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Problem 7-31 continued

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Problem 8.1

An information packet contains 200 bits. This packet is transmitted over a communications channel where the probability of error for each bit is 10-3. What is the probability that the packet is received error-free?

Solution Recognizing that the number of errors has a binomial distribution over the sequence of 200 bits, let x represent the number of errors with p = 0.001 and n = 200. Then the probability of no errors is

P ⎡⎢⎣ x = 0⎤⎥⎦ = (1 − p )

n

= (1 − .001)

200

= .999 200 = 0.82

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page…8-1

Problem 8.2 Suppose the packet of the Problem 8.1 includes an error-correcting code that can correct up to three errors located anywhere in the packet. What is the probability that a particular packet is received in error in this case? Solution The probability of a packet error is equal to the probability of more than three bit errors. This is equivalent to 1 minus the probability of 0, 1, 2, or 3 errors: 1 − P[x ≤ 3] = 1 − (P[x = 0] + P[x = 1] + P[x = 2] + P[x = 3]) ⎛n⎞ ⎛n⎞ ⎛n⎞ = 1 − (1 − p) n − ⎜⎜ ⎟⎟ p(1 − p )n−1 − ⎜⎜ ⎟⎟ p 2 (1 − p )n −2 − ⎜⎜ ⎟⎟ p 3 (1 − p )n −3 ⎝3 ⎠ ⎝2⎠ ⎝1 ⎠ n(n − 1) 2 n(n − 1)(n − 2) 3 ⎤ ⎡ p (1 − p ) + p ⎥ = 1 − (1 − p )n−3 ⎢(1 − p )3 + np(1 − p )2 + 2 6 ⎣ ⎦ = 5.5 × 10 −5

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page…8-2

Problem 8.3 Continuing with Example 8.6, find the following conditional probabilities: P[X=0|Y=1] and P[X =1|Y=0].

Solution From Bayes’ Rule

[

]

[

]

P X = 0Y = 1 =

[

]

P Y = 1 X = 0 P[ X = 0 ]

P[Y = 1] pp0 = pp0 + (1 − p ) p1

P X = 1Y = 0 = =

[

]

P Y = 0 X = 1 P[X = 1] P[Y = 0]

pp1 pp1 + (1 − p ) p0

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page…8-3

Problem 8.4 Consider a binary symmetric channel for which the conditional probability of error p = 10-4, and symbols 0 and 1 occur with equal probability. Calculate the following probabilities: a) The probability of receiving symbol 0. b) The probability of receiving symbol 1. c) The probability that symbol 0 was sent, given that symbol 0 is received d) The probability that symbol 1 was sent, given that symbol 0 is received.

Solution (a) P[Y = 0] = P[Y = 0 | X = 0]P[ X = 0] + P[Y = 0 | X = 1]P[ X = 1] = (1 − p ) p0 + pp1 = .9999 1 + .0001 1 2 2 =1 2

(b)

P[Y = 1] = 1 − P[Y = 0] =1

(c)

2

From Eq.(8.30) P [X = 0 Y = 0 ] = =

(1 − p ) p0 (1 − p ) p0 + pp1

(1 − 10 ) (1 − 10 ) + 10 −4

−4

= 1 − 10

(d)

1

1

2

2

−4 1

2

−4

From Prob. 8.3 P [X = 1Y = 0 ] = =

pp1 pp1 + (1 − p ) p0 10 −4

10 −4 12 1 + (1 − 10 − 4 ) 1 2 2

= 10 −4

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page…8-4

Problem 8.5 Determine the mean and variance of a random variable that is uniformly distributed between a and b.

Solution The mean of the uniform distribution is given by ∞

µ = E[X ] =

∫ xf

X

( x)dx

−∞ b

=∫ x a

1 dx b−a b

x2 = 2(b − a ) a b2 − a 2 2(b − a ) b+a = 2 =

The variance is given by

[

] ∫ (x − µ)

E (X − µ ) = 2

∞

2

f X ( x)dx

−∞

=∫

b

(x − µ )2 dx

b−a 3 3 1 (b − µ ) (a − µ ) = − b−a 3 3 a

If we substitute µ =

[

]

E (X − µ ) = 2

b+a then 2

3 3 1 ⎡ (b − a ) (a − b ) ⎤ − ⎢ ⎥ b − a ⎣ 24 24 ⎦

2 ( b − a) =

12

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page…8-5

Problem 8.6 Let X be a random variable and let Y = (X-µX)/σX. What is the mean and variance of the random variable Y? Solution

⎡ X − µ X ⎤ E[X ] − µ X 0 E[Y ] = E ⎢ = =0 ⎥= σX σX ⎣ σX ⎦

[ ]

E(Y − µ Y ) = E Y 2

2

⎛ X − µX = E⎜⎜ ⎝ σX

E( X − µ X )

2

=

σX2

⎞ ⎟⎟ ⎠

2

σX2 = =1 σX2

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page…8-6

Problem 8.7 What is the probability density function of the random variable Y of Example 8.8? Sketch this density function.

Solution From Example 8.8, the distribution of Y is ⎧0 ⎪ −1 ⎪ 2π − 2 cos ( y ) FY ( y ) = ⎨ 2π ⎪ ⎪⎩1

y < −1 | y |< 1 y >1

Thus, the density of Y is given by ⎧0 ⎪ dFY ( y ) ⎪ 1 =⎨ 2 dy ⎪π 1 − y ⎪0 ⎩

y < −1 | y |< 1 y >1

This density is sketched in the following figure.

fY(y)

1

π

y -1

1

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page…8-7

Problem 8.8 Show that the mean and variance of a Gaussian random variable X with the density 2 function given by Eq. (8.48) are µ and σ X . X

Solution Consider the difference E[X]-µX: E[X ] − µ X =

∞

∫

(x − µ X ) exp⎧− (x − µ X )2 ⎫dx ⎨ ⎩

2π σ X

−∞

2σ X

2

⎬ ⎭

Let y = x − µ X and substitute E[ X ] − µ X = ∫

∞

y 2π σ X

−∞

2 ⎛ ⎞ dy exp⎜⎜ − y 2 ⎟ 2σ X ⎟⎠ ⎝

=0

since integrand has odd symmetry. This implies E[X ] = µ X . With this result Var( X ) = E( x − µ X )

2

=∫

∞

−∞

(x − µ X )2 exp⎧ − (x − µ X )2 ⎫dx ⎨ ⎩

2π σ X

2σ X

2

⎬ ⎭

In this case let y=

x − µX

σX

and making the substitution, we obtain Var ( X ) = σ X

2

∫

∞

−∞

⎧− y2 ⎫ y2 exp⎨ ⎬dy 2π ⎩ 2 ⎭

Recalling the integration-by-parts, i.e., ∫ udv = uv − ∫ vdu , let u = y and

⎛ − y2 ⎞ ⎟⎟dy . Then dv = y exp⎜⎜ 2 ⎝ ⎠ Continued on next slide

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page…8-8

Problem 8.8 continued Var ( X ) = σ X

2

(− ) y exp⎛ − y 2 2π

⎜ ⎝

∞

⎛ y2 ⎞ 1 ⎞ 2 ∞ ⎜⎜ − ⎟⎟dy + σ exp X ∫ 2 ⎟⎠ −∞ 2π ⎝ 2 ⎠ −∞

= 0 + σ X •1 2

=σX

2

where the second integral is one since it is integral of the normalized Gaussian probability density.

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page…8-9

Problem 8.9 Show that for a Gaussian random variable X with mean µX and variance σ X2 the transformation Y = (X - µX)/σX, converts X to a normalized Gaussian random variable.

Solution x − µX Let y = . Then

σX

E[Y ] =

1 2π

⎛ y2 ⎞ y exp ⎜ − 2 ⎟dy ∫−∞ ⎠ ⎝ ∞

=0

by the odd symmetry of the integrand. If E[Y] = 0, then from the definition of Y, E[X] = µX. In a similar fashion

[ ]

EY2 =

1 2π

⎛ y2 ⎞ 2 ⎜⎜ − ⎟⎟dy exp y ∫−∞ ⎝ 2 ⎠ ∞

∞

⎧ y2 ⎫ (−) y 1 = ⋅ exp⎨− ⎬ + 2π 2π ⎩ 2 ⎭ −∞ =1

2 ⎞ ⎛ exp⎜ − y ⎟dy 2 −∞ ⎠ ⎝

∫

∞

where we use integration by parts as in Problem 8.8. This result implies ⎛ x − µX E ⎜⎜ ⎝ σX

2

⎞ ⎟⎟ = 1 ⎠

and hence E( x − µ X ) = σ X2 2

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page…8-10

Problem 8.10 Determine the mean and variance of the sum of five independent uniformly-distributed random variables on the interval from -1 to +1. Solution Let Xi be the individual uniformly distributed random variables for i = 1,..,5, and let Y be the random variable representing the sum: 5

Y = ∑ Xi i =1

Since Xi has zero mean and Var(Xi) = 1/3 (see Problem 8.5), we have 5

E[Y ] = ∑ E[X i ] = 0 i =1

and

[

] [ ]

Var (Y ) = E (Y − µY ) = E Y 2 2

[

= E (∑ X i )

2

]

[ ]

5

[

= ∑ E X i2 + ∑ E X i X j i =1

]

i≠ j

Since the Xi are independent, we may write this as

( )

[ ]

Var(Y ) = 5 1 + ∑ E[ X i ]E X j 3 = 5 +0 3 5 = 3

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page…8-11

Problem 8.11 A random process is defined by the function

X (t , θ ) = A cos(2πft + θ ) where A and f are constants, and θ is uniformly distributed over the interval 0 to 2π. Is X stationary to the first order?

Solution Denote

Y = X (t1 , θ ) = A cos(2πft1 + θ ) for any t1. From Problem 8.7, the distribution of Y and therefore of X for any t1 is

⎧0 ⎪ ⎪ 2π − 2 cos −1 ( y / A) FX (t1 ) ( y ) = ⎨ 2π ⎪ ⎪⎩1

y < −A | y |< A y>A

Since the distribution is independent of t it is stationary to first order.

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page…8-12

Problem 8.12 Show that a random process that is stationary to the second order is also stationary to the first order.

Solution

Let the distribution F be stationary to second order FX (t1 ) X (t2 ) ( x1 , x2 ) = FX ( t1 +τ ) X ( t2 +τ ) ( x1 , x2 )

Then, FX ( t1 ) X (t2 ) ( x1 , ∞ ) = FX (t1 ) (x1 )

= FX ( t1 +τ ) X ( t2 +τ ) ( x1 , ∞ ) = FX ( t1 +τ ) ( x1 )

Thus the first order distributions are stationary as well.

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page…8-13

Problem 8.13 Let X(t) be a random process defined by

X (t ) = A cos(2πft ) where A is uniformly distributed between 0 and 1, and f is constant. Determine the autocorrelation function of X. Is X wide-sense stationary?

Solution

[ ] = E[A ][cos(2πf (t

E[X (t1 )X (t 2 )] = E A 2 cos(2πft1 ) cos(2πft 2 ) 2

1

[ ]

− t 2 )) + cos 2πf (t1 + t 2 )]

1

x3 1 E A = ∫ x dx = = 0 3 0 3 2

1

2

Since the autocorrelation function depends on t1 + t 2 as well as t1 − t 2 , the process is not wide-sense stationary.

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page…8-14

Problem 8.14 A discrete-time random process {Yn: n = …,-1,0,1,2, …} is defined by Yn = α 0 Z n + α 1 Z n −1 where {Zn} is a random process with autocorrelation function RZ (n) = σ

[

]

δ (n) . What is the

2

autocorrelation function RY ( n, m) = E Yn Ym ? Is the process {Yn} wide-sense stationary?

Solution

We implicitly assume that Zn is stationary and has a constant mean µZ. Then the mean of Yn is given by

E[Yn ] = α 0 E[Z n ] + α1E[ Z n−1 ] = (α 0 + α1 )µ Z

The autocorrelation of Y is given by E[YnYm ] = E[(α 0 Z n + α 0 Z n−1 )(α 0 Z m +α 1Z m−1 )]

= α 02 E[Z n Z m ] + α1α 0 E[Z n Z m−1 ] + α 0α1E[Z n−1 Z m ] + α12 E[Z m−1 Z n−1 ]

= α 02σ 2δ (n − m ) + α1α 0σ 2δ (m − 1 − n ) + α 0α1σ 2δ (n − 1 − m ) + α12δ (m − 1 − (n − 1))

(

)

= α 02 + α12 σ 2δ (n − m ) + α 0α1σ 2 [δ (n − m − 1) + δ (m − n − 1)]

Since the autocorrelation only depends on the time difference n-m, the process is widesense stationary with

(

)

RY (n) = α 02 + α12 σ 2δ (n) + α 0α1σ 2 (δ (n − 1) + δ (n + 1) )

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page…8-15

Problem 8.15 For the discrete-time process of Problem 8.14, use the discrete Fourier transform to approximate the corresponding spectrum. That is, N −1

S Y (k ) = ∑ RY (n)W kn n=0

If the sampling in the time domain is at n/Ts where n = 0, 1, 2, …, N-1. What frequency does k correspond to?

Solution Let

β 0 = (α 02 + α12 )σ 2 and β1 = α 0α1σ 2 . Then N −1

SY (k ) = ∑ [β 0δ (n ) + β1 (δ (n − 1) + δ (n + 1))] W kn n =0

(

= β 0W 0 + β1 W −k + W + k

)

+ j 2πk ⎛ − jN2πk ⎞ ⎜ = β 0 + β1 ⎜ e + e N ⎟⎟ ⎝ ⎠ ⎛ 2πk ⎞ = β 0 + 2 β1 cos⎜ ⎟ ⎝ N ⎠

The term SY (k ) corresponds to frequency

kf s 1 where f S = . TS N

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page…8-16

Problem 8.16 Is the discrete-time process {Yn: n = 1,2,…} defined by: Y0 = 0 and Yn +1 = αYn + Wn , a Gaussian process, if Wn is Gaussian?

Solution (Proof by mathematical induction.) The first term Y1 = αY0 + W0 is Gaussian since Y0 = 0 and W0 are Gaussian. The second term Y2 = αY1 + W1 is Gaussian since Y1 and W1 are Gaussian. Assume Yn is Gaussian. Then Yn +1 = αYn + Wn is Gaussian since Yn and Wn are both Gaussian.

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page…8-18

Problem 8.17 A discrete-time white noise process {Wn} has an autocorrelation function given by RW(n) = N0δ(n). (a) Using the discrete Fourier transform, determine the power spectral density of {Wn}. (b) The white noise process is passed through a discrete-time filter having a discretefrequency response H (k ) =

1 − (αW k ) N 1 − αW k

where, for a N-point discrete Fourier transform, W = exp{j2π/N}. What is the spectrum of the filter output?

Solution The spectrum of the discrete white noise process is N −1

S (k ) = ∑ R(n ) W nk n =0

N −1

= ∑ N 0δ (n )W nk n =0

= N0

The spectrum of the process after filtering is SY (k ) = H (k ) S (k ) 2

1 − (αW k ) N = N0 1 − αW k

2

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page…8-19

Problem 8.18 Consider a deck of 52 cards, divided into four different suits, with 13 cards in each suit ranging from the two up through the ace. Assume that all the cards are equally likely to be drawn. (a) Suppose that a single card is drawn from a full deck. What is the probability that this card is the ace of diamonds? What is the probability that the single card drawn is an ace of any one of the four suits? (b) Suppose that two cards are drawn from the full deck. What is the probability that the cards drawn are an ace and a king, not necessarily the same suit? What if they are of the same suit?

Solution (a)

P[Ace of diamonds] = P[Any ace] =

1 52

1 13

(b)

P[Ace and king ] = P[Ace on first draw ]P[King on second] + P[King on first draw ]P[Ace on seco 1 4 1 4 = × + × 13 51 13 51 8 = 663 1 1 1 1 P[Ace and king of same suit ] = × + × 13 51 13 51 1 = 663

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page…8-20

Problem 8.19

Suppose a player has one red die and one white die. How many outcomes are possible in the random experiment of tossing the two dice? Suppose the dice are indistinguishable, how many outcomes are possible?

Solution The number of possible outcomes is 6 × 6 = 36 , if distinguishable. If the die are indistinguishable then the outcomes are (11) (12)…(16) (22)(23)…(26) (33)(34)…(36) (44)(45)(46) (55)(56) (66) And the number of possible outcomes are 21.

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page…8-21

Problem 8.20 Refer to Problem 8.19. (a) What is the probability of throwing a red 5 and a white 2? (b) If the dice are indistinguishable, what is the probability of throwing a sum of 7? If they are distinguishable, what is this probability?

Solution (a) P[Red 5 and white 2] =

1 1 1 × = 6 6 36

(b) The probability of the sum does not depend upon whether the die are distinguishable or not. If we consider the distinguishable case the possible outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) so 6 1 P[sum of 7] = = 36 6

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page…8-22

Problem 8.21 Consider a random variable X that is uniformly distributed between the values of 0 and 1 with probability ¼ takes on the value 1 with probability ¼ and is uniformly distributed between values 1 and 2 with probability ½ . Determine the distribution function of the random variable X. Solution ⎧ ⎪0 ⎪x ⎪ 4 ⎪ FX ( x) = ⎨ 1 2 ⎪ ⎪ 1 + 1 ( x − 1) ⎪2 2 ⎪1 ⎩

x≤0 0 < x 2

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page…8-24

Problem 8.22 Consider a random variable X defined by the double-exponential density where a and b are constants.

f X ( x) = a exp(− b x )

−∞ < x Y] = P[Y > X]. If we only consider the case X > Y, then we have the conditions: 0 < X < T and 0 < Y < X < τ+Y. Combining these conditions we have Y < X < min(T, τ+Y). Consequently, T

min (T ,τ + y )

0

y

P[ X − Y < τ ] = ∫

∫f

min (T ,τ + y )

T

=∫

∫ y

0

=

X

1 T2

( x) f Y ( y )dx dy

2

⎛1⎞ ⎜ ⎟ dx dy ⎝T ⎠

T

∫ {min(T ,τ + y ) − y}dy 0

Combining the two terms of the integrand, P[ X − Y < τ ] =

T

1 min(T − y, τ )dy T 2 ∫0 T

⎛ ⎞ 1 y2 ⎜ , τy ⎟⎟ = 2 min⎜ Ty − 2 T ⎝ ⎠0 ⎛1 τ ⎞ = min⎜ , ⎟ ⎝2 T ⎠

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page…8-54

Problem 8.45 A telegraph system (an early version of digital communications) transmits either a dot or dash signal. Assume the transmission properties are such that 2/5 of the dots and 1/3 of the dashes are received incorrectly. Suppose the ratio of transmitted dots to transmitted dashes is 5 to 3. What is the probability that a received signal as the transmitted if: a) The received signal is a dot? b) The received signal is a dash?

Solution (a) Let X represent the transmitted signal and Y represent the received signal. Then by application of Bayes’ rule

P(Y = dot ) = P( X = dot | No error )P( No dot error) + P( X = dash | error )P(dash error ) =5 3 + 3 1 8 5 8 3 =3 +1 =1 8 8 2

( ) ( )( )

(b) Similarly, P[Y = dash ] = P[ X = dash | no error]P(no dash error ) + P( X = dot )P[dot error] = 3 ⋅2 + 5 2 8 3 8 5 =2 +2 = 1 8 8 2

[ ]

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page…8-55

Problem 8.46 Four radio signals are emitted successively. The probability of reception for each of them is independent of the reception of the others and equal, respectively, 0.1, 0.2, 0.3 and 0.4. Find the probability that k signals will be received where k = 1, 2, 3, 4. Solution For one successful reception, the probability is given by the sum of the probabilities of the four mutually exclusive cases P = p1 (1 − p2 )(1 − p3 )(1 − p4 ) +

(1 − p1 ) p2 (1 − p3 )(1 − p4 ) + (1 − p1 )(1 − p2 ) p3 (1 − p4 ) + (1 − p1 )(1 − p2 )(1 − p3 ) p4

= .1 ⋅ .8 ⋅ .7 ⋅ .6 + .9 ⋅ .2 ⋅ .7 ⋅ .6 + .9 ⋅ .8 ⋅ .3 ⋅ .6 + .9 ⋅ .8 ⋅ .7 ⋅ .4 = 0.4404

For k = 2, there six mutually exclusive cases P = p1 p2 (1 − p3 )(1 − p4 ) +

p1 (1 − p2 ) p3 (1 − p4 ) + p1 (1 − p2 )(1 − p3 ) p4 +

(1 − p1 ) p2 p3 (1 − p4 ) + (1 − p1 ) p2 (1 − p3 ) p4 + (1 − p1 )(1 − p2 ) p3 p4

= .1 ⋅ .2 ⋅ .7 ⋅ .6 + .1 ⋅ .8 ⋅ .3 ⋅ .6 + .1 ⋅ .8 ⋅ .7 ⋅ .4 + .9 ⋅ .2 ⋅ .3 ⋅ .6 + .9 ⋅ .2 ⋅ .7 ⋅ .4 + .9 ⋅ .8 ⋅ .3 ⋅ .4 = 0.2144 For k =3 there are four mutually exclusive cases P = p1 p2 p3 (1 − p4 ) +

p1 (1 − p2 ) p3 p4 + p1 p2 (1 − p3 ) p4 +

(1 − p1 ) p2 p3 p4 = .1 ⋅ .2 ⋅ .3 ⋅ .6 + .1 ⋅ .8 ⋅ .3 ⋅ .4 + .1 ⋅ .2 ⋅ .7 ⋅ .4 + .9 ⋅ .2 ⋅ .3 ⋅ .4 = 0.0404

For k = 4 there is only one term

P = p1 p2 p3 p4 = .1 ⋅ .2 ⋅ .3 ⋅ .4 = 0.0024

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page…8-56

Problem 8.47 In a computer-communication network, the arrival time τ between messages is modeled with an exponential distribution function, having the density

⎧ 1 −λτ ⎪ e f T (τ ) = ⎨ λ ⎪⎩0

τ ≥0 otherwise

a) What is the mean time between messages with this distribution? b) What is the variance in this time between messages?

Solution (Typo in problem statement, should read fT(τ)=(1/λ)exp(-τ/λ) for τ>0) (a) The mean time between messages is ∞

E[T ] = ∫ τf T (τ )dτ 0

∞

τ exp(− τ / λ )dτ λ 0

=∫

∞

= − τ exp(− τ / λ ) 0 + ∫ exp(− τ / λ )dτ ∞

0

∞

= 0 − λ exp(−τ / λ ) 0 =λ

where the third line follows by integration by parts. (b) To compute the variance, we first determine the second moment of T ∞

[ ] = ∫τ

ET

2

2

f T (τ )dτ

0

∞

τ2 exp(− τ / λ )dτ λ 0

=∫

∞

= − τ 2 exp(− τ / λ ) + 2∫ τ exp(− τ / λ )dτ ∞ 0

= 0 + 2λE[T ]

0

= 2λ2

Continued on next slide

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page…8-57

Problem 8.47 continued

The variance is then given by the difference of the second moment and the first moment squared (see Problem 8.23)

[ ]

Var (T ) = E T 2 − (E[T ])

2

= 2λ2 − λ2 = λ2

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page…8-58

Problem 8.48 If X has a density fX(x), find the density of Y where a) Y = aX + b for constants a and b. 2 b) Y = X . c) Y = X , assuming X is a non-negative random variable. Solution (a) If Y = aX + b , using the results of Section 8.3 for Y = g(X)

dg −1 ( y ) f Y ( y ) = f X (g ( y ) ) dy −1

⎛ y −b⎞ 1 = fX ⎜ ⎟ ⎝ a ⎠a (b) If Y = X 2 , then

( (

)

(

fY ( y ) = f X − y + f X +

⎛ 1 ⎞ ⎟ y ⎜ ⎜2 y ⎟ ⎝ ⎠

))

(c) If Y = X , then we must assume X is positive valued so, this is a one-to-one mapping and

( )

fY ( y ) = f X y 2 ⋅ 2 y

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page…8-59

Problem 8.49 Let X and Y be independent random variables with densities fX(x) and fY(y), respectively. Show that the random variable Z = X+Y has a density given by z

f Z ( z) =

∫f

Y

( z − s ) f X ( s )ds

−∞

Hint:

P[Z ≤ z ] = P[ X ≤ z, Y ≤ z − X ]

Solution (Typo in problem statement - should be “positive” independent random variables)

Using the hint, we have that FZ(z) = P[Z ≤ z] and z z−x

FZ ( z ) =

∫∫f

X

( x) f Y ( y )dydx

− ∞ −∞

To differentiate this result with respect to z, we use the fact that if b

g ( z ) = ∫ h( x, z )dx a

then db da ∂g ( z ) ∂ = ∫ h( x, z )dx + h(b, z ) − h( a, z ) dz dz dz ∂z a b

(1)

Inspecting FZ(z), we identify h(x,z) z−x

h ( x, z ) =

∫f

X

( x) f Y ( y )dy

−∞

and a = -∞ and b = z. We then obtain d FZ ( z ) dz z −( −∞ ) z z−z ⎤ ⎡ d z− x dz ( ) ( ) ( ) ( ) f x f y dy dx f z f y dy f X (−∞) f Y ( y )dy ⋅ 0 + − = ∫⎢ ⎥ X Y ∫ X Y dz −∫∞ dz −∫∞ −∞ ⎣ −∞ ⎦

f Z ( z) =

⎡d ∫−∞⎢⎣ dz z

=

z− x

∫f

−∞

X

⎤ ( x) f Y ( y )dy ⎥ dx ⎦

Continued on next slide

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page…8-60

Problem 8-49 continued

where the second term of the second line is zero since the random variables are positive, and the third term is zero due to the factor zero. Applying the differentiation rule a second time, we obtain

z

f Z ( z) =

⎡

∫ ⎢⎣0 + f

X

( x) f Y ( z − x)

−∞

d ( z − x) d (−∞) ⎤ dx − f X ( x) f Y (−∞) dz dz ⎥⎦

z

=

∫f

X

( x) f Y ( z − x)dx

−∞

which is the desired result.

An alternative solution is the following: we note that P[Z ≤ z | X = x] = P[X + Y ≤ z | X = x]

= P[x + Y ≤ z | X = x ] = P[x + Y ≤ z ] = P[Y ≤ z − x]

where the third equality follows from the independence of X and Y. By differentiating both sides with respect to z, we see that f Z | X ( z | x) = f Y ( z − x) By the properties of conditional densities f Z , X ( z , x) = f X ( x) f Z | X ( z | x) = f X ( x) f Y ( z − x) Integrating to form the marginal distribution, we have ∞

f Z ( z) =

∫f

X

( x) f Y ( z − x)dx

−∞

If Y is a positive random variable then fY(z-x) is zero for x > z and the desired result follows.

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page…8-61

Problem 8.50 Find the spectral density SZ(f) if

Z (t ) = X (t )Y (t ) where X(t) and Y(t) are independent zero-mean random processes with

RX (τ ) = a1e

−α1 τ

and

RY (τ ) = a2 e

−α 2 τ

.

Solution The autocorrelation of Z(t) is given by

RZ (τ ) = E[Z (t )Z (t + τ )]

= E[ X (t )X (t + τ )Y (t )Y (t + τ )] = E[ X (t ) X (t + τ )]E[Y (t )Y (t + τ )] = RX (τ )RY (τ )

By the Wiener-Khintchine relations, the spectrum of Z(t) is given by S Z ( f ) = F −1 [R X (τ )RY (τ )]

= F −1 [a1a2 exp(− (α1 + α 2 )τ =

2a1a2 (α1 + α 2 ) 2 (α1 + α 2 ) 2 + (2πf )

)]

where the last line follows from the Fourier transform of the double-sided exponential (See Example 2.3).

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page…8-62

Problem 8.51 Consider a random process X(t) defined by X (t ) = sin (2πf c t ) where the frequency fc is a random variable uniformly distributed over the interval [0,W]. Show that X(t) is nonstationary. Hint: Examine specific sample functions of the random process X(t) for, say, the frequencies W/4, W/2, and W.

Solution To be stationary to first order implies that the mean value of the process X(t) must be constant and independent of t. In this case, E[ X (t )] = E[sin (2πf c t )] W

=

1 sin (2πwt )dw W ∫0

− cos(2πwt ) = 2πWt 0

W

=

1 − cos(2πWt ) 2πWt

This mean value clearly depends on t, and thus the process X(t) is nonstationary.

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page…8-63

Problem 8.52 The oscillators used in communication systems are not ideal but often suffer from a distortion known as phase noise. Such an oscillator may be modeled by the random process Y (t ) = A cos(2πf c t + φ (t ) ) where

φ (t )

φ (t )

is a slowly varying random process. Describe and justify the conditions on the random process

such that Y(t) is wide-sense stationary.

Solution The first condition for wide-sense stationary process is a constant mean. Consider t = t0, then E[Y (t 0 )] = E[ A cos(2πf c t0 + φ (t0 ) )]

In general, the function cos θ takes from values -1 to +1 when θ varies from 0 to 2π. In this case θ corresponds to 2πfct0 + φ(t0). If φ(t0) varies only by a small amount then θ will be biased toward the point 2πfct0 + E[φ(t0)], and the mean value of E[Y(t0)] will depend upon the choice of t0. However, if φ(t0) is uniformly distributed over [0, 2π] then 2πfct0 + φ(t0) will be uniformly distributed over [0, 2π] when considered modulo 2π, and the mean E[Y(t0)] will be zero and will not depend upon t0. Thus the first requirement is that φ(t) must be uniformly distributed over [0,2π] for all t. The second condition for a wide-sense stationary Y(t) is that the autocorrelation depends only upon the time difference E[Y (t1 )Y (t 2 )] = E[A cos(2πf c t1 + φ (t1 ) )A cos(2πf c t 2 + φ (t 2 ) )] A2 E[cos(2πf c (t1 + t 2 ) + φ (t1 ) + φ (t 2 ) ) + cos(2πf c (t1 − t 2 ) + φ (t1 ) − φ (t 2 ) )] = 2

where we have used the relation cos A cos B = 1 2 (cos( A + B) + cos( A − B) ) . In general, this correlation does not depend solely on the time difference t2-t1. However, if we assume: We first note that if φ(t1) and φ(t2) are both uniformly distributed over [0,2π] then so is ψ = φ (t1 ) + φ (t 2 ) (modulo 2π), and E[cos(2πf c (t1 + t 2 ) + ψ )] =

1 2π

2π

∫ cos(2πf 0

c

(t1 + t 2 ) + ψ )dψ

(1)

=0

Continued on next slide

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page…8-64

Problem 8.52 continued

We consider next the term RY(t1,t2)= E[cos(2πf c (t1 − t 2 ) + φ (t1 ) − φ (t 2 ) )] and three special cases: (a) if ∆t = t1-t2 is small then φ (t1 ) ≈ φ (t 2 ) since φ(t) is a slowly varying process, and A2 cos(2πf c (t1 − t 2 ) ) 2 (b) if ∆t is large then φ(t1) and φ(t2) should be approximately independent and φ (t1 ) − φ (t 2 ) would be approximately uniformly distributed over [0,2π]. In this case RY (t1 , t 2 ) ≈ 0 using the argument of Eq. (1). RY (t1 , t 2 ) =

(c) for intermediate values of ∆t, we require that

φ (t1 ) − φ (t 2 ) ≈ g (t1 − t 2 ) for some arbitrary function g(t). Under these conditions the random process Y(t) will be wide-sense stationary.

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page…8-65

Problem 8.53 A baseband signal is disturbed by a noise process N(t) as shown by

X (t ) = A sin (0.3πt ) + N (t ) where N(t) is a stationary Gaussian process of zero mean and variance σ2. (a) What are the density functions of the random variables X1 and X2 where

X 1 = X (t ) t =1 X 2 = X (t ) t =2 (b) The noise process N(t) has an autocorrelation function given by

RN (τ ) = σ 2 exp(− τ What is the joint density function of X1 and X2, that is,

)

f X1 , X 2 ( x1 , x2 ) ?

Solution (a) The random variable X1 has a mean E[ X (t1 )] = E[ A sin (0.3π ) + N (t1 )]

= A sin(0.3π ) + E[N (t1 )] = A sin (0.3π )

Since X1 is equal to N(t1) plus a constant, the variance of X1 is the same as that of N(t1). In addition, since N(t1) is a Gaussian random variable, X1 is also Gaussian with a density given by 1 f X1 ( x ) = exp{− ( x − µ1 ) / 2σ 2 } 2π σ where µ1 = E[ X (t1 )] . By a similar argument, the density function of X2 is f X 2 ( x) =

{

1 exp − ( x − µ 2 ) / 2σ 2 2π σ

}

where µ 2 = A sin(0.6π ) .

Continued on next slide

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page…8-66

Problem 8-53 continued

(b) First note that since the mean of X(t) is not constant, X(t) is not a stationary random process. However, X(t) is still a Gaussian random process, so the joint distribution of N Gaussian random variables may be written as Eq. (8.90). For the case of N = 2, this equation reduces to f X (x) =

1 2π Λ

1/ 2

{

}

exp − (x − µ)Λ−1 (x − µ)T / 2

where Λ is the 2x2 covariance matrix. Recall that cov(X1,X2) =E[(X1-µ1)(X2-µ2)], so that ⎡cov( X 1 , X 1 ) cov( X 1 , X 2 ) ⎤ Λ=⎢ ⎥ ⎣cov( X 2 , X 1 ) cov( X 2 , X 2 )⎦ ⎡ R (0) RN (1) ⎤ =⎢ N ⎥ ⎣ RN (1) RN (0)⎦ ⎡ σ2 σ 2 exp(−1)⎤ =⎢ 2 ⎥ σ2 ⎦ ⎣σ exp(− 1)

If we let ρ = exp(-1) then

Λ = σ 4 (1 − ρ 2 ) and

Λ−1 =

⎡ 1 1 2 ⎢ σ (1 − ρ ) ⎣− ρ 2

− ρ⎤ 1 ⎥⎦

Making these substitutions into the above expression, we obtain upon simplification f X1 , X 2 ( x1 , x2 ) =

⎧ ( x − µ1 ) 2 + ( x2 − µ 2 ) 2 − 2 ρ ( x1 − µ1 )( x2 − µ 2 ) ⎫ exp⎨− 1 ⎬ 2σ 2 (1 − ρ 2 ) 1− ρ2 ⎩ ⎭

1 2πσ 2

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page…8-67

Problem 9.1 In practice, we often cannot measure the signal by itself but must measure the signal plus noise. Explain how the SNR would be calculated in this case. Solution Let r(t) = s(t) + n(t) be the received signal plus noise. Assuming the signal is independent of the noise, we have that the received power is

[

R0 = E r 2 (t )

[

]

= E (s (t ) + n(t ) )

2

]

[ ] [ ] = E[s (t )] + 2E[s (t )]E[n(t )] + E[n (t )] = E s 2 (t ) + 2E[s (t )n(t )] + E n 2 (t ) 2

2

= S +0+ N where S is the signal power and N is the average noise power. We then measure the noise alone

[

R1 = E n 2 (t )

]

=N

and the SNR is given by

SNR =

R0 − R1 R1

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Problem 9.2 A DSC-SC modulated signal is transmitted over a noisy channel, having a noise spectral density N0/2 of 2x10-17 watts per hertz. The message bandwidth is 4 kHz and the carrier frequency is 200 kHz. Assume the average received power of the signal is -80 dBm. Determine the post-detection signal-to-noise ratio of the receiver. Solution From Eq. (9.23), the post-detection SNR of DSB-SC is SNR DSB post =

Ac2 P 2 N 0W

Ac2 P = −80 dBm = 10 −11 watts. With a message bandwidth 2 of 4 kHz, the post-detection SNR is The average received power is

SNR

DSB post

10−11 = = 62.5 ~ 18.0 dB (4 ×10−17 )4000

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Problem 9.3. For the same received signal power, compare the post-detection SNRs of DSB-SC with coherent detection and envelope detection with ka = 0.2 and 0.4. Assume the average message power is P = 1. Solution DSB

From Eq. (9.23), the post-detection SNR of DSB-SC with received power DSB SNR post =

Ac2 P 2 N 0W

DSB

(

)

Ac2 2 1 + k a P is 2

AM

From Eq. (9.30), the post-detection SNR of AM with received power AM SNR post =

2

Ac P is 2

Ac2 k a2 P 2 N 0W

AM

So, by equating the transmit powers for DSB-Sc and AM, we obtain DSB

Ac2 P AM Ac2 1 + ka 2 P = 2 2 AM 2 DSB 2 Ac Ac P ⇒ = 2 2 1 + ka 2 P

(

)

Substituting this result into the expression for the post-detection SNR of AM,

SNR AM post =

Ac 2 P ⎛ ka 2 P ⎞ DSB ⎜ ⎟ = SNR post ∆ 2 N 0W ⎝ 1 + ka 2 P ⎠

DSB

Where the factor ∆ is 2

∆=

ka P 2

1+ k a P

With ka= 0.2 and P= 1, the AM SNR is a factor

2 ( .2 ) ∆=

With ka = 0.4 and P = 1, the AM SNR is a factor ∆ =

1.04

= .04 less.

(.4)2 1 + .16

=

.16 ≈ 0.14 less. 1.16

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Problem 9.4. In practice, there is an arbitrary phase θ in Eq. (9.24). How will this affect the results of Section 9.5.2? Solution Envelope detection is insensitive to a phase offset.

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Problem 9.5. The message signal of Problem 9.2 having a bandwidth W of 4 kHz is transmitted over the same noisy channel having a noise spectral density N0/2 of 2x10-17 watts per hertz using single-sideband modulation. If the average received power of the signal is -80 dBm, what is the post-detection signal-to-noise ratio of the receiver? Compare the transmission bandwidth of the SSB receiver to that of the DSB-SC receiver. Solution From Eq. (9.23) SNR SSB post =

with

Ac2 P 2 N 0W

Ac 2 P = −80 dBm , W = 4 kHz , and N 0 = 4 × 10 −17 . The 2

SNR SSB post = 18 dB The transmission bandwidth of SSB is 4 kHz, half of the 8 kHz used with DSB-SC.

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Problem 9.6 The signal m(t) = cos(2000πt) is transmitted by means of frequency modulation. If the frequency sensitivity kf is 2 kHz per volt, what is the Carson’s rule bandwidth of the FM signal. If the pre-detection SNR is 17 dB, calculate the postdetection SNR. Assume the FM demodulator includes an ideal low-pass filter with bandwidth 3.1 kHz. Solution The Carson Rule bandwidth is BT = 2(k f A + f m ) = 2(2(1) + 2 ) = 8 kHz . Then from Eq.(9.59), SNR FM post =

3 Ac2 k 2f P 2 N 0W 3

=

Ac2 2 N 0 BT

⎛ 3k 2f P ⎞ BT ⎟ ⎜⎜ 3 ⎟ W ⎝ ⎠

We observed that the first factor is the pre-detection SNR, and we may write this as 2 1 ⎛ ⎞ FM 3 ⋅ 2 2 ⎜ ⎟ SNR FM SNR 8 = post pre ⎜ ( 3.1)3 ⎟ ⎝ ⎠ FM = SNR pre ×1.61

~ 19.2 dB (There is an error in the answer given in the text.)

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Problem 9.7 Compute the post-detection SNR in the lower channel for Example 9.2 and compare to the upper channel. Solution

The SNR of lower channel is, from Eq. (9.59) SNR FM post =

3 Ac 2 k f 2 ( P / 2) 2 N 0W 3

where we have assumed that half the power is in the lower channel. Using the approximation to Carson’s Rule BT = 2(k f P 1 / 2 + D ) ≈ 2k f P

1

2

= 200 kHz , that is,

k 2f P = BT2 / 4 this expression becomes SNR

FM post

Ac 2 = 2 N 0 BT

3 ( BT / 2 ) 2 W

3 ⎛ BT ⎞ = SNR FM pre ⎜ ⎟ 8⎝ W ⎠

2

3

With a pre-detection SNR of 12 dB, we determine the post-detection SNR as follows SNR

FM post

= SNR

FM pre

3 ⎛ 200 ⎞ ⎜ ⎟ 8 ⎝ 19 ⎠

3

= 1012 /10 × 0.375 × (10.53)3 = 6.94 × 103 ~ 38.4 dB

(The answer in the text for the lower channel is off by factor 0.5 or 3 dB.) For the upper channel, Example 9.2 indicates this result should be scaled by 2/52 and 3 2 FM 3 ⎛ 200 ⎞ SNR FM = SNR post pre ⎜ ⎟ 8 ⎝ 19 ⎠ 52 ~ 24.3 dB So the upper channel is 10log10(52/2) ≈ 14.1 dB worse than lower channel.

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Problem 9.8 An FM system has a pre-detection SNR of 15 dB. If the transmission bandwidth is 30 MHz and the message bandwidth is 6 MHz, what is the post-detection SNR? Suppose the system includes pre-emphasis and de-emphasis filters as described by Eqs. (9.63) and (9.64). What is the post-detection SNR if the f3dB of the de-emphasis filter is 800 kHz? Solution From Eq. (9.59), (see Problem 9.7), the post-detection SNR without pre-emphasis is 3

3 ⎛ BT ⎞ ⎜ ⎟ 4⎝W ⎠ ~ 15 dB + 19.7 dB = 34.7 dB

FM SNR FM post = SNR pre

From Eq. (9.65), the pre-emphasis improvement is

( 6 / 0.8) I= 3 ⎡⎣( 6 / 0.8 ) − tan −1 ( 6 / 0.8) ) ⎤⎦ 3

= 23.2 ~ 13.6 dB With this improvement the post-detection SNR with pre-emphasis is 48.3 dB.

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Problem 9.9 A sample function x(t ) = Ac cos(2πf c t ) + w(t )

is applied to a low-pass RC filter. The amplitude Ac and frequency fc of the sinusoidal component are constant, and w(t) is white noise of zero mean and power spectral density N0/2. Find an expression for the output signal-to-noise ratio with the sinusoidal component of x(t) regarded as the signal of interest. Solution The noise variance is proportional to the noise bandwidth of the filter so from Example 8.16,

[

]

E n 2 (t ) = B N N 0 =

1 N0 4 RC

and the signal power is Ac2 / 2 fir a sinusoid, so the signal-to-noise ratio is given by 2

SNR =

2

Ac 2 A RC = c N0 ⎛ N ⎞ 2⎜ 0 ⎟ ⎝ 4 RC ⎠

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Problem 9.10 A DSC-SC modulated signal is transmitted over a noisy channel, with the power spectral density of the noise as shown in Fig. 9.19. The message bandwidth is 4 kHz and the carrier frequency is 200 kHz. Assume the average received power of the signal is -80 dBm, determine the output signal-to-noise ratio of the receiver. Solution From Fig. 9.19, the noise power spectral density ate 200 kHz is approximately 5x10-19 W/Hz. Using this value for N0/2 (we are assuming the noise spectral density is approximately flat across a bandwidth of 4 kHz), the post-detection SNR is given by

SNR =

Ac2 P 2 N 0W

10−11 = 4 × 103 × 5 × 10−19 = 5 × 103 ~ 37 dB where we have used the fact that the received power is -80 dBm implies that Ac2 P / 2 = 10 −11 watts .

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Problem 9.11 Derive an expression for the post-detection signal-to-noise ratio of the coherent receiver of Fig. 9.6, assuming that the modulated signal s(t) is produced by sinusoidal modulating wave m(t ) = Am cos(2πf m t )

Perform your calculation for the following two receiver types: (a) Coherent DSB-SC receiver (b) Coherent SSB receiver. Assume the message bandwidth is fm. Evaluate these expressions if the received signal strength is 100 picowatts, the noise spectral density is 10-15 watts per hertz, and fm is 3 kHz. Solution (a) The post-detection SNR of the DSB detector is

SNR DSB =

Ac2 P A2 A2 = c m 2 N 0W 4 N 0 f m

(b) The post-detection SNR of the SSB detector is SNR SSB =

Ac2 P A2 A2 = c m 4 N 0W 8 N 0 f m

Although the SNR of the SSB system is half of the DSB-SC SNR, note that the SSB system only transmits half as much power.

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Problem 9.12 Evaluate the autocorrelation function of the in-phase and quadrature components of narrowband noise at the coherent detector input for the DSB-SC system. Assume the band-pass noise spectral density is SN(f) = N0/2 for |f-fc| < BT. Solution From Eg. (8.98), the in-phase power spectral density is (see Section 8.11) S N I ( f ) = S NQ ( f )

⎪⎧ S ( f − f c ) + S N ( f + f c ) =⎨ N ⎪⎩0 ⎧⎪ N =⎨ 0 ⎪⎩0

f < BT / 2

otherwise

f < BT / 2

otherwise

From Example 8.13, the autocorrelation function corresponding to this power spectral density is RNQ (τ ) = RN I (τ ) = N 0 BT sinc ( BTτ )

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Problem 9.13 Assume a message signal m(t) has power spectral density

⎧ f ⎪a SM ( f ) = ⎨ W ⎪0 ⎩

f ≤W

otherwise

where a and W are constants. Find the expression for post-detection SNR of the receiver when (a) The signal is transmitted by DSB-SC. (b) The signal is transmitted by envelope modulation with ka = 0.3. (c) The signal is transmitted with frequency modulation with kf = 500 hertz per volt. Assume that white Gaussian noise of zero mean and power spectral density N0/2 is added to the signal at the receiver input. Solution

(a) with DSB-SC modulation and detection, the post-detection SNR is given by SNR DSB =

Ac2 P 2 N 0W

For the given message spectrum, the power is ∞

P=

∫S

M

( f )df

−∞

W

= 2∫ a 0

f df W

= aW

where we have used the even-symmetry of the message spectrum on the second line. Consequently, the post-detection SNR is SNR

DSB

Ac2 a = 2N 0

(b) for envelope detection with ka = 0.3, the post-detection SNR is

Continued on next slide

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Problem 9.13 continued

SNR AM =

Ac2 ka2 P 2 N 0W

=

Ac2 a 2 ka 2 N0

Ac2 a = 0.09 2 N0 (c) for frequency modulation and detection with kf = 500 Hz/V, the post-detection SNR is

SNR FM =

3 Ac2 k 2f P 2 N 0W 3

Ac2 a ⎛ k f 3⎜ = 2 N 0 ⎜⎝ W

⎞ ⎟⎟ ⎠

2

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Problem 9.14 A 10 kilowatt transmitter amplitude modulates a carrier with a tone m(t) = sin(2000πt), using 50 percent modulation. Propagation losses between the transmitter and the receiver attenuate the signal by 90 dB. The receiver has a front-end noise N0 = -113 dBW/Hz and includes a bandpass filter BT = 2W = 10 kHz. What is the post-detection signal-to-noise ratio, assuming the receiver uses an envelope detector? Solution If the output of a 10 kW transmitter is attenuated by 90 dB through propagation, then the received signal level R is

R = 10 4 × 10 −90 / 10

(1)

= 10 −5 watts For an amplitude modulated signal, this received power corresponds to R=

Ac2 ( 1 + k a2 P ) 2

(2)

From Eq. (9.30), the post-detection SNR of an AM receiver using envelope detection is 2

SNR AM post =

2

AC k a P 2 N 0W

Substituting for ka, P, and Ac2 / 2 (obtained from Eq. (2)), we find SNR AM post =

ka2 P R 1 + ka2 P N 0W

10−5 0.25 × 0.5 × 1 + 0.25 × 0.5 (5 × 10−12 )(5 × 103 ) = 44.4 =

where ka = 0.5 and P = 0.5.

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Problem 9.15 The average noise power per unit bandwidth measured at the front end of an AM receiver is 10-6 watts per Hz. The modulating signal is sinusoidal, with a carrier power of 80 watts and a sideband power of 10 watts per sideband. The message bandwidth is 4 kHz. Assuming the use of an envelope detector in the receiver, determine the output signal-to-noise ratio of the system. By how many decibels is this system inferior to DSB-SC modulation system? Solution For this AM system, the carrier power is 80 watts, that is, 2

AC = 80 watts 2

(1)

and the total sideband power is 20 watts, that is, 2 AC 2 k a P = 20 watts 2

(2)

Comparing Eq.s (1) and (2), we determine that k a2 P = detection SNR of the AM system is 2

1

4

. Consequently, that post-

2

AC k a P 2 N 0W

AM SNR post =

20 10 × 4000 = 5000 ~ 37 dB =

−6

For the corresponding DSB system the post detection SNR is given by SNR DSB post = =

1 + k a2 P SNR AM post k a2 P 1 + 14 1

4

= 5 × SNR AM post ~ 7dB higher

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Problem 9.16 An AM receiver, operating with a sinusoidal modulating wave and 80% modulation, has a post-detection signal-to-noise ratio of 30 dB. What is the corresponding pre-detection signal-to-noise ratio? Solution We are given that ka = 0.80, and for sinusoidal modulation P = 0.5. A post-detection SNR of 30 dB corresponds to an absolute SNR of 1000. From Eq.(9.30), 2

SNR

AM post

2

A k P = C a 2 N 0W

1000 =

Ac2 (0.8) 2 0.5 2 N 0W

Re-arranging this equation, we obtain Ac2 = 3125 2 N 0W From Eq. (9.26 ) the pre-detection SNR is given by 2

AM SNR pre =

(

2

AC 1 + k a P 2 N 0 BT

)

Ac2 = 1 + k a2 P 2 N 0 (2W )

(

(

)

3125 1 + (0.8) 2 0.5 2 = 2062.5 =

)

where we have assumed that BT = 2W. This pre-detection SNR is equivalent to approximately 36 dB.

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Problem 9.17. The signal m(t ) = cos(400π t ) is transmitted via FM. There is an ideal band-pass filter passing 100 ≤ |f| ≤ 300 at the discriminator output. Calculate the postdetection SNR given that kf = 1 kHz per volt, and the pre-detection SNR is 500. Use Carson’s rule to estimate the pre-detection bandwidth. Solution We begin by estimating the Carson’s rule bandwidth

BT = 2(k f A + f m )

= 2(1000(1) + 200 ) = 2400 Hz

We are given that the pre-detection SNR is 500. From Section 9.7 this implies FM = SNR pre

500 =

Ac2 2 N 0 BT Ac2 1 2 N 0 2400

Re-arranging this equation, we obtain Ac2 = 1.2 × 10 6 Hz 2N 0 The nuance in this problem is that the post-detection filter is not ideal with unity gain from 0 to W and zero for higher frequencies. Consequently, we must re-evaluate the postdetection noise using Eq. (9.58) 300 ⎤ N 0 ⎡ −100 2 Avg. post - detection noise power = 2 ⎢ ∫ f df + ∫ f 2 df ⎥ Ac ⎣ −300 100 ⎦ 2N 0 = 300 3 − 100 3 3 Ac2

[

=

]

2N 0 2.6 × 10 7 3 Ac2

The post-detection SNR then becomes

Continued on next slide

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Problem 9.17 continued

SNR

FM post

=

3 Ac2 k 2f P

(

2 N 0 2.6 × 10 7

⎛ A2 = 3⎜⎜ c ⎝ 2N 0

)

2 ⎞ kf P ⎟⎟ 7 ⎠ 2.6 × 10

(

= 3 1.2 × 10 6

) (1000) 0.5 2

2.6 × 10 7

= 69230.8 where we have used the fact that kf = 1000 Hz/V and P = 0.5 watts. In decibels, the postdetection SNR is 48.4 dB.

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Problem 9.18. Suppose that the spectrum of a modulating signal occupies the frequency band f1 ≤ f ≤ f 2 . To accommodate this signal, the receiver of an FM system (without

pre-emphasis) uses an ideal band-pass filter connected to the output of the frequency discriminator; the filter passes frequencies in the interval f1 ≤ f ≤ f 2 . Determine the output signal-to-noise ratio and figure of merit of the system in the presence of additive white noise at the receiver input. Solution Since the post detection filter is no longer an ideal brickwall filter, we must revert to Eq. (9.58) to compute the post-detection noise power. For this scenario (similar to Problem 9.17)

Avg. post - detection noise power = =

N0 Ac2

⎡ − f1 2 ⎢ ∫ f df + ⎢⎣ − f 2

[

2N 0 3 f 2 − f 13 3 Ac2

f2

∫

f1

⎤ f 2 df ⎥ ⎥⎦

]

Since the average output power is still k 2f P , the post detection SNR is given by FM = SNR post

3 Ac2 k 2f P

(

2 N 0 f 23 − f13

)

For comparison purposes, the reference SNR is SNRref

Ac2 = 2 N 0 ( f 2 − f1 )

The corresponding figure of merit is Figure of merit = = =

SNR FM post SNR ref 3 Ac2 k 2f P

(

2 N 0 f 23 − f 13

)

Ac2 2 N 0 ( f 2 − f1 )

3k 2f P f 22 + f 2 f 1 + f 12

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Problem 9.19. An FM system, operating at a pre-detection SNR of 14 dB, requires a post-detection SNR of 30 dB, and has a message power of 1 watt and bandwidth of 50 kHz. Using Carson’s rule, estimate what the transmission bandwidth of the system must be. Suppose this system includes pre-emphasis and de-emphasis network with f3dB of 10 kHz. What transmission bandwidth is required in this case? Solution We are given the pre-detection SNR of 14 dB (~25.1), so Ac2 FM SNR pre = = 25.1 2 N 0 BT

and the post-detection SNR of 30 dB (~1000), so SNR

FM post

=

3 Ac2 k 2f P 2 N 0W 3

= 1000

Combining these two expressions, we obtain FM SNR post FM SNR pre

=

3k 2f PBT W3

= 39.8

Approximating the Carson’s rule for general modulation BT = 2 ( k f P1/ 2 + W ) ≈ 2k f P1/ 2 , and if we replace k 2f P with BT2 / 4 in this last equation, we obtain FM SNR post FM SNR pre

≈

3BT3 = 39.8 4W 3

Upon substituting W = 50 kHz, this last equation yields BT = 187.9 kHz.

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Problem 9.20. Assume that the narrowband noise n(t) is Gaussian and its power spectral density SN(f) is symmetric about the midband frequency fc. Show that the in-phase and quadrature components of n(t) are statistically independent. Solution The narrowband noise n(t) can be expressed as:

n(t ) = nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) = Re ⎡⎣ z (t )e j 2π fct ⎤⎦

,

where nI(t) and nQ(t) are in-phase and quadrature components of n(t), respectively. The term z(t) is called the complex envelope of n(t). The noise n(t) has the power spectral density SN(f) that may be represented as shown below

We shall denote Rnn (τ ) , RnI nI (τ ) and RnQ nQ (τ ) as autocorrelation functions of n(t), nI(t) and nQ(t), respectively. Then Rnn (τ ) = E [ n(t )n(t + τ ) ]

{

}

= E ⎡⎣ nI (t ) cos(2π f ct ) − nQ (t ) sin(2π f c t ) ⎤⎦ ⋅ ⎡⎣ nI (t + τ ) cos(2π f c (t + τ )) − nQ (t + τ ) sin(2π f c (t + τ )) ⎤⎦ 1 1 = ⎡⎣ RnI nI (τ ) + RnQ nQ (τ ) ⎤⎦ cos(2π f cτ ) + ⎡⎣ RnI nI (τ ) − RnQ nQ (τ ) ⎤⎦ cos(2π f c (2t + τ )) 2 2 1 1 − ⎡⎣ RnQ nI (τ ) − RnI nQ (τ ) ⎤⎦ sin(2π f cτ ) − ⎡⎣ RnQ nI (τ ) + RnI nQ (τ ) ⎤⎦ sin(2π f c (2t + τ )) 2 2 Since n(t) is stationary, the right-hand side of the above equation must be independent of t, this implies RnI nI (τ ) = RnQ nQ (τ ) (1)

RnI nQ (τ ) = − RnQ nI (τ )

(2)

Continued on next slide

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Problem 9.20 continued Substituting the above two equations into the expression for Rnn(τ), we have

Rnn (τ ) = RnI nI (τ ) cos(2π f cτ ) − RnQ nI (τ ) sin(2π f cτ )

(3)

The autocorrelation function of the complex envelope z (t ) = nI (t ) + jnQ (t ) is

Rzz (τ ) = E ⎡⎣ z* (t ) z (t + τ ) ⎤⎦

(4)

= 2 RnI nI (τ ) + j 2 RnQ nI (τ ) From the bandpass to low-pass transformation of Section 3.8, the spectrum of the complex envelope z is given bye ⎧⎪ S ( f + f c ) SZ ( f ) = ⎨ N ⎪⎩0

f > − fc otherwise

Since SN(f) is symmetric about fc, SZ(f) is symmetric about f = 0. Consequently, the inverse Fourier transform of SZ(f) = Rzz(τ) must be real. Since Rzz (τ ) is real valued, based on Eq. (4), we have

RnQ nI (τ ) = 0 , which means the in-phase and quadrature components of n(t) are uncorrelated. Since the in-phase and quadrature components are also Gaussian, this implies that they are also statistically independent.

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Problem 9.21. Suppose that the receiver bandpass-filter magnitude response H BP ( f )

has symmetry about ± f c and noise bandwidth BT. From the properties of narrowband noise described in Section 8.11, what is the spectral density SN(f) of the in-phase and quadrature components of the narrowband noise n(t) at the output of the filter? Show that the autocorrelation of n(t) is RN (τ ) = ρ (τ ) cos(2π f cτ ) where ρ (τ ) = F −1 [ S N ( f )] ; justify the approximation ρ (τ ) ≈ 1 for τ < 1 / BT . Solution Let the noise spectral density of the bandpass process be SH(f) then N0 2 H BP ( f ) 2 From Section 8.11, the power spectral densities of the in-phase and quadrature components are given by SH ( f ) =

⎧⎪ S ( f − f c ) + S H ( f + f c ), SN ( f ) = ⎨ H ⎪⎩0,

f ≤ BT / 2 otherwise

.

Since the spectrum SH(f) is symmetric about fc, , the spectral density of the in-phase and quadrature components is ⎧⎪ H ( f − f c ) 2 N 0 S N ( f ) = ⎨ BP ⎪⎩0

f < BT / 2

(1)

otherwise

Note that if |HBP(f)| is symmetric about fc then |HBP(f-fc)| will be symmetric about 0. Consequently, the power spectral densities of the in-phase and quadrature components are symmetric about the origin. This implies that the corresponding autocorrelation functions are real valued (since they are related by the inverse Fourier transform). In Problem 9.20, we shown that if the autocorrelation function of the in-phase component is real valued then autocorrelation of n(t) is RN (τ ) = RnI nI (τ ) cos(2π f cτ ) . If we denote 2 ρ (τ ) = Rn n (τ ) = F −1 [ S N ( f )] = NO F −1 ⎡ H BP ( f − f c ) ⎤ I I

⎣

⎦

Continued on next slide

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Problem 9.21 continued

then the autocorrelation of the bandpass noise is R N (τ ) = ρ (τ ) cos(2πf c t ) For τ 1/ BT (there is a typo in the text), we have

ρ (τ ) =

∞

∫S

N

( f ) exp ( − j 2π f τ ) df

−∞ ∞

= ∫ S N ( f ) cos ( 2π f τ ) df 0

due to the real even-symmetric nature of SN(f). If the signal has noise bandwidth BT then

ρ (τ ) ≈

BT

∫S

N

( f ) cos ( 2π f τ ) df

N

( f ) cos ( 0 ) df

N

( f )df

0

≈

BT

∫S 0

=

BT

∫S 0

= a constant where the second line follows from the assumption that τ 1/ BT . With suitable scaling the constant can be set to one.

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Problem 9.22. Assume that, in the DSB-SC demodulator of Fig. 9.6, there is a phase error φ in the synchronized oscillator such that its output is cos(2π f c t + φ ) . Find an expression for the coherent detector output and show that the post-detection SNR is reduced by the factor cos 2 φ . Solution The signal at the input to the coherent detector of Fig. 9.6 is x(t) where

x(t ) = s(t ) + nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) = Ac m(t ) cos(2π f ct ) + nI (t ) cos(2π f c t ) − nQ (t )sin(2π f ct ) The output of mixer2 in Fig. 9.6 is v(t ) = x(t ) cos(2π f c t + φ )

= [ Ac m(t ) + nI (t ) ] cos(2π f c t ) cos(2π f c t + φ ) − nQ (t ) sin(2π f c t ) cos(2π f c t + φ ) =

1 1 1 1 [ Ac m(t ) + nI (t )] cos φ + nQ (t ) sin φ + [ Ac m(t ) + nI (t )] cos(4π f ct + φ ) − nQ (t ) sin(4π f ct + φ ) 2 2 2 2

With the higher frequency components will be eliminated by the low pass filter, the received message at the output of the low-pass filter is y (t ) =

1 1 1 Ac m(t ) cos φ + nI (t ) cos φ + nQ (t ) sin φ 2 2 2

To compute the post-detection SNR we note that the average output message power in this last expression is 1 2 Ac P cos 2 φ 4

and the average output noise power is 1 1 1 ⋅ 2 N 0W cos 2 φ + ⋅ 2 N 0W sin 2 φ = ⋅ 2 N 0W 4 4 4

[

] [

]

where E n I2 (t ) = E nQ2 (t ) = N 0W . Consequently, the post-detection SNR is SNR =

1/ 4 Ac2 P cos 2 φ Ac2 P cos 2 φ = 1/ 4 ⋅ 2 N 0W 2 N 0W

Compared with (9.23), the above post-detection SNR is reduced by a factor of cos 2 φ .

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Problem 9.23. In a receiver using coherent detection, the sinusoidal wave generated by the local oscillator suffers from a phase error θ(t) with respect to the carrier wave cos(2π f c t ) . Assuming that θ(t) is a zero-mean Gaussian process of variance σ θ2 and that most of the time the maximum value of θ(t) is small compared to unity, find the meansquare error of the receiver output for DSB-SC modulation. The mean-square error is defined as the expected value of the squared difference between the receiver output and message signal component of a synchronous receiver output. Solution Based on the solution of Problem 9.22, we have the DSB-SC demodulator output is y (t ) =

1 1 1 Ac m(t ) cos [θ (t ) ] + nI (t ) cos [θ (t ) ] + nQ (t ) sin [θ (t ) ] 2 2 2

Recall from Section 9. that the output of a synchronous receiver is 1 1 Ac m(t ) + nI (t ) 2 2

The mean-square error (MSE) is defined by 2 ⎡⎛ 1 ⎞ ⎤ MSE = E ⎢⎜ y (t ) − Ac m(t ) ⎟ ⎥ 2 ⎠ ⎥⎦ ⎢⎣⎝

Substituting the above expression for y(t), the mean-square error is 2 ⎡⎡ 1 1 1 ⎤ ⎤ MSE = E ⎢ ⎢ Ac m(t ) ⎡⎣cos (θ (t ) ) − 1⎤⎦ + nI (t ) cos (θ (t ) ) + nQ (t ) sin (θ (t ) ) ⎥ ⎥ 2 2 ⎦ ⎥⎦ ⎢⎣ ⎣ 2 2 A2 1 1 = c E ⎢⎡ m 2 (t ) ⎡⎣cos (θ (t ) ) − 1⎤⎦ ⎥⎤ + E ⎡⎣ nI2 (t ) cos 2 (θ (t ) ) ⎤⎦ + E ⎡⎣ nQ2 (t ) sin 2 (θ (t ) ) ⎤⎦ ⎦ 4 4 ⎣ 4

where we have used the independence of m(t), nI(t), nQ(t), and θ(t) and the fact that E [ nI (t ) ] = E ⎡⎣ nQ (t ) ⎤⎦ = 0 to eliminate the cross terms.

Continued on next slide

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Problem 9.23 continued 2 Ac2 1 1 E ⎡⎣ m 2 (t ) ⎤⎦ E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + E ⎡⎣ nI2 (t ) ⎤⎦ E ⎡⎣cos 2 (θ (t ) ) ⎤⎦ + E ⎡⎣ nQ2 (t ) ⎤⎦ E ⎡⎣sin 2 (θ (t ) ⎤⎦ MSE = ⎣ ⎦ 4 4 4 2 2 AP 1 1 = c E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + N 0WE ⎡⎣cos 2 (θ (t ) ) ⎤⎦ + N 0WE ⎡⎣sin 2 (θ (t ) ⎤⎦ ⎣ ⎦ 4 4 4 2 2 AP NW = c E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + 0 ⎣ ⎦ 4 2

[

] [

]

where we have used the equivalences of E[m2(t)] = P, and E n I2 (t ) = E nQ2 (t ) = 2 N 0W . The last line uses the fact that cos (θ(t))+sin (θ(t)) = 1. If we now use the relation that 1-cos A = 2sin2(A/2), this expression becomes 2

2

⎡ ⎛ θ (t ) ⎞ ⎤ N 0W MSE = Ac2 PE ⎢sin 4 ⎜ ⎟⎥ + 2 ⎝ 2 ⎠⎦ ⎣ Since the maximum value of θ(t) > 2W . Solution We modify Eq. (9.58) to include the effects of a non-ideal post-detection filter in order to estimate the average post-detection noise power: N0 Ac2

∫

W

−W

f 2 | H BP ( f ) |2 df = =

N0 Ac2

∫

2 N0 Ac2

W

−W

∫

W

0

f2⋅

1 df 1 + ( f / W )4

f2⋅

1 df 1 + ( f / W )4

This can be evaluated by a partial fraction expansion of the integrand but for simplicity, we appeal to the formula:

x 2 dx 1 ∫ a + bx4 = 4bk

⎡1 x 2 − 2kx + 2k 2 2kx ⎤ + tan −1 2 log , ⎢2 2 2 x + 2kx + 2k 2k − x 2 ⎥⎦ ⎣

ab > 0, k =

4

a 2b

Using this result, we get the average post-detection noise power is Avg. post-detection noise power =

2 N0 W 3 ⋅ Ac2 4 2

⎡ ⎤ N 0W 3 2− 2 log 0.42 π + = ⎢ ⎥ Ac2 2+ 2 ⎣ ⎦

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Problem 9.25. Consider a communication system with a transmission loss of 100 dB and a noise density of 10-14 W/Hz at the receiver input. If the average message power is P = 1 watt and the bandwidth is 10 kHz, find the average transmitter power (in kilowatts) required for a post-detection SNR of 40 dB or better when the modulation is: (a) AM with ka = 1; repeat the calculation for ka = 0.1. (b) FM with kf = 10, 50 and 100 kHz per volt. In the FM case, check for threshold limitations by confirming that the pre-detection SNR is greater that 12 dB.

Solution (a) In the AM case, the post detection SNR is given by SNR AM post =

Ac2 k a2 P 2 N oW

Ac2 k a2 (1) 10 = 2(2 × 10 −14 )(10 4 ) 4

Ac2 k a2 = 2 × 10 −6 2 where an SNR of 40 dB corresponds to 104 absolute and N0/2 = 10-14 W/Hz. For the different values of ka ka = 1 ⇒

Ac2 = 4 × 10 −6

k a = 0.1 ⇒

Ac2 = 4 × 10 −4

Average modulated signal power at the input of the detector is

1 2 Ac (1 + ka2 P ) . 2

1 2 Ac (1 + ka2 P ) = 4 × 10−6 2 1 2 ka = 0.1 ⇒ Ac (1 + ka2 P ) = 2.02 × 10−4 2 ka = 1 ⇒

The transmitted power is 100dB (1010) greater than the received signal power so ka = 1 ⇒ transmitted power = 4 × 10 4 = 40 kW ka = 0.1 ⇒ transmitted power = 2.02 × 106 = 2020 kW

Continued on next slide

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Problem 9.25 continued (b) In the FM case, the post detection SNR is SNR

FM post

=

10 = 4

Ac2 k 2f 2

3 Ac2 k 2f P 2 N oW 3 3 Ac2 k 2f (1) 2(2 × 10 −14 )(10 4 ) 3

= 0.667 × 10 2

For the different values of ka A c2 k f = 10 kHz/V ⇒ = 0.667 × 10 −6 2 A2 k f = 50 kHz/V ⇒ c = 26.667 × 10 −9 2 A c2 k f = 100 kHz/V ⇒ = 0.667 × 10 −8 2 The transmitted power is 100dB (1010) greater than the received signal power so kf = 10 kHz/V ⇒ transmitted power = 0.667x104 W = 6.67 kW kf = 50 kHz/V ⇒ transmitted power = 26.667x101 W = 0.27 kW kf = 100 kHz/V ⇒ transmitted power = 0.667x102 W = 0.07 kW To check the pre-detection SNR, we note that it is given by : Ac2 Ac2 FM SNR pre = = 2 N 0 BT 4 N 0 (k f P1/ 2 + W ) where from Carson’s rule BT = 2(k f P1/ 2 + W ) . From the above Ac2 = FM = SNR pre

4 × 102 , so 3k 2f

4 × 102 102 = 3k 2f × 4 N 0 (k f P1/ 2 + W ) 3k 2f × 2 × 10−14 ( k f + 104 )

For the different values of kf, the pre-detection SNR is FM k f = 10kHz ⇒ SNR pre = 104 /12 = 29dB > 12dB FM k f = 50kHz ⇒ SNR pre = 11.11 = 10.45dB < 12dB FM k f = 100kHz ⇒ SNRpre = 1.515 = 1.8dB < 12dB

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Problem 9.25 continued Therefore, for kf = 50 kHz and 100 kHz, the pre-detection SNR is too low and the transmitter power would have to be increased by 1.55 dB and 10.2 dB, respectively.

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Problem 9.26 In this experiment we investigate the performance of amplitude modulation in noise. The MatLab script for this AM experiment is provided in Appendix 8 and simulates envelope modulation by a sine wave with a modulation index of 0.3, adds noise, and then envelope detects the message. Using this script: (a) Plot the envelope modulated signal. (b) Using the supporting function “spectra”, plot its spectrum. (c) Plot the envelope detected signal before low-pass filtering. (d) Compare the post-detection SNR to theory. Using the Matlab script given in Appendix 7 we obtain the following plots (a) By inserting the statements plot(t,AM) xlabel('Time') ylabel('Amplitude') at the end of Modulator section of the code, we obtain the following plot of the envelope modulated signal: 2 1.5 1

Amplitude

0.5 0 -0.5 -1 -1.5 -2

0

0.2

0.4

0.6

0.8

1 Time

1.2

1.4

1.6

1.8

2

Continued on next slide

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Problem 9.26 continued (b) The provided script simulates 2 seconds of the AM signal. Since the modulating signal is only 2 Hz, this is not a sufficient signal length to accurately estimate the spectrum. We extend the simulation to 200 seconds by modifying the statement t = [0:1/Fs:200]; To plot the spectrum, we insert the following statements after the AM section [P,F] = spectrum(AM,4096,0,4096,Fs); plot(F,10*log10(P(:,1))) xlabel('Frequency') ylabel('Spectrum') We use the large FFT size of 4096 to provide sufficient frequency resolution. (The resolution is Fs (1000 Hz) divided by the FFT size. We plot the spectrum of decibels because it more clearly shows the sideband components. With a linear plot, and this low modulation index, the sideband components would be difficult to see. The following figure enlarges the plot around the carrier frequency of 100 Hz.

30 20

Spectrum

10 0 -10 -20 -30 -40 -50 85

90

95 100 Frequency

105

(c) To plot the envelope-detected signal before low-pass filtering, we insert the statements (Decrease the time duration to 2 seconds to speed up processing for this part.) plot(AM_rec) xlabel('Time samples') ylabel('Amplitude') The following plot is obtained and illustrates the tracking of the envelope detector. Continued on next slide

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Problem 9.26 continued

Amplitude

0.5

0

-0.5

0

500

1000 Time samples

1500

2000

(d) To compare the simulated post detection SNR to theory. Create a loop around the main body of the simulation by adding the following statements for kk = 1:15 SNRdBr = 10 + 2*kk …. PreSNR(kk) = 20*log10(std(RxAM)/std(RxAMn-RxAM)); No(kk) = 2*sigma^2/Fs; …. SNRdBpost(kk) = 10*log10(C/error); W = 50; P = 0.5; Theory(kk) = 10*log10 ( A^2*ka^2*0.5 / (2*No(kk)*W)); end plot(PreSNR, SNRdBpost) hold on, plot(PreSNR, Theory,'g'); The results are shown in the following chart. Continued on next slide

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Problem 9.26 continued

35

30

Post-detection SNR (dB)

25

20 Simulation 15 Theory 10

5

0 10

15

20

25 Pre-detection SNR (dB)

30

35

40

These results indicate that the simulation is performing slightly better than theory? Why? As an exercise try adjusting either the frequency of the message tone or the decay of the envelope detector and compare the results.

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Problem 9.27. In this computer experiment, we investigate the performance of FM in noise. Using the Matlab script for the FM experiment provided in Appendix 8: (a) Plot the spectrum of the baseband FM phasor. (b) Plot the spectrum of the band-pass FM plus noise. (c) Plot the spectrum of the detected signal prior to low-pass filtering. (d) Plot the spectrum of the detected signal after low pass filtering. (e) Compare pre-detection and post-detection SNRs for an FM receiver. In the following parts (a) through (d), set the initial CNdB value to 13 dB in order to be operating above the FM threshold. (a) By inserting the following statements after the definition of FM, we obtain the baseband spectrum [P,F] = spectrum(FM,4096,0,4096,Fs); plot(F,P(:,1)) xlabel('Frequency (Hz)') ylabel('Spectrum') An enlarged snapshot of the spectrum near 0 Hz is shown here. It shows the tones at the regular spacing that one would expect with FM tone modulation. Note that initial plot shows the “negative frequency” portion of the spectrum just below Fs = 500 Hz. This is due to the nature of the FFT and the sampling process.

140

120

Spectrum

100

80

60

40

20

0

0

5

10

15

20 25 30 Frequency (Hz)

35

40

45

Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 9.27 continued (b) The spectrum of the bandpass FM plus noise is obtained by inserting the statements [P,F] = spectrum((FM+Noise).*Carrier,4096,0,4096,Fs); plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum') An expanded view of the result around the carrier frequency of 50 Hz is shown below. The spectrum has been plotted on a decibel scale to show both the FM tone spectrum and the noise pedestal.

20

Spectrum(dB)

0

-20

-40

-60

-80

-100

10

20

30

40 50 60 Frequency (Hz)

70

80

90

100

(c) To plot the spectrum of the noisy signal before low-pass filtering, we insert the following statements in the FM discriminator function, prior to the low pass filter [P,F] = spectrum(BBdec,1024,0,1024,Fsample/4) plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum(dB)') Continued on next slide

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Problem 9.27 continued The following plot is obtained when expanded near the origin. We plot the spectrum in decibels in order to show the noise and the non-flat nature of its spectrum more clearly. The decibel scale also illustrates some low-level distortion that has been introduced by the demodulation process as exhibited by the small second harmonic at 2 Hz and the low dc level.

60

Spectrum(dB)

50

40

30

20

10

0

5

10

15 Frequency (Hz)

20

25

30

(d) To plot the spectrum of the noisy signal before low-pass filtering, we insert the following statements in the FM discriminator function, after the low-pass filter [P,F] = spectrum(Message,1024,0,1024,Fsample/4) plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum(dB)') The following plot is obtained when expanded near the origin. Again we plot the spectrum in decibels in order to show the noise and, in this case, the effect of the lowpass filtering. The low-pass filtering does not affect the distortion introduced by the demodulator in the passband. Continued on next slide

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Problem 9.27 continued

60

Spectrum(dB)

40

20

0

-20

-40

5

10

15 Frequency (Hz)

20

25

(a) Running the code as provided produces the following comparison of the postdetection and pre-detection SNR.

45

40

Post-detection SNR (dB)

35

30

25

20

15

10

0

5

10

15

20

25

C/N (dB)

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Problem 10.1. Let H 0 be the event that a 0 is transmitted and let R0 be the event that a 0 is received. Define H1 and R1, similarly for a 1. Express the BER in terms of the probability of these events when: (a) The probability of a 1 error is the same as the probability of a 0 error. (b) The probability of a 1 being transmitted is not the same as the probability of a 0 being transmitted. Solution In both cases, the probability of error may be expressed as

P[error ] = P(R0 H 1 )P(H 1 ) + P(R1 H 0 )P(H 0 )

(1)

(a) The BER is the same as the P[error] and with P(R0|H1) = P(R1/H0) = p then P[error ] = p[P( H 1 ) + P(H 0 )] = p since P(H1)+P(H0) = 1. (b) With P(H0) ≠ P(H1), the answer is given by the general result of Eq. (1).

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Problem 10.2. Suppose that in Eq. (10.4), r(t) represents a complex baseband signal instead of a real signal. What would be the ideal choice for g(t) in this case? Justify your answer. Solution Inspecting the Schwarz inequality of Eq. (10.12), we see that equality is achieved with

g (T − t ) = cs* ( t ) if s(t) is complex.

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⎡ α (t − T / 2) ⎤ Problem 10.3 If g (t ) = c rect ⎢ , determine c such g(t) satisfies Eq. (10.10) T ⎣ ⎦⎥ where α > 1.

Solution From the definition of the rect(.) function,

⎛ α (t − T / 2) ⎞ g (t ) = c rect ⎜ ⎟ T ⎝ ⎠ t − T / 2 < T /(2α )

⎧⎪c =⎨ ⎪⎩0

otherwise

Substituting this into Eq. (10.10) 2

T = ∫ g ( t ) dt T

0

= c2 ∫

T / 2 +T /(2α )

T / 2 −T /(2α )

12 dt

= c 2T / α

And so c = α .

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Problem 10.4. Show that with on-off signaling, the probability of a Type II error in Eq.(10.23) is given by ⎛γ ⎞ P[Y > γ | H 0 ] = Q ⎜ ⎟ ⎝σ ⎠ Solution A Type II error probability is

1 P[Y > γ | H 0 ] = 2πσ Let s =

y

σ

+∞

∫γ

⎛ y2 ⎞ exp ⎜ − 2 ⎟dy ⎝ 2σ ⎠

, and then

P[Y > γ | H 0 ] =

1 2π

⎛ s2 ⎞ ⎛γ ⎞ exp ∫γ /σ ⎜⎝ − 2 ⎟⎠ds = Q ⎜⎝ σ ⎟⎠ +∞

using the definition of the Q-function given in Section 8.4.

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Problem 10.5 Prove the property of root-raised cosine pulse shape p(t) given by Eq. (10.32), using the following steps: (a) If R ( f ) is the Fourier transform representation of p(t), what is the Fourier transform representation of p(t-lT)? ∞

(b) What is the Fourier transform of q (τ ) =

∫ p(τ − t ) p(t − lT )dt ? What spectral

−∞

shape does it have? (c) What q(τ)? What is q(kT)? Use these results to show that Eq. (10.32) holds. Solution (a) From the time-shifting property of Fourier transforms (see Section 2.2 ), we have that

F[ p(t − lT )] = R( f ) exp(− j 2πflT ) (b) From the convolution property of Fourier transforms (See Section 2.2) we have that Q ( f ) = F[q (τ )] = F[ p (t )]F[ p (t − lT )]

= R 2 ( f ) exp(− j 2πflT )

(c) Since R(f) is the root-raised cosine spectrum, R2(f) is the raised cosine spectrum and so q(τ) corresponds to a raised cosine pulse. In particular, using the time-shifting property of inverse Fourier transforms q (τ ) = m (τ − lT )

where m(τ) is the raised cosine pulse shape. Using the properties of the raised cosine pulse shape (see Section 6.4) q(kT ) = m(kT − lT ) ⎧1 =⎨ ⎩0 = δ (k − l )

k =l k ≠l

and Eq. (10.32) holds.

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Problem 10.6 Compare the transmission bandwidth required for binary PAM and BPSK modulation, if both signals have a data rate of 9600 bps and use root-raised cosine pulse spectrum with a roll-off factor of 0.5. Solution

1+ β , 2T where β is the roll-off factor (0.5) and T is the symbol duration (1/9600 sec). Therefore, BT = (1+0.5)×9600 = 14.4 kHz.

For BPSK modulation (bandpass signal), the transmission bandwidth is BT = 2 ×

For binary PAM modulation (baseband signal), BT =

1+ β = 7.2 kHz. 2T

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Problem 10.7 Sketch a block diagram of a transmission system including both transmitter and receiver for BPSK modulation with root-raised cosine pulse shaping. Solution

The BPSK transmitter with root-raised cosine pulse shaping is shown in (a), and the corresponding BPSK receiver is shown in (b). (a)

Data

Modulated Impulse Train

s (t ) = ± m(t ) sin( 2πf c t )

Root Raised Cosine Filter

sin(2πf c t )

(b) Data

Root Raised Cosine Filter

r (t )

T-spaced Sample Clock

sin(2πf c t )

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Problem 10.8 Show that the integral of the high frequency term in Eq. (10.53) is approximately zero. Solution Consider the integral over the period from 0 to T of the high frequency term in Eq. (10.53):

∫

T

0

AC 2 A2 cos ( 4π f c t + 2φ (t ) ) dt = c sin ( 4π f ct + 2φ (t ) ) 2 8π f c

T 0

=

Ac 2 ⎡sin ( 4π f cT + 2φ (T ) ) − sin ( 2φ ( 0 ) ) ⎤⎦ 8π f c ⎣

<

Ac2 4π f c

where the first line follows since φ(t) is constant over a symbol interval. By the bandpass assumption fc >> 1, so this last line is small.

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Problem 10.9. Use Eqs. (10.61), (10.64), and (10.66) to show that N1 and N2 are uncorrelated and therefore independent Gaussian random variables. Compute the variance of N1-N2. Solution The correlation of N1 and N2 is

E ( N1 N 2 ) = E ⎡ 2∫ ⎣⎢ 0

T

= 2∫

T

0

=2

∫

T

0

w( s ) w(t ) cos ( 2π f1t ) cos ( 2π f 2 s ) dsdt ⎤ ⎦⎥

∫ E [ w(s)w(t )] cos ( 2π f t ) cos ( 2π f s ) dsdt T

1

0

N0 2

2

∫ ∫ δ ( t − s ) cos ( 2π f t ) cos ( 2π f s ) ds dt 1

2

= N 0 ∫ cos ( 2π f1t ) cos ( 2π f 2t ) dt T

0

=0 where the last line follows from Eq.(10.61). Since N1 and N2 are uncorrelated 2 2 2 E ⎡( N1 − N 2 ) ⎤ = E ⎡( N1 ) ⎤ + 2E [ N1 N 2 ] + E ⎡( N 2 ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2 2 = E ⎡( N1 ) ⎤ + E ⎡( N 2 ) ⎤ ⎣ ⎦ ⎣ ⎦

The variance of the N1 term is E ( N1 N1 ) = E ⎡ 2∫ ⎣⎢ 0

T

= 2∫

T

0

=2

∫

T

0

w( s ) w(t ) cos ( 2π f1t ) cos ( 2π f1s ) dsdt ⎤ ⎦⎥

∫ E [ w(s)w(t )] cos ( 2π f t ) cos ( 2π f s ) dsdt T

1

0

N0 2

1

∫ ∫ δ ( t − s ) cos ( 2π f t ) cos ( 2π f s ) ds dt 1

1

= N 0 ∫ cos 2 ( 2π f 0t ) dt T

0

Using the double angle formula 2cos2θ = 1+2cosθ, we have N T 2 E ⎡( N1 ) ⎤ = 0 ∫ (1 + cos 4π ft ) dt ⎣ ⎦ 2 0 NT = 0 2

The derivation of the variance of N2 is similar and the combined variance is N0T.

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Problem 10.10. Plot the BER performance of differential BPSK and compare the results to Fig. 10.16. Solution The bit error probability of differential BPSK is (Eq. (10.75))

⎛ E ⎞ PeDPSK = 0.5exp ⎜ − b ⎟ . ⎝ N0 ⎠ The following Matlab script plots this performance EbNodB=[0:0.25:12]; EbNo = 10.^(EbNodB/10); BER = 0.5*exp(-EbNo); semilogy(EbNodB,BER) grid xlabel('Eb/No (dB)') ylabel('BER of DPSK') axis([0 20 1E-7 0.1]) This script produces the following plot. -1

10

-2

10

-3

BER of DPSK

10

-4

10

-5

10

-6

10

-7

10

0

2

4

6

8

10 12 Eb/No (dB)

14

16

18

20

The performance of DPSK is slightly worse than BPSK and QPSK. The relative loss with DPSK is less than 1 dB at Eb/N0 of 8 dB and higher. The loss at lower Eb/N0 ratios is greater. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 10.11. A communication system that transmits single isolated pulses is subject to multipath such that, if the transmitted pulse is p(t) of length T, the received signal is s (t ) = p (t ) + α p (t − τ )

Assuming that α and τ are known, determine the optimum receiver filter for signal in the presence of white Gaussian noise of power spectral density N0/2. What is the postdetection SNR at the output of this filter? Solution We first note that the pulse is non-zero over the interval 0 ≤ t ≤ T + τ . From Section 10.2 the appropriate linear receiver is

Y =∫

T +τ

0

g (T + τ − u )r (u )du

and the optimum choice for g(t) is

g (T + τ − t ) = c ( p(t ) + α p(t − τ ) ) where c is chosen such that T +τ

∫

g (t ) dt = T + τ 2

0

With this filtering arrangement, if follows from the modified Eq. (10.9) that E ⎡⎣ N 2 ⎤⎦ =

N 0 (T + t ) 2

Continued on next slide

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Problem 10.11 continued The corresponding signal level S is T +τ

S =c

∫

g (T − t ) ( p (t ) + α p (t + τ ) ) dt

0

T +τ

=c

∫ ( p(t ) + α p(t + τ ) )

2

dt

0

= T +τ which follows from the normalization properties of c. The received signal to noise is then

S2 T +τ SNR = = 2 E ⎡⎣ N ⎤⎦ N 0 / 2 Although the units on this expression may appear unusual, note that the units of N0 are (volt)2/Hz = (volt)2-sec. The units of the numerator are also (volt)2-sec, although the (volt)2 has been suppressed. Consequently, the SNR is dimensionless, as it should be.

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Problem 10.12. The impulse response corresponding to a root-raised cosine spectrum, normalized to satisfy Eq.(10.10), is given by ⎡ (1 + α )πt ⎤ T ⎡ (1 − α )πt ⎤ cos ⎢ sin ⎢ + ⎥ ⎥⎦ 4α ⎣ T ⎦ 4αt ⎣ T g (t ) = 2 π ⎛ 4αt ⎞ 1− ⎜ ⎟ ⎝ T ⎠

where T = 1/2B0 is the symbol period and α is the roll-off factor. Obtain a discrete-time representation of this impulse response by sampling it at t = 0.1nT for integer n such that –3T < t < 3T. Numerically approximate match filtering (e.g. with Matlab) by performing the discrete-time convolution 60

q k = 0 .1 ∑ g n g k − n n = −60

where gn = g(0.1nT). What is the value of qk = q(0.1kT) for k = ±20, ±10, and 0? Solution A Matlab script for this problem is shown below. Note the starting time of -3.01 is used to avoid divide-by-zero problems. Using the filter function is just one way the discrete convolution can be performed. alpha = 0.5; B0 = 0.5; T = 1/(2*B0); t = [-3.01: 0.1 :3] * T; %-- root raised cosine impulse response g = cos( (1+alpha)*pi*t/T) + (T/4/alpha) ./ t .* sin( (1alpha)*pi*t/T); g = g ./ (1 - (4*alpha*t/T).^2 ); g = 4*alpha/pi * g; %--- discrete convolution ----q = 0.1*filter(g,1, [g zeros(1,60)]); tp = [-6.01:0.1:6] * T; stem(tp,q) xlabel('Time (T)') ylabel('Amplitude') axis([-4 4 -.2 1]), grid on Continued on next slide

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Problem 10.12 continued The plot of qk is shown below for α = 0.5 . At k = ±20 and ±10, the amplitude is approximately zero. At k = 0 the amplitude is 1.

0.8

Amplitude

0.6

0.4

0.2

0

-0.2 -4

-3

-2

-1

0 Time (T)

1

2

3

4

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Problem 10.13. Determine the discrete-time autocorrelation function of the noise sequence {Nk} defined by Eq. (10.34) ∞

N k = ∫ p(kT − t ) w(t )dt −∞

where w(t) is a white Gaussian noise process and the pulse p(t) corresponds to a rootraised cosine spectrum. How are the noise samples corresponding to adjacent bit intervals related? Solution The autocorrelation function of the noise at samples spaced by T is

RN (n) = E [ N k N k + n ] ∞ ∞ = E ⎡ ∫ p (kT − t ) w(t )dt ⋅ ∫ p ((k + n)T − s) w( s )ds ⎤ −∞ ⎣⎢ −∞ ⎦⎥

=∫

∞

=∫

∞

∫

∞

∫

∞

−∞ −∞

−∞ −∞

p(kT − t ) p((k + n)T − s )E [ w(t ) w( s) ] dtds p(kT − t ) p((k + n)T − s )

N0 δ (t − s)dtds 2

where we have interchanged integration and expectation on the third line, and the fourth line follows from the uncorrelated properties of the white noise. We next apply the sifting property of the delta function to obtain ∞

RN (n) = ∫ p (kT − t ) p ((k + n)T − t ) −∞

N0 dt 2

N0 ∞ p (kT − t ) p (t − (k + n)T )dt 2 ∫−∞ N = 0 δ (n) 2 =

where the second line follows from the even symmetry property of the raised cosine pulse, and third line follows from Eq. (10.32). Therefore, noise samples corresponding to adjacent bit intervals are not correlated.

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Problem 10.14. Draw the Gray-encoded constellation (signal-space diagram) for 16QAM and for 64-QAM. Can you suggest a constellation for 32-QAM? Solution A general hint for Gray encoding is to (a) first Gray encode two bits and assign one pair of the resulting encoding to each quadrant. (b) Gray encode the remaining bits within one of the quadrants. (c) obtain the Gray encodings for the remaining quadrants by reflecting the result across the in-phase and quadrature axes. 16-QAM constellation:

64-QAM constellation:

Continued on next slide

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Problem 10.14 continued

32-QAM constellation: (There does not appear to be a Gray encoding for 32-QAM)

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Problem 10.15. Write the defining equation for a QAM-modulated signal. Based on the discussion of QPSK and multi-level PAM, draw the block diagram for a coherent QAM receiver. Solution The QAM modulated signal can be defined as: s (t ) = ∑ ⎡⎣bkI h(t − kT ) cos(2π f c t ) + bkQ h(t − kT ) sin(2π f c t ) ⎤⎦ , k

where bkI , bkQ are different modulation levels on the I and Q channels, respectively. T is the QAM symbol duration, h(t) is the pulse shape and is nonzero during 0 ≤ t < T, and fc is the carrier frequency. The block diagram for a coherent QAM receiver is

cos(2π f c t )

sin(2π f ct )

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Problem 10.16. Show that if T is a multiple of the period of fc, then the terms sin(2π f c t ) and cos(2π f c t ) are orthogonal over the interval [t0 , T + t0 ] . Solution

∫

T + t0

t0

sin(2π f c t ) cos(2π f c t ) dt = ∫

T + t0

t0

=

1 8π f c

=−

1 sin(4π f c t )dt 2

[ − cos(4π f ct )] Tt +t

1 8π f c

0

0

[cos(4π fc (t0 + T )) − cos(4π fct0 )]

−1 sin(4π f c t0 + 2π f cT ) ⋅ sin(2π f cT ) 4π f c where we have used the equivalence cosA - cosB = 2sin[(A+B)/2]sin[(B-A)/2)]. If T is a multiple of the period of fc, then fcT = integer, and sin(2πf c T ) = 0 . =

Therefore,

∫

t 0 +T

t0

sin( 2πf c t ) cos( 2πf c t )dt = 0 . That is, sin( 2πf c t ) and cos(2πf c t ) are

orthogonal over the interval [t0, t0+T].

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Problem 10.17. For a rectangular pulse shape, by how much does null-to-null transmission bandwidth increase, if the transmission rate is increased by a factor of three? Solution Without loss of generality, consider the baseband BPSK signal: s (t ) = ∑ bk h(t − kT ), k

where T is the symbol duration, bk = +1 or -1 for transmitted 1 or 0, respectively. The pulse h(t) is rectangular, ⎛t −T /2⎞ h(t ) = rect ⎜ ⎟. ⎝ T ⎠

The Fourier transform H(f) of h(t) is H ( f ) = Tsinc( fT ) ⋅ e − j 2π fT / 2 sin(π fT ) − jπ fT . =T e π fT Inspecting a plot of the sinc function, we see the null-to-null transmission bandwidth of H(f) is B = 2/T. When the transmission rate is increased by a factor three, we have the new symbol duration T ′ = T / 3 . The null-to-null bandwidth B′ = 2 / T ′ = 3B , increased by a factor of 3.

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Problem 10.18. Under the bandpass assumptions, determine the conditions under which the two signals cos(2π f 0t ) and cos(2π f1t ) are orthogonal over the interval from 0 to T. Solution For two signals to be orthogonal over the interval from 0 to T, they must satisfy

∫

T

0

cos(2πf 0 t ) cos(2πf1t )dt = 0 .

To verify this we perform the integration as follows: T 1 T π f t π f t dt = cos(2 ) cos(2 ) [cos(2π ( f0 + f1 )t ) + cos(2π ( f0 − f1 )t )] dt 0 1 ∫0 2 ∫0 1 1 = sin(2π ( f 0 + f1 )) T0 + sin(2π ( f 0 − f1 )) 4π ( f 0 + f1 ) 4π ( f 0 − f1 )

=

T 0

1 1 sin(2π ( f 0 + f1 )T ) + sin(2π ( f 0 − f1 )T ) 4π ( f 0 + f1 ) 4π ( f 0 − f1 )

By the bandpass assumption (f0+f1) >> 1 so the first term in the last line is negligible. For the second term to be zero it must satisfy 2π ( f 0 − f 1 )T = nπ where n is an integer. This implies that (f0 - f1)= n/2T.

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Problem 10.19. Encode the sequence 1101 with a Hamming (7,4) block code. Solution Coded bit sequence c = x ⋅ G , where G is defined by (10.89).

⎡1000101 ⎤ ⎢0100111 ⎥ ⎥ c = [1101] ⋅ ⎢ ⎢0010110 ⎥ ⎢ ⎥ ⎣0001011 ⎦ = [1101001]

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Problem 10.20. The Hamming (7,4) encoded sequence 1001000 was received. If the number of transmission errors is less than two, what was the transmitted sequence? Solution The syndrome of the received sequence is S = R H where H is defined by (10.92). S = R⋅H ⎡101 ⎤ ⎢111 ⎥ ⎢ ⎥ ⎢110 ⎥ ⎢ ⎥ = [1001000] ⋅ ⎢011 ⎥ ⎢100 ⎥ ⎢ ⎥ ⎢010 ⎥ ⎢001 ⎥ ⎣ ⎦ = [110]

Based on Table 10.4, the error vector E = [0010000]. The transmitted sequence is E ⊕ R = [1011000].

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Problem 10.21. A Hamming (15,11) block code is applied to a BPSK transmission scheme. Compare the block error rate performance of the uncoded and coded systems. Explain how this would differ if the modulation strategy was QPSK.

Solution 1) For the uncoded system, the probability of a bit error with BPSK is ⎛ 2 Eb ⎞ ⎟ Pe = Q⎜ ⎜ N ⎟ 0 ⎠ ⎝ The probability of a block error with block length of 15 bits, assuming independent errors is: Pbuncoded = 1 − (1 − Pe )15

2) For the coded system, with a (15,11) Hamming code, the probability of block error is ⎛ 15 ⎞ Pbcoded = 1 − (1 − Pe′)15 − ⎜ ⎟ (1 − Pe′)14 Pe′ , ⎝1 ⎠ where Pe′ is the bit error probability of coded bit, since the code can correct a single bit error. The probability of bit error in this case is: ⎛ 2 Ec Pe′ = Q ⎜ ⎜ N 0 ⎝

⎞ ⎟⎟ , ⎠

where Ec is the coded bit energy, and Ec = 11/15Eb. Therefore ⎛ 22 Eb Pe′ = Q ⎜ ⎜ 15 N 0 ⎝

⎞ ⎟⎟ ⎠

To compare the block error probabilities of uncoded and coded systems, we use Matlab to plot the block error rate curves for Pbuncoded and Pbcoded versus Eb/N0 (dB), as shown below

Continued on next slide

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Problem 10.21 continued -1

10

-2

10

-3

Block Error Rate

10

Block of 15 uncoded bits

Hamming(15,11)

-4

10

-5

10

-6

10

-7

10

0

2

4

6

8

10 12 Eb/No (dB)

14

16

18

20

The Matlab script that generates the above plot is EbNodB=[0:0.25:12]; EbNo = 10.^(EbNodB/10); Pe = 0.5*erfc(sqrt(EbNo)); Puncoded = 1 - (1-Pe).^15; EcNo = 11/15 * EbNo; Peprime = 0.5*erfc(sqrt(EcNo)); Pcoded = 1 - (1-Peprime).^15 - 15*(1-Peprime).^14.*Peprime; semilogy(EbNodB,Puncoded) grid xlabel('Eb/No (dB)') ylabel('Block Error Rate') axis([0 20 1E-7 0.1]) hold on, semilogy(EbNodB,Pcoded,'g'), hold off

3) Since for QPSK modulation, bit error probabilities of uncoded bits Pe and coded bits Pe' are unchanged compared with BPSK modulation, the block error probabilities of two systems are also the same as those of BPSK modulation.

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Problem 10.22. Show that the choice γ = µ / 2 minimizes the probability of error given by Eq. (10.26). Hint: The Q-function is continuously differentiable. Solution From (10.26), we have the average probability of error as: Pe (γ ) =

1 ⎛ µ −γ ⎞ 1 ⎛ γ ⎞ Q⎜ ⎟ + Q⎜ ⎟ 2 ⎝ σ ⎠ 2 ⎝σ ⎠

Recall the definition of Q-function:

Q ( x) =

1

=

1

∫

+∞

2π (let u = − s ) x

2π

∫

−x

−∞

exp(− s 2 / 2)ds

exp(−u 2 / 2)du

So the derivative is given by −1 dQ( x) = exp(− x 2 / 2) ≤ 0 dx 2π Substituting this result into the definition of Pe(γ) we obtain ⎛ ( µ − γ ) 2 ⎞ −1 1 −1 ⎛ γ2 ⎞ 1 dPe (γ ) 1 −1 = ⋅ ⋅ exp ⎜ − ⋅ + ⋅ ⋅ exp ⎟ ⎜− 2 ⎟⋅ dγ 2 2π 2σ 2 ⎠ σ 2 2π ⎝ ⎝ 2σ ⎠ σ = Setting

1 2 2πσ

⎧⎪ ⎛ ( µ − γ )2 ⎞ ⎛ γ 2 ⎞ ⎫⎪ − − exp exp ⎨ ⎜ ⎟ ⎜ − 2 ⎟⎬ 2σ 2 ⎠ ⎪⎩ ⎝ ⎝ 2σ ⎠ ⎪⎭

dPe (γ ) = 0 implies dγ

⎛ ( µ − γ )2 ⎞ ⎛ γ2 ⎞ = exp ⎜ − exp ⎟ ⎜− 2 ⎟ 2σ 2 ⎠ ⎝ ⎝ 2σ ⎠ 2 2 (µ − γ ) = γ γ = µ/2

Continued on next slide

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Problem 10.22 continued

Checking the second derivative, we have ⎡ 2( µ − γ ) ⎛ ( µ − γ ) 2 ⎞ 2γ ⎛ γ 2 ⎞⎤ d 2 Pe (γ ) 1 = ⋅ − + exp exp ⎢ ⎜ ⎟ ⎜ − 2 ⎟⎥ 2 d 2γ 2σ 2 ⎠ 2σ 2 2 2πσ ⎣ 2σ ⎝ ⎝ 2σ ⎠ ⎦ >0 when γ = µ/2. Therefore at γ = µ / 2 , Pe (γ ) has a minimum value.

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Problem 10.23. For M-ary PAM, (a) Show that the formula for probability of error, namely, ⎛ M −1 ⎞ ⎛ A ⎞ Pe = 2 ⎜ ⎟Q⎜ ⎟ ⎝ M ⎠ ⎝σ ⎠ holds for M = 2, 3, and 4. By mathematical induction, show that it holds for all M.

(b) Show the formula for average power, namely, ( M 2 − 1) A2 P= 3 holds for M = 2, and 3. Show it holds for all M. Solution (a) M-ary PAM with the separation between nearest neighbours as 2A. Assume that all M symbols are equally transmitted.

(i) For M=2, we have the result given in the text for binary PAM ⎛ A⎞ Pe2 PAM = Q ⎜ ⎟ ⎝σ ⎠ M −1 ⎛ A ⎞ =2 Q⎜ ⎟ M ⎝σ ⎠

for M = 2. (ii) For M = 3, the constellation is: 1 1 P e = P [ y > − A | (−2 A) is transmitted ] + P [ y > A or y < − A | 0 is transmitted ] 3 3 1 + P [ y < A | (2 A) is transmitted ] 3 ⎛ ( y + 2 A) 2 ⎞ ⎛ y2 ⎞ 1 ∞ 1 1 ∞ 1 = ∫ + dy exp ⎜ − exp ⎟ ⎜ − 2 ⎟dy 3 − A 2πσ 2σ 2 ⎠ 3 ∫A 2πσ ⎝ ⎝ 2σ ⎠ ⎛ y2 ⎞ ⎛ ( y − 2 A) 2 ⎞ 1 −A 1 1 A 1 + ∫ exp ⎜ − 2 ⎟dy + ∫ exp ⎜ − ⎟dy 3 −∞ 2πσ 3 −∞ 2πσ 2σ 2 ⎠ ⎝ 2σ ⎠ ⎝ 4 ⎛ A⎞ = Q⎜ ⎟ 3 ⎝σ ⎠

Continued on next slide

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Problem 10.23 continued

From the formula Pe = Pe =

2( M − 1) ⎛ A ⎞ 4 ⎛ A⎞ Q⎜ ⎟ , when M=3, Pe = Q⎜ ⎟ . Thus the formula M 3 ⎝σ ⎠ ⎝σ ⎠

2( M − 1) ⎛ A ⎞ Q⎜ ⎟ holds for M = 3. M ⎝σ ⎠

(iii) For M = 4, the constellation is: - 3A

-A

+A

+ 3A

1 1 P [ y > −2 A | (−3 A) is transmitted ] + P [ y < −2 A or y > 0 | − A is transmitted ] 4 4 1 1 + P [ y < 0 or y > 2 A | + A is transmitted ] + P [ y < 2 A | (3 A) is transmitted ] 4 4 2 ⎛ ( y + 3 A) ⎞ 1 ∞ 1 = ∫ exp ⎜ − ⎟dy 4 −2 A 2πσ 2σ 2 ⎠ ⎝

P e=

⎡1 ∞ 1 ⎛ ( y + A )2 ⎞ ⎛ ( y + A)2 ⎞ ⎤ 1 −2 A 1 + 2⎢ ∫ exp ⎜ − exp ⎜ − ⎟dy ⎥ ⎟dy + ∫ 2 2 −∞ ⎜ ⎟ ⎜ ⎟ ⎥ σ σ 2 4 2 πσ 2 ⎢⎣ 4 0 2πσ ⎝ ⎠ ⎦ ⎝ ⎠ 2 ⎛ ( y − 3 A) ⎞ 1 A 1 exp ⎜ − + ∫ ⎟dy 4 −∞ 2πσ 2σ 2 ⎠ ⎝ 6 ⎛ A⎞ = Q⎜ ⎟ 4 ⎝σ ⎠ where the factor 2 in the third last line, comes from the symmetry of the second and third 2( M − 1) ⎛ A ⎞ terms of the first equation. From the formula Pe = Q⎜ ⎟ , when M=4, M ⎝σ ⎠ 6 ⎛ A⎞ 2( M − 1) ⎛ A ⎞ Pe = Q ⎜ ⎟ . Thus the formula Pe = Q⎜ ⎟ holds for M = 4. 4 ⎝σ ⎠ M ⎝σ ⎠

(iv) Assume that the formula of Pe holds for (M-1)-ary PAM. By mathematical induction, we need to show it also holds for M-ary PAM. The (M-1)-ary PAM constellation may be illustrated as shown: K

Continued on next slide

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Problem 10.23 continued

By adding one point P2 on the (M-1)-ary PAM constellation, which has the distance 2A from point P1, we obtain M-ary PAM constellation as follows (in practice, the average or dc level may be adjusted as well but this has no effect on the symbol error rate): K

Since error probabilities of P1 symbol on the (M-1)-ary PAM is the same as that of P2 point on the M-ary PAM, the error probability of M-ary PAM is PeM − ary =

M − 1 ( M −1)− ary 1 Pe + ⋅ symbol error prob. of P1 symbol on M -ary (1) M M

where 1/M is the probability that P1 is transmitted and (M-1)/M is the probability that one of the other constellation points is transmitted. The probability of error formula for (M-1)-ary PAM is given by ( M − 2) ⎛ A ⎞ (2) Pe( M −1) − ary = 2 Q⎜ ⎟ . ( M − 1) ⎝ σ ⎠ The symbol error rate of P1 symbol on M-ary PAM is PP1 =

1 P [ y < ( µ − A), or y > ( µ + A) | P1 is transmitted ] M

where µ is the signal level of P1 symbol.

⎛ ( y − µ )2 ⎞ 1 ∫−∞ exp ⎜⎝ − 2σ 2 ⎟⎠ dy + M 2 ⎛ A⎞ = Q⎜ ⎟ M ⎝σ ⎠

1 PP1 = M

µ−A

⎛ ( y − µ )2 ⎞ ∫µ + A exp ⎜⎝ − 2σ 2 ⎟⎠ dy +∞

(3)

Substituting Eqs. (2) and (3) into (1), we obtain the symbol error probability of M-ary PAM M −1 M −2 ⎛ A⎞ 2 ⎛ A⎞ ⋅2⋅ Q⎜ ⎟ + Q⎜ ⎟ M M −1 ⎝ σ ⎠ M ⎝ σ ⎠ . M −1 ⎛ A ⎞ Q⎜ ⎟ =2 M ⎝σ ⎠

PeM − PAM =

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Problem 10.23 continued

The formula holds for M-ary PAM. Therefore, by mathematical induction, the formula holds for all M.

(b) To compute the average symbol power we note: i) For M = 2, the average symbol power is A2 and the formula P =

( M 2 − 1) A2 holds for 3

M=2. ii) For M = 3, the average symbol energy is P=

1 8 (2 A) 2 + 02 + (2 A) 2 ) = A2 . ( 3 3

( M 2 − 1) A2 holds for M=3. The formula P = 3 iii) For general even M, the M-ary PAM constellation points are

{−(M − 1) A,L, −3 A, − A, A,3 A,L, (M − 1) A} . The average symbol energy is P=

2 ⎡⎣ ( M − 1) 2 + ( M − 3) 2 + L + 32 + 1⎤⎦ M

2 A2 = M

A2

M /2

∑ (2k − 1)

2

k =1

M /2 M /2 2A ⎡ 2 M /2 2 ⎤ 2 k 4 k 1⎥ − + . ∑ ∑ ∑ ⎢ M ⎣ k =1 k =1 k =1 ⎦ 2 A2 ⎡ M ( M / 2 + 1)( M + 1) M ( M / 2 + 1) M ⎤ 4 = −4 + ⎥ ⎢ 26 22 2⎦ M ⎣

=

2

( M 2 − 1) A2 = 3 where we have used the summation formulas of Appendix 6. iv) For general odd M, the M-ary PAM constellation points are

{−(M − 1) A,L, −2 A, 0, 2 A,L (M − 1) A} . Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 10.23 continued The average symbol energy is 2 ⎡ ( M − 1) 2 + ( M − 3) 2 + L + 22 ⎤⎦ 2 P= ⎣ A M 2 2 ⎤ 2 A2 2 ⎡ ⎛ M − 1 ⎞ ⎛ M − 3 ⎞ 2 2 ⎢⎜ = ⎟ +⎜ ⎟ + ... + 1 ⎥ M ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ =

8 A2 M

( M −1) / 2

∑

k2

k =1

8 A ( M − 1)( M + 1)( M ) 226 M 2 2 ( M − 1) A = 3 =

2

where the fourth line uses the summation formula found in Appendix 6.

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Problem 10.24. Consider binary FSK transmission where ( f1 − f 2 )T is not an integer. (a) What is the mean output of the upper correlator of Fig. 10.12, if a 1 is transmitted? What is the mean output of the lower correlator? (b) Are the random variables N1 and N2 independent under these conditions? What is the variance of N1 – N2? (c) Describe the properties of the random variable D of Fig. 10.12 in this case.

Solution: (a) If a 1 is transmitted,

r (t ) = Ac cos(2πf 1t ) + n(t )

where n(t) is a narrow band Gaussian noise. The output of the upper correlator is Y1: T

Y1 = ∫ r (t ) 2 cos(2πf 1t )dt 0

=∫

T

2 Ac cos(2πf1t ) cos(2πf 1t )dt + ∫

0

≅

T

0

1 2

Ac T + ∫

T

0

2n(t ) cos(2πf1t )dt

2n(t ) cos(2πf1t )dt

The expected value of Y1 is E[Y1 ] =

1 AcT , since n(t) has zero mean. 2

The output of the lower correlator is Y2: T

Y2 = ∫ r (t ) 2 cos(2πf 2 t )dt 0

=∫

T

2 Ac cos(2πf1t ) cos(2πf 2 t )dt + ∫

0

= ≅

T

0

Ac 2 Ac

∫

T

0

∫ 2

T

0

cos(2π ( f 1 + f 2 )t )dt +

Ac 2

cos(2π ( f 1 − f 2 )t )dt + ∫

T

0

∫

T

0

2n(t ) cos(2πf 2 t )dt T

cos(2π ( f1 − f 2 )t )dt + 2 ∫ n(t ) cos(2πf 2 t )dt 0

2n(t ) cos(2πf 2 t )dt

Continued on next slide

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Problem 10.24 continued where the first term of the third line is negligible due to the bandpass assumption. The expected value of Y2 is A T E[Y2 ] = c ∫ cos(2π ( f1 − f 2 )t )dt 2 0 A 1 sin ⎡⎣ 2π ( f1 − f 2 ) t ⎤⎦ T0 = c ⋅ 2 2π ( f1 − f 2 ) Ac sin(2π ( f1 − f 2 ) T ) 2 2π ( f1 − f 2 )

=

which clearly differs from the orthogonal case. (b) The random variables N1 and N2 are given by

N1 = ∫

T

0 T

N2 = ∫

0

2n(t ) cos(2πf1t )dt 2n(t ) cos(2πf 2 t )dt

Since n(t) is a Gaussian process, both N1 and N2 are Gaussian. To show N1 and N2 are correlated consider T T E[ N1 N 2 ] = E ⎡ ∫ n(t ) cos(2π f1t )dt ⋅ ∫ n(τ ) cos(2π f 2τ )dτ ⎤ 0 ⎣⎢ 0 ⎦⎥

=∫ =∫

T

T

0

∫

T

T

0

0

∫

0

E[n(t )n(τ )]cos(2π f1t ) cos(2π f 2τ )dtdτ N0 δ (t − τ ) cos(2π f1t ) cos(2π f 2τ )dtdτ 2

N0 T cos(2π f1t ) cos(2π f 2t )dt 2 ∫0 N T = 0 ∫ [ cos(2π ( f1 + f 2 )t ) + cos(2π ( f1 − f 2 )t ) ] dt 4 0 N sin ( 2π ( f1 + f 2 )t ) T sin ( 2π ( f1 − f 2 )t ) T = 0 0 + 0 4 2π ( f1 + f 2 ) 2π ( f1 − f 2 ) =

≅

N0 sinc(2( f1 − f 2 )T ) 4

Continued on next slide

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Problem 10.24 continued

where the first term of the second last line is assumed negligible due to the bandpass assumption. Since N1 and N2 are correlated, they are not independent. The variance of (N1-N2) is

var[ N1 − N 2 ] = var[ N1 ] + var[ N 2 ] − 2E[ N1 N 2 ] = N0 −

N0 sinc ( 2( f1 − f 2 )T ) 2

(c) The random variable D is Gaussian with zero mean and variance var[N1-N2].

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Problem 10.25. Show that the noise variance of the in-phase component nI(t) of the band-pass noise is the same as the band-pass noise n(t) variance; that is, for a band-pass noise bandwidth BN E ⎡⎣ nI2 (t ) ⎤⎦ = N 0 BN Solution Recall the spectra of narrowband noise n(t) and its in-phase component nI (t ) shown in ∞

Figure 8.23. The variance of a random process x(t ) = Rx (0) = ∫ X ( f )df , where X(f) is −∞

the power spectral density of x(t). Therefore, Var[n(t )] = E ⎡⎣ n 2 (t ) ⎤⎦ ∞

= ∫ S N ( f )df −∞

N0 ⋅ 2B 2 = N0 ⋅ 2B = 2⋅

Where we have used the fact that for a bandpass signal BT = 2B, that is twice the lowpass bandwidth. Similarly, the variance of the in-phase noise is Var[nI (t )] = E ⎡⎣ nI2 (t ) ⎤⎦ ∞

= ∫ S nI ( f )df −∞

= N0 ⋅ 2B

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Problem 10.26 In this problem, we investigate the effects when transmit and receive filters do not combine to form an ISI-free pulse shape. To be specific, data is transmitted at baseband using binary PAM with an exponential pulse shape g(t)=exp(-t/T)u(t) where T is the symbol period (see Example 2.2). The receiver detects the data using an integrate-and-dump detector. (a)

With data represented as ±1, what is magnitude of the signal component at the output of the detector.

(b)

What is the worst case magnitude of the intersymbol interference at the output of the detector. (Assume the data stream has infinite length.) Using the value obtained in part (a) as a reference, by what percentage is the eye opening reduced by this interference.

(c)

What is the rms magnitude of the intersymbol interference at the output of the detector? If this interference is treated as equivalent to noise, what is the equivalent signal-to-noise ratio at the output of the detector? Comment on how this would affect bit error rate performance of this system when there is also receiver noise present. (Typo in problem statement, there should be minus sign in exponential.) Solution (a) For a data pulse g (t ) = A exp( −t / T )u (t )

where A is the binary PAM symbol (±1). The desired output of an integrate-and-dump filter in the nth symbol period is ( n +1)T

∫

Gn =

g (t − nT )dt

nT T

= ∫ An exp(−t / T )dt 0

= AnT (1 − exp(−1) ) If the data is either ±1, then magnitude of the output is T(1-e-1). (b) In the nth symbol period the received signal is y (t ) =

∞

∑ A g (t − kT )

k =−∞

n

Continued on next slide

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Problem 10.26 continued The output of the detection filter in the nth symbol period is ( n +1)T

Yn =

∫

y (t )dt

nT

=

( n +1)T

∞

nT

k =−∞

∫ ∑A

( n +1)T

=

∫

k

exp ( −(t − kT ) / T )dt ∞ ( n +1)T

An exp ( −(t − nT ) / T ) dt + ∑ k =1

nT

∫

An − k exp {−(t − (n − k )T ) / T } dt

nT

where, due to the causality of the pulse shape, the symbols An+1 and later due not cause intersymbol interference into symbol An. The first term in the above is the desired signal and the second term is the intersymbol interference. By letting s = t – (n-k)T, we can express this interference as ∞ ( k +1)T

Jn = ∑ k =1

∫

An − k exp ( −t / T ) dt

kT

∞

= ∑ An − k T ( exp(− k ) − exp(−(k + 1)) ) k =1

where each term in the summation corresponds to the interference caused by a previous symbol. For worst case interference we assume that all of the An-k have the same sign. Then this worst case interference is given by ∞

J n = ∑ An − k T ( exp(− k ) − exp(−(k + 1)) ) k =1

∞

≤ T (1 − exp(−1) ) ∑ exp(− k ) k =1

To simplify the notation, we let α = exp(-1). Then ∞

J

max n

= T (1 − α ) ∑ α k k =1

= T (1 − α )

α 1−α

= αT Comparing this worst case interference to the desired signal level Gn, the eye-opening is reduced by J nmax Tα × 100 = × 100 = 58% Gn T (1 − α )

Continued on next slide

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Problem 10.26 continued (c) From part (b), we found that kth preceding symbol contributes an interference

I nk = An − k (1 − α ) α k The total interference is ∞

J n = ∑ I nk k =1 ∞

= ∑ An − k T (1 − α )α k k =1

Since all symbol intervals are equivalent, we drop the subscript n on Jn. The mean value of this interference is E[J] = 0 since E[An-k] = 0. The variance of this interference is Var ( J ) = E ⎡⎣ J 2 ⎤⎦ ∞

= ∑ E ⎡⎣ An2− k ⎤⎦ T 2 (1 − α ) α 2 k 2

k =1

α2 1−α 2 2 1−α = (α T ) 1+ α

= T 2 (1 − α ) 2

where we have assumed the symbols are independent so that E[AiAj] = 0 if i ≠ j. The rms interference is given by the square root of the variance so

1−α 1+ α = 0.25T

J rms = α T

which is clearly less than the worst case interference Jmax. If we represent the signal power by S, the noise power by N, then the equivalent signalto-noise ratio taking account of the intersymbol interference is S 2 N + J rms The intersymbol interference will further degrade performance. In fact, if the worst case interference is large enough such that the eye closes, it will result in a lower limit on the bit error rate regardless of how little noise there is. SNR =

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Problem 10.27. A BPSK signal is applied to a matched-filter receiver that lacks perfect phase synchronization with the transmitter. Specifically, it is supplied with a local carrier whose phase differs from that of the carrier used in the transmitter by φ radians. (a) Determine the effect of the phase error φ on the average probability of error of this receiver. (b) As a check on the formula derived in part (a), show that when the phase error is zero the formula reduces to the same form as in Eq. (10.44). Solution (a) With BPSK, assume the transmitted signal is (10.36): N

s (t ) = Ac ∑ bk h(t − kT ) cos(2π f c t ) , k =0

⎛ t −T / 2 ⎞ where bk = +1 for a 1 and bk = -1 for a 0, h(t) is the rectangular pulse rect ⎜ ⎟. ⎝ T ⎠

The received signal is x(t ) = s (t ) + n(t ) N

= Ac ∑ bk h(t − kT ) cos(2π f c t ) + nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) k =0

The receiver matched filter is the integrate-and-dump filter. The output for the kth symbol after down-conversion with phase error φ and match filtering is: Yk = ∫

kT

( k −1)T

x(t ) cos(2π f c t + φ )dt

[ Acbk + nI (t )] cos(2π f ct ) cos(2π fct + φ )dt − ∫( k −1)T nQ (t ) sin(2π f ct ) cos(2π fct + φ )dt ( k −1)T

=∫

kT

kT

kT 1 1 [ Ac bk + nI (t )][cos φ + cos(4π f c t + φ )]dt − ∫ n (t )[sin(4π f c t + φ ) + sin(−φ )]dt ( k −1)T 2 ( k −1)T 2 Q T 1 kT 1 kT nI (t ) cos φ dt + ∫ nQ (t ) sin φ dt ≅ Ac bk cos φ + ∫ 2 2 ( k −1)T 2 ( k −1)T T = Ac bk cos φ + N k 2

=∫

kT

where we define Nk =

1 kT 1 kT nI (t ) cos φ dt + ∫ nQ (t ) sin φ dt ∫ ( − 1) k T 2 2 ( k −1)T

Continued on next slide

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Problem 10.27 continued

The random variable Nk has zero mean and variance var[ N k ] = cos 2 φ =

Let µ =

N 0T NT + sin 2 φ 0 4 4

N 0T =σ2 4

T Ac cos φ . Then the probability of bit error Pe is 2

Pe = P[bk = 1]P[Yk < 0 | bk = 1] + P[bk = −1]P[Yk > 0 | bk = −1] ⎧ ( y − µ )2 ⎫ ⎧ ( y + µ )2 ⎫ 1 0 1 1 +∞ 1 exp dy exp − + ⎨ ⎬ ⎨− ⎬dy 2 ∫−∞ 2πσ 2 ∫0 σ2 ⎭ σ2 ⎭ 2πσ ⎩ ⎩ ⎛µ⎞ = Q⎜ ⎟ ⎝σ ⎠

=

with Eb =

⎛ 2 Eb cos φ Ac2T 1 T , µ = Ac cos φ , σ = N 0T , we have Pe = Q ⎜⎜ 2 2 2 N0 ⎝

⎛ 2 Eb (b) When the phase error φ=0, Pe = Q ⎜⎜ ⎝ N0

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ , as the same as Eq. (10.44). ⎠

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Problem 10.28. A binary FSK system transmits data at the rate of 2.5 megabits per second. During the course of transmission, white Gaussian noise of zero mean and power spectral density 10-20 watts per hertz is added to the signal. In the absence of noise, the amplitude of the received signal is 1 µV across 50 ohm impedance. Determine the average probability of error assuming coherent detection of the binary FSK signal. Solution The average probability of error for coherent FSK is ⎛ Eb ⎞ Pe = Q ⎜⎜ ⎟⎟ ⎝ N0 ⎠

from Eq. (10.68). For this example, we have noise power spectral density is N 0 = 2 × 10 −20 watts / Hz

and the energy per bit is Eb =

1 Ac2T , 2 R

In the text, we have nominally assumed the resistance is 1 ohm and omitted it. In this problem we use the resistance of R = 50 ohms. The symbol duration is 1 seconds and the amplitude of received signal is Ac = 1µV. Therefore, T= 2.5 × 106

1 1×10−12 1 × Eb = × 2 50 2.5 ×106 = 4 ×10−21 watts / Hz Substituting the above values into the expression for Pe and we have the probability of error is Pe = Q ( 0.2 ) ≅ 0.26

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Problem 10.29. One of the simplest forms of forward error correction code is the repetition code. With an N-repetition code, the same bit is sent N times, and the decoder decides in favor of the bit that is detected on the majority of trials (assuming N is odd). For a BPSK transmission scheme, determine the BER performance of a 3-repetition code. Solution With 3-repetition code, the decoder will output the correct bit if there are one or fewer errors in the 3-bit code. Thus, assuming bit errors are independent, the bit error rate is ⎛ 3⎞ Pbcoded = (1 − Pe )3 + ⎜ ⎟ Pe (1 − Pe ) 2 , ⎝1 ⎠ = (1 − Pe ) 2 (1 + 2 Pe ) where Pe is the bit error rate of channel bit. With BPSK, the formula for bit error probability is ⎛ 2 Ec ⎞ Pe = Q ⎜⎜ ⎟⎟ ⎝ N0 ⎠ , ⎛ 2 Eb ⎞ = Q ⎜⎜ ⎟⎟ ⎝ 3N 0 ⎠ since ratio of channel bit energy to information bit energy is given by Ec = 1/3Eb. Therefore, the bit error probability of the 3-repetition code is coded b

P

⎛ ⎛ 2 Eb = ⎜1 − Q ⎜⎜ ⎜ ⎝ 3N0 ⎝

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

2

⎛ ⎛ 2 Eb ⎜ 1 + 2Q ⎜⎜ ⎜ ⎝ 3N 0 ⎝

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

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Problem 10.30 In this experiment, we simulate the performance of bipolar signalling in additive white Gaussian noise. The Matlab script included in Appendix 7 for this experiment: • generates a random sequence with rectangular pulse shaping • adds Gaussian noise • detects the data with a simulated integrate-and-dump detector With this Matlab script (a) Compute the spectrum of the transmitted signal and compare to the theoretical. (b) Explain the computation of the noise variance given an Eb/N0 ratio. (c) Confirm the theoretically predicted bit error rate for Eb/N0 from 0 to 10 dB. Solution (a) The provided script plots the simulated spectrum before noise is added. If we add the statement hold on, plot(F, abs(2*sinc(F)).^2,'g'), hold off at the same point, we obtain the following comparison graph. The two graphs agree reasonably well. There are two reasons for the differences observed with the simulated spectrum. The first is the relatively short random sequence used for generating the plot and the second is an aliasing effect. 4.5 4 3.5

Spectrum

3 2.5 2 1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1 1.2 Frequency

1.4

1.6

1.8

2

Continued on next slide

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Problem 10.30 continued

(b) The calculation of the noise variance in a discrete time simulation proceeds as follows. We are given the sampling rate Fs and the required Eb/N0 to simulation. We then note that ∞

Eb =

∫

2

p (t ) dt (1)

−∞

≈ ∑ pk Ts 2

where p(t) is the pulse shape, {pk} is its sample version and Ts = 1/Fs is the sample interval. On the other hand, if generate noise of variance σ2, due to Nyquist considerations this can only be distributed over a bandwidth Fs, thus the noise spectral density is N0 σ 2 = 2 Fs s

(2)

Re-arranging Eq. (2) and substituting Eq. (1) and the knowns, we have

σ2 =

N0 Fs 2 −1

⎛F =⎜ s ⎝ 2

⎞ ⎛ Eb ⎞ ⎟ ⎜ N ⎟ Eb ⎠⎝ 0 ⎠

⎛F =⎜ s ⎝ 2

⎞ ⎛ Eb ⎞ ⎟⎜ N ⎟ ⎠⎝ 0 ⎠

1⎛ E ⎞ = ⎜ b⎟ 2 ⎝ N0 ⎠

−1

−1

∑p

k

∑p

2

Ts

2

k

which agrees with what is used in the script (except that in the script we have suppressed Fs and Ts, knowing they would cancel).

Continued on next slide

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Problem 10.30 continued

(c) To compute the bit error rate for 0 to 10 dB, we add the following statements around the provided script for kk = 0:10 Eb_N0 = 10^(kk/10); Nbits = 100000; % increase for higher Eb/N0 …(provided script) BER(kk+1) = Nerrs/Nbits end semilogy([0:10], BER) xlabel('Eb/No (dB)') ylabel('BER') grid on hold on, semilogy([0:10], 0.4*erfc(sqrt(10.^([0:10]/10))),'g') The following plot is then produced by the Matlab script which shows good agreement between theory and simulation. -1

10

Simulation

-2

10

Theory -3

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 10.31 In this experiment, we simulate the performance of bipolar signalling in additive white Gaussian noise but with root-raised-cosine pulse shaping. A Matlab script is included in Appendix 7 for doing this. With this simulation: (a) Compute the spectrum of the transmitted signal and compare to the theoretical. Also compare to the transmit spectrum with rectangular pulse shaping (b) Plot the eye diagram of the received signal under no noise conditions. Explain the relationship of the eye opening to bit error rate performance. (c) Confirm the theoretically predicted bit error rate for Eb/N0 from 0 to 10 dB. Solution (a) We compare the spectra by inserting the following statements prior to noise being added to the signal [P,F] = spectrum(S,256,0,Hanning(256),Fs); plot(F,P(:,1)); midpt = floor(length(F)/2); hold on, plot(F, abs([(1+cos(pi*F(1:midpt)))/2; 0*F(midpt+1:end)]),'g'), hold off xlabel('Frequency'), ylabel('Spectrum') The comparison plot is shown below. 1.4

1.2

Spectrum

1

0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1 1.2 Frequency

1.4

1.6

1.8

2

(b) To plot the eye diagram we eliminate the noise by setting Eb/N0 to a high value Eb_N0 = 2000; Then running the Matlab script produces the following eye diagram. Continued on next slide

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Problem 10.31 continued 1.5

1

Amplitude

0.5

0

-0.5

-1

-1.5 -1

-0.8

-0.6

-0.4

-0.2 0 0.2 Symbol periods

0.4

0.6

0.8

1

(c) We simulate the bit error rate by commenting out the plotting statements and adding a set of statements similar to those used in Problem 10.30. -1

10

Simulation

-2

10

Theory

-3

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 10.32 In this experiment, we simulate the effect of various mismatches in the communication system and their effect on performance. In particular, modify the MatLab scripts of the two preceding problems to: (a) Simulate the performance of a system using rectangular pulse shaping at the transmitter and raised cosine pulse shaping at the receiver. Comment on the performance degradation. (b) In the case of matched root-raised cosine filtering, include a complex phase rotation exp(jθ) in the channel. Plot the resulting eye diagram for θ being the equivalent of 5, 10, 20, and 45°. Compare to the case of 0°. Do likewise for the BER performance. What modification to the theoretical BER formula would accurately model this behaviour? Solution (a) We can create this mismatch by inserting the statements: pulseTx = ones(1,Fs); pulseRx = [ 0.0064 0.0000 -0.0101 0.0000 0.0182 -0.0000 -0.0424 ... 0.0000 0.2122 0.5000 0.6367 0.5000 0.2122 -0.0000 ... -0.0424 0.0000 0.0182 -0.0000 -0.0101 0.0000 0.0064 ]; Delay = floor((length(pulseTx)-1)/2 + (length(pulseRx)-1)/2 + 1); Eb = sum(pulseTx.^2); And by modifying the statements S = filter(pulseTx,1,[b_delta zeros(1,Delay)]); De = filter(pulseRx,1,R); Then we obtain the performance shown below. Part of the loss seen is due to the filter mismatch but part of it is also due to a timing error; with the arrangement of the simulation the optimum sampling point for the data falls between the discrete samples. This sampling time loss could be recovered by interpolation.

Continued on next slide

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Problem 10.32 continued -1

10

Simulation

-2

10

-3

Theory

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

(b) Implementation of the phase rotation requires simulation of the complete complex baseband. To do this we must modify the channel portion of the simulation to the following %--- add Gaussian noise ---Noise = sqrt(N0/2)*(randn(size(S))+j*randn(size(S))); R = S + Noise; R = R*exp(j*10/180*pi); R = real(R); Where we have now included the quadrature component of the noise. Note the receiver only uses the in-phase portion (real part) of the signal to characterize this degradation. The resulting performance for rotations of 10, 20 and 45° are shown below. Note that the 45° rotation results in a 3 dB loss in performance.

Continued on next slide

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Problem 10.32 continued 0

10

-1

10

BER

45 degrees

20 deg.

-2

10

10 deg.

Theory

-3

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 11.1 What is the root-mean-square voltage across a 10 Mega-ohm resistor at room temperature if measured over a 1 GHz bandwidth? What is the available noise power? Solution Following Example 11.2, the available noise power is PN = kTBN = 1.38 ×10 − 23 × 290 ×10 9 = 4 ×10 −12 watts

The root-mean-square voltage across a 10 mega-ohm resistor is

Vrms = PN R = 4 × 10 −12 × 10 7 = 6.3 millivolts

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Problem 11.2 What is the available noise power over 1 MHz due to shot noise from a junction diode that has a voltage differential of 0.7 volts and carries average current of 0.1 milliamperes, if the current source of the Norton equivalent circuit has a resistance of 250 ohms? Solution From Eq. (11.9), the saturation current for a junction diode is given by

IS =

I ⎛ qV ⎞ exp⎜ ⎟ −1 ⎝ RT ⎠

= 1.8 × 10 −12 I Consequently, the noise contribution from the saturation current may be ignored. From Eq. (11.10) the expected current variance is then E ⎡⎣ I shot 2 ⎤⎦ = 2q ( I + 2 I s ) BN ≈ 2qIBN

(

) (

)

= 2 × 1.6 ×10−19 × 0.1× 10−3 × 106 = 3.2 × 10−17 Amp 2

The corresponding noise power with an equivalent resistance of 250 ohms is

[

2

= 8 × 10

−15

]

PN = E I shot R watts

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Problem 11.3 An electronic device has a noise figure of 10 dB. What is the equivalent noise temperature? Solution From Eq. (11.7), the equivalent noise temperature is

Te = T0 (F − 1)

= 290(10 − 1)

= 2610 o K

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Problem 11.4. The device of Problem 11.3 has a gain of 17 dB and is connected to a spectrum analyzer. If the input to the device has an equivalent temperature of 290°K and the spectrum analyzer is noiseless, express the measured power spectral density in dBm/Hz. If the spectrum analyzer has a noise figure of 25 dB, what is the measured power spectral density in this case? Solution For the device of Problem 11.3, the total output noise is, from Eq. (11.15), N = Gk (T + Te ) BN = GkTFBN

(

)(

)

(

)

= 1017 /10 1.38 × 10−23 ( 290 ) 1010 /10 BN = 2.0 ×10−18 BN

The noise spectral density at the device output is approximately N = 2.0 × 10−18 W/Hz BN ~ -147 dBm/Hz

If the spectrum analyzer has a noise figure of 25 dB, then we must use the results of the following section. Specifically, Eq. (11.21), to obtain the total noise figure of F = F1 +

(

F2 G1

)

= 1010 /10 +

1025 /10 1017 /10

= 16.3 ~ 12.1 dB

Since the overall noise figure is increased 2.1 dB by the spectrum analyzer, the noise spectral density at the spectrum analyzer output is (assuming unity gain for the spectrum analyzer) -144.9 dBm/Hz. (There is an error in the second answer given in the text.)

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Problem 11.5 A broadcast television receiver consists of an antenna with a noise temperature of 290°K and a pre-amplifier with a gain of 20 dB and a noise figure of 9 dB. A second-stage amplifier in the receiver provides another 20 dB of gain and has a noise figure of 20 dB. What is the noise figure of the overall system? Solution From Eq (11.21), after converting from decibels to absolute F = F1 +

F2 − 1 F3 − 1 + G1 G1G2

7.94 − 1 99 + 1 100 = 2 + 6.94 + .99 = 9.93 = 2+

Converting this result back to decibels, the overall noise figure is 9.97 dB.

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Problem 11.6 A satellite antenna has a diameter of 4.6 meters and operates at 12 GHz. What is the antenna gain if the aperture efficiency is 60%? If the same antenna was used at 4 GHz what would be the corresponding gain? Solution From Eq (11.25), the antenna gain is

G=

4πAeff

λ2

The effective area is given by Eq.(11.24) Aeff = ηA =η

πd 2

4 = 9.97 m 2

where the efficiency is 60% and the diameter is 4.6 meters. At 12 GHz, the wavelength λ = c/f = 0.025 meters. Consequently, the antenna gain is G = 2000458.7 ~ 53.0 dB

With a transmission frequency of 4 GHz, the wavelength λ = 0.075 m and the antenna gain is G = 22273.2 ~ 43.5 dB

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Problem 11.7 A satellite at a distance of 40,000 kilometers transmits a signal at 12 GHz with an EIRP of 10 watts towards a 4.6 meter antenna that has an aperture efficiency of 60%. What is the received signal level at the antenna output? Solution From Eq.(11.32), the path loss due to free-space transmission of a 12 GHz signal over 40,000 kilometers is

⎛ 4πr ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠ ⎛ 4π 40 × 10 6 ⎞ ⎟⎟ = 20 log10 ⎜⎜ ⎝ 0.025 ⎠ ~ 206.1 dB Substituting this result in Eq (11.29), the received power is PR = EIRP − LP + GR = 10 dBW − 206.1 dB + 53.0 dB = −143 dBW

where we have used the antenna gain of 53 dB from Problem 11.6.

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Problem 11.8 The antenna of Problem 11.7 has a noise temperature of 70°K and is directly connected to a receiver with an equivalent noise temperature of 50°K and a gain of 60 dB. What is the system noise temperature? If the transmitted signal has a bandwidth of 100 kHz, what is the carrier-to-noise ratio? If the digital signal has a bit rate of 150 kbps, what is the Eb/N0? Solution From Eq. (11.22), the combined system noise temperature is

Trx 1 o = 70 + 50 o

Ts = Tant +

= 120 o K where the electrical gain of the antenna is 1. For a bandwidth of 100 kHz the available noise power is N = kTs B = 1.38 × 10 −23 ⋅ 120 ⋅ 10 5 = 1.66 × 10 −16 watts ~ −157.8 dBW

Comparing to the result for Problem 11.7, we have that the C/N is 14.8 dB. To convert the C/N0 to an Eb/N0, we use the formula E C = b ×R N0 N0

where the bit rate R relates the energy per bit Eb to the power C. In decibels, ⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠ dB − Hz ⎝ N 0 ⎠ dB

Continued on next slide

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Problem 11.8 continued

Re-arranging, we have

⎛ Eb ⎞ ⎛ C ⎞ − 10 log10 R ⎜ ⎟ =⎜ ⎟ ⎝ N 0 ⎠dB ⎝ N 0 ⎠ dB − Hz ⎛C⎞ = ⎜ ⎟ + 10 log BN − 10 log10 R ⎝ N ⎠ dB = 14.8 + 50 − 51.8 = 13 dB

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Problem 11.9 Transmitting and receiving antennas for a 4 GHz signal are located on top of 20 meter towers separated by 2 kilometers. For free-space propagation, what is the maximum height permitted for an object located midway between the two towers? Solution The radius of the first Fresnel zone with d1 = d2 = 1 kilometer and λ = c/f = 0.075 m is

h= =

λd1d 2 d1 + d 2

(.075)(1000)(1000) 2000

= 6.1 m Consequently, the maximum height of intermediate object is 20 m – 6.1m = 13.9 m, if we require free-space propagation conditions.

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Problem 11.10 A measurement campaign indicates that the median path loss at 900 MHz in a suburban area may be modeled with a path-loss exponent of 2.9. What is the median path loss at a distance of 3 kilometers using this model? How does this loss compare to the free-space loss at the same distance? Solution From Eq (11.37), the free-space path loss at one meter, with a transmission frequency of 900 MHz, is 2

⎛ λ ⎞ ⎛ .333 ⎞ β0 = ⎜ ⎟ =⎜ ⎟ = .000704 ⎝ 4π r0 ⎠ ⎝ 4π ⎠ 2

with r0 = 1 m. From Eq.(11.37), the path loss with the terrestrial propagation model is

β0 PR = PT ( r / r0 )n =

.0007

( 3000 )

2.9

= 5.8 × 10−14 ~ − 132 dB The free-space loss over the same distance is given by

PR ⎛ λ ⎞ =⎜ ⎟ PT ⎝ 4π r ⎠

2

= 7.8 × 10−11 ~ − 101.1 dB or 31 dB less.

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Problem 11.11 Express the true median of the Rayleigh distribution as a fraction of the Rrms value? What is the decibel error in the approximation Rmedian ≈ Rrms? Solution The median of the distribution satisfies P [ R < r ] = 0.5 . Consequently, from Eq. (11.38) we have that the median r satisfies

⎧− r2 ⎫ 1 − exp⎨ 2 ⎬ = 1 2 ⎩ Rrms ⎭ Solving for r we obtain r2 − 2 = ln 1 2 Rrms r = Rrms ln 1

2

= 0.83Rrms

Consequently, there is a 20log10(0.83) =1.62 dB error when using the rms value of the amplitude instead of the median value.

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Problem 11.12 Compute the noise spectral density in watts per hertz of: (a) an ideal resistor at nominal temperature of 290°K; (b) an amplifier with an equivalent noise temperature of 22,000°K. Solution (a) From Eq. (11.19), the noise power spectral density is N 0 = kTe = 1.38 × 10−23 × 290 = 4.0 × 10−21 W/Hz

(b) From Eq. (11.19), the noise power spectral density is N 0 = kTe = 1.38 × 10−23 × 22000 = 3.04 × 10−19 W/Hz

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Problem 11.13 For the two cases of Problem 11.12, compute the pre-detection SNR when the received signal power is: (a) -60 dBm and the receive bandwidth is 1 MHz; (b) -90 dBm and the receive bandwidth is 30 kHz. Express the answers in both absolute terms and decibels. Solution (a) The signal power is obtained by converting -60 dBm to watts S = 10( −60 /10) = 10−6 mW = 10−9 W

The noise power from the ideal resistor is from Eq. (11.13) N = kTe BN = 4.0 × 10−21 × (106 ) = 4.0 × 10−15 W The SNR is the ratio of the two

SNR =

S 10−9 = = 2.5 ×105 ~ 54 dB N 4.0 ×10−15

A similar calculation for the amplifier of the previous problem results in 10−9 S = 2.94 × 103 ~ 34.7 dB SNR = = −19 6 N 3.04 ×10 ×10

(b) The signal power is obtained by converting -90 dBm to watts S = 10( −90 /10) = 10−9 mW = 10−12 W

The noise power from the ideal resistor is from Eq. (11.13)

Continued on next slide

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Problem 11.13 continued

N = kTe BN = 4.0 × 10−21 × (30 ×103 ) = 1.2 ×10−16 W The SNR is the ratio of the two 10−12 S = 8.3 × 103 ~ 39.2 dB SNR = = −16 N 1.2 ×10 A similar calculation for the amplifier of the previous problem results in

SNR =

S 10−12 = = 1.1×102 ~ 20.4 dB −19 3 N 3.04 ×10 × 30 ×10

(

)

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Problem 11.14 A wireless local area network transmits a signal that has a noise bandwidth of approximately 6 MHz. If the signal strength at the receiver input terminals is –90 dBm and the receiver noise figure is 8 dB, what is the pre-detection signal-to-noise ratio? Solution The signal power is obtained by converting -90 dBm to watts S = 10( −90 /10) = 10−9 mW = 10−12 W

The noise power with an 8 dB noise figure F is from Eqs. (11.15) and (11.16) N = kT0 FB = 1.38 × 10−23 × (290) × 108/10 × (6 × 106 ) = 1.52 × 10−13 W The pre-detection SNR is the ratio of the two S 10−12 SNR = = = 6.6 ~ 8.2 dB N 1.52 ×10−13

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Problem 11.15 A communications receiver includes a whip antenna whose noise temperature is approximately that of the Earth, that is, 290°K. The receiver pre-amplifier has a noise figure of 4 dB and a gain of 25 dB. What is the equivalent noise temperature of the antenna and the pre-amplifier? What is the combined noise figure? Solution (a) Following Example 11.4, the combined noise temperature of the antenna and preamplifier is, from Eq. (11.17) Tsys = Tant + Tamp

= 290 + 290( F − 1) = 728K

(b) From Eq. (11.16), the combined noise figure is Fcomb =

T + Te 290 + 728 = = 3.51 ~ 5.45dB T 290

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Problem 11.16 A parabolic antenna with a diameter of 0.75 meters is used to receive a 12 GHz satellite signal. What is the gain in decibels of this antenna? Assume the antenna efficiency is 60%. Solution From Eq.(11.25), the antenna gain is

GR =

4π Aeff

(1)

λ2

The signal wavelength is λ = c f = 3 × 108 12 × 109 = 0.025m , and the effective area is Aeff = η

πd2 4

= 0.60

π (.75) 2 4

Substituting these two results into Eq. (1), the antenna gain is GR =

4π

( 0.025)

× ( 0.6 ) 2

π (0.75) 2 4

= 5.33 × 103 ~ 37.3dB

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Problem 11.17 If the system noise temperature of a satellite receiver is 300°K, what is the required received signal strength to produce a C/N0 of 80 dB? Solution (There is a typo in problem statement, the units should be “dB-Hz”.) From Eq. (11.19), the noise power spectral density is N 0 = kTs = 1.38 × 10−23 × (300) = 4.14 ×10−21 W/Hz ~ -203.8 dBW/Hz

In decibels, the carrier to noise density is given by

( C / N0 )dB − Hz = (C )dBW − ( N0 )dBW − Hz 80 = ( C )dBW − (−203.8) Solving for C, we obtain C = -123.8 dBW = -93.8 dBm.

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Problem 11.18 If a satellite is 40,000 km from the antenna of Problem 11.16, what satellite EIRP will produce a signal strength of –110 dBm at the antenna terminals? Assume the transmission frequency is 12 GHz. Solution The received power is given by Eq. (11.29) PR = EIRP + GR − LP

(1)

where all quantities are in decibels. From Problem 11.16, the antenna gain is GR = 37.3 dB. The free-space path loss is given by Eq. (11.32) ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

From Problem 11.16, the wavelength is λ = 0.025m at 12 GHz. So, at a distance r = 40,000 km, the path loss is

⎛ 4π (40000 ×103 ⎞ LP = 20 log10 ⎜ ⎟ = 206.1 dB 0.025 ⎝ ⎠ Substituting these in Eq.(1) with a received power of -110 dBm, we obtain −110 dBm = EIRP + 37.3 dB − 206.1dB

Solving this equation, we find the require EIRP is 58.8 dBm.

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Problem 11.19 Antennas are placed on two 35-meter office towers that are separated by ten kilometers. What is the minimum height of a building between the two towers that would disturb the assumption of free-space propagation? Solution From Eq. (11.35), the radius of the first Fresnel zone is

h=

λ d1d 2 d1 + d 2

This radius is maximized midway between the two towers and must be kept clear to approximate free-space propagation. With d1 = d2 = 5km, the radius in meters is h = 2500λ = 50 λ

The maximum building height (in meters) is

b = 35 − h = 35 − 50 λ For example, at a transmission frequency of 4 GHz, the maximum height is b = 21.3 m.

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Problem 11.20 If a receiver has a sensitivity of –90 dBm and a 12 dB noise figure what is minimum pre-detection signal-to-noise ratio of an 8 kHz signal? Solution The noise in an 8 kHz bandwidth for a receiver with an 8 dB noise figure is, from Eqs. (11.15) and (11.16),

N = kT0 FB

(

)

= 1.38 × 10−23 × (290) × 1012 /10 × (8 × 103 ) = 5.07 × 10−16 W The receiver sensitivity is defined as the minimum received signal power that will provide a demodulated signal with acceptable performance, thus the minimum signal power is S = -90 dBm ~ 10-12 W. The minimum pre-detection SNR is the ratio of the two S 10−12 SNR = = = 1.97 × 103 ~ 32.9 dB −16 N 5.07 ×10

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Problem 11.21 A satellite antenna is installed on the tail of an aircraft and has a noise temperature of 100°K. The antenna is connected by a coaxial cable to a low-noise amplifier in the equipment bay at the front of the aircraft. The cable causes 2 dB attenuation of the signal. The low-noise amplifier has a gain of 60 dB and a noise temperature of 120°K. What is the system noise temperature? Where would a better place for the low-noise amplifier be? Solution Following Example 11.4, the system noise temperature is

Ts = Tant + = 100 +

Tcable Tamp + Gant Gcable 290 120 + 1 .631

= 580K where we have used the facts that the antenna does not provide any electrical gain, thus Gant = 1; and the fact the fact that cable causes a 2 dB loss so Gcable = 10-2/10 = 0.631. Locating the low-noise amplifier in the tail of the aircraft, close to the antenna would be a better system design. With the amplifier in the antenna tail, the system noise temperature would be approximately 220 K.

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Problem 11.22 A wireless local area network transmitter radiates 200 milliwatts. Experimentation indicates that the path loss may be accurately described by

Lp = 31 + 33 log10(r) where the path loss is in decibels and r is the range in meters. If the minimum receiver sensitivity is –85 dBm, what is the range of the transmitter? Solution Since the problem says nothing about the transmit and receive antennas, we shall assume they are omni-directional with a gain of 0 dB. In this case, the Friis equation (in decibels) for the received signal strength reduces to

PR = PT − LP

= PT − ( 31 + 33log10 r )

(1)

With a transmit power of 200 mW, equivalent to 23 dBm, and a minimum signal strength of -85 dBm, Eq. (1) becomes −85 = 23 − ( 31 + 33log10 r )

Solving this equation for the maximum range, we find r is 215.4 meters.

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Problem 11.23 A mobile radio transmits 30 watts and the median path loss may be approximated by

Lp = 69 + 31 log10(r) where the path loss is in decibels and r is the range in kilometers. If the receiver sensitivity is -110 dBm and 12 dB of margin must be included to compensate for variations about the median path loss, what is the range of the transmitter? Solution Since the problem says nothing about the transmit and receive antennas, we shall assume they are omni-directional with a gain of 0 dB. In this case, the Friis equation for the received signal strength reduces to

PR = PT − LP − L0

= PT − ( 69 + 31log10 r ) − L0

(1)

where L0 represents the required margin. A transmit power of 30 W is equivalent to 14.8 dBW, and a minimum signal strength of -110 dBm is equivalent to -140 dBW. Thus, Eq. (1) becomes −140 = 14.8 − ( 69 + 31log10 r ) − 12

Solving this equation for the maximum range, we find r is 240.2 kilometers. In practice, the range will likely be somewhat less than this due to the curvature of the earth and depending on the height of the base station antenna.

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Problem 11.24 A cellular telephone transmits 600 milliwatts of power. If the receiver sensitivity is –90 dBm, what would the range of the telephone be under free space propagation? Assume the transmitting and receiving antennas have unity gain and the transmissions are at 900 MHz. If propagation conditions actually show a path-loss exponent of 3.1 with a fixed loss β = 36 dB, what would the range be in this case? Solution (a) The Friis equation for the received power in decibels is PR = PT + GR + GT − LP

(1)

where the antenna gains are GR = GT = 0 dB. The transmit power of 600 mW is equivalent to or 27.8 dBm. For free-space propagation, the path loss is ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

At 900 MHz, the wavelength is λ = c f = 3 × 108 900 × 106 = 0.33 m . Making these substitutions, we have ⎛ 4π r ⎞ −90 = 27.8 + 0 + 0 − 20 log10 ⎜ ⎟ ⎝ 0.33 ⎠

Solving this equation for the maximum range, we find the r is 20.4 kilometers. (b) In this case, the Friis equation still applies but the path loss is given by Eq. (11.37) −1

⎛ 10−36 /10 ⎞ LP = ⎜ 3.1 ⎟ ⎝ r ⎠ ~ 36 + 31log10 ( r ) Substituting the second line into Eq. (1), we have −90 = 27.8 + 0 + 0 − (36 + 31log10 r )

Solving this equation for the maximum range, we find that r is 435 meters. Clearly, the propagation conditions can make a huge difference on the range.

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Problem 11.25 A line-of-sight 10-kilometer radio link is required to transmit data at a rate of 1 megabit per second at a center frequency of 4 GHz. The transmitter uses an antenna with 10 dB gain and QPSK modulation with a root-raised cosine pulse shape spectrum having a roll-off factor of 0.5. The receiver also has an antenna with 10 dB gain and has a system noise temperature of 900 K. What is the minimum transmit power required to achieve a bit error rate of 10-5? Solution From the BER performance of QPSK in Fig. 10.16, we find that a BER of 10-5 implies an Eb/N0 of 9.5 dB is required. From this, we obtain the required C/N0 using knowledge of the transmission rate R = 1 Mbps.

⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠dB − Hz ⎝ N 0 ⎠dB = 69.5 dB-Hz

The system noise temperature of 900 K implies N 0 = kTe = 1.38 × 10−23 × 900 = 1.24 × 10−20 W/Hz ~ −199.1 dBW/Hz

Using this information, the received power level may be calculated from PR = C ⎛ C ⎞ =⎜ + ( N 0 )dBW − Hz ⎟ ⎝ N 0 ⎠dB − Hz = 69.5 + ( −199.1) = −129.6 dBW

We now appeal to the decibel form of the Friis equation: PR = PT + GR + GT − LP

(1)

where the antenna gains are GR = GT = 10 dB. Since the problem sight says line-of-sight transmission, we shall assume free-space propagation, and the path loss is

Continued on next slide

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Problem 11.25 continued ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

At 4 GHz, the wavelength is λ = c f = 3 × 108 4 × 109 = 0.075 m . Making all these substitutions into Eq. (1) with a range r = 10 km, we obtain

⎛ 4π 10 ×103 ⎞ −129.6 = PT + 10 + 10 − 20 log10 ⎜ ⎟ ⎝ 0.075 ⎠ Solving this equation for the transmitted power, we find that the required PT is -25.1 dBW or 4.9 dBm.

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Problem 11.26 A land-mobile radio transmits 128 kbps at a frequency of 700 MHz. The transmitter uses an omni-directional antenna and 16-QAM modulation with a root-raised cosine pulse spectrum having a roll-off of 0.4. The receiver has an antenna with 3 dB gain and a noise figure of 6 dB. If the path loss between the transmitter and receiver is given by LP (r ) = 30 + 28log10 (r ) dB

where r is in meters, what is the maximum range at which the bit error rate of 10-4 may be achieved? Solution From the BER performance of 16-QAM in Fig. 10.16, we find that a BER of 10-4 implies an Eb/N0 of 13 dB. From this, we obtain the C/N0 by using knowledge of the transmission rate R = 128 kbps.

⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠dB − Hz ⎝ N 0 ⎠dB = 64.1 dB-Hz The noise figure of 6 dB implies N 0 = kFT0

(

)

= 1.38 × 10−23 × 106 /10 × ( 290 ) = 1.59 × 10

−20

W/Hz

~ −198.0 dBW/Hz

and the received power level is ⎛ C ⎞ + ( N 0 )dBW − Hz PR = C = ⎜ ⎟ ⎝ N 0 ⎠ dB − Hz = 64.1 + ( −198.0 ) = −133.9 dBW

We now appeal to the decibel form of the Friis equation: PR = PT + GR + GT − LP

(1)

Continued on next slide

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Problem 11.26 continued

where the antenna gains are GR = GT = 0 dB. The path loss is LP = 30 + 28log10 ( r )

Making all these substitutions into Eq. (1), we obtain −133.9 = PT + 0 + 0 − ( 30 + 28log10 r )

or

r = 10(

PT +103.9 ) / 28

In the following figure, we plot the range in kilometres versus the transmit power in dBW. 20 18 16

Range (km)

14 12 10 8 6 4 2 0 -20

-15

-10

-5

0

5

10

15

Transmit Power (dBW)

For example, with a transmit power of 10 W or 10 dBW, we find that range is 11.7 km.

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Problem 8.1

An information packet contains 200 bits. This packet is transmitted over a communications channel where the probability of error for each bit is 10-3. What is the probability that the packet is received error-free?

Solution Recognizing that the number of errors has a binomial distribution over the sequence of 200 bits, let x represent the number of errors with p = 0.001 and n = 200. Then the probability of no errors is

P ⎡⎢⎣ x = 0⎤⎥⎦ = (1 − p )

n

= (1 − .001)

200

= .999 200 = 0.82

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page…8-1

Problem 8.2 Suppose the packet of the Problem 8.1 includes an error-correcting code that can correct up to three errors located anywhere in the packet. What is the probability that a particular packet is received in error in this case? Solution The probability of a packet error is equal to the probability of more than three bit errors. This is equivalent to 1 minus the probability of 0, 1, 2, or 3 errors: 1 − P[x ≤ 3] = 1 − (P[x = 0] + P[x = 1] + P[x = 2] + P[x = 3]) ⎛n⎞ ⎛n⎞ ⎛n⎞ = 1 − (1 − p) n − ⎜⎜ ⎟⎟ p(1 − p )n−1 − ⎜⎜ ⎟⎟ p 2 (1 − p )n −2 − ⎜⎜ ⎟⎟ p 3 (1 − p )n −3 ⎝3 ⎠ ⎝2⎠ ⎝1 ⎠ n(n − 1) 2 n(n − 1)(n − 2) 3 ⎤ ⎡ p (1 − p ) + p ⎥ = 1 − (1 − p )n−3 ⎢(1 − p )3 + np(1 − p )2 + 2 6 ⎣ ⎦ = 5.5 × 10 −5

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page…8-2

Problem 8.3 Continuing with Example 8.6, find the following conditional probabilities: P[X=0|Y=1] and P[X =1|Y=0].

Solution From Bayes’ Rule

[

]

[

]

P X = 0Y = 1 =

[

]

P Y = 1 X = 0 P[ X = 0 ]

P[Y = 1] pp0 = pp0 + (1 − p ) p1

P X = 1Y = 0 = =

[

]

P Y = 0 X = 1 P[X = 1] P[Y = 0]

pp1 pp1 + (1 − p ) p0

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page…8-3

Problem 8.4 Consider a binary symmetric channel for which the conditional probability of error p = 10-4, and symbols 0 and 1 occur with equal probability. Calculate the following probabilities: a) The probability of receiving symbol 0. b) The probability of receiving symbol 1. c) The probability that symbol 0 was sent, given that symbol 0 is received d) The probability that symbol 1 was sent, given that symbol 0 is received.

Solution (a) P[Y = 0] = P[Y = 0 | X = 0]P[ X = 0] + P[Y = 0 | X = 1]P[ X = 1] = (1 − p ) p0 + pp1 = .9999 1 + .0001 1 2 2 =1 2

(b)

P[Y = 1] = 1 − P[Y = 0] =1

(c)

2

From Eq.(8.30) P [X = 0 Y = 0 ] = =

(1 − p ) p0 (1 − p ) p0 + pp1

(1 − 10 ) (1 − 10 ) + 10 −4

−4

= 1 − 10

(d)

1

1

2

2

−4 1

2

−4

From Prob. 8.3 P [X = 1Y = 0 ] = =

pp1 pp1 + (1 − p ) p0 10 −4

10 −4 12 1 + (1 − 10 − 4 ) 1 2 2

= 10 −4

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page…8-4

Problem 8.5 Determine the mean and variance of a random variable that is uniformly distributed between a and b.

Solution The mean of the uniform distribution is given by ∞

µ = E[X ] =

∫ xf

X

( x)dx

−∞ b

=∫ x a

1 dx b−a b

x2 = 2(b − a ) a b2 − a 2 2(b − a ) b+a = 2 =

The variance is given by

[

] ∫ (x − µ)

E (X − µ ) = 2

∞

2

f X ( x)dx

−∞

=∫

b

(x − µ )2 dx

b−a 3 3 1 (b − µ ) (a − µ ) = − b−a 3 3 a

If we substitute µ =

[

]

E (X − µ ) = 2

b+a then 2

3 3 1 ⎡ (b − a ) (a − b ) ⎤ − ⎢ ⎥ b − a ⎣ 24 24 ⎦

2 ( b − a) =

12

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page…8-5

Problem 8.6 Let X be a random variable and let Y = (X-µX)/σX. What is the mean and variance of the random variable Y? Solution

⎡ X − µ X ⎤ E[X ] − µ X 0 E[Y ] = E ⎢ = =0 ⎥= σX σX ⎣ σX ⎦

[ ]

E(Y − µ Y ) = E Y 2

2

⎛ X − µX = E⎜⎜ ⎝ σX

E( X − µ X )

2

=

σX2

⎞ ⎟⎟ ⎠

2

σX2 = =1 σX2

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page…8-6

Problem 8.7 What is the probability density function of the random variable Y of Example 8.8? Sketch this density function.

Solution From Example 8.8, the distribution of Y is ⎧0 ⎪ −1 ⎪ 2π − 2 cos ( y ) FY ( y ) = ⎨ 2π ⎪ ⎪⎩1

y < −1 | y |< 1 y >1

Thus, the density of Y is given by ⎧0 ⎪ dFY ( y ) ⎪ 1 =⎨ 2 dy ⎪π 1 − y ⎪0 ⎩

y < −1 | y |< 1 y >1

This density is sketched in the following figure.

fY(y)

1

π

y -1

1

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page…8-7

Problem 8.8 Show that the mean and variance of a Gaussian random variable X with the density 2 function given by Eq. (8.48) are µ and σ X . X

Solution Consider the difference E[X]-µX: E[X ] − µ X =

∞

∫

(x − µ X ) exp⎧− (x − µ X )2 ⎫dx ⎨ ⎩

2π σ X

−∞

2σ X

2

⎬ ⎭

Let y = x − µ X and substitute E[ X ] − µ X = ∫

∞

y 2π σ X

−∞

2 ⎛ ⎞ dy exp⎜⎜ − y 2 ⎟ 2σ X ⎟⎠ ⎝

=0

since integrand has odd symmetry. This implies E[X ] = µ X . With this result Var( X ) = E( x − µ X )

2

=∫

∞

−∞

(x − µ X )2 exp⎧ − (x − µ X )2 ⎫dx ⎨ ⎩

2π σ X

2σ X

2

⎬ ⎭

In this case let y=

x − µX

σX

and making the substitution, we obtain Var ( X ) = σ X

2

∫

∞

−∞

⎧− y2 ⎫ y2 exp⎨ ⎬dy 2π ⎩ 2 ⎭

Recalling the integration-by-parts, i.e., ∫ udv = uv − ∫ vdu , let u = y and

⎛ − y2 ⎞ ⎟⎟dy . Then dv = y exp⎜⎜ 2 ⎝ ⎠ Continued on next slide

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page…8-8

Problem 8.8 continued Var ( X ) = σ X

2

(− ) y exp⎛ − y 2 2π

⎜ ⎝

∞

⎛ y2 ⎞ 1 ⎞ 2 ∞ ⎜⎜ − ⎟⎟dy + σ exp X ∫ 2 ⎟⎠ −∞ 2π ⎝ 2 ⎠ −∞

= 0 + σ X •1 2

=σX

2

where the second integral is one since it is integral of the normalized Gaussian probability density.

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page…8-9

Problem 8.9 Show that for a Gaussian random variable X with mean µX and variance σ X2 the transformation Y = (X - µX)/σX, converts X to a normalized Gaussian random variable.

Solution x − µX Let y = . Then

σX

E[Y ] =

1 2π

⎛ y2 ⎞ y exp ⎜ − 2 ⎟dy ∫−∞ ⎠ ⎝ ∞

=0

by the odd symmetry of the integrand. If E[Y] = 0, then from the definition of Y, E[X] = µX. In a similar fashion

[ ]

EY2 =

1 2π

⎛ y2 ⎞ 2 ⎜⎜ − ⎟⎟dy exp y ∫−∞ ⎝ 2 ⎠ ∞

∞

⎧ y2 ⎫ (−) y 1 = ⋅ exp⎨− ⎬ + 2π 2π ⎩ 2 ⎭ −∞ =1

2 ⎞ ⎛ exp⎜ − y ⎟dy 2 −∞ ⎠ ⎝

∫

∞

where we use integration by parts as in Problem 8.8. This result implies ⎛ x − µX E ⎜⎜ ⎝ σX

2

⎞ ⎟⎟ = 1 ⎠

and hence E( x − µ X ) = σ X2 2

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page…8-10

Problem 8.10 Determine the mean and variance of the sum of five independent uniformly-distributed random variables on the interval from -1 to +1. Solution Let Xi be the individual uniformly distributed random variables for i = 1,..,5, and let Y be the random variable representing the sum: 5

Y = ∑ Xi i =1

Since Xi has zero mean and Var(Xi) = 1/3 (see Problem 8.5), we have 5

E[Y ] = ∑ E[X i ] = 0 i =1

and

[

] [ ]

Var (Y ) = E (Y − µY ) = E Y 2 2

[

= E (∑ X i )

2

]

[ ]

5

[

= ∑ E X i2 + ∑ E X i X j i =1

]

i≠ j

Since the Xi are independent, we may write this as

( )

[ ]

Var(Y ) = 5 1 + ∑ E[ X i ]E X j 3 = 5 +0 3 5 = 3

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page…8-11

Problem 8.11 A random process is defined by the function

X (t , θ ) = A cos(2πft + θ ) where A and f are constants, and θ is uniformly distributed over the interval 0 to 2π. Is X stationary to the first order?

Solution Denote

Y = X (t1 , θ ) = A cos(2πft1 + θ ) for any t1. From Problem 8.7, the distribution of Y and therefore of X for any t1 is

⎧0 ⎪ ⎪ 2π − 2 cos −1 ( y / A) FX (t1 ) ( y ) = ⎨ 2π ⎪ ⎪⎩1

y < −A | y |< A y>A

Since the distribution is independent of t it is stationary to first order.

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page…8-12

Problem 8.12 Show that a random process that is stationary to the second order is also stationary to the first order.

Solution

Let the distribution F be stationary to second order FX (t1 ) X (t2 ) ( x1 , x2 ) = FX ( t1 +τ ) X ( t2 +τ ) ( x1 , x2 )

Then, FX ( t1 ) X (t2 ) ( x1 , ∞ ) = FX (t1 ) (x1 )

= FX ( t1 +τ ) X ( t2 +τ ) ( x1 , ∞ ) = FX ( t1 +τ ) ( x1 )

Thus the first order distributions are stationary as well.

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page…8-13

Problem 8.13 Let X(t) be a random process defined by

X (t ) = A cos(2πft ) where A is uniformly distributed between 0 and 1, and f is constant. Determine the autocorrelation function of X. Is X wide-sense stationary?

Solution

[ ] = E[A ][cos(2πf (t

E[X (t1 )X (t 2 )] = E A 2 cos(2πft1 ) cos(2πft 2 ) 2

1

[ ]

− t 2 )) + cos 2πf (t1 + t 2 )]

1

x3 1 E A = ∫ x dx = = 0 3 0 3 2

1

2

Since the autocorrelation function depends on t1 + t 2 as well as t1 − t 2 , the process is not wide-sense stationary.

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page…8-14

Problem 8.14 A discrete-time random process {Yn: n = …,-1,0,1,2, …} is defined by Yn = α 0 Z n + α 1 Z n −1 where {Zn} is a random process with autocorrelation function RZ (n) = σ

[

]

δ (n) . What is the

2

autocorrelation function RY ( n, m) = E Yn Ym ? Is the process {Yn} wide-sense stationary?

Solution

We implicitly assume that Zn is stationary and has a constant mean µZ. Then the mean of Yn is given by

E[Yn ] = α 0 E[Z n ] + α1E[ Z n−1 ] = (α 0 + α1 )µ Z

The autocorrelation of Y is given by E[YnYm ] = E[(α 0 Z n + α 0 Z n−1 )(α 0 Z m +α 1Z m−1 )]

= α 02 E[Z n Z m ] + α1α 0 E[Z n Z m−1 ] + α 0α1E[Z n−1 Z m ] + α12 E[Z m−1 Z n−1 ]

= α 02σ 2δ (n − m ) + α1α 0σ 2δ (m − 1 − n ) + α 0α1σ 2δ (n − 1 − m ) + α12δ (m − 1 − (n − 1))

(

)

= α 02 + α12 σ 2δ (n − m ) + α 0α1σ 2 [δ (n − m − 1) + δ (m − n − 1)]

Since the autocorrelation only depends on the time difference n-m, the process is widesense stationary with

(

)

RY (n) = α 02 + α12 σ 2δ (n) + α 0α1σ 2 (δ (n − 1) + δ (n + 1) )

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page…8-15

Problem 8.15 For the discrete-time process of Problem 8.14, use the discrete Fourier transform to approximate the corresponding spectrum. That is, N −1

S Y (k ) = ∑ RY (n)W kn n=0

If the sampling in the time domain is at n/Ts where n = 0, 1, 2, …, N-1. What frequency does k correspond to?

Solution Let

β 0 = (α 02 + α12 )σ 2 and β1 = α 0α1σ 2 . Then N −1

SY (k ) = ∑ [β 0δ (n ) + β1 (δ (n − 1) + δ (n + 1))] W kn n =0

(

= β 0W 0 + β1 W −k + W + k

)

+ j 2πk ⎛ − jN2πk ⎞ ⎜ = β 0 + β1 ⎜ e + e N ⎟⎟ ⎝ ⎠ ⎛ 2πk ⎞ = β 0 + 2 β1 cos⎜ ⎟ ⎝ N ⎠

The term SY (k ) corresponds to frequency

kf s 1 where f S = . TS N

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page…8-16

Problem 8.16 Is the discrete-time process {Yn: n = 1,2,…} defined by: Y0 = 0 and Yn +1 = αYn + Wn , a Gaussian process, if Wn is Gaussian?

Solution (Proof by mathematical induction.) The first term Y1 = αY0 + W0 is Gaussian since Y0 = 0 and W0 are Gaussian. The second term Y2 = αY1 + W1 is Gaussian since Y1 and W1 are Gaussian. Assume Yn is Gaussian. Then Yn +1 = αYn + Wn is Gaussian since Yn and Wn are both Gaussian.

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page…8-18

Problem 8.17 A discrete-time white noise process {Wn} has an autocorrelation function given by RW(n) = N0δ(n). (a) Using the discrete Fourier transform, determine the power spectral density of {Wn}. (b) The white noise process is passed through a discrete-time filter having a discretefrequency response H (k ) =

1 − (αW k ) N 1 − αW k

where, for a N-point discrete Fourier transform, W = exp{j2π/N}. What is the spectrum of the filter output?

Solution The spectrum of the discrete white noise process is N −1

S (k ) = ∑ R(n ) W nk n =0

N −1

= ∑ N 0δ (n )W nk n =0

= N0

The spectrum of the process after filtering is SY (k ) = H (k ) S (k ) 2

1 − (αW k ) N = N0 1 − αW k

2

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page…8-19

Problem 8.18 Consider a deck of 52 cards, divided into four different suits, with 13 cards in each suit ranging from the two up through the ace. Assume that all the cards are equally likely to be drawn. (a) Suppose that a single card is drawn from a full deck. What is the probability that this card is the ace of diamonds? What is the probability that the single card drawn is an ace of any one of the four suits? (b) Suppose that two cards are drawn from the full deck. What is the probability that the cards drawn are an ace and a king, not necessarily the same suit? What if they are of the same suit?

Solution (a)

P[Ace of diamonds] = P[Any ace] =

1 52

1 13

(b)

P[Ace and king ] = P[Ace on first draw ]P[King on second] + P[King on first draw ]P[Ace on seco 1 4 1 4 = × + × 13 51 13 51 8 = 663 1 1 1 1 P[Ace and king of same suit ] = × + × 13 51 13 51 1 = 663

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page…8-20

Problem 8.19

Suppose a player has one red die and one white die. How many outcomes are possible in the random experiment of tossing the two dice? Suppose the dice are indistinguishable, how many outcomes are possible?

Solution The number of possible outcomes is 6 × 6 = 36 , if distinguishable. If the die are indistinguishable then the outcomes are (11) (12)…(16) (22)(23)…(26) (33)(34)…(36) (44)(45)(46) (55)(56) (66) And the number of possible outcomes are 21.

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page…8-21

Problem 8.20 Refer to Problem 8.19. (a) What is the probability of throwing a red 5 and a white 2? (b) If the dice are indistinguishable, what is the probability of throwing a sum of 7? If they are distinguishable, what is this probability?

Solution (a) P[Red 5 and white 2] =

1 1 1 × = 6 6 36

(b) The probability of the sum does not depend upon whether the die are distinguishable or not. If we consider the distinguishable case the possible outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1) so 6 1 P[sum of 7] = = 36 6

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page…8-22

Problem 8.21 Consider a random variable X that is uniformly distributed between the values of 0 and 1 with probability ¼ takes on the value 1 with probability ¼ and is uniformly distributed between values 1 and 2 with probability ½ . Determine the distribution function of the random variable X. Solution ⎧ ⎪0 ⎪x ⎪ 4 ⎪ FX ( x) = ⎨ 1 2 ⎪ ⎪ 1 + 1 ( x − 1) ⎪2 2 ⎪1 ⎩

x≤0 0 < x 2

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page…8-24

Problem 8.22 Consider a random variable X defined by the double-exponential density where a and b are constants.

f X ( x) = a exp(− b x )

−∞ < x Y] = P[Y > X]. If we only consider the case X > Y, then we have the conditions: 0 < X < T and 0 < Y < X < τ+Y. Combining these conditions we have Y < X < min(T, τ+Y). Consequently, T

min (T ,τ + y )

0

y

P[ X − Y < τ ] = ∫

∫f

min (T ,τ + y )

T

=∫

∫ y

0

=

X

1 T2

( x) f Y ( y )dx dy

2

⎛1⎞ ⎜ ⎟ dx dy ⎝T ⎠

T

∫ {min(T ,τ + y ) − y}dy 0

Combining the two terms of the integrand, P[ X − Y < τ ] =

T

1 min(T − y, τ )dy T 2 ∫0 T

⎛ ⎞ 1 y2 ⎜ , τy ⎟⎟ = 2 min⎜ Ty − 2 T ⎝ ⎠0 ⎛1 τ ⎞ = min⎜ , ⎟ ⎝2 T ⎠

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page…8-54

Problem 8.45 A telegraph system (an early version of digital communications) transmits either a dot or dash signal. Assume the transmission properties are such that 2/5 of the dots and 1/3 of the dashes are received incorrectly. Suppose the ratio of transmitted dots to transmitted dashes is 5 to 3. What is the probability that a received signal as the transmitted if: a) The received signal is a dot? b) The received signal is a dash?

Solution (a) Let X represent the transmitted signal and Y represent the received signal. Then by application of Bayes’ rule

P(Y = dot ) = P( X = dot | No error )P( No dot error) + P( X = dash | error )P(dash error ) =5 3 + 3 1 8 5 8 3 =3 +1 =1 8 8 2

( ) ( )( )

(b) Similarly, P[Y = dash ] = P[ X = dash | no error]P(no dash error ) + P( X = dot )P[dot error] = 3 ⋅2 + 5 2 8 3 8 5 =2 +2 = 1 8 8 2

[ ]

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page…8-55

Problem 8.46 Four radio signals are emitted successively. The probability of reception for each of them is independent of the reception of the others and equal, respectively, 0.1, 0.2, 0.3 and 0.4. Find the probability that k signals will be received where k = 1, 2, 3, 4. Solution For one successful reception, the probability is given by the sum of the probabilities of the four mutually exclusive cases P = p1 (1 − p2 )(1 − p3 )(1 − p4 ) +

(1 − p1 ) p2 (1 − p3 )(1 − p4 ) + (1 − p1 )(1 − p2 ) p3 (1 − p4 ) + (1 − p1 )(1 − p2 )(1 − p3 ) p4

= .1 ⋅ .8 ⋅ .7 ⋅ .6 + .9 ⋅ .2 ⋅ .7 ⋅ .6 + .9 ⋅ .8 ⋅ .3 ⋅ .6 + .9 ⋅ .8 ⋅ .7 ⋅ .4 = 0.4404

For k = 2, there six mutually exclusive cases P = p1 p2 (1 − p3 )(1 − p4 ) +

p1 (1 − p2 ) p3 (1 − p4 ) + p1 (1 − p2 )(1 − p3 ) p4 +

(1 − p1 ) p2 p3 (1 − p4 ) + (1 − p1 ) p2 (1 − p3 ) p4 + (1 − p1 )(1 − p2 ) p3 p4

= .1 ⋅ .2 ⋅ .7 ⋅ .6 + .1 ⋅ .8 ⋅ .3 ⋅ .6 + .1 ⋅ .8 ⋅ .7 ⋅ .4 + .9 ⋅ .2 ⋅ .3 ⋅ .6 + .9 ⋅ .2 ⋅ .7 ⋅ .4 + .9 ⋅ .8 ⋅ .3 ⋅ .4 = 0.2144 For k =3 there are four mutually exclusive cases P = p1 p2 p3 (1 − p4 ) +

p1 (1 − p2 ) p3 p4 + p1 p2 (1 − p3 ) p4 +

(1 − p1 ) p2 p3 p4 = .1 ⋅ .2 ⋅ .3 ⋅ .6 + .1 ⋅ .8 ⋅ .3 ⋅ .4 + .1 ⋅ .2 ⋅ .7 ⋅ .4 + .9 ⋅ .2 ⋅ .3 ⋅ .4 = 0.0404

For k = 4 there is only one term

P = p1 p2 p3 p4 = .1 ⋅ .2 ⋅ .3 ⋅ .4 = 0.0024

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page…8-56

Problem 8.47 In a computer-communication network, the arrival time τ between messages is modeled with an exponential distribution function, having the density

⎧ 1 −λτ ⎪ e f T (τ ) = ⎨ λ ⎪⎩0

τ ≥0 otherwise

a) What is the mean time between messages with this distribution? b) What is the variance in this time between messages?

Solution (Typo in problem statement, should read fT(τ)=(1/λ)exp(-τ/λ) for τ>0) (a) The mean time between messages is ∞

E[T ] = ∫ τf T (τ )dτ 0

∞

τ exp(− τ / λ )dτ λ 0

=∫

∞

= − τ exp(− τ / λ ) 0 + ∫ exp(− τ / λ )dτ ∞

0

∞

= 0 − λ exp(−τ / λ ) 0 =λ

where the third line follows by integration by parts. (b) To compute the variance, we first determine the second moment of T ∞

[ ] = ∫τ

ET

2

2

f T (τ )dτ

0

∞

τ2 exp(− τ / λ )dτ λ 0

=∫

∞

= − τ 2 exp(− τ / λ ) + 2∫ τ exp(− τ / λ )dτ ∞ 0

= 0 + 2λE[T ]

0

= 2λ2

Continued on next slide

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page…8-57

Problem 8.47 continued

The variance is then given by the difference of the second moment and the first moment squared (see Problem 8.23)

[ ]

Var (T ) = E T 2 − (E[T ])

2

= 2λ2 − λ2 = λ2

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page…8-58

Problem 8.48 If X has a density fX(x), find the density of Y where a) Y = aX + b for constants a and b. 2 b) Y = X . c) Y = X , assuming X is a non-negative random variable. Solution (a) If Y = aX + b , using the results of Section 8.3 for Y = g(X)

dg −1 ( y ) f Y ( y ) = f X (g ( y ) ) dy −1

⎛ y −b⎞ 1 = fX ⎜ ⎟ ⎝ a ⎠a (b) If Y = X 2 , then

( (

)

(

fY ( y ) = f X − y + f X +

⎛ 1 ⎞ ⎟ y ⎜ ⎜2 y ⎟ ⎝ ⎠

))

(c) If Y = X , then we must assume X is positive valued so, this is a one-to-one mapping and

( )

fY ( y ) = f X y 2 ⋅ 2 y

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page…8-59

Problem 8.49 Let X and Y be independent random variables with densities fX(x) and fY(y), respectively. Show that the random variable Z = X+Y has a density given by z

f Z ( z) =

∫f

Y

( z − s ) f X ( s )ds

−∞

Hint:

P[Z ≤ z ] = P[ X ≤ z, Y ≤ z − X ]

Solution (Typo in problem statement - should be “positive” independent random variables)

Using the hint, we have that FZ(z) = P[Z ≤ z] and z z−x

FZ ( z ) =

∫∫f

X

( x) f Y ( y )dydx

− ∞ −∞

To differentiate this result with respect to z, we use the fact that if b

g ( z ) = ∫ h( x, z )dx a

then db da ∂g ( z ) ∂ = ∫ h( x, z )dx + h(b, z ) − h( a, z ) dz dz dz ∂z a b

(1)

Inspecting FZ(z), we identify h(x,z) z−x

h ( x, z ) =

∫f

X

( x) f Y ( y )dy

−∞

and a = -∞ and b = z. We then obtain d FZ ( z ) dz z −( −∞ ) z z−z ⎤ ⎡ d z− x dz ( ) ( ) ( ) ( ) f x f y dy dx f z f y dy f X (−∞) f Y ( y )dy ⋅ 0 + − = ∫⎢ ⎥ X Y ∫ X Y dz −∫∞ dz −∫∞ −∞ ⎣ −∞ ⎦

f Z ( z) =

⎡d ∫−∞⎢⎣ dz z

=

z− x

∫f

−∞

X

⎤ ( x) f Y ( y )dy ⎥ dx ⎦

Continued on next slide

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page…8-60

Problem 8-49 continued

where the second term of the second line is zero since the random variables are positive, and the third term is zero due to the factor zero. Applying the differentiation rule a second time, we obtain

z

f Z ( z) =

⎡

∫ ⎢⎣0 + f

X

( x) f Y ( z − x)

−∞

d ( z − x) d (−∞) ⎤ dx − f X ( x) f Y (−∞) dz dz ⎥⎦

z

=

∫f

X

( x) f Y ( z − x)dx

−∞

which is the desired result.

An alternative solution is the following: we note that P[Z ≤ z | X = x] = P[X + Y ≤ z | X = x]

= P[x + Y ≤ z | X = x ] = P[x + Y ≤ z ] = P[Y ≤ z − x]

where the third equality follows from the independence of X and Y. By differentiating both sides with respect to z, we see that f Z | X ( z | x) = f Y ( z − x) By the properties of conditional densities f Z , X ( z , x) = f X ( x) f Z | X ( z | x) = f X ( x) f Y ( z − x) Integrating to form the marginal distribution, we have ∞

f Z ( z) =

∫f

X

( x) f Y ( z − x)dx

−∞

If Y is a positive random variable then fY(z-x) is zero for x > z and the desired result follows.

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page…8-61

Problem 8.50 Find the spectral density SZ(f) if

Z (t ) = X (t )Y (t ) where X(t) and Y(t) are independent zero-mean random processes with

RX (τ ) = a1e

−α1 τ

and

RY (τ ) = a2 e

−α 2 τ

.

Solution The autocorrelation of Z(t) is given by

RZ (τ ) = E[Z (t )Z (t + τ )]

= E[ X (t )X (t + τ )Y (t )Y (t + τ )] = E[ X (t ) X (t + τ )]E[Y (t )Y (t + τ )] = RX (τ )RY (τ )

By the Wiener-Khintchine relations, the spectrum of Z(t) is given by S Z ( f ) = F −1 [R X (τ )RY (τ )]

= F −1 [a1a2 exp(− (α1 + α 2 )τ =

2a1a2 (α1 + α 2 ) 2 (α1 + α 2 ) 2 + (2πf )

)]

where the last line follows from the Fourier transform of the double-sided exponential (See Example 2.3).

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page…8-62

Problem 8.51 Consider a random process X(t) defined by X (t ) = sin (2πf c t ) where the frequency fc is a random variable uniformly distributed over the interval [0,W]. Show that X(t) is nonstationary. Hint: Examine specific sample functions of the random process X(t) for, say, the frequencies W/4, W/2, and W.

Solution To be stationary to first order implies that the mean value of the process X(t) must be constant and independent of t. In this case, E[ X (t )] = E[sin (2πf c t )] W

=

1 sin (2πwt )dw W ∫0

− cos(2πwt ) = 2πWt 0

W

=

1 − cos(2πWt ) 2πWt

This mean value clearly depends on t, and thus the process X(t) is nonstationary.

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page…8-63

Problem 8.52 The oscillators used in communication systems are not ideal but often suffer from a distortion known as phase noise. Such an oscillator may be modeled by the random process Y (t ) = A cos(2πf c t + φ (t ) ) where

φ (t )

φ (t )

is a slowly varying random process. Describe and justify the conditions on the random process

such that Y(t) is wide-sense stationary.

Solution The first condition for wide-sense stationary process is a constant mean. Consider t = t0, then E[Y (t 0 )] = E[ A cos(2πf c t0 + φ (t0 ) )]

In general, the function cos θ takes from values -1 to +1 when θ varies from 0 to 2π. In this case θ corresponds to 2πfct0 + φ(t0). If φ(t0) varies only by a small amount then θ will be biased toward the point 2πfct0 + E[φ(t0)], and the mean value of E[Y(t0)] will depend upon the choice of t0. However, if φ(t0) is uniformly distributed over [0, 2π] then 2πfct0 + φ(t0) will be uniformly distributed over [0, 2π] when considered modulo 2π, and the mean E[Y(t0)] will be zero and will not depend upon t0. Thus the first requirement is that φ(t) must be uniformly distributed over [0,2π] for all t. The second condition for a wide-sense stationary Y(t) is that the autocorrelation depends only upon the time difference E[Y (t1 )Y (t 2 )] = E[A cos(2πf c t1 + φ (t1 ) )A cos(2πf c t 2 + φ (t 2 ) )] A2 E[cos(2πf c (t1 + t 2 ) + φ (t1 ) + φ (t 2 ) ) + cos(2πf c (t1 − t 2 ) + φ (t1 ) − φ (t 2 ) )] = 2

where we have used the relation cos A cos B = 1 2 (cos( A + B) + cos( A − B) ) . In general, this correlation does not depend solely on the time difference t2-t1. However, if we assume: We first note that if φ(t1) and φ(t2) are both uniformly distributed over [0,2π] then so is ψ = φ (t1 ) + φ (t 2 ) (modulo 2π), and E[cos(2πf c (t1 + t 2 ) + ψ )] =

1 2π

2π

∫ cos(2πf 0

c

(t1 + t 2 ) + ψ )dψ

(1)

=0

Continued on next slide

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page…8-64

Problem 8.52 continued

We consider next the term RY(t1,t2)= E[cos(2πf c (t1 − t 2 ) + φ (t1 ) − φ (t 2 ) )] and three special cases: (a) if ∆t = t1-t2 is small then φ (t1 ) ≈ φ (t 2 ) since φ(t) is a slowly varying process, and A2 cos(2πf c (t1 − t 2 ) ) 2 (b) if ∆t is large then φ(t1) and φ(t2) should be approximately independent and φ (t1 ) − φ (t 2 ) would be approximately uniformly distributed over [0,2π]. In this case RY (t1 , t 2 ) ≈ 0 using the argument of Eq. (1). RY (t1 , t 2 ) =

(c) for intermediate values of ∆t, we require that

φ (t1 ) − φ (t 2 ) ≈ g (t1 − t 2 ) for some arbitrary function g(t). Under these conditions the random process Y(t) will be wide-sense stationary.

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page…8-65

Problem 8.53 A baseband signal is disturbed by a noise process N(t) as shown by

X (t ) = A sin (0.3πt ) + N (t ) where N(t) is a stationary Gaussian process of zero mean and variance σ2. (a) What are the density functions of the random variables X1 and X2 where

X 1 = X (t ) t =1 X 2 = X (t ) t =2 (b) The noise process N(t) has an autocorrelation function given by

RN (τ ) = σ 2 exp(− τ What is the joint density function of X1 and X2, that is,

)

f X1 , X 2 ( x1 , x2 ) ?

Solution (a) The random variable X1 has a mean E[ X (t1 )] = E[ A sin (0.3π ) + N (t1 )]

= A sin(0.3π ) + E[N (t1 )] = A sin (0.3π )

Since X1 is equal to N(t1) plus a constant, the variance of X1 is the same as that of N(t1). In addition, since N(t1) is a Gaussian random variable, X1 is also Gaussian with a density given by 1 f X1 ( x ) = exp{− ( x − µ1 ) / 2σ 2 } 2π σ where µ1 = E[ X (t1 )] . By a similar argument, the density function of X2 is f X 2 ( x) =

{

1 exp − ( x − µ 2 ) / 2σ 2 2π σ

}

where µ 2 = A sin(0.6π ) .

Continued on next slide

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page…8-66

Problem 8-53 continued

(b) First note that since the mean of X(t) is not constant, X(t) is not a stationary random process. However, X(t) is still a Gaussian random process, so the joint distribution of N Gaussian random variables may be written as Eq. (8.90). For the case of N = 2, this equation reduces to f X (x) =

1 2π Λ

1/ 2

{

}

exp − (x − µ)Λ−1 (x − µ)T / 2

where Λ is the 2x2 covariance matrix. Recall that cov(X1,X2) =E[(X1-µ1)(X2-µ2)], so that ⎡cov( X 1 , X 1 ) cov( X 1 , X 2 ) ⎤ Λ=⎢ ⎥ ⎣cov( X 2 , X 1 ) cov( X 2 , X 2 )⎦ ⎡ R (0) RN (1) ⎤ =⎢ N ⎥ ⎣ RN (1) RN (0)⎦ ⎡ σ2 σ 2 exp(−1)⎤ =⎢ 2 ⎥ σ2 ⎦ ⎣σ exp(− 1)

If we let ρ = exp(-1) then

Λ = σ 4 (1 − ρ 2 ) and

Λ−1 =

⎡ 1 1 2 ⎢ σ (1 − ρ ) ⎣− ρ 2

− ρ⎤ 1 ⎥⎦

Making these substitutions into the above expression, we obtain upon simplification f X1 , X 2 ( x1 , x2 ) =

⎧ ( x − µ1 ) 2 + ( x2 − µ 2 ) 2 − 2 ρ ( x1 − µ1 )( x2 − µ 2 ) ⎫ exp⎨− 1 ⎬ 2σ 2 (1 − ρ 2 ) 1− ρ2 ⎩ ⎭

1 2πσ 2

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page…8-67

Problem 9.1 In practice, we often cannot measure the signal by itself but must measure the signal plus noise. Explain how the SNR would be calculated in this case. Solution Let r(t) = s(t) + n(t) be the received signal plus noise. Assuming the signal is independent of the noise, we have that the received power is

[

R0 = E r 2 (t )

[

]

= E (s (t ) + n(t ) )

2

]

[ ] [ ] = E[s (t )] + 2E[s (t )]E[n(t )] + E[n (t )] = E s 2 (t ) + 2E[s (t )n(t )] + E n 2 (t ) 2

2

= S +0+ N where S is the signal power and N is the average noise power. We then measure the noise alone

[

R1 = E n 2 (t )

]

=N

and the SNR is given by

SNR =

R0 − R1 R1

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Problem 9.2 A DSC-SC modulated signal is transmitted over a noisy channel, having a noise spectral density N0/2 of 2x10-17 watts per hertz. The message bandwidth is 4 kHz and the carrier frequency is 200 kHz. Assume the average received power of the signal is -80 dBm. Determine the post-detection signal-to-noise ratio of the receiver. Solution From Eq. (9.23), the post-detection SNR of DSB-SC is SNR DSB post =

Ac2 P 2 N 0W

Ac2 P = −80 dBm = 10 −11 watts. With a message bandwidth 2 of 4 kHz, the post-detection SNR is The average received power is

SNR

DSB post

10−11 = = 62.5 ~ 18.0 dB (4 ×10−17 )4000

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Problem 9.3. For the same received signal power, compare the post-detection SNRs of DSB-SC with coherent detection and envelope detection with ka = 0.2 and 0.4. Assume the average message power is P = 1. Solution DSB

From Eq. (9.23), the post-detection SNR of DSB-SC with received power DSB SNR post =

Ac2 P 2 N 0W

DSB

(

)

Ac2 2 1 + k a P is 2

AM

From Eq. (9.30), the post-detection SNR of AM with received power AM SNR post =

2

Ac P is 2

Ac2 k a2 P 2 N 0W

AM

So, by equating the transmit powers for DSB-Sc and AM, we obtain DSB

Ac2 P AM Ac2 1 + ka 2 P = 2 2 AM 2 DSB 2 Ac Ac P ⇒ = 2 2 1 + ka 2 P

(

)

Substituting this result into the expression for the post-detection SNR of AM,

SNR AM post =

Ac 2 P ⎛ ka 2 P ⎞ DSB ⎜ ⎟ = SNR post ∆ 2 N 0W ⎝ 1 + ka 2 P ⎠

DSB

Where the factor ∆ is 2

∆=

ka P 2

1+ k a P

With ka= 0.2 and P= 1, the AM SNR is a factor

2 ( .2 ) ∆=

With ka = 0.4 and P = 1, the AM SNR is a factor ∆ =

1.04

= .04 less.

(.4)2 1 + .16

=

.16 ≈ 0.14 less. 1.16

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Problem 9.4. In practice, there is an arbitrary phase θ in Eq. (9.24). How will this affect the results of Section 9.5.2? Solution Envelope detection is insensitive to a phase offset.

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Problem 9.5. The message signal of Problem 9.2 having a bandwidth W of 4 kHz is transmitted over the same noisy channel having a noise spectral density N0/2 of 2x10-17 watts per hertz using single-sideband modulation. If the average received power of the signal is -80 dBm, what is the post-detection signal-to-noise ratio of the receiver? Compare the transmission bandwidth of the SSB receiver to that of the DSB-SC receiver. Solution From Eq. (9.23) SNR SSB post =

with

Ac2 P 2 N 0W

Ac 2 P = −80 dBm , W = 4 kHz , and N 0 = 4 × 10 −17 . The 2

SNR SSB post = 18 dB The transmission bandwidth of SSB is 4 kHz, half of the 8 kHz used with DSB-SC.

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Problem 9.6 The signal m(t) = cos(2000πt) is transmitted by means of frequency modulation. If the frequency sensitivity kf is 2 kHz per volt, what is the Carson’s rule bandwidth of the FM signal. If the pre-detection SNR is 17 dB, calculate the postdetection SNR. Assume the FM demodulator includes an ideal low-pass filter with bandwidth 3.1 kHz. Solution The Carson Rule bandwidth is BT = 2(k f A + f m ) = 2(2(1) + 2 ) = 8 kHz . Then from Eq.(9.59), SNR FM post =

3 Ac2 k 2f P 2 N 0W 3

=

Ac2 2 N 0 BT

⎛ 3k 2f P ⎞ BT ⎟ ⎜⎜ 3 ⎟ W ⎝ ⎠

We observed that the first factor is the pre-detection SNR, and we may write this as 2 1 ⎛ ⎞ FM 3 ⋅ 2 2 ⎜ ⎟ SNR FM SNR 8 = post pre ⎜ ( 3.1)3 ⎟ ⎝ ⎠ FM = SNR pre ×1.61

~ 19.2 dB (There is an error in the answer given in the text.)

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Problem 9.7 Compute the post-detection SNR in the lower channel for Example 9.2 and compare to the upper channel. Solution

The SNR of lower channel is, from Eq. (9.59) SNR FM post =

3 Ac 2 k f 2 ( P / 2) 2 N 0W 3

where we have assumed that half the power is in the lower channel. Using the approximation to Carson’s Rule BT = 2(k f P 1 / 2 + D ) ≈ 2k f P

1

2

= 200 kHz , that is,

k 2f P = BT2 / 4 this expression becomes SNR

FM post

Ac 2 = 2 N 0 BT

3 ( BT / 2 ) 2 W

3 ⎛ BT ⎞ = SNR FM pre ⎜ ⎟ 8⎝ W ⎠

2

3

With a pre-detection SNR of 12 dB, we determine the post-detection SNR as follows SNR

FM post

= SNR

FM pre

3 ⎛ 200 ⎞ ⎜ ⎟ 8 ⎝ 19 ⎠

3

= 1012 /10 × 0.375 × (10.53)3 = 6.94 × 103 ~ 38.4 dB

(The answer in the text for the lower channel is off by factor 0.5 or 3 dB.) For the upper channel, Example 9.2 indicates this result should be scaled by 2/52 and 3 2 FM 3 ⎛ 200 ⎞ SNR FM = SNR post pre ⎜ ⎟ 8 ⎝ 19 ⎠ 52 ~ 24.3 dB So the upper channel is 10log10(52/2) ≈ 14.1 dB worse than lower channel.

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Problem 9.8 An FM system has a pre-detection SNR of 15 dB. If the transmission bandwidth is 30 MHz and the message bandwidth is 6 MHz, what is the post-detection SNR? Suppose the system includes pre-emphasis and de-emphasis filters as described by Eqs. (9.63) and (9.64). What is the post-detection SNR if the f3dB of the de-emphasis filter is 800 kHz? Solution From Eq. (9.59), (see Problem 9.7), the post-detection SNR without pre-emphasis is 3

3 ⎛ BT ⎞ ⎜ ⎟ 4⎝W ⎠ ~ 15 dB + 19.7 dB = 34.7 dB

FM SNR FM post = SNR pre

From Eq. (9.65), the pre-emphasis improvement is

( 6 / 0.8) I= 3 ⎡⎣( 6 / 0.8 ) − tan −1 ( 6 / 0.8) ) ⎤⎦ 3

= 23.2 ~ 13.6 dB With this improvement the post-detection SNR with pre-emphasis is 48.3 dB.

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Problem 9.9 A sample function x(t ) = Ac cos(2πf c t ) + w(t )

is applied to a low-pass RC filter. The amplitude Ac and frequency fc of the sinusoidal component are constant, and w(t) is white noise of zero mean and power spectral density N0/2. Find an expression for the output signal-to-noise ratio with the sinusoidal component of x(t) regarded as the signal of interest. Solution The noise variance is proportional to the noise bandwidth of the filter so from Example 8.16,

[

]

E n 2 (t ) = B N N 0 =

1 N0 4 RC

and the signal power is Ac2 / 2 fir a sinusoid, so the signal-to-noise ratio is given by 2

SNR =

2

Ac 2 A RC = c N0 ⎛ N ⎞ 2⎜ 0 ⎟ ⎝ 4 RC ⎠

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Problem 9.10 A DSC-SC modulated signal is transmitted over a noisy channel, with the power spectral density of the noise as shown in Fig. 9.19. The message bandwidth is 4 kHz and the carrier frequency is 200 kHz. Assume the average received power of the signal is -80 dBm, determine the output signal-to-noise ratio of the receiver. Solution From Fig. 9.19, the noise power spectral density ate 200 kHz is approximately 5x10-19 W/Hz. Using this value for N0/2 (we are assuming the noise spectral density is approximately flat across a bandwidth of 4 kHz), the post-detection SNR is given by

SNR =

Ac2 P 2 N 0W

10−11 = 4 × 103 × 5 × 10−19 = 5 × 103 ~ 37 dB where we have used the fact that the received power is -80 dBm implies that Ac2 P / 2 = 10 −11 watts .

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Problem 9.11 Derive an expression for the post-detection signal-to-noise ratio of the coherent receiver of Fig. 9.6, assuming that the modulated signal s(t) is produced by sinusoidal modulating wave m(t ) = Am cos(2πf m t )

Perform your calculation for the following two receiver types: (a) Coherent DSB-SC receiver (b) Coherent SSB receiver. Assume the message bandwidth is fm. Evaluate these expressions if the received signal strength is 100 picowatts, the noise spectral density is 10-15 watts per hertz, and fm is 3 kHz. Solution (a) The post-detection SNR of the DSB detector is

SNR DSB =

Ac2 P A2 A2 = c m 2 N 0W 4 N 0 f m

(b) The post-detection SNR of the SSB detector is SNR SSB =

Ac2 P A2 A2 = c m 4 N 0W 8 N 0 f m

Although the SNR of the SSB system is half of the DSB-SC SNR, note that the SSB system only transmits half as much power.

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Problem 9.12 Evaluate the autocorrelation function of the in-phase and quadrature components of narrowband noise at the coherent detector input for the DSB-SC system. Assume the band-pass noise spectral density is SN(f) = N0/2 for |f-fc| < BT. Solution From Eg. (8.98), the in-phase power spectral density is (see Section 8.11) S N I ( f ) = S NQ ( f )

⎪⎧ S ( f − f c ) + S N ( f + f c ) =⎨ N ⎪⎩0 ⎧⎪ N =⎨ 0 ⎪⎩0

f < BT / 2

otherwise

f < BT / 2

otherwise

From Example 8.13, the autocorrelation function corresponding to this power spectral density is RNQ (τ ) = RN I (τ ) = N 0 BT sinc ( BTτ )

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Problem 9.13 Assume a message signal m(t) has power spectral density

⎧ f ⎪a SM ( f ) = ⎨ W ⎪0 ⎩

f ≤W

otherwise

where a and W are constants. Find the expression for post-detection SNR of the receiver when (a) The signal is transmitted by DSB-SC. (b) The signal is transmitted by envelope modulation with ka = 0.3. (c) The signal is transmitted with frequency modulation with kf = 500 hertz per volt. Assume that white Gaussian noise of zero mean and power spectral density N0/2 is added to the signal at the receiver input. Solution

(a) with DSB-SC modulation and detection, the post-detection SNR is given by SNR DSB =

Ac2 P 2 N 0W

For the given message spectrum, the power is ∞

P=

∫S

M

( f )df

−∞

W

= 2∫ a 0

f df W

= aW

where we have used the even-symmetry of the message spectrum on the second line. Consequently, the post-detection SNR is SNR

DSB

Ac2 a = 2N 0

(b) for envelope detection with ka = 0.3, the post-detection SNR is

Continued on next slide

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Problem 9.13 continued

SNR AM =

Ac2 ka2 P 2 N 0W

=

Ac2 a 2 ka 2 N0

Ac2 a = 0.09 2 N0 (c) for frequency modulation and detection with kf = 500 Hz/V, the post-detection SNR is

SNR FM =

3 Ac2 k 2f P 2 N 0W 3

Ac2 a ⎛ k f 3⎜ = 2 N 0 ⎜⎝ W

⎞ ⎟⎟ ⎠

2

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Problem 9.14 A 10 kilowatt transmitter amplitude modulates a carrier with a tone m(t) = sin(2000πt), using 50 percent modulation. Propagation losses between the transmitter and the receiver attenuate the signal by 90 dB. The receiver has a front-end noise N0 = -113 dBW/Hz and includes a bandpass filter BT = 2W = 10 kHz. What is the post-detection signal-to-noise ratio, assuming the receiver uses an envelope detector? Solution If the output of a 10 kW transmitter is attenuated by 90 dB through propagation, then the received signal level R is

R = 10 4 × 10 −90 / 10

(1)

= 10 −5 watts For an amplitude modulated signal, this received power corresponds to R=

Ac2 ( 1 + k a2 P ) 2

(2)

From Eq. (9.30), the post-detection SNR of an AM receiver using envelope detection is 2

SNR AM post =

2

AC k a P 2 N 0W

Substituting for ka, P, and Ac2 / 2 (obtained from Eq. (2)), we find SNR AM post =

ka2 P R 1 + ka2 P N 0W

10−5 0.25 × 0.5 × 1 + 0.25 × 0.5 (5 × 10−12 )(5 × 103 ) = 44.4 =

where ka = 0.5 and P = 0.5.

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Problem 9.15 The average noise power per unit bandwidth measured at the front end of an AM receiver is 10-6 watts per Hz. The modulating signal is sinusoidal, with a carrier power of 80 watts and a sideband power of 10 watts per sideband. The message bandwidth is 4 kHz. Assuming the use of an envelope detector in the receiver, determine the output signal-to-noise ratio of the system. By how many decibels is this system inferior to DSB-SC modulation system? Solution For this AM system, the carrier power is 80 watts, that is, 2

AC = 80 watts 2

(1)

and the total sideband power is 20 watts, that is, 2 AC 2 k a P = 20 watts 2

(2)

Comparing Eq.s (1) and (2), we determine that k a2 P = detection SNR of the AM system is 2

1

4

. Consequently, that post-

2

AC k a P 2 N 0W

AM SNR post =

20 10 × 4000 = 5000 ~ 37 dB =

−6

For the corresponding DSB system the post detection SNR is given by SNR DSB post = =

1 + k a2 P SNR AM post k a2 P 1 + 14 1

4

= 5 × SNR AM post ~ 7dB higher

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Problem 9.16 An AM receiver, operating with a sinusoidal modulating wave and 80% modulation, has a post-detection signal-to-noise ratio of 30 dB. What is the corresponding pre-detection signal-to-noise ratio? Solution We are given that ka = 0.80, and for sinusoidal modulation P = 0.5. A post-detection SNR of 30 dB corresponds to an absolute SNR of 1000. From Eq.(9.30), 2

SNR

AM post

2

A k P = C a 2 N 0W

1000 =

Ac2 (0.8) 2 0.5 2 N 0W

Re-arranging this equation, we obtain Ac2 = 3125 2 N 0W From Eq. (9.26 ) the pre-detection SNR is given by 2

AM SNR pre =

(

2

AC 1 + k a P 2 N 0 BT

)

Ac2 = 1 + k a2 P 2 N 0 (2W )

(

(

)

3125 1 + (0.8) 2 0.5 2 = 2062.5 =

)

where we have assumed that BT = 2W. This pre-detection SNR is equivalent to approximately 36 dB.

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Problem 9.17. The signal m(t ) = cos(400π t ) is transmitted via FM. There is an ideal band-pass filter passing 100 ≤ |f| ≤ 300 at the discriminator output. Calculate the postdetection SNR given that kf = 1 kHz per volt, and the pre-detection SNR is 500. Use Carson’s rule to estimate the pre-detection bandwidth. Solution We begin by estimating the Carson’s rule bandwidth

BT = 2(k f A + f m )

= 2(1000(1) + 200 ) = 2400 Hz

We are given that the pre-detection SNR is 500. From Section 9.7 this implies FM = SNR pre

500 =

Ac2 2 N 0 BT Ac2 1 2 N 0 2400

Re-arranging this equation, we obtain Ac2 = 1.2 × 10 6 Hz 2N 0 The nuance in this problem is that the post-detection filter is not ideal with unity gain from 0 to W and zero for higher frequencies. Consequently, we must re-evaluate the postdetection noise using Eq. (9.58) 300 ⎤ N 0 ⎡ −100 2 Avg. post - detection noise power = 2 ⎢ ∫ f df + ∫ f 2 df ⎥ Ac ⎣ −300 100 ⎦ 2N 0 = 300 3 − 100 3 3 Ac2

[

=

]

2N 0 2.6 × 10 7 3 Ac2

The post-detection SNR then becomes

Continued on next slide

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Problem 9.17 continued

SNR

FM post

=

3 Ac2 k 2f P

(

2 N 0 2.6 × 10 7

⎛ A2 = 3⎜⎜ c ⎝ 2N 0

)

2 ⎞ kf P ⎟⎟ 7 ⎠ 2.6 × 10

(

= 3 1.2 × 10 6

) (1000) 0.5 2

2.6 × 10 7

= 69230.8 where we have used the fact that kf = 1000 Hz/V and P = 0.5 watts. In decibels, the postdetection SNR is 48.4 dB.

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Problem 9.18. Suppose that the spectrum of a modulating signal occupies the frequency band f1 ≤ f ≤ f 2 . To accommodate this signal, the receiver of an FM system (without

pre-emphasis) uses an ideal band-pass filter connected to the output of the frequency discriminator; the filter passes frequencies in the interval f1 ≤ f ≤ f 2 . Determine the output signal-to-noise ratio and figure of merit of the system in the presence of additive white noise at the receiver input. Solution Since the post detection filter is no longer an ideal brickwall filter, we must revert to Eq. (9.58) to compute the post-detection noise power. For this scenario (similar to Problem 9.17)

Avg. post - detection noise power = =

N0 Ac2

⎡ − f1 2 ⎢ ∫ f df + ⎢⎣ − f 2

[

2N 0 3 f 2 − f 13 3 Ac2

f2

∫

f1

⎤ f 2 df ⎥ ⎥⎦

]

Since the average output power is still k 2f P , the post detection SNR is given by FM = SNR post

3 Ac2 k 2f P

(

2 N 0 f 23 − f13

)

For comparison purposes, the reference SNR is SNRref

Ac2 = 2 N 0 ( f 2 − f1 )

The corresponding figure of merit is Figure of merit = = =

SNR FM post SNR ref 3 Ac2 k 2f P

(

2 N 0 f 23 − f 13

)

Ac2 2 N 0 ( f 2 − f1 )

3k 2f P f 22 + f 2 f 1 + f 12

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Problem 9.19. An FM system, operating at a pre-detection SNR of 14 dB, requires a post-detection SNR of 30 dB, and has a message power of 1 watt and bandwidth of 50 kHz. Using Carson’s rule, estimate what the transmission bandwidth of the system must be. Suppose this system includes pre-emphasis and de-emphasis network with f3dB of 10 kHz. What transmission bandwidth is required in this case? Solution We are given the pre-detection SNR of 14 dB (~25.1), so Ac2 FM SNR pre = = 25.1 2 N 0 BT

and the post-detection SNR of 30 dB (~1000), so SNR

FM post

=

3 Ac2 k 2f P 2 N 0W 3

= 1000

Combining these two expressions, we obtain FM SNR post FM SNR pre

=

3k 2f PBT W3

= 39.8

Approximating the Carson’s rule for general modulation BT = 2 ( k f P1/ 2 + W ) ≈ 2k f P1/ 2 , and if we replace k 2f P with BT2 / 4 in this last equation, we obtain FM SNR post FM SNR pre

≈

3BT3 = 39.8 4W 3

Upon substituting W = 50 kHz, this last equation yields BT = 187.9 kHz.

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Problem 9.20. Assume that the narrowband noise n(t) is Gaussian and its power spectral density SN(f) is symmetric about the midband frequency fc. Show that the in-phase and quadrature components of n(t) are statistically independent. Solution The narrowband noise n(t) can be expressed as:

n(t ) = nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) = Re ⎡⎣ z (t )e j 2π fct ⎤⎦

,

where nI(t) and nQ(t) are in-phase and quadrature components of n(t), respectively. The term z(t) is called the complex envelope of n(t). The noise n(t) has the power spectral density SN(f) that may be represented as shown below

We shall denote Rnn (τ ) , RnI nI (τ ) and RnQ nQ (τ ) as autocorrelation functions of n(t), nI(t) and nQ(t), respectively. Then Rnn (τ ) = E [ n(t )n(t + τ ) ]

{

}

= E ⎡⎣ nI (t ) cos(2π f ct ) − nQ (t ) sin(2π f c t ) ⎤⎦ ⋅ ⎡⎣ nI (t + τ ) cos(2π f c (t + τ )) − nQ (t + τ ) sin(2π f c (t + τ )) ⎤⎦ 1 1 = ⎡⎣ RnI nI (τ ) + RnQ nQ (τ ) ⎤⎦ cos(2π f cτ ) + ⎡⎣ RnI nI (τ ) − RnQ nQ (τ ) ⎤⎦ cos(2π f c (2t + τ )) 2 2 1 1 − ⎡⎣ RnQ nI (τ ) − RnI nQ (τ ) ⎤⎦ sin(2π f cτ ) − ⎡⎣ RnQ nI (τ ) + RnI nQ (τ ) ⎤⎦ sin(2π f c (2t + τ )) 2 2 Since n(t) is stationary, the right-hand side of the above equation must be independent of t, this implies RnI nI (τ ) = RnQ nQ (τ ) (1)

RnI nQ (τ ) = − RnQ nI (τ )

(2)

Continued on next slide

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Problem 9.20 continued Substituting the above two equations into the expression for Rnn(τ), we have

Rnn (τ ) = RnI nI (τ ) cos(2π f cτ ) − RnQ nI (τ ) sin(2π f cτ )

(3)

The autocorrelation function of the complex envelope z (t ) = nI (t ) + jnQ (t ) is

Rzz (τ ) = E ⎡⎣ z* (t ) z (t + τ ) ⎤⎦

(4)

= 2 RnI nI (τ ) + j 2 RnQ nI (τ ) From the bandpass to low-pass transformation of Section 3.8, the spectrum of the complex envelope z is given bye ⎧⎪ S ( f + f c ) SZ ( f ) = ⎨ N ⎪⎩0

f > − fc otherwise

Since SN(f) is symmetric about fc, SZ(f) is symmetric about f = 0. Consequently, the inverse Fourier transform of SZ(f) = Rzz(τ) must be real. Since Rzz (τ ) is real valued, based on Eq. (4), we have

RnQ nI (τ ) = 0 , which means the in-phase and quadrature components of n(t) are uncorrelated. Since the in-phase and quadrature components are also Gaussian, this implies that they are also statistically independent.

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Problem 9.21. Suppose that the receiver bandpass-filter magnitude response H BP ( f )

has symmetry about ± f c and noise bandwidth BT. From the properties of narrowband noise described in Section 8.11, what is the spectral density SN(f) of the in-phase and quadrature components of the narrowband noise n(t) at the output of the filter? Show that the autocorrelation of n(t) is RN (τ ) = ρ (τ ) cos(2π f cτ ) where ρ (τ ) = F −1 [ S N ( f )] ; justify the approximation ρ (τ ) ≈ 1 for τ < 1 / BT . Solution Let the noise spectral density of the bandpass process be SH(f) then N0 2 H BP ( f ) 2 From Section 8.11, the power spectral densities of the in-phase and quadrature components are given by SH ( f ) =

⎧⎪ S ( f − f c ) + S H ( f + f c ), SN ( f ) = ⎨ H ⎪⎩0,

f ≤ BT / 2 otherwise

.

Since the spectrum SH(f) is symmetric about fc, , the spectral density of the in-phase and quadrature components is ⎧⎪ H ( f − f c ) 2 N 0 S N ( f ) = ⎨ BP ⎪⎩0

f < BT / 2

(1)

otherwise

Note that if |HBP(f)| is symmetric about fc then |HBP(f-fc)| will be symmetric about 0. Consequently, the power spectral densities of the in-phase and quadrature components are symmetric about the origin. This implies that the corresponding autocorrelation functions are real valued (since they are related by the inverse Fourier transform). In Problem 9.20, we shown that if the autocorrelation function of the in-phase component is real valued then autocorrelation of n(t) is RN (τ ) = RnI nI (τ ) cos(2π f cτ ) . If we denote 2 ρ (τ ) = Rn n (τ ) = F −1 [ S N ( f )] = NO F −1 ⎡ H BP ( f − f c ) ⎤ I I

⎣

⎦

Continued on next slide

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Problem 9.21 continued

then the autocorrelation of the bandpass noise is R N (τ ) = ρ (τ ) cos(2πf c t ) For τ 1/ BT (there is a typo in the text), we have

ρ (τ ) =

∞

∫S

N

( f ) exp ( − j 2π f τ ) df

−∞ ∞

= ∫ S N ( f ) cos ( 2π f τ ) df 0

due to the real even-symmetric nature of SN(f). If the signal has noise bandwidth BT then

ρ (τ ) ≈

BT

∫S

N

( f ) cos ( 2π f τ ) df

N

( f ) cos ( 0 ) df

N

( f )df

0

≈

BT

∫S 0

=

BT

∫S 0

= a constant where the second line follows from the assumption that τ 1/ BT . With suitable scaling the constant can be set to one.

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Problem 9.22. Assume that, in the DSB-SC demodulator of Fig. 9.6, there is a phase error φ in the synchronized oscillator such that its output is cos(2π f c t + φ ) . Find an expression for the coherent detector output and show that the post-detection SNR is reduced by the factor cos 2 φ . Solution The signal at the input to the coherent detector of Fig. 9.6 is x(t) where

x(t ) = s(t ) + nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) = Ac m(t ) cos(2π f ct ) + nI (t ) cos(2π f c t ) − nQ (t )sin(2π f ct ) The output of mixer2 in Fig. 9.6 is v(t ) = x(t ) cos(2π f c t + φ )

= [ Ac m(t ) + nI (t ) ] cos(2π f c t ) cos(2π f c t + φ ) − nQ (t ) sin(2π f c t ) cos(2π f c t + φ ) =

1 1 1 1 [ Ac m(t ) + nI (t )] cos φ + nQ (t ) sin φ + [ Ac m(t ) + nI (t )] cos(4π f ct + φ ) − nQ (t ) sin(4π f ct + φ ) 2 2 2 2

With the higher frequency components will be eliminated by the low pass filter, the received message at the output of the low-pass filter is y (t ) =

1 1 1 Ac m(t ) cos φ + nI (t ) cos φ + nQ (t ) sin φ 2 2 2

To compute the post-detection SNR we note that the average output message power in this last expression is 1 2 Ac P cos 2 φ 4

and the average output noise power is 1 1 1 ⋅ 2 N 0W cos 2 φ + ⋅ 2 N 0W sin 2 φ = ⋅ 2 N 0W 4 4 4

[

] [

]

where E n I2 (t ) = E nQ2 (t ) = N 0W . Consequently, the post-detection SNR is SNR =

1/ 4 Ac2 P cos 2 φ Ac2 P cos 2 φ = 1/ 4 ⋅ 2 N 0W 2 N 0W

Compared with (9.23), the above post-detection SNR is reduced by a factor of cos 2 φ .

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Problem 9.23. In a receiver using coherent detection, the sinusoidal wave generated by the local oscillator suffers from a phase error θ(t) with respect to the carrier wave cos(2π f c t ) . Assuming that θ(t) is a zero-mean Gaussian process of variance σ θ2 and that most of the time the maximum value of θ(t) is small compared to unity, find the meansquare error of the receiver output for DSB-SC modulation. The mean-square error is defined as the expected value of the squared difference between the receiver output and message signal component of a synchronous receiver output. Solution Based on the solution of Problem 9.22, we have the DSB-SC demodulator output is y (t ) =

1 1 1 Ac m(t ) cos [θ (t ) ] + nI (t ) cos [θ (t ) ] + nQ (t ) sin [θ (t ) ] 2 2 2

Recall from Section 9. that the output of a synchronous receiver is 1 1 Ac m(t ) + nI (t ) 2 2

The mean-square error (MSE) is defined by 2 ⎡⎛ 1 ⎞ ⎤ MSE = E ⎢⎜ y (t ) − Ac m(t ) ⎟ ⎥ 2 ⎠ ⎥⎦ ⎢⎣⎝

Substituting the above expression for y(t), the mean-square error is 2 ⎡⎡ 1 1 1 ⎤ ⎤ MSE = E ⎢ ⎢ Ac m(t ) ⎡⎣cos (θ (t ) ) − 1⎤⎦ + nI (t ) cos (θ (t ) ) + nQ (t ) sin (θ (t ) ) ⎥ ⎥ 2 2 ⎦ ⎥⎦ ⎢⎣ ⎣ 2 2 A2 1 1 = c E ⎢⎡ m 2 (t ) ⎡⎣cos (θ (t ) ) − 1⎤⎦ ⎥⎤ + E ⎡⎣ nI2 (t ) cos 2 (θ (t ) ) ⎤⎦ + E ⎡⎣ nQ2 (t ) sin 2 (θ (t ) ) ⎤⎦ ⎦ 4 4 ⎣ 4

where we have used the independence of m(t), nI(t), nQ(t), and θ(t) and the fact that E [ nI (t ) ] = E ⎡⎣ nQ (t ) ⎤⎦ = 0 to eliminate the cross terms.

Continued on next slide

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Problem 9.23 continued 2 Ac2 1 1 E ⎡⎣ m 2 (t ) ⎤⎦ E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + E ⎡⎣ nI2 (t ) ⎤⎦ E ⎡⎣cos 2 (θ (t ) ) ⎤⎦ + E ⎡⎣ nQ2 (t ) ⎤⎦ E ⎡⎣sin 2 (θ (t ) ⎤⎦ MSE = ⎣ ⎦ 4 4 4 2 2 AP 1 1 = c E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + N 0WE ⎡⎣cos 2 (θ (t ) ) ⎤⎦ + N 0WE ⎡⎣sin 2 (θ (t ) ⎤⎦ ⎣ ⎦ 4 4 4 2 2 AP NW = c E ⎡⎢(1 − cos (θ (t ) ) ) ⎤⎥ + 0 ⎣ ⎦ 4 2

[

] [

]

where we have used the equivalences of E[m2(t)] = P, and E n I2 (t ) = E nQ2 (t ) = 2 N 0W . The last line uses the fact that cos (θ(t))+sin (θ(t)) = 1. If we now use the relation that 1-cos A = 2sin2(A/2), this expression becomes 2

2

⎡ ⎛ θ (t ) ⎞ ⎤ N 0W MSE = Ac2 PE ⎢sin 4 ⎜ ⎟⎥ + 2 ⎝ 2 ⎠⎦ ⎣ Since the maximum value of θ(t) > 2W . Solution We modify Eq. (9.58) to include the effects of a non-ideal post-detection filter in order to estimate the average post-detection noise power: N0 Ac2

∫

W

−W

f 2 | H BP ( f ) |2 df = =

N0 Ac2

∫

2 N0 Ac2

W

−W

∫

W

0

f2⋅

1 df 1 + ( f / W )4

f2⋅

1 df 1 + ( f / W )4

This can be evaluated by a partial fraction expansion of the integrand but for simplicity, we appeal to the formula:

x 2 dx 1 ∫ a + bx4 = 4bk

⎡1 x 2 − 2kx + 2k 2 2kx ⎤ + tan −1 2 log , ⎢2 2 2 x + 2kx + 2k 2k − x 2 ⎥⎦ ⎣

ab > 0, k =

4

a 2b

Using this result, we get the average post-detection noise power is Avg. post-detection noise power =

2 N0 W 3 ⋅ Ac2 4 2

⎡ ⎤ N 0W 3 2− 2 log 0.42 π + = ⎢ ⎥ Ac2 2+ 2 ⎣ ⎦

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Problem 9.25. Consider a communication system with a transmission loss of 100 dB and a noise density of 10-14 W/Hz at the receiver input. If the average message power is P = 1 watt and the bandwidth is 10 kHz, find the average transmitter power (in kilowatts) required for a post-detection SNR of 40 dB or better when the modulation is: (a) AM with ka = 1; repeat the calculation for ka = 0.1. (b) FM with kf = 10, 50 and 100 kHz per volt. In the FM case, check for threshold limitations by confirming that the pre-detection SNR is greater that 12 dB.

Solution (a) In the AM case, the post detection SNR is given by SNR AM post =

Ac2 k a2 P 2 N oW

Ac2 k a2 (1) 10 = 2(2 × 10 −14 )(10 4 ) 4

Ac2 k a2 = 2 × 10 −6 2 where an SNR of 40 dB corresponds to 104 absolute and N0/2 = 10-14 W/Hz. For the different values of ka ka = 1 ⇒

Ac2 = 4 × 10 −6

k a = 0.1 ⇒

Ac2 = 4 × 10 −4

Average modulated signal power at the input of the detector is

1 2 Ac (1 + ka2 P ) . 2

1 2 Ac (1 + ka2 P ) = 4 × 10−6 2 1 2 ka = 0.1 ⇒ Ac (1 + ka2 P ) = 2.02 × 10−4 2 ka = 1 ⇒

The transmitted power is 100dB (1010) greater than the received signal power so ka = 1 ⇒ transmitted power = 4 × 10 4 = 40 kW ka = 0.1 ⇒ transmitted power = 2.02 × 106 = 2020 kW

Continued on next slide

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Problem 9.25 continued (b) In the FM case, the post detection SNR is SNR

FM post

=

10 = 4

Ac2 k 2f 2

3 Ac2 k 2f P 2 N oW 3 3 Ac2 k 2f (1) 2(2 × 10 −14 )(10 4 ) 3

= 0.667 × 10 2

For the different values of ka A c2 k f = 10 kHz/V ⇒ = 0.667 × 10 −6 2 A2 k f = 50 kHz/V ⇒ c = 26.667 × 10 −9 2 A c2 k f = 100 kHz/V ⇒ = 0.667 × 10 −8 2 The transmitted power is 100dB (1010) greater than the received signal power so kf = 10 kHz/V ⇒ transmitted power = 0.667x104 W = 6.67 kW kf = 50 kHz/V ⇒ transmitted power = 26.667x101 W = 0.27 kW kf = 100 kHz/V ⇒ transmitted power = 0.667x102 W = 0.07 kW To check the pre-detection SNR, we note that it is given by : Ac2 Ac2 FM SNR pre = = 2 N 0 BT 4 N 0 (k f P1/ 2 + W ) where from Carson’s rule BT = 2(k f P1/ 2 + W ) . From the above Ac2 = FM = SNR pre

4 × 102 , so 3k 2f

4 × 102 102 = 3k 2f × 4 N 0 (k f P1/ 2 + W ) 3k 2f × 2 × 10−14 ( k f + 104 )

For the different values of kf, the pre-detection SNR is FM k f = 10kHz ⇒ SNR pre = 104 /12 = 29dB > 12dB FM k f = 50kHz ⇒ SNR pre = 11.11 = 10.45dB < 12dB FM k f = 100kHz ⇒ SNRpre = 1.515 = 1.8dB < 12dB

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Problem 9.25 continued Therefore, for kf = 50 kHz and 100 kHz, the pre-detection SNR is too low and the transmitter power would have to be increased by 1.55 dB and 10.2 dB, respectively.

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Problem 9.26 In this experiment we investigate the performance of amplitude modulation in noise. The MatLab script for this AM experiment is provided in Appendix 8 and simulates envelope modulation by a sine wave with a modulation index of 0.3, adds noise, and then envelope detects the message. Using this script: (a) Plot the envelope modulated signal. (b) Using the supporting function “spectra”, plot its spectrum. (c) Plot the envelope detected signal before low-pass filtering. (d) Compare the post-detection SNR to theory. Using the Matlab script given in Appendix 7 we obtain the following plots (a) By inserting the statements plot(t,AM) xlabel('Time') ylabel('Amplitude') at the end of Modulator section of the code, we obtain the following plot of the envelope modulated signal: 2 1.5 1

Amplitude

0.5 0 -0.5 -1 -1.5 -2

0

0.2

0.4

0.6

0.8

1 Time

1.2

1.4

1.6

1.8

2

Continued on next slide

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Problem 9.26 continued (b) The provided script simulates 2 seconds of the AM signal. Since the modulating signal is only 2 Hz, this is not a sufficient signal length to accurately estimate the spectrum. We extend the simulation to 200 seconds by modifying the statement t = [0:1/Fs:200]; To plot the spectrum, we insert the following statements after the AM section [P,F] = spectrum(AM,4096,0,4096,Fs); plot(F,10*log10(P(:,1))) xlabel('Frequency') ylabel('Spectrum') We use the large FFT size of 4096 to provide sufficient frequency resolution. (The resolution is Fs (1000 Hz) divided by the FFT size. We plot the spectrum of decibels because it more clearly shows the sideband components. With a linear plot, and this low modulation index, the sideband components would be difficult to see. The following figure enlarges the plot around the carrier frequency of 100 Hz.

30 20

Spectrum

10 0 -10 -20 -30 -40 -50 85

90

95 100 Frequency

105

(c) To plot the envelope-detected signal before low-pass filtering, we insert the statements (Decrease the time duration to 2 seconds to speed up processing for this part.) plot(AM_rec) xlabel('Time samples') ylabel('Amplitude') The following plot is obtained and illustrates the tracking of the envelope detector. Continued on next slide

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Problem 9.26 continued

Amplitude

0.5

0

-0.5

0

500

1000 Time samples

1500

2000

(d) To compare the simulated post detection SNR to theory. Create a loop around the main body of the simulation by adding the following statements for kk = 1:15 SNRdBr = 10 + 2*kk …. PreSNR(kk) = 20*log10(std(RxAM)/std(RxAMn-RxAM)); No(kk) = 2*sigma^2/Fs; …. SNRdBpost(kk) = 10*log10(C/error); W = 50; P = 0.5; Theory(kk) = 10*log10 ( A^2*ka^2*0.5 / (2*No(kk)*W)); end plot(PreSNR, SNRdBpost) hold on, plot(PreSNR, Theory,'g'); The results are shown in the following chart. Continued on next slide

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Problem 9.26 continued

35

30

Post-detection SNR (dB)

25

20 Simulation 15 Theory 10

5

0 10

15

20

25 Pre-detection SNR (dB)

30

35

40

These results indicate that the simulation is performing slightly better than theory? Why? As an exercise try adjusting either the frequency of the message tone or the decay of the envelope detector and compare the results.

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Problem 9.27. In this computer experiment, we investigate the performance of FM in noise. Using the Matlab script for the FM experiment provided in Appendix 8: (a) Plot the spectrum of the baseband FM phasor. (b) Plot the spectrum of the band-pass FM plus noise. (c) Plot the spectrum of the detected signal prior to low-pass filtering. (d) Plot the spectrum of the detected signal after low pass filtering. (e) Compare pre-detection and post-detection SNRs for an FM receiver. In the following parts (a) through (d), set the initial CNdB value to 13 dB in order to be operating above the FM threshold. (a) By inserting the following statements after the definition of FM, we obtain the baseband spectrum [P,F] = spectrum(FM,4096,0,4096,Fs); plot(F,P(:,1)) xlabel('Frequency (Hz)') ylabel('Spectrum') An enlarged snapshot of the spectrum near 0 Hz is shown here. It shows the tones at the regular spacing that one would expect with FM tone modulation. Note that initial plot shows the “negative frequency” portion of the spectrum just below Fs = 500 Hz. This is due to the nature of the FFT and the sampling process.

140

120

Spectrum

100

80

60

40

20

0

0

5

10

15

20 25 30 Frequency (Hz)

35

40

45

Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 9.27 continued (b) The spectrum of the bandpass FM plus noise is obtained by inserting the statements [P,F] = spectrum((FM+Noise).*Carrier,4096,0,4096,Fs); plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum') An expanded view of the result around the carrier frequency of 50 Hz is shown below. The spectrum has been plotted on a decibel scale to show both the FM tone spectrum and the noise pedestal.

20

Spectrum(dB)

0

-20

-40

-60

-80

-100

10

20

30

40 50 60 Frequency (Hz)

70

80

90

100

(c) To plot the spectrum of the noisy signal before low-pass filtering, we insert the following statements in the FM discriminator function, prior to the low pass filter [P,F] = spectrum(BBdec,1024,0,1024,Fsample/4) plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum(dB)') Continued on next slide

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Problem 9.27 continued The following plot is obtained when expanded near the origin. We plot the spectrum in decibels in order to show the noise and the non-flat nature of its spectrum more clearly. The decibel scale also illustrates some low-level distortion that has been introduced by the demodulation process as exhibited by the small second harmonic at 2 Hz and the low dc level.

60

Spectrum(dB)

50

40

30

20

10

0

5

10

15 Frequency (Hz)

20

25

30

(d) To plot the spectrum of the noisy signal before low-pass filtering, we insert the following statements in the FM discriminator function, after the low-pass filter [P,F] = spectrum(Message,1024,0,1024,Fsample/4) plot(F,10*log10(P(:,1))) xlabel('Frequency (Hz)') ylabel('Spectrum(dB)') The following plot is obtained when expanded near the origin. Again we plot the spectrum in decibels in order to show the noise and, in this case, the effect of the lowpass filtering. The low-pass filtering does not affect the distortion introduced by the demodulator in the passband. Continued on next slide

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Problem 9.27 continued

60

Spectrum(dB)

40

20

0

-20

-40

5

10

15 Frequency (Hz)

20

25

(a) Running the code as provided produces the following comparison of the postdetection and pre-detection SNR.

45

40

Post-detection SNR (dB)

35

30

25

20

15

10

0

5

10

15

20

25

C/N (dB)

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Problem 10.1. Let H 0 be the event that a 0 is transmitted and let R0 be the event that a 0 is received. Define H1 and R1, similarly for a 1. Express the BER in terms of the probability of these events when: (a) The probability of a 1 error is the same as the probability of a 0 error. (b) The probability of a 1 being transmitted is not the same as the probability of a 0 being transmitted. Solution In both cases, the probability of error may be expressed as

P[error ] = P(R0 H 1 )P(H 1 ) + P(R1 H 0 )P(H 0 )

(1)

(a) The BER is the same as the P[error] and with P(R0|H1) = P(R1/H0) = p then P[error ] = p[P( H 1 ) + P(H 0 )] = p since P(H1)+P(H0) = 1. (b) With P(H0) ≠ P(H1), the answer is given by the general result of Eq. (1).

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Problem 10.2. Suppose that in Eq. (10.4), r(t) represents a complex baseband signal instead of a real signal. What would be the ideal choice for g(t) in this case? Justify your answer. Solution Inspecting the Schwarz inequality of Eq. (10.12), we see that equality is achieved with

g (T − t ) = cs* ( t ) if s(t) is complex.

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⎡ α (t − T / 2) ⎤ Problem 10.3 If g (t ) = c rect ⎢ , determine c such g(t) satisfies Eq. (10.10) T ⎣ ⎦⎥ where α > 1.

Solution From the definition of the rect(.) function,

⎛ α (t − T / 2) ⎞ g (t ) = c rect ⎜ ⎟ T ⎝ ⎠ t − T / 2 < T /(2α )

⎧⎪c =⎨ ⎪⎩0

otherwise

Substituting this into Eq. (10.10) 2

T = ∫ g ( t ) dt T

0

= c2 ∫

T / 2 +T /(2α )

T / 2 −T /(2α )

12 dt

= c 2T / α

And so c = α .

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Problem 10.4. Show that with on-off signaling, the probability of a Type II error in Eq.(10.23) is given by ⎛γ ⎞ P[Y > γ | H 0 ] = Q ⎜ ⎟ ⎝σ ⎠ Solution A Type II error probability is

1 P[Y > γ | H 0 ] = 2πσ Let s =

y

σ

+∞

∫γ

⎛ y2 ⎞ exp ⎜ − 2 ⎟dy ⎝ 2σ ⎠

, and then

P[Y > γ | H 0 ] =

1 2π

⎛ s2 ⎞ ⎛γ ⎞ exp ∫γ /σ ⎜⎝ − 2 ⎟⎠ds = Q ⎜⎝ σ ⎟⎠ +∞

using the definition of the Q-function given in Section 8.4.

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Problem 10.5 Prove the property of root-raised cosine pulse shape p(t) given by Eq. (10.32), using the following steps: (a) If R ( f ) is the Fourier transform representation of p(t), what is the Fourier transform representation of p(t-lT)? ∞

(b) What is the Fourier transform of q (τ ) =

∫ p(τ − t ) p(t − lT )dt ? What spectral

−∞

shape does it have? (c) What q(τ)? What is q(kT)? Use these results to show that Eq. (10.32) holds. Solution (a) From the time-shifting property of Fourier transforms (see Section 2.2 ), we have that

F[ p(t − lT )] = R( f ) exp(− j 2πflT ) (b) From the convolution property of Fourier transforms (See Section 2.2) we have that Q ( f ) = F[q (τ )] = F[ p (t )]F[ p (t − lT )]

= R 2 ( f ) exp(− j 2πflT )

(c) Since R(f) is the root-raised cosine spectrum, R2(f) is the raised cosine spectrum and so q(τ) corresponds to a raised cosine pulse. In particular, using the time-shifting property of inverse Fourier transforms q (τ ) = m (τ − lT )

where m(τ) is the raised cosine pulse shape. Using the properties of the raised cosine pulse shape (see Section 6.4) q(kT ) = m(kT − lT ) ⎧1 =⎨ ⎩0 = δ (k − l )

k =l k ≠l

and Eq. (10.32) holds.

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Problem 10.6 Compare the transmission bandwidth required for binary PAM and BPSK modulation, if both signals have a data rate of 9600 bps and use root-raised cosine pulse spectrum with a roll-off factor of 0.5. Solution

1+ β , 2T where β is the roll-off factor (0.5) and T is the symbol duration (1/9600 sec). Therefore, BT = (1+0.5)×9600 = 14.4 kHz.

For BPSK modulation (bandpass signal), the transmission bandwidth is BT = 2 ×

For binary PAM modulation (baseband signal), BT =

1+ β = 7.2 kHz. 2T

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Problem 10.7 Sketch a block diagram of a transmission system including both transmitter and receiver for BPSK modulation with root-raised cosine pulse shaping. Solution

The BPSK transmitter with root-raised cosine pulse shaping is shown in (a), and the corresponding BPSK receiver is shown in (b). (a)

Data

Modulated Impulse Train

s (t ) = ± m(t ) sin( 2πf c t )

Root Raised Cosine Filter

sin(2πf c t )

(b) Data

Root Raised Cosine Filter

r (t )

T-spaced Sample Clock

sin(2πf c t )

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Problem 10.8 Show that the integral of the high frequency term in Eq. (10.53) is approximately zero. Solution Consider the integral over the period from 0 to T of the high frequency term in Eq. (10.53):

∫

T

0

AC 2 A2 cos ( 4π f c t + 2φ (t ) ) dt = c sin ( 4π f ct + 2φ (t ) ) 2 8π f c

T 0

=

Ac 2 ⎡sin ( 4π f cT + 2φ (T ) ) − sin ( 2φ ( 0 ) ) ⎤⎦ 8π f c ⎣

<

Ac2 4π f c

where the first line follows since φ(t) is constant over a symbol interval. By the bandpass assumption fc >> 1, so this last line is small.

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Problem 10.9. Use Eqs. (10.61), (10.64), and (10.66) to show that N1 and N2 are uncorrelated and therefore independent Gaussian random variables. Compute the variance of N1-N2. Solution The correlation of N1 and N2 is

E ( N1 N 2 ) = E ⎡ 2∫ ⎣⎢ 0

T

= 2∫

T

0

=2

∫

T

0

w( s ) w(t ) cos ( 2π f1t ) cos ( 2π f 2 s ) dsdt ⎤ ⎦⎥

∫ E [ w(s)w(t )] cos ( 2π f t ) cos ( 2π f s ) dsdt T

1

0

N0 2

2

∫ ∫ δ ( t − s ) cos ( 2π f t ) cos ( 2π f s ) ds dt 1

2

= N 0 ∫ cos ( 2π f1t ) cos ( 2π f 2t ) dt T

0

=0 where the last line follows from Eq.(10.61). Since N1 and N2 are uncorrelated 2 2 2 E ⎡( N1 − N 2 ) ⎤ = E ⎡( N1 ) ⎤ + 2E [ N1 N 2 ] + E ⎡( N 2 ) ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2 2 = E ⎡( N1 ) ⎤ + E ⎡( N 2 ) ⎤ ⎣ ⎦ ⎣ ⎦

The variance of the N1 term is E ( N1 N1 ) = E ⎡ 2∫ ⎣⎢ 0

T

= 2∫

T

0

=2

∫

T

0

w( s ) w(t ) cos ( 2π f1t ) cos ( 2π f1s ) dsdt ⎤ ⎦⎥

∫ E [ w(s)w(t )] cos ( 2π f t ) cos ( 2π f s ) dsdt T

1

0

N0 2

1

∫ ∫ δ ( t − s ) cos ( 2π f t ) cos ( 2π f s ) ds dt 1

1

= N 0 ∫ cos 2 ( 2π f 0t ) dt T

0

Using the double angle formula 2cos2θ = 1+2cosθ, we have N T 2 E ⎡( N1 ) ⎤ = 0 ∫ (1 + cos 4π ft ) dt ⎣ ⎦ 2 0 NT = 0 2

The derivation of the variance of N2 is similar and the combined variance is N0T.

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Problem 10.10. Plot the BER performance of differential BPSK and compare the results to Fig. 10.16. Solution The bit error probability of differential BPSK is (Eq. (10.75))

⎛ E ⎞ PeDPSK = 0.5exp ⎜ − b ⎟ . ⎝ N0 ⎠ The following Matlab script plots this performance EbNodB=[0:0.25:12]; EbNo = 10.^(EbNodB/10); BER = 0.5*exp(-EbNo); semilogy(EbNodB,BER) grid xlabel('Eb/No (dB)') ylabel('BER of DPSK') axis([0 20 1E-7 0.1]) This script produces the following plot. -1

10

-2

10

-3

BER of DPSK

10

-4

10

-5

10

-6

10

-7

10

0

2

4

6

8

10 12 Eb/No (dB)

14

16

18

20

The performance of DPSK is slightly worse than BPSK and QPSK. The relative loss with DPSK is less than 1 dB at Eb/N0 of 8 dB and higher. The loss at lower Eb/N0 ratios is greater. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 10.11. A communication system that transmits single isolated pulses is subject to multipath such that, if the transmitted pulse is p(t) of length T, the received signal is s (t ) = p (t ) + α p (t − τ )

Assuming that α and τ are known, determine the optimum receiver filter for signal in the presence of white Gaussian noise of power spectral density N0/2. What is the postdetection SNR at the output of this filter? Solution We first note that the pulse is non-zero over the interval 0 ≤ t ≤ T + τ . From Section 10.2 the appropriate linear receiver is

Y =∫

T +τ

0

g (T + τ − u )r (u )du

and the optimum choice for g(t) is

g (T + τ − t ) = c ( p(t ) + α p(t − τ ) ) where c is chosen such that T +τ

∫

g (t ) dt = T + τ 2

0

With this filtering arrangement, if follows from the modified Eq. (10.9) that E ⎡⎣ N 2 ⎤⎦ =

N 0 (T + t ) 2

Continued on next slide

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Problem 10.11 continued The corresponding signal level S is T +τ

S =c

∫

g (T − t ) ( p (t ) + α p (t + τ ) ) dt

0

T +τ

=c

∫ ( p(t ) + α p(t + τ ) )

2

dt

0

= T +τ which follows from the normalization properties of c. The received signal to noise is then

S2 T +τ SNR = = 2 E ⎡⎣ N ⎤⎦ N 0 / 2 Although the units on this expression may appear unusual, note that the units of N0 are (volt)2/Hz = (volt)2-sec. The units of the numerator are also (volt)2-sec, although the (volt)2 has been suppressed. Consequently, the SNR is dimensionless, as it should be.

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Problem 10.12. The impulse response corresponding to a root-raised cosine spectrum, normalized to satisfy Eq.(10.10), is given by ⎡ (1 + α )πt ⎤ T ⎡ (1 − α )πt ⎤ cos ⎢ sin ⎢ + ⎥ ⎥⎦ 4α ⎣ T ⎦ 4αt ⎣ T g (t ) = 2 π ⎛ 4αt ⎞ 1− ⎜ ⎟ ⎝ T ⎠

where T = 1/2B0 is the symbol period and α is the roll-off factor. Obtain a discrete-time representation of this impulse response by sampling it at t = 0.1nT for integer n such that –3T < t < 3T. Numerically approximate match filtering (e.g. with Matlab) by performing the discrete-time convolution 60

q k = 0 .1 ∑ g n g k − n n = −60

where gn = g(0.1nT). What is the value of qk = q(0.1kT) for k = ±20, ±10, and 0? Solution A Matlab script for this problem is shown below. Note the starting time of -3.01 is used to avoid divide-by-zero problems. Using the filter function is just one way the discrete convolution can be performed. alpha = 0.5; B0 = 0.5; T = 1/(2*B0); t = [-3.01: 0.1 :3] * T; %-- root raised cosine impulse response g = cos( (1+alpha)*pi*t/T) + (T/4/alpha) ./ t .* sin( (1alpha)*pi*t/T); g = g ./ (1 - (4*alpha*t/T).^2 ); g = 4*alpha/pi * g; %--- discrete convolution ----q = 0.1*filter(g,1, [g zeros(1,60)]); tp = [-6.01:0.1:6] * T; stem(tp,q) xlabel('Time (T)') ylabel('Amplitude') axis([-4 4 -.2 1]), grid on Continued on next slide

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Problem 10.12 continued The plot of qk is shown below for α = 0.5 . At k = ±20 and ±10, the amplitude is approximately zero. At k = 0 the amplitude is 1.

0.8

Amplitude

0.6

0.4

0.2

0

-0.2 -4

-3

-2

-1

0 Time (T)

1

2

3

4

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Problem 10.13. Determine the discrete-time autocorrelation function of the noise sequence {Nk} defined by Eq. (10.34) ∞

N k = ∫ p(kT − t ) w(t )dt −∞

where w(t) is a white Gaussian noise process and the pulse p(t) corresponds to a rootraised cosine spectrum. How are the noise samples corresponding to adjacent bit intervals related? Solution The autocorrelation function of the noise at samples spaced by T is

RN (n) = E [ N k N k + n ] ∞ ∞ = E ⎡ ∫ p (kT − t ) w(t )dt ⋅ ∫ p ((k + n)T − s) w( s )ds ⎤ −∞ ⎣⎢ −∞ ⎦⎥

=∫

∞

=∫

∞

∫

∞

∫

∞

−∞ −∞

−∞ −∞

p(kT − t ) p((k + n)T − s )E [ w(t ) w( s) ] dtds p(kT − t ) p((k + n)T − s )

N0 δ (t − s)dtds 2

where we have interchanged integration and expectation on the third line, and the fourth line follows from the uncorrelated properties of the white noise. We next apply the sifting property of the delta function to obtain ∞

RN (n) = ∫ p (kT − t ) p ((k + n)T − t ) −∞

N0 dt 2

N0 ∞ p (kT − t ) p (t − (k + n)T )dt 2 ∫−∞ N = 0 δ (n) 2 =

where the second line follows from the even symmetry property of the raised cosine pulse, and third line follows from Eq. (10.32). Therefore, noise samples corresponding to adjacent bit intervals are not correlated.

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Problem 10.14. Draw the Gray-encoded constellation (signal-space diagram) for 16QAM and for 64-QAM. Can you suggest a constellation for 32-QAM? Solution A general hint for Gray encoding is to (a) first Gray encode two bits and assign one pair of the resulting encoding to each quadrant. (b) Gray encode the remaining bits within one of the quadrants. (c) obtain the Gray encodings for the remaining quadrants by reflecting the result across the in-phase and quadrature axes. 16-QAM constellation:

64-QAM constellation:

Continued on next slide

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Problem 10.14 continued

32-QAM constellation: (There does not appear to be a Gray encoding for 32-QAM)

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Problem 10.15. Write the defining equation for a QAM-modulated signal. Based on the discussion of QPSK and multi-level PAM, draw the block diagram for a coherent QAM receiver. Solution The QAM modulated signal can be defined as: s (t ) = ∑ ⎡⎣bkI h(t − kT ) cos(2π f c t ) + bkQ h(t − kT ) sin(2π f c t ) ⎤⎦ , k

where bkI , bkQ are different modulation levels on the I and Q channels, respectively. T is the QAM symbol duration, h(t) is the pulse shape and is nonzero during 0 ≤ t < T, and fc is the carrier frequency. The block diagram for a coherent QAM receiver is

cos(2π f c t )

sin(2π f ct )

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Problem 10.16. Show that if T is a multiple of the period of fc, then the terms sin(2π f c t ) and cos(2π f c t ) are orthogonal over the interval [t0 , T + t0 ] . Solution

∫

T + t0

t0

sin(2π f c t ) cos(2π f c t ) dt = ∫

T + t0

t0

=

1 8π f c

=−

1 sin(4π f c t )dt 2

[ − cos(4π f ct )] Tt +t

1 8π f c

0

0

[cos(4π fc (t0 + T )) − cos(4π fct0 )]

−1 sin(4π f c t0 + 2π f cT ) ⋅ sin(2π f cT ) 4π f c where we have used the equivalence cosA - cosB = 2sin[(A+B)/2]sin[(B-A)/2)]. If T is a multiple of the period of fc, then fcT = integer, and sin(2πf c T ) = 0 . =

Therefore,

∫

t 0 +T

t0

sin( 2πf c t ) cos( 2πf c t )dt = 0 . That is, sin( 2πf c t ) and cos(2πf c t ) are

orthogonal over the interval [t0, t0+T].

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Problem 10.17. For a rectangular pulse shape, by how much does null-to-null transmission bandwidth increase, if the transmission rate is increased by a factor of three? Solution Without loss of generality, consider the baseband BPSK signal: s (t ) = ∑ bk h(t − kT ), k

where T is the symbol duration, bk = +1 or -1 for transmitted 1 or 0, respectively. The pulse h(t) is rectangular, ⎛t −T /2⎞ h(t ) = rect ⎜ ⎟. ⎝ T ⎠

The Fourier transform H(f) of h(t) is H ( f ) = Tsinc( fT ) ⋅ e − j 2π fT / 2 sin(π fT ) − jπ fT . =T e π fT Inspecting a plot of the sinc function, we see the null-to-null transmission bandwidth of H(f) is B = 2/T. When the transmission rate is increased by a factor three, we have the new symbol duration T ′ = T / 3 . The null-to-null bandwidth B′ = 2 / T ′ = 3B , increased by a factor of 3.

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Problem 10.18. Under the bandpass assumptions, determine the conditions under which the two signals cos(2π f 0t ) and cos(2π f1t ) are orthogonal over the interval from 0 to T. Solution For two signals to be orthogonal over the interval from 0 to T, they must satisfy

∫

T

0

cos(2πf 0 t ) cos(2πf1t )dt = 0 .

To verify this we perform the integration as follows: T 1 T π f t π f t dt = cos(2 ) cos(2 ) [cos(2π ( f0 + f1 )t ) + cos(2π ( f0 − f1 )t )] dt 0 1 ∫0 2 ∫0 1 1 = sin(2π ( f 0 + f1 )) T0 + sin(2π ( f 0 − f1 )) 4π ( f 0 + f1 ) 4π ( f 0 − f1 )

=

T 0

1 1 sin(2π ( f 0 + f1 )T ) + sin(2π ( f 0 − f1 )T ) 4π ( f 0 + f1 ) 4π ( f 0 − f1 )

By the bandpass assumption (f0+f1) >> 1 so the first term in the last line is negligible. For the second term to be zero it must satisfy 2π ( f 0 − f 1 )T = nπ where n is an integer. This implies that (f0 - f1)= n/2T.

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Problem 10.19. Encode the sequence 1101 with a Hamming (7,4) block code. Solution Coded bit sequence c = x ⋅ G , where G is defined by (10.89).

⎡1000101 ⎤ ⎢0100111 ⎥ ⎥ c = [1101] ⋅ ⎢ ⎢0010110 ⎥ ⎢ ⎥ ⎣0001011 ⎦ = [1101001]

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Problem 10.20. The Hamming (7,4) encoded sequence 1001000 was received. If the number of transmission errors is less than two, what was the transmitted sequence? Solution The syndrome of the received sequence is S = R H where H is defined by (10.92). S = R⋅H ⎡101 ⎤ ⎢111 ⎥ ⎢ ⎥ ⎢110 ⎥ ⎢ ⎥ = [1001000] ⋅ ⎢011 ⎥ ⎢100 ⎥ ⎢ ⎥ ⎢010 ⎥ ⎢001 ⎥ ⎣ ⎦ = [110]

Based on Table 10.4, the error vector E = [0010000]. The transmitted sequence is E ⊕ R = [1011000].

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Problem 10.21. A Hamming (15,11) block code is applied to a BPSK transmission scheme. Compare the block error rate performance of the uncoded and coded systems. Explain how this would differ if the modulation strategy was QPSK.

Solution 1) For the uncoded system, the probability of a bit error with BPSK is ⎛ 2 Eb ⎞ ⎟ Pe = Q⎜ ⎜ N ⎟ 0 ⎠ ⎝ The probability of a block error with block length of 15 bits, assuming independent errors is: Pbuncoded = 1 − (1 − Pe )15

2) For the coded system, with a (15,11) Hamming code, the probability of block error is ⎛ 15 ⎞ Pbcoded = 1 − (1 − Pe′)15 − ⎜ ⎟ (1 − Pe′)14 Pe′ , ⎝1 ⎠ where Pe′ is the bit error probability of coded bit, since the code can correct a single bit error. The probability of bit error in this case is: ⎛ 2 Ec Pe′ = Q ⎜ ⎜ N 0 ⎝

⎞ ⎟⎟ , ⎠

where Ec is the coded bit energy, and Ec = 11/15Eb. Therefore ⎛ 22 Eb Pe′ = Q ⎜ ⎜ 15 N 0 ⎝

⎞ ⎟⎟ ⎠

To compare the block error probabilities of uncoded and coded systems, we use Matlab to plot the block error rate curves for Pbuncoded and Pbcoded versus Eb/N0 (dB), as shown below

Continued on next slide

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Problem 10.21 continued -1

10

-2

10

-3

Block Error Rate

10

Block of 15 uncoded bits

Hamming(15,11)

-4

10

-5

10

-6

10

-7

10

0

2

4

6

8

10 12 Eb/No (dB)

14

16

18

20

The Matlab script that generates the above plot is EbNodB=[0:0.25:12]; EbNo = 10.^(EbNodB/10); Pe = 0.5*erfc(sqrt(EbNo)); Puncoded = 1 - (1-Pe).^15; EcNo = 11/15 * EbNo; Peprime = 0.5*erfc(sqrt(EcNo)); Pcoded = 1 - (1-Peprime).^15 - 15*(1-Peprime).^14.*Peprime; semilogy(EbNodB,Puncoded) grid xlabel('Eb/No (dB)') ylabel('Block Error Rate') axis([0 20 1E-7 0.1]) hold on, semilogy(EbNodB,Pcoded,'g'), hold off

3) Since for QPSK modulation, bit error probabilities of uncoded bits Pe and coded bits Pe' are unchanged compared with BPSK modulation, the block error probabilities of two systems are also the same as those of BPSK modulation.

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Problem 10.22. Show that the choice γ = µ / 2 minimizes the probability of error given by Eq. (10.26). Hint: The Q-function is continuously differentiable. Solution From (10.26), we have the average probability of error as: Pe (γ ) =

1 ⎛ µ −γ ⎞ 1 ⎛ γ ⎞ Q⎜ ⎟ + Q⎜ ⎟ 2 ⎝ σ ⎠ 2 ⎝σ ⎠

Recall the definition of Q-function:

Q ( x) =

1

=

1

∫

+∞

2π (let u = − s ) x

2π

∫

−x

−∞

exp(− s 2 / 2)ds

exp(−u 2 / 2)du

So the derivative is given by −1 dQ( x) = exp(− x 2 / 2) ≤ 0 dx 2π Substituting this result into the definition of Pe(γ) we obtain ⎛ ( µ − γ ) 2 ⎞ −1 1 −1 ⎛ γ2 ⎞ 1 dPe (γ ) 1 −1 = ⋅ ⋅ exp ⎜ − ⋅ + ⋅ ⋅ exp ⎟ ⎜− 2 ⎟⋅ dγ 2 2π 2σ 2 ⎠ σ 2 2π ⎝ ⎝ 2σ ⎠ σ = Setting

1 2 2πσ

⎧⎪ ⎛ ( µ − γ )2 ⎞ ⎛ γ 2 ⎞ ⎫⎪ − − exp exp ⎨ ⎜ ⎟ ⎜ − 2 ⎟⎬ 2σ 2 ⎠ ⎪⎩ ⎝ ⎝ 2σ ⎠ ⎪⎭

dPe (γ ) = 0 implies dγ

⎛ ( µ − γ )2 ⎞ ⎛ γ2 ⎞ = exp ⎜ − exp ⎟ ⎜− 2 ⎟ 2σ 2 ⎠ ⎝ ⎝ 2σ ⎠ 2 2 (µ − γ ) = γ γ = µ/2

Continued on next slide

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Problem 10.22 continued

Checking the second derivative, we have ⎡ 2( µ − γ ) ⎛ ( µ − γ ) 2 ⎞ 2γ ⎛ γ 2 ⎞⎤ d 2 Pe (γ ) 1 = ⋅ − + exp exp ⎢ ⎜ ⎟ ⎜ − 2 ⎟⎥ 2 d 2γ 2σ 2 ⎠ 2σ 2 2 2πσ ⎣ 2σ ⎝ ⎝ 2σ ⎠ ⎦ >0 when γ = µ/2. Therefore at γ = µ / 2 , Pe (γ ) has a minimum value.

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Problem 10.23. For M-ary PAM, (a) Show that the formula for probability of error, namely, ⎛ M −1 ⎞ ⎛ A ⎞ Pe = 2 ⎜ ⎟Q⎜ ⎟ ⎝ M ⎠ ⎝σ ⎠ holds for M = 2, 3, and 4. By mathematical induction, show that it holds for all M.

(b) Show the formula for average power, namely, ( M 2 − 1) A2 P= 3 holds for M = 2, and 3. Show it holds for all M. Solution (a) M-ary PAM with the separation between nearest neighbours as 2A. Assume that all M symbols are equally transmitted.

(i) For M=2, we have the result given in the text for binary PAM ⎛ A⎞ Pe2 PAM = Q ⎜ ⎟ ⎝σ ⎠ M −1 ⎛ A ⎞ =2 Q⎜ ⎟ M ⎝σ ⎠

for M = 2. (ii) For M = 3, the constellation is: 1 1 P e = P [ y > − A | (−2 A) is transmitted ] + P [ y > A or y < − A | 0 is transmitted ] 3 3 1 + P [ y < A | (2 A) is transmitted ] 3 ⎛ ( y + 2 A) 2 ⎞ ⎛ y2 ⎞ 1 ∞ 1 1 ∞ 1 = ∫ + dy exp ⎜ − exp ⎟ ⎜ − 2 ⎟dy 3 − A 2πσ 2σ 2 ⎠ 3 ∫A 2πσ ⎝ ⎝ 2σ ⎠ ⎛ y2 ⎞ ⎛ ( y − 2 A) 2 ⎞ 1 −A 1 1 A 1 + ∫ exp ⎜ − 2 ⎟dy + ∫ exp ⎜ − ⎟dy 3 −∞ 2πσ 3 −∞ 2πσ 2σ 2 ⎠ ⎝ 2σ ⎠ ⎝ 4 ⎛ A⎞ = Q⎜ ⎟ 3 ⎝σ ⎠

Continued on next slide

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Problem 10.23 continued

From the formula Pe = Pe =

2( M − 1) ⎛ A ⎞ 4 ⎛ A⎞ Q⎜ ⎟ , when M=3, Pe = Q⎜ ⎟ . Thus the formula M 3 ⎝σ ⎠ ⎝σ ⎠

2( M − 1) ⎛ A ⎞ Q⎜ ⎟ holds for M = 3. M ⎝σ ⎠

(iii) For M = 4, the constellation is: - 3A

-A

+A

+ 3A

1 1 P [ y > −2 A | (−3 A) is transmitted ] + P [ y < −2 A or y > 0 | − A is transmitted ] 4 4 1 1 + P [ y < 0 or y > 2 A | + A is transmitted ] + P [ y < 2 A | (3 A) is transmitted ] 4 4 2 ⎛ ( y + 3 A) ⎞ 1 ∞ 1 = ∫ exp ⎜ − ⎟dy 4 −2 A 2πσ 2σ 2 ⎠ ⎝

P e=

⎡1 ∞ 1 ⎛ ( y + A )2 ⎞ ⎛ ( y + A)2 ⎞ ⎤ 1 −2 A 1 + 2⎢ ∫ exp ⎜ − exp ⎜ − ⎟dy ⎥ ⎟dy + ∫ 2 2 −∞ ⎜ ⎟ ⎜ ⎟ ⎥ σ σ 2 4 2 πσ 2 ⎢⎣ 4 0 2πσ ⎝ ⎠ ⎦ ⎝ ⎠ 2 ⎛ ( y − 3 A) ⎞ 1 A 1 exp ⎜ − + ∫ ⎟dy 4 −∞ 2πσ 2σ 2 ⎠ ⎝ 6 ⎛ A⎞ = Q⎜ ⎟ 4 ⎝σ ⎠ where the factor 2 in the third last line, comes from the symmetry of the second and third 2( M − 1) ⎛ A ⎞ terms of the first equation. From the formula Pe = Q⎜ ⎟ , when M=4, M ⎝σ ⎠ 6 ⎛ A⎞ 2( M − 1) ⎛ A ⎞ Pe = Q ⎜ ⎟ . Thus the formula Pe = Q⎜ ⎟ holds for M = 4. 4 ⎝σ ⎠ M ⎝σ ⎠

(iv) Assume that the formula of Pe holds for (M-1)-ary PAM. By mathematical induction, we need to show it also holds for M-ary PAM. The (M-1)-ary PAM constellation may be illustrated as shown: K

Continued on next slide

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Problem 10.23 continued

By adding one point P2 on the (M-1)-ary PAM constellation, which has the distance 2A from point P1, we obtain M-ary PAM constellation as follows (in practice, the average or dc level may be adjusted as well but this has no effect on the symbol error rate): K

Since error probabilities of P1 symbol on the (M-1)-ary PAM is the same as that of P2 point on the M-ary PAM, the error probability of M-ary PAM is PeM − ary =

M − 1 ( M −1)− ary 1 Pe + ⋅ symbol error prob. of P1 symbol on M -ary (1) M M

where 1/M is the probability that P1 is transmitted and (M-1)/M is the probability that one of the other constellation points is transmitted. The probability of error formula for (M-1)-ary PAM is given by ( M − 2) ⎛ A ⎞ (2) Pe( M −1) − ary = 2 Q⎜ ⎟ . ( M − 1) ⎝ σ ⎠ The symbol error rate of P1 symbol on M-ary PAM is PP1 =

1 P [ y < ( µ − A), or y > ( µ + A) | P1 is transmitted ] M

where µ is the signal level of P1 symbol.

⎛ ( y − µ )2 ⎞ 1 ∫−∞ exp ⎜⎝ − 2σ 2 ⎟⎠ dy + M 2 ⎛ A⎞ = Q⎜ ⎟ M ⎝σ ⎠

1 PP1 = M

µ−A

⎛ ( y − µ )2 ⎞ ∫µ + A exp ⎜⎝ − 2σ 2 ⎟⎠ dy +∞

(3)

Substituting Eqs. (2) and (3) into (1), we obtain the symbol error probability of M-ary PAM M −1 M −2 ⎛ A⎞ 2 ⎛ A⎞ ⋅2⋅ Q⎜ ⎟ + Q⎜ ⎟ M M −1 ⎝ σ ⎠ M ⎝ σ ⎠ . M −1 ⎛ A ⎞ Q⎜ ⎟ =2 M ⎝σ ⎠

PeM − PAM =

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Problem 10.23 continued

The formula holds for M-ary PAM. Therefore, by mathematical induction, the formula holds for all M.

(b) To compute the average symbol power we note: i) For M = 2, the average symbol power is A2 and the formula P =

( M 2 − 1) A2 holds for 3

M=2. ii) For M = 3, the average symbol energy is P=

1 8 (2 A) 2 + 02 + (2 A) 2 ) = A2 . ( 3 3

( M 2 − 1) A2 holds for M=3. The formula P = 3 iii) For general even M, the M-ary PAM constellation points are

{−(M − 1) A,L, −3 A, − A, A,3 A,L, (M − 1) A} . The average symbol energy is P=

2 ⎡⎣ ( M − 1) 2 + ( M − 3) 2 + L + 32 + 1⎤⎦ M

2 A2 = M

A2

M /2

∑ (2k − 1)

2

k =1

M /2 M /2 2A ⎡ 2 M /2 2 ⎤ 2 k 4 k 1⎥ − + . ∑ ∑ ∑ ⎢ M ⎣ k =1 k =1 k =1 ⎦ 2 A2 ⎡ M ( M / 2 + 1)( M + 1) M ( M / 2 + 1) M ⎤ 4 = −4 + ⎥ ⎢ 26 22 2⎦ M ⎣

=

2

( M 2 − 1) A2 = 3 where we have used the summation formulas of Appendix 6. iv) For general odd M, the M-ary PAM constellation points are

{−(M − 1) A,L, −2 A, 0, 2 A,L (M − 1) A} . Continued on next slide Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 10.23 continued The average symbol energy is 2 ⎡ ( M − 1) 2 + ( M − 3) 2 + L + 22 ⎤⎦ 2 P= ⎣ A M 2 2 ⎤ 2 A2 2 ⎡ ⎛ M − 1 ⎞ ⎛ M − 3 ⎞ 2 2 ⎢⎜ = ⎟ +⎜ ⎟ + ... + 1 ⎥ M ⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎦⎥ =

8 A2 M

( M −1) / 2

∑

k2

k =1

8 A ( M − 1)( M + 1)( M ) 226 M 2 2 ( M − 1) A = 3 =

2

where the fourth line uses the summation formula found in Appendix 6.

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Problem 10.24. Consider binary FSK transmission where ( f1 − f 2 )T is not an integer. (a) What is the mean output of the upper correlator of Fig. 10.12, if a 1 is transmitted? What is the mean output of the lower correlator? (b) Are the random variables N1 and N2 independent under these conditions? What is the variance of N1 – N2? (c) Describe the properties of the random variable D of Fig. 10.12 in this case.

Solution: (a) If a 1 is transmitted,

r (t ) = Ac cos(2πf 1t ) + n(t )

where n(t) is a narrow band Gaussian noise. The output of the upper correlator is Y1: T

Y1 = ∫ r (t ) 2 cos(2πf 1t )dt 0

=∫

T

2 Ac cos(2πf1t ) cos(2πf 1t )dt + ∫

0

≅

T

0

1 2

Ac T + ∫

T

0

2n(t ) cos(2πf1t )dt

2n(t ) cos(2πf1t )dt

The expected value of Y1 is E[Y1 ] =

1 AcT , since n(t) has zero mean. 2

The output of the lower correlator is Y2: T

Y2 = ∫ r (t ) 2 cos(2πf 2 t )dt 0

=∫

T

2 Ac cos(2πf1t ) cos(2πf 2 t )dt + ∫

0

= ≅

T

0

Ac 2 Ac

∫

T

0

∫ 2

T

0

cos(2π ( f 1 + f 2 )t )dt +

Ac 2

cos(2π ( f 1 − f 2 )t )dt + ∫

T

0

∫

T

0

2n(t ) cos(2πf 2 t )dt T

cos(2π ( f1 − f 2 )t )dt + 2 ∫ n(t ) cos(2πf 2 t )dt 0

2n(t ) cos(2πf 2 t )dt

Continued on next slide

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Problem 10.24 continued where the first term of the third line is negligible due to the bandpass assumption. The expected value of Y2 is A T E[Y2 ] = c ∫ cos(2π ( f1 − f 2 )t )dt 2 0 A 1 sin ⎡⎣ 2π ( f1 − f 2 ) t ⎤⎦ T0 = c ⋅ 2 2π ( f1 − f 2 ) Ac sin(2π ( f1 − f 2 ) T ) 2 2π ( f1 − f 2 )

=

which clearly differs from the orthogonal case. (b) The random variables N1 and N2 are given by

N1 = ∫

T

0 T

N2 = ∫

0

2n(t ) cos(2πf1t )dt 2n(t ) cos(2πf 2 t )dt

Since n(t) is a Gaussian process, both N1 and N2 are Gaussian. To show N1 and N2 are correlated consider T T E[ N1 N 2 ] = E ⎡ ∫ n(t ) cos(2π f1t )dt ⋅ ∫ n(τ ) cos(2π f 2τ )dτ ⎤ 0 ⎣⎢ 0 ⎦⎥

=∫ =∫

T

T

0

∫

T

T

0

0

∫

0

E[n(t )n(τ )]cos(2π f1t ) cos(2π f 2τ )dtdτ N0 δ (t − τ ) cos(2π f1t ) cos(2π f 2τ )dtdτ 2

N0 T cos(2π f1t ) cos(2π f 2t )dt 2 ∫0 N T = 0 ∫ [ cos(2π ( f1 + f 2 )t ) + cos(2π ( f1 − f 2 )t ) ] dt 4 0 N sin ( 2π ( f1 + f 2 )t ) T sin ( 2π ( f1 − f 2 )t ) T = 0 0 + 0 4 2π ( f1 + f 2 ) 2π ( f1 − f 2 ) =

≅

N0 sinc(2( f1 − f 2 )T ) 4

Continued on next slide

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Problem 10.24 continued

where the first term of the second last line is assumed negligible due to the bandpass assumption. Since N1 and N2 are correlated, they are not independent. The variance of (N1-N2) is

var[ N1 − N 2 ] = var[ N1 ] + var[ N 2 ] − 2E[ N1 N 2 ] = N0 −

N0 sinc ( 2( f1 − f 2 )T ) 2

(c) The random variable D is Gaussian with zero mean and variance var[N1-N2].

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Problem 10.25. Show that the noise variance of the in-phase component nI(t) of the band-pass noise is the same as the band-pass noise n(t) variance; that is, for a band-pass noise bandwidth BN E ⎡⎣ nI2 (t ) ⎤⎦ = N 0 BN Solution Recall the spectra of narrowband noise n(t) and its in-phase component nI (t ) shown in ∞

Figure 8.23. The variance of a random process x(t ) = Rx (0) = ∫ X ( f )df , where X(f) is −∞

the power spectral density of x(t). Therefore, Var[n(t )] = E ⎡⎣ n 2 (t ) ⎤⎦ ∞

= ∫ S N ( f )df −∞

N0 ⋅ 2B 2 = N0 ⋅ 2B = 2⋅

Where we have used the fact that for a bandpass signal BT = 2B, that is twice the lowpass bandwidth. Similarly, the variance of the in-phase noise is Var[nI (t )] = E ⎡⎣ nI2 (t ) ⎤⎦ ∞

= ∫ S nI ( f )df −∞

= N0 ⋅ 2B

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Problem 10.26 In this problem, we investigate the effects when transmit and receive filters do not combine to form an ISI-free pulse shape. To be specific, data is transmitted at baseband using binary PAM with an exponential pulse shape g(t)=exp(-t/T)u(t) where T is the symbol period (see Example 2.2). The receiver detects the data using an integrate-and-dump detector. (a)

With data represented as ±1, what is magnitude of the signal component at the output of the detector.

(b)

What is the worst case magnitude of the intersymbol interference at the output of the detector. (Assume the data stream has infinite length.) Using the value obtained in part (a) as a reference, by what percentage is the eye opening reduced by this interference.

(c)

What is the rms magnitude of the intersymbol interference at the output of the detector? If this interference is treated as equivalent to noise, what is the equivalent signal-to-noise ratio at the output of the detector? Comment on how this would affect bit error rate performance of this system when there is also receiver noise present. (Typo in problem statement, there should be minus sign in exponential.) Solution (a) For a data pulse g (t ) = A exp( −t / T )u (t )

where A is the binary PAM symbol (±1). The desired output of an integrate-and-dump filter in the nth symbol period is ( n +1)T

∫

Gn =

g (t − nT )dt

nT T

= ∫ An exp(−t / T )dt 0

= AnT (1 − exp(−1) ) If the data is either ±1, then magnitude of the output is T(1-e-1). (b) In the nth symbol period the received signal is y (t ) =

∞

∑ A g (t − kT )

k =−∞

n

Continued on next slide

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Problem 10.26 continued The output of the detection filter in the nth symbol period is ( n +1)T

Yn =

∫

y (t )dt

nT

=

( n +1)T

∞

nT

k =−∞

∫ ∑A

( n +1)T

=

∫

k

exp ( −(t − kT ) / T )dt ∞ ( n +1)T

An exp ( −(t − nT ) / T ) dt + ∑ k =1

nT

∫

An − k exp {−(t − (n − k )T ) / T } dt

nT

where, due to the causality of the pulse shape, the symbols An+1 and later due not cause intersymbol interference into symbol An. The first term in the above is the desired signal and the second term is the intersymbol interference. By letting s = t – (n-k)T, we can express this interference as ∞ ( k +1)T

Jn = ∑ k =1

∫

An − k exp ( −t / T ) dt

kT

∞

= ∑ An − k T ( exp(− k ) − exp(−(k + 1)) ) k =1

where each term in the summation corresponds to the interference caused by a previous symbol. For worst case interference we assume that all of the An-k have the same sign. Then this worst case interference is given by ∞

J n = ∑ An − k T ( exp(− k ) − exp(−(k + 1)) ) k =1

∞

≤ T (1 − exp(−1) ) ∑ exp(− k ) k =1

To simplify the notation, we let α = exp(-1). Then ∞

J

max n

= T (1 − α ) ∑ α k k =1

= T (1 − α )

α 1−α

= αT Comparing this worst case interference to the desired signal level Gn, the eye-opening is reduced by J nmax Tα × 100 = × 100 = 58% Gn T (1 − α )

Continued on next slide

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Problem 10.26 continued (c) From part (b), we found that kth preceding symbol contributes an interference

I nk = An − k (1 − α ) α k The total interference is ∞

J n = ∑ I nk k =1 ∞

= ∑ An − k T (1 − α )α k k =1

Since all symbol intervals are equivalent, we drop the subscript n on Jn. The mean value of this interference is E[J] = 0 since E[An-k] = 0. The variance of this interference is Var ( J ) = E ⎡⎣ J 2 ⎤⎦ ∞

= ∑ E ⎡⎣ An2− k ⎤⎦ T 2 (1 − α ) α 2 k 2

k =1

α2 1−α 2 2 1−α = (α T ) 1+ α

= T 2 (1 − α ) 2

where we have assumed the symbols are independent so that E[AiAj] = 0 if i ≠ j. The rms interference is given by the square root of the variance so

1−α 1+ α = 0.25T

J rms = α T

which is clearly less than the worst case interference Jmax. If we represent the signal power by S, the noise power by N, then the equivalent signalto-noise ratio taking account of the intersymbol interference is S 2 N + J rms The intersymbol interference will further degrade performance. In fact, if the worst case interference is large enough such that the eye closes, it will result in a lower limit on the bit error rate regardless of how little noise there is. SNR =

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Problem 10.27. A BPSK signal is applied to a matched-filter receiver that lacks perfect phase synchronization with the transmitter. Specifically, it is supplied with a local carrier whose phase differs from that of the carrier used in the transmitter by φ radians. (a) Determine the effect of the phase error φ on the average probability of error of this receiver. (b) As a check on the formula derived in part (a), show that when the phase error is zero the formula reduces to the same form as in Eq. (10.44). Solution (a) With BPSK, assume the transmitted signal is (10.36): N

s (t ) = Ac ∑ bk h(t − kT ) cos(2π f c t ) , k =0

⎛ t −T / 2 ⎞ where bk = +1 for a 1 and bk = -1 for a 0, h(t) is the rectangular pulse rect ⎜ ⎟. ⎝ T ⎠

The received signal is x(t ) = s (t ) + n(t ) N

= Ac ∑ bk h(t − kT ) cos(2π f c t ) + nI (t ) cos(2π f c t ) − nQ (t ) sin(2π f c t ) k =0

The receiver matched filter is the integrate-and-dump filter. The output for the kth symbol after down-conversion with phase error φ and match filtering is: Yk = ∫

kT

( k −1)T

x(t ) cos(2π f c t + φ )dt

[ Acbk + nI (t )] cos(2π f ct ) cos(2π fct + φ )dt − ∫( k −1)T nQ (t ) sin(2π f ct ) cos(2π fct + φ )dt ( k −1)T

=∫

kT

kT

kT 1 1 [ Ac bk + nI (t )][cos φ + cos(4π f c t + φ )]dt − ∫ n (t )[sin(4π f c t + φ ) + sin(−φ )]dt ( k −1)T 2 ( k −1)T 2 Q T 1 kT 1 kT nI (t ) cos φ dt + ∫ nQ (t ) sin φ dt ≅ Ac bk cos φ + ∫ 2 2 ( k −1)T 2 ( k −1)T T = Ac bk cos φ + N k 2

=∫

kT

where we define Nk =

1 kT 1 kT nI (t ) cos φ dt + ∫ nQ (t ) sin φ dt ∫ ( − 1) k T 2 2 ( k −1)T

Continued on next slide

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Problem 10.27 continued

The random variable Nk has zero mean and variance var[ N k ] = cos 2 φ =

Let µ =

N 0T NT + sin 2 φ 0 4 4

N 0T =σ2 4

T Ac cos φ . Then the probability of bit error Pe is 2

Pe = P[bk = 1]P[Yk < 0 | bk = 1] + P[bk = −1]P[Yk > 0 | bk = −1] ⎧ ( y − µ )2 ⎫ ⎧ ( y + µ )2 ⎫ 1 0 1 1 +∞ 1 exp dy exp − + ⎨ ⎬ ⎨− ⎬dy 2 ∫−∞ 2πσ 2 ∫0 σ2 ⎭ σ2 ⎭ 2πσ ⎩ ⎩ ⎛µ⎞ = Q⎜ ⎟ ⎝σ ⎠

=

with Eb =

⎛ 2 Eb cos φ Ac2T 1 T , µ = Ac cos φ , σ = N 0T , we have Pe = Q ⎜⎜ 2 2 2 N0 ⎝

⎛ 2 Eb (b) When the phase error φ=0, Pe = Q ⎜⎜ ⎝ N0

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ , as the same as Eq. (10.44). ⎠

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Problem 10.28. A binary FSK system transmits data at the rate of 2.5 megabits per second. During the course of transmission, white Gaussian noise of zero mean and power spectral density 10-20 watts per hertz is added to the signal. In the absence of noise, the amplitude of the received signal is 1 µV across 50 ohm impedance. Determine the average probability of error assuming coherent detection of the binary FSK signal. Solution The average probability of error for coherent FSK is ⎛ Eb ⎞ Pe = Q ⎜⎜ ⎟⎟ ⎝ N0 ⎠

from Eq. (10.68). For this example, we have noise power spectral density is N 0 = 2 × 10 −20 watts / Hz

and the energy per bit is Eb =

1 Ac2T , 2 R

In the text, we have nominally assumed the resistance is 1 ohm and omitted it. In this problem we use the resistance of R = 50 ohms. The symbol duration is 1 seconds and the amplitude of received signal is Ac = 1µV. Therefore, T= 2.5 × 106

1 1×10−12 1 × Eb = × 2 50 2.5 ×106 = 4 ×10−21 watts / Hz Substituting the above values into the expression for Pe and we have the probability of error is Pe = Q ( 0.2 ) ≅ 0.26

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Problem 10.29. One of the simplest forms of forward error correction code is the repetition code. With an N-repetition code, the same bit is sent N times, and the decoder decides in favor of the bit that is detected on the majority of trials (assuming N is odd). For a BPSK transmission scheme, determine the BER performance of a 3-repetition code. Solution With 3-repetition code, the decoder will output the correct bit if there are one or fewer errors in the 3-bit code. Thus, assuming bit errors are independent, the bit error rate is ⎛ 3⎞ Pbcoded = (1 − Pe )3 + ⎜ ⎟ Pe (1 − Pe ) 2 , ⎝1 ⎠ = (1 − Pe ) 2 (1 + 2 Pe ) where Pe is the bit error rate of channel bit. With BPSK, the formula for bit error probability is ⎛ 2 Ec ⎞ Pe = Q ⎜⎜ ⎟⎟ ⎝ N0 ⎠ , ⎛ 2 Eb ⎞ = Q ⎜⎜ ⎟⎟ ⎝ 3N 0 ⎠ since ratio of channel bit energy to information bit energy is given by Ec = 1/3Eb. Therefore, the bit error probability of the 3-repetition code is coded b

P

⎛ ⎛ 2 Eb = ⎜1 − Q ⎜⎜ ⎜ ⎝ 3N0 ⎝

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

2

⎛ ⎛ 2 Eb ⎜ 1 + 2Q ⎜⎜ ⎜ ⎝ 3N 0 ⎝

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

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Problem 10.30 In this experiment, we simulate the performance of bipolar signalling in additive white Gaussian noise. The Matlab script included in Appendix 7 for this experiment: • generates a random sequence with rectangular pulse shaping • adds Gaussian noise • detects the data with a simulated integrate-and-dump detector With this Matlab script (a) Compute the spectrum of the transmitted signal and compare to the theoretical. (b) Explain the computation of the noise variance given an Eb/N0 ratio. (c) Confirm the theoretically predicted bit error rate for Eb/N0 from 0 to 10 dB. Solution (a) The provided script plots the simulated spectrum before noise is added. If we add the statement hold on, plot(F, abs(2*sinc(F)).^2,'g'), hold off at the same point, we obtain the following comparison graph. The two graphs agree reasonably well. There are two reasons for the differences observed with the simulated spectrum. The first is the relatively short random sequence used for generating the plot and the second is an aliasing effect. 4.5 4 3.5

Spectrum

3 2.5 2 1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1 1.2 Frequency

1.4

1.6

1.8

2

Continued on next slide

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Problem 10.30 continued

(b) The calculation of the noise variance in a discrete time simulation proceeds as follows. We are given the sampling rate Fs and the required Eb/N0 to simulation. We then note that ∞

Eb =

∫

2

p (t ) dt (1)

−∞

≈ ∑ pk Ts 2

where p(t) is the pulse shape, {pk} is its sample version and Ts = 1/Fs is the sample interval. On the other hand, if generate noise of variance σ2, due to Nyquist considerations this can only be distributed over a bandwidth Fs, thus the noise spectral density is N0 σ 2 = 2 Fs s

(2)

Re-arranging Eq. (2) and substituting Eq. (1) and the knowns, we have

σ2 =

N0 Fs 2 −1

⎛F =⎜ s ⎝ 2

⎞ ⎛ Eb ⎞ ⎟ ⎜ N ⎟ Eb ⎠⎝ 0 ⎠

⎛F =⎜ s ⎝ 2

⎞ ⎛ Eb ⎞ ⎟⎜ N ⎟ ⎠⎝ 0 ⎠

1⎛ E ⎞ = ⎜ b⎟ 2 ⎝ N0 ⎠

−1

−1

∑p

k

∑p

2

Ts

2

k

which agrees with what is used in the script (except that in the script we have suppressed Fs and Ts, knowing they would cancel).

Continued on next slide

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Problem 10.30 continued

(c) To compute the bit error rate for 0 to 10 dB, we add the following statements around the provided script for kk = 0:10 Eb_N0 = 10^(kk/10); Nbits = 100000; % increase for higher Eb/N0 …(provided script) BER(kk+1) = Nerrs/Nbits end semilogy([0:10], BER) xlabel('Eb/No (dB)') ylabel('BER') grid on hold on, semilogy([0:10], 0.4*erfc(sqrt(10.^([0:10]/10))),'g') The following plot is then produced by the Matlab script which shows good agreement between theory and simulation. -1

10

Simulation

-2

10

Theory -3

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 10.31 In this experiment, we simulate the performance of bipolar signalling in additive white Gaussian noise but with root-raised-cosine pulse shaping. A Matlab script is included in Appendix 7 for doing this. With this simulation: (a) Compute the spectrum of the transmitted signal and compare to the theoretical. Also compare to the transmit spectrum with rectangular pulse shaping (b) Plot the eye diagram of the received signal under no noise conditions. Explain the relationship of the eye opening to bit error rate performance. (c) Confirm the theoretically predicted bit error rate for Eb/N0 from 0 to 10 dB. Solution (a) We compare the spectra by inserting the following statements prior to noise being added to the signal [P,F] = spectrum(S,256,0,Hanning(256),Fs); plot(F,P(:,1)); midpt = floor(length(F)/2); hold on, plot(F, abs([(1+cos(pi*F(1:midpt)))/2; 0*F(midpt+1:end)]),'g'), hold off xlabel('Frequency'), ylabel('Spectrum') The comparison plot is shown below. 1.4

1.2

Spectrum

1

0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1 1.2 Frequency

1.4

1.6

1.8

2

(b) To plot the eye diagram we eliminate the noise by setting Eb/N0 to a high value Eb_N0 = 2000; Then running the Matlab script produces the following eye diagram. Continued on next slide

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Problem 10.31 continued 1.5

1

Amplitude

0.5

0

-0.5

-1

-1.5 -1

-0.8

-0.6

-0.4

-0.2 0 0.2 Symbol periods

0.4

0.6

0.8

1

(c) We simulate the bit error rate by commenting out the plotting statements and adding a set of statements similar to those used in Problem 10.30. -1

10

Simulation

-2

10

Theory

-3

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 10.32 In this experiment, we simulate the effect of various mismatches in the communication system and their effect on performance. In particular, modify the MatLab scripts of the two preceding problems to: (a) Simulate the performance of a system using rectangular pulse shaping at the transmitter and raised cosine pulse shaping at the receiver. Comment on the performance degradation. (b) In the case of matched root-raised cosine filtering, include a complex phase rotation exp(jθ) in the channel. Plot the resulting eye diagram for θ being the equivalent of 5, 10, 20, and 45°. Compare to the case of 0°. Do likewise for the BER performance. What modification to the theoretical BER formula would accurately model this behaviour? Solution (a) We can create this mismatch by inserting the statements: pulseTx = ones(1,Fs); pulseRx = [ 0.0064 0.0000 -0.0101 0.0000 0.0182 -0.0000 -0.0424 ... 0.0000 0.2122 0.5000 0.6367 0.5000 0.2122 -0.0000 ... -0.0424 0.0000 0.0182 -0.0000 -0.0101 0.0000 0.0064 ]; Delay = floor((length(pulseTx)-1)/2 + (length(pulseRx)-1)/2 + 1); Eb = sum(pulseTx.^2); And by modifying the statements S = filter(pulseTx,1,[b_delta zeros(1,Delay)]); De = filter(pulseRx,1,R); Then we obtain the performance shown below. Part of the loss seen is due to the filter mismatch but part of it is also due to a timing error; with the arrangement of the simulation the optimum sampling point for the data falls between the discrete samples. This sampling time loss could be recovered by interpolation.

Continued on next slide

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Problem 10.32 continued -1

10

Simulation

-2

10

-3

Theory

BER

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

(b) Implementation of the phase rotation requires simulation of the complete complex baseband. To do this we must modify the channel portion of the simulation to the following %--- add Gaussian noise ---Noise = sqrt(N0/2)*(randn(size(S))+j*randn(size(S))); R = S + Noise; R = R*exp(j*10/180*pi); R = real(R); Where we have now included the quadrature component of the noise. Note the receiver only uses the in-phase portion (real part) of the signal to characterize this degradation. The resulting performance for rotations of 10, 20 and 45° are shown below. Note that the 45° rotation results in a 3 dB loss in performance.

Continued on next slide

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Problem 10.32 continued 0

10

-1

10

BER

45 degrees

20 deg.

-2

10

10 deg.

Theory

-3

10

-4

10

-5

10

-6

10

0

1

2

3

4

5 6 Eb/No (dB)

7

8

9

10

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Problem 11.1 What is the root-mean-square voltage across a 10 Mega-ohm resistor at room temperature if measured over a 1 GHz bandwidth? What is the available noise power? Solution Following Example 11.2, the available noise power is PN = kTBN = 1.38 ×10 − 23 × 290 ×10 9 = 4 ×10 −12 watts

The root-mean-square voltage across a 10 mega-ohm resistor is

Vrms = PN R = 4 × 10 −12 × 10 7 = 6.3 millivolts

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Problem 11.2 What is the available noise power over 1 MHz due to shot noise from a junction diode that has a voltage differential of 0.7 volts and carries average current of 0.1 milliamperes, if the current source of the Norton equivalent circuit has a resistance of 250 ohms? Solution From Eq. (11.9), the saturation current for a junction diode is given by

IS =

I ⎛ qV ⎞ exp⎜ ⎟ −1 ⎝ RT ⎠

= 1.8 × 10 −12 I Consequently, the noise contribution from the saturation current may be ignored. From Eq. (11.10) the expected current variance is then E ⎡⎣ I shot 2 ⎤⎦ = 2q ( I + 2 I s ) BN ≈ 2qIBN

(

) (

)

= 2 × 1.6 ×10−19 × 0.1× 10−3 × 106 = 3.2 × 10−17 Amp 2

The corresponding noise power with an equivalent resistance of 250 ohms is

[

2

= 8 × 10

−15

]

PN = E I shot R watts

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Problem 11.3 An electronic device has a noise figure of 10 dB. What is the equivalent noise temperature? Solution From Eq. (11.7), the equivalent noise temperature is

Te = T0 (F − 1)

= 290(10 − 1)

= 2610 o K

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Problem 11.4. The device of Problem 11.3 has a gain of 17 dB and is connected to a spectrum analyzer. If the input to the device has an equivalent temperature of 290°K and the spectrum analyzer is noiseless, express the measured power spectral density in dBm/Hz. If the spectrum analyzer has a noise figure of 25 dB, what is the measured power spectral density in this case? Solution For the device of Problem 11.3, the total output noise is, from Eq. (11.15), N = Gk (T + Te ) BN = GkTFBN

(

)(

)

(

)

= 1017 /10 1.38 × 10−23 ( 290 ) 1010 /10 BN = 2.0 ×10−18 BN

The noise spectral density at the device output is approximately N = 2.0 × 10−18 W/Hz BN ~ -147 dBm/Hz

If the spectrum analyzer has a noise figure of 25 dB, then we must use the results of the following section. Specifically, Eq. (11.21), to obtain the total noise figure of F = F1 +

(

F2 G1

)

= 1010 /10 +

1025 /10 1017 /10

= 16.3 ~ 12.1 dB

Since the overall noise figure is increased 2.1 dB by the spectrum analyzer, the noise spectral density at the spectrum analyzer output is (assuming unity gain for the spectrum analyzer) -144.9 dBm/Hz. (There is an error in the second answer given in the text.)

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Problem 11.5 A broadcast television receiver consists of an antenna with a noise temperature of 290°K and a pre-amplifier with a gain of 20 dB and a noise figure of 9 dB. A second-stage amplifier in the receiver provides another 20 dB of gain and has a noise figure of 20 dB. What is the noise figure of the overall system? Solution From Eq (11.21), after converting from decibels to absolute F = F1 +

F2 − 1 F3 − 1 + G1 G1G2

7.94 − 1 99 + 1 100 = 2 + 6.94 + .99 = 9.93 = 2+

Converting this result back to decibels, the overall noise figure is 9.97 dB.

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Problem 11.6 A satellite antenna has a diameter of 4.6 meters and operates at 12 GHz. What is the antenna gain if the aperture efficiency is 60%? If the same antenna was used at 4 GHz what would be the corresponding gain? Solution From Eq (11.25), the antenna gain is

G=

4πAeff

λ2

The effective area is given by Eq.(11.24) Aeff = ηA =η

πd 2

4 = 9.97 m 2

where the efficiency is 60% and the diameter is 4.6 meters. At 12 GHz, the wavelength λ = c/f = 0.025 meters. Consequently, the antenna gain is G = 2000458.7 ~ 53.0 dB

With a transmission frequency of 4 GHz, the wavelength λ = 0.075 m and the antenna gain is G = 22273.2 ~ 43.5 dB

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Problem 11.7 A satellite at a distance of 40,000 kilometers transmits a signal at 12 GHz with an EIRP of 10 watts towards a 4.6 meter antenna that has an aperture efficiency of 60%. What is the received signal level at the antenna output? Solution From Eq.(11.32), the path loss due to free-space transmission of a 12 GHz signal over 40,000 kilometers is

⎛ 4πr ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠ ⎛ 4π 40 × 10 6 ⎞ ⎟⎟ = 20 log10 ⎜⎜ ⎝ 0.025 ⎠ ~ 206.1 dB Substituting this result in Eq (11.29), the received power is PR = EIRP − LP + GR = 10 dBW − 206.1 dB + 53.0 dB = −143 dBW

where we have used the antenna gain of 53 dB from Problem 11.6.

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Problem 11.8 The antenna of Problem 11.7 has a noise temperature of 70°K and is directly connected to a receiver with an equivalent noise temperature of 50°K and a gain of 60 dB. What is the system noise temperature? If the transmitted signal has a bandwidth of 100 kHz, what is the carrier-to-noise ratio? If the digital signal has a bit rate of 150 kbps, what is the Eb/N0? Solution From Eq. (11.22), the combined system noise temperature is

Trx 1 o = 70 + 50 o

Ts = Tant +

= 120 o K where the electrical gain of the antenna is 1. For a bandwidth of 100 kHz the available noise power is N = kTs B = 1.38 × 10 −23 ⋅ 120 ⋅ 10 5 = 1.66 × 10 −16 watts ~ −157.8 dBW

Comparing to the result for Problem 11.7, we have that the C/N is 14.8 dB. To convert the C/N0 to an Eb/N0, we use the formula E C = b ×R N0 N0

where the bit rate R relates the energy per bit Eb to the power C. In decibels, ⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠ dB − Hz ⎝ N 0 ⎠ dB

Continued on next slide

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Problem 11.8 continued

Re-arranging, we have

⎛ Eb ⎞ ⎛ C ⎞ − 10 log10 R ⎜ ⎟ =⎜ ⎟ ⎝ N 0 ⎠dB ⎝ N 0 ⎠ dB − Hz ⎛C⎞ = ⎜ ⎟ + 10 log BN − 10 log10 R ⎝ N ⎠ dB = 14.8 + 50 − 51.8 = 13 dB

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Problem 11.9 Transmitting and receiving antennas for a 4 GHz signal are located on top of 20 meter towers separated by 2 kilometers. For free-space propagation, what is the maximum height permitted for an object located midway between the two towers? Solution The radius of the first Fresnel zone with d1 = d2 = 1 kilometer and λ = c/f = 0.075 m is

h= =

λd1d 2 d1 + d 2

(.075)(1000)(1000) 2000

= 6.1 m Consequently, the maximum height of intermediate object is 20 m – 6.1m = 13.9 m, if we require free-space propagation conditions.

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Problem 11.10 A measurement campaign indicates that the median path loss at 900 MHz in a suburban area may be modeled with a path-loss exponent of 2.9. What is the median path loss at a distance of 3 kilometers using this model? How does this loss compare to the free-space loss at the same distance? Solution From Eq (11.37), the free-space path loss at one meter, with a transmission frequency of 900 MHz, is 2

⎛ λ ⎞ ⎛ .333 ⎞ β0 = ⎜ ⎟ =⎜ ⎟ = .000704 ⎝ 4π r0 ⎠ ⎝ 4π ⎠ 2

with r0 = 1 m. From Eq.(11.37), the path loss with the terrestrial propagation model is

β0 PR = PT ( r / r0 )n =

.0007

( 3000 )

2.9

= 5.8 × 10−14 ~ − 132 dB The free-space loss over the same distance is given by

PR ⎛ λ ⎞ =⎜ ⎟ PT ⎝ 4π r ⎠

2

= 7.8 × 10−11 ~ − 101.1 dB or 31 dB less.

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Problem 11.11 Express the true median of the Rayleigh distribution as a fraction of the Rrms value? What is the decibel error in the approximation Rmedian ≈ Rrms? Solution The median of the distribution satisfies P [ R < r ] = 0.5 . Consequently, from Eq. (11.38) we have that the median r satisfies

⎧− r2 ⎫ 1 − exp⎨ 2 ⎬ = 1 2 ⎩ Rrms ⎭ Solving for r we obtain r2 − 2 = ln 1 2 Rrms r = Rrms ln 1

2

= 0.83Rrms

Consequently, there is a 20log10(0.83) =1.62 dB error when using the rms value of the amplitude instead of the median value.

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Problem 11.12 Compute the noise spectral density in watts per hertz of: (a) an ideal resistor at nominal temperature of 290°K; (b) an amplifier with an equivalent noise temperature of 22,000°K. Solution (a) From Eq. (11.19), the noise power spectral density is N 0 = kTe = 1.38 × 10−23 × 290 = 4.0 × 10−21 W/Hz

(b) From Eq. (11.19), the noise power spectral density is N 0 = kTe = 1.38 × 10−23 × 22000 = 3.04 × 10−19 W/Hz

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Problem 11.13 For the two cases of Problem 11.12, compute the pre-detection SNR when the received signal power is: (a) -60 dBm and the receive bandwidth is 1 MHz; (b) -90 dBm and the receive bandwidth is 30 kHz. Express the answers in both absolute terms and decibels. Solution (a) The signal power is obtained by converting -60 dBm to watts S = 10( −60 /10) = 10−6 mW = 10−9 W

The noise power from the ideal resistor is from Eq. (11.13) N = kTe BN = 4.0 × 10−21 × (106 ) = 4.0 × 10−15 W The SNR is the ratio of the two

SNR =

S 10−9 = = 2.5 ×105 ~ 54 dB N 4.0 ×10−15

A similar calculation for the amplifier of the previous problem results in 10−9 S = 2.94 × 103 ~ 34.7 dB SNR = = −19 6 N 3.04 ×10 ×10

(b) The signal power is obtained by converting -90 dBm to watts S = 10( −90 /10) = 10−9 mW = 10−12 W

The noise power from the ideal resistor is from Eq. (11.13)

Continued on next slide

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Problem 11.13 continued

N = kTe BN = 4.0 × 10−21 × (30 ×103 ) = 1.2 ×10−16 W The SNR is the ratio of the two 10−12 S = 8.3 × 103 ~ 39.2 dB SNR = = −16 N 1.2 ×10 A similar calculation for the amplifier of the previous problem results in

SNR =

S 10−12 = = 1.1×102 ~ 20.4 dB −19 3 N 3.04 ×10 × 30 ×10

(

)

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Problem 11.14 A wireless local area network transmits a signal that has a noise bandwidth of approximately 6 MHz. If the signal strength at the receiver input terminals is –90 dBm and the receiver noise figure is 8 dB, what is the pre-detection signal-to-noise ratio? Solution The signal power is obtained by converting -90 dBm to watts S = 10( −90 /10) = 10−9 mW = 10−12 W

The noise power with an 8 dB noise figure F is from Eqs. (11.15) and (11.16) N = kT0 FB = 1.38 × 10−23 × (290) × 108/10 × (6 × 106 ) = 1.52 × 10−13 W The pre-detection SNR is the ratio of the two S 10−12 SNR = = = 6.6 ~ 8.2 dB N 1.52 ×10−13

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Problem 11.15 A communications receiver includes a whip antenna whose noise temperature is approximately that of the Earth, that is, 290°K. The receiver pre-amplifier has a noise figure of 4 dB and a gain of 25 dB. What is the equivalent noise temperature of the antenna and the pre-amplifier? What is the combined noise figure? Solution (a) Following Example 11.4, the combined noise temperature of the antenna and preamplifier is, from Eq. (11.17) Tsys = Tant + Tamp

= 290 + 290( F − 1) = 728K

(b) From Eq. (11.16), the combined noise figure is Fcomb =

T + Te 290 + 728 = = 3.51 ~ 5.45dB T 290

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Problem 11.16 A parabolic antenna with a diameter of 0.75 meters is used to receive a 12 GHz satellite signal. What is the gain in decibels of this antenna? Assume the antenna efficiency is 60%. Solution From Eq.(11.25), the antenna gain is

GR =

4π Aeff

(1)

λ2

The signal wavelength is λ = c f = 3 × 108 12 × 109 = 0.025m , and the effective area is Aeff = η

πd2 4

= 0.60

π (.75) 2 4

Substituting these two results into Eq. (1), the antenna gain is GR =

4π

( 0.025)

× ( 0.6 ) 2

π (0.75) 2 4

= 5.33 × 103 ~ 37.3dB

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Problem 11.17 If the system noise temperature of a satellite receiver is 300°K, what is the required received signal strength to produce a C/N0 of 80 dB? Solution (There is a typo in problem statement, the units should be “dB-Hz”.) From Eq. (11.19), the noise power spectral density is N 0 = kTs = 1.38 × 10−23 × (300) = 4.14 ×10−21 W/Hz ~ -203.8 dBW/Hz

In decibels, the carrier to noise density is given by

( C / N0 )dB − Hz = (C )dBW − ( N0 )dBW − Hz 80 = ( C )dBW − (−203.8) Solving for C, we obtain C = -123.8 dBW = -93.8 dBm.

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Problem 11.18 If a satellite is 40,000 km from the antenna of Problem 11.16, what satellite EIRP will produce a signal strength of –110 dBm at the antenna terminals? Assume the transmission frequency is 12 GHz. Solution The received power is given by Eq. (11.29) PR = EIRP + GR − LP

(1)

where all quantities are in decibels. From Problem 11.16, the antenna gain is GR = 37.3 dB. The free-space path loss is given by Eq. (11.32) ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

From Problem 11.16, the wavelength is λ = 0.025m at 12 GHz. So, at a distance r = 40,000 km, the path loss is

⎛ 4π (40000 ×103 ⎞ LP = 20 log10 ⎜ ⎟ = 206.1 dB 0.025 ⎝ ⎠ Substituting these in Eq.(1) with a received power of -110 dBm, we obtain −110 dBm = EIRP + 37.3 dB − 206.1dB

Solving this equation, we find the require EIRP is 58.8 dBm.

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Problem 11.19 Antennas are placed on two 35-meter office towers that are separated by ten kilometers. What is the minimum height of a building between the two towers that would disturb the assumption of free-space propagation? Solution From Eq. (11.35), the radius of the first Fresnel zone is

h=

λ d1d 2 d1 + d 2

This radius is maximized midway between the two towers and must be kept clear to approximate free-space propagation. With d1 = d2 = 5km, the radius in meters is h = 2500λ = 50 λ

The maximum building height (in meters) is

b = 35 − h = 35 − 50 λ For example, at a transmission frequency of 4 GHz, the maximum height is b = 21.3 m.

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Problem 11.20 If a receiver has a sensitivity of –90 dBm and a 12 dB noise figure what is minimum pre-detection signal-to-noise ratio of an 8 kHz signal? Solution The noise in an 8 kHz bandwidth for a receiver with an 8 dB noise figure is, from Eqs. (11.15) and (11.16),

N = kT0 FB

(

)

= 1.38 × 10−23 × (290) × 1012 /10 × (8 × 103 ) = 5.07 × 10−16 W The receiver sensitivity is defined as the minimum received signal power that will provide a demodulated signal with acceptable performance, thus the minimum signal power is S = -90 dBm ~ 10-12 W. The minimum pre-detection SNR is the ratio of the two S 10−12 SNR = = = 1.97 × 103 ~ 32.9 dB −16 N 5.07 ×10

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Problem 11.21 A satellite antenna is installed on the tail of an aircraft and has a noise temperature of 100°K. The antenna is connected by a coaxial cable to a low-noise amplifier in the equipment bay at the front of the aircraft. The cable causes 2 dB attenuation of the signal. The low-noise amplifier has a gain of 60 dB and a noise temperature of 120°K. What is the system noise temperature? Where would a better place for the low-noise amplifier be? Solution Following Example 11.4, the system noise temperature is

Ts = Tant + = 100 +

Tcable Tamp + Gant Gcable 290 120 + 1 .631

= 580K where we have used the facts that the antenna does not provide any electrical gain, thus Gant = 1; and the fact the fact that cable causes a 2 dB loss so Gcable = 10-2/10 = 0.631. Locating the low-noise amplifier in the tail of the aircraft, close to the antenna would be a better system design. With the amplifier in the antenna tail, the system noise temperature would be approximately 220 K.

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Problem 11.22 A wireless local area network transmitter radiates 200 milliwatts. Experimentation indicates that the path loss may be accurately described by

Lp = 31 + 33 log10(r) where the path loss is in decibels and r is the range in meters. If the minimum receiver sensitivity is –85 dBm, what is the range of the transmitter? Solution Since the problem says nothing about the transmit and receive antennas, we shall assume they are omni-directional with a gain of 0 dB. In this case, the Friis equation (in decibels) for the received signal strength reduces to

PR = PT − LP

= PT − ( 31 + 33log10 r )

(1)

With a transmit power of 200 mW, equivalent to 23 dBm, and a minimum signal strength of -85 dBm, Eq. (1) becomes −85 = 23 − ( 31 + 33log10 r )

Solving this equation for the maximum range, we find r is 215.4 meters.

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Problem 11.23 A mobile radio transmits 30 watts and the median path loss may be approximated by

Lp = 69 + 31 log10(r) where the path loss is in decibels and r is the range in kilometers. If the receiver sensitivity is -110 dBm and 12 dB of margin must be included to compensate for variations about the median path loss, what is the range of the transmitter? Solution Since the problem says nothing about the transmit and receive antennas, we shall assume they are omni-directional with a gain of 0 dB. In this case, the Friis equation for the received signal strength reduces to

PR = PT − LP − L0

= PT − ( 69 + 31log10 r ) − L0

(1)

where L0 represents the required margin. A transmit power of 30 W is equivalent to 14.8 dBW, and a minimum signal strength of -110 dBm is equivalent to -140 dBW. Thus, Eq. (1) becomes −140 = 14.8 − ( 69 + 31log10 r ) − 12

Solving this equation for the maximum range, we find r is 240.2 kilometers. In practice, the range will likely be somewhat less than this due to the curvature of the earth and depending on the height of the base station antenna.

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Problem 11.24 A cellular telephone transmits 600 milliwatts of power. If the receiver sensitivity is –90 dBm, what would the range of the telephone be under free space propagation? Assume the transmitting and receiving antennas have unity gain and the transmissions are at 900 MHz. If propagation conditions actually show a path-loss exponent of 3.1 with a fixed loss β = 36 dB, what would the range be in this case? Solution (a) The Friis equation for the received power in decibels is PR = PT + GR + GT − LP

(1)

where the antenna gains are GR = GT = 0 dB. The transmit power of 600 mW is equivalent to or 27.8 dBm. For free-space propagation, the path loss is ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

At 900 MHz, the wavelength is λ = c f = 3 × 108 900 × 106 = 0.33 m . Making these substitutions, we have ⎛ 4π r ⎞ −90 = 27.8 + 0 + 0 − 20 log10 ⎜ ⎟ ⎝ 0.33 ⎠

Solving this equation for the maximum range, we find the r is 20.4 kilometers. (b) In this case, the Friis equation still applies but the path loss is given by Eq. (11.37) −1

⎛ 10−36 /10 ⎞ LP = ⎜ 3.1 ⎟ ⎝ r ⎠ ~ 36 + 31log10 ( r ) Substituting the second line into Eq. (1), we have −90 = 27.8 + 0 + 0 − (36 + 31log10 r )

Solving this equation for the maximum range, we find that r is 435 meters. Clearly, the propagation conditions can make a huge difference on the range.

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Problem 11.25 A line-of-sight 10-kilometer radio link is required to transmit data at a rate of 1 megabit per second at a center frequency of 4 GHz. The transmitter uses an antenna with 10 dB gain and QPSK modulation with a root-raised cosine pulse shape spectrum having a roll-off factor of 0.5. The receiver also has an antenna with 10 dB gain and has a system noise temperature of 900 K. What is the minimum transmit power required to achieve a bit error rate of 10-5? Solution From the BER performance of QPSK in Fig. 10.16, we find that a BER of 10-5 implies an Eb/N0 of 9.5 dB is required. From this, we obtain the required C/N0 using knowledge of the transmission rate R = 1 Mbps.

⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠dB − Hz ⎝ N 0 ⎠dB = 69.5 dB-Hz

The system noise temperature of 900 K implies N 0 = kTe = 1.38 × 10−23 × 900 = 1.24 × 10−20 W/Hz ~ −199.1 dBW/Hz

Using this information, the received power level may be calculated from PR = C ⎛ C ⎞ =⎜ + ( N 0 )dBW − Hz ⎟ ⎝ N 0 ⎠dB − Hz = 69.5 + ( −199.1) = −129.6 dBW

We now appeal to the decibel form of the Friis equation: PR = PT + GR + GT − LP

(1)

where the antenna gains are GR = GT = 10 dB. Since the problem sight says line-of-sight transmission, we shall assume free-space propagation, and the path loss is

Continued on next slide

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Problem 11.25 continued ⎛ 4π r ⎞ LP = 20 log10 ⎜ ⎟ ⎝ λ ⎠

At 4 GHz, the wavelength is λ = c f = 3 × 108 4 × 109 = 0.075 m . Making all these substitutions into Eq. (1) with a range r = 10 km, we obtain

⎛ 4π 10 ×103 ⎞ −129.6 = PT + 10 + 10 − 20 log10 ⎜ ⎟ ⎝ 0.075 ⎠ Solving this equation for the transmitted power, we find that the required PT is -25.1 dBW or 4.9 dBm.

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Problem 11.26 A land-mobile radio transmits 128 kbps at a frequency of 700 MHz. The transmitter uses an omni-directional antenna and 16-QAM modulation with a root-raised cosine pulse spectrum having a roll-off of 0.4. The receiver has an antenna with 3 dB gain and a noise figure of 6 dB. If the path loss between the transmitter and receiver is given by LP (r ) = 30 + 28log10 (r ) dB

where r is in meters, what is the maximum range at which the bit error rate of 10-4 may be achieved? Solution From the BER performance of 16-QAM in Fig. 10.16, we find that a BER of 10-4 implies an Eb/N0 of 13 dB. From this, we obtain the C/N0 by using knowledge of the transmission rate R = 128 kbps.

⎛ C ⎞ ⎛E ⎞ = ⎜ b ⎟ + 10 log10 R ⎜ ⎟ ⎝ N 0 ⎠dB − Hz ⎝ N 0 ⎠dB = 64.1 dB-Hz The noise figure of 6 dB implies N 0 = kFT0

(

)

= 1.38 × 10−23 × 106 /10 × ( 290 ) = 1.59 × 10

−20

W/Hz

~ −198.0 dBW/Hz

and the received power level is ⎛ C ⎞ + ( N 0 )dBW − Hz PR = C = ⎜ ⎟ ⎝ N 0 ⎠ dB − Hz = 64.1 + ( −198.0 ) = −133.9 dBW

We now appeal to the decibel form of the Friis equation: PR = PT + GR + GT − LP

(1)

Continued on next slide

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Problem 11.26 continued

where the antenna gains are GR = GT = 0 dB. The path loss is LP = 30 + 28log10 ( r )

Making all these substitutions into Eq. (1), we obtain −133.9 = PT + 0 + 0 − ( 30 + 28log10 r )

or

r = 10(

PT +103.9 ) / 28

In the following figure, we plot the range in kilometres versus the transmit power in dBW. 20 18 16

Range (km)

14 12 10 8 6 4 2 0 -20

-15

-10

-5

0

5

10

15

Transmit Power (dBW)

For example, with a transmit power of 10 W or 10 dBW, we find that range is 11.7 km.

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