Problem 7.7 : We assume that P (x i )=1/3, i =1, 2, 3. Then P (y 1 )= 1 ((1 p)+p) = P (y j )=1/3, j=2, 3. Hence : and similarly.

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1 (b) We note that the above capacity is the same to the capacity of the binary symmetric channel. Indeed, if we considerthe grouping of the output symbols into a = {y 1,y 2 } and b = {y 3,y 4 } we get a binary symmetric channel, with transition probabilities: P (a x 1 )=P (y 1 x 1 )+P (y 2 x 1 )= (1 p), P(a x 2 )=p, etc. Problem 7.7 : We assume that P (x i )=1/3, i =1, 2, 3. Then P (y 1 )= 1 ((1 p)+p) = P (y j )=1/3, j=2, 3. Hence : and similarly I(x 1 ; Y ) = 3 j=1 P (y j x 1 )log P (y j x 1 ) P (y j ) = log3+p log p +(1 p) log(1 p) =(1 p)log (1 p) 1/3 + p log p 1/3 and the same is true for I(x 2 ; Y ), I(x 3 ; Y ). Thus, equiprobable input symbols achieve the channel capacity : C =log3+p log p +(1 p)log(1 p) bits/symbol sent Problem 7.8 : (a) the probability that a codeword transmitted over the BSC is received correctly, is equal to the probability that all R bits are received correctly. Since each bit transmission is independent from the others : P (correct codeword) = (1 p) R (b) P ( at least one bit error in the codeword) = 1 P (correct codeword) = 1 (1 p) R (c) P (n e or less errors in R bits) = n e i=1 ( ) R p i (1 p) R i i (d) For R =5,p=.1, n e =5: (1 p) R = (1 p) R =.49 ne i=1 ( R i ) p i (1 p) R i =1 (1 p) R =.49 14

2 where Q 1, Q 2 are the transition probability matrices of channel 1 and channel 2 respectively. We have assumed that the output space of both channels has been augmented by adding two new symbols so that the size of the matrices Q, Q 1 and Q 2 is the same. The transition probabilities to these newly added output symbols is equal to zero. Using the fact that the function I(p; Q) is a convex function in Q we obtain : C = max p I(X; Y )=max I(p; Q) = max I(p; 1 p 2 Q Q 2) p 1 2 max I(p; Q 1)+ 1 p 2 max I(p; Q 2) p = 1 2 C C 2 Since Q 1 and Q 2 are different, the inequality is strict. Hence : C< 1 2 C C 2 Problem 7.14 : The capacity of a channel is : C =max p(x) I(X; Y )=max p(x) Since in general H(X Y ) andh(y X), we obtain : [H(Y ) H(Y X)] = max[h(x) H(X Y )] p(x) C min{max[h(y )], max[h(x)]} However, the maximum of H(X) is attained when X is uniformly distributed, in which case max[h(x)] = log X. Similarly : max[h(y )] = log Y and by substituting in the previous inequality, we obtain C min{max[h(y )], max[h(x)]} =min{log Y, log X } = min{log M,log N} Problem 7.15 : (a) Let q be the probability of the input symbol, and therefore (1 q) the probability of the input symbol 1. Then : H(Y X) = x P (x)h(y X = x) 145

3 = qh(y X =)+(1 q)h(y X =1) = (1 q)h(y X =1)=(1 q) The probability mass function of the output symbols is : P (Y =) = qp(y = X =)+(1 q)p (Y = X =1) = q +(1 q)(1 ɛ) =1 ɛ + qɛ P (Y =1) = (1 q)ɛ = ɛ qɛ Hence : C =max[h(ɛ qɛ) (1 q)] q To find the probability q that achieves the maximum, we set the derivative of C with respect to q equal to. Thus : Therefore : ϑc ϑq ==+ɛ log 2(ɛ qɛ) ɛ log 2 (1 ɛ + qɛ) log 2 and the channel capacity is ɛ qɛ 1 ɛ + qɛ = ɛ C = H 2 ɛ 1+2 ɛ = q = ɛ +2 ɛ (ɛ 1) ɛ(1 + 2 ɛ ) 2 ɛ ɛ(1 + 2 ɛ ) (b) If ɛ, then using L Hospital s rule we find that lim =, lim 2 ɛ = ɛ ɛ ɛ ɛ and therefore lim C(ɛ) =H() = ɛ If ɛ =.5, then =1andC = H( 1) 2 = In this case the probability of the input 5 5 symbol is q = ɛ +2 ɛ (ɛ 1) (.5 1) = = 3 ɛ(1 + 2 ɛ ).5 (1 +.25) 5 If ɛ =1,thenC = H(.5) = 1. The input distribution that achieves capacity is P () = P (1) =

4 (c) The following figure shows the topology of the cascade channels. If we start at the input labeled, then the output will be. If however we transmit a 1, then the output will be zero with probability P (Y = X =1) = (1 ɛ)+ɛ(1 ɛ)+ɛ 2 (1 ɛ)+ = (1 ɛ)(1 + ɛ + ɛ 2 + ) = 1 ɛ 1 ɛn =1 ɛn 1 ɛ Thus, the resulting system is equivalent to a Z channel with ɛ 1 = ɛ n ɛ 1 ɛ 1 ɛ ɛ ɛ ɛ 1 (d) As n, ɛ n and the capacity of the channel goes to. Problem 7.16 : The SNR is : SNR = Thus the capacity of the channel is : C = W log 2 (1 + 2P N 2W = P 2W = =14 P N W )=16 log 2 (1 + 1) bits/sec Problem 7.17 : The capacity of the additive white Gaussian channel is : C = 1 ( 2 log 1+ P ) N W Forthe nonwhite Gaussian noise channel, although the noise poweris equal to the noise powerin the white Gaussian noise channel, the capacity is higher, The reason is that since noise samples are correlated, knowledge of the previous noise samples provides partial information on the future noise samples and therefore reduces their effective variance. 147

5 Problem 5.12 : The correlation of the two signals in binary FSK is: ρ = sin(2π ft) 2π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero. Thus: and therefore : ϑρ ϑ f ==cos(2π ft)2πt 2π ft 2π ft =tan(2π ft) sin(2π ft) (2π ft) 2πT 2 Solving numerically (or graphically) the equation x = tan(x), we obtain x = Thus, 2π ft = = f =.7151 T and the value of ρ is We know that the probability of error can be expressed in terms of the distance d 12 between the signal points, as : P e = Q d N where the distance between the two signal points is : d 2 12 =2E b(1 ρ) and therefore : [ ] 2Eb (1 ρ) Eb P e = Q = Q 2N N Problem 5.13 : (a) It is straightforward to see that : Set I : Four level PAM Set II : Orthogonal Set III : Biorthogonal 86

6 Problem 5.18 : For binary phase modulation, the error probability is [ ] 2Eb A2 T P 2 = Q = Q With P 2 =1 6 we find from tables that A2 T =4.74 = A 2 T = N N If the data rate is 1 Kbps, then the bit interval is T =1 4 and therefore, the signal amplitude is A = = Similarly we find that when the rate is 1 5 bps and 1 6 bps, the required amplitude of the signal is A = and A = respectively. N Problem 5.19 : (a) The PDF of the noise n is : p(n) = λ 2 e λ n where λ = 2 σ The optimal receiver uses the criterion : p(r A) p(r A) = e λ[ r A r+a ] A > < A 1= r A > < A The average probability of error is : P (e) = 1 2 P (e A)+1 2 P (e A) = 1 2 = 1 2 = λ 4 f(r A)dr λ 2 e λ r A dr A e λ x dx + λ 4 = 1 2 e λa = 1 2 e 2A σ A f(r A)dr λ 2 e λ r+a dr e λ x dx 92

7 (b) The variance of the noise is : Hence, the SNR is: σn 2 = λ 2 = λ and the probability of error is given by: For P (e) =1 5 we obtain: e λ x x 2 dx e λx x 2 dx = λ 2! λ 3 = 2 λ 2 = σ2 SNR = A2 σ 2 P (e) = 1 2 e 2SNR ln(2 1 5 )= 2SNR = SNR = = db If the noise was Gaussian, then the probability of error for antipodal signalling is: [ ] 2Eb P (e) =Q = Q [ SNR ] N where SNR is the signal to noise ratio at the output of the matched filter. With P (e) =1 5 we find SNR =4.26 and therefore SNR = = db. Thus the required signal to noise ratio is 5 db less when the additive noise is Gaussian. Problem 5.2 : The constellation of Fig. P5-2(a) has four points at a distance 2A from the origin and four points at a distance 2 2A. Thus, the average transmitted power of the constellation is: P a = 1 [ 4 (2A) 2 +4 (2 2A) 2] =6A 2 8 The second constellation has four points at a distance 7A from the origin, two points at a distance 3A and two points at a distance A. Thus, the average transmitted power of the second constellation is: P b = 1 [ 4 ( 7A) 2 +2 ( 3A) 2 +2A 2] = A2 Since P b <P a the second constellation is more power efficient. 93

ρ = sin(2π ft) 2π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero.

ρ = sin(2π ft) 2π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero. Problem 5.1 : The correlation of the two signals in binary FSK is: ρ = sin(π ft) π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero. Thus: ϑρ