Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13 pps

9.99 9.991 9.992 9.993 9.994 9.995 9.996 9.997 9.998 9.999 10 x10 -7 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Frequency (f) Tc(f) 2) The following figure is a plot of the amplitude characteristics of the RC filter, |C(f)|. The values of the vertical axis indicate that |C(f)| can be considered constant for frequencies up to 2000 Hz. Since the same is true for the envelope delay, we conclude that a lowpass signal of bandwidth ∆f = 1 KHz will not be distorted if it passes the RC filter. 0.999 1 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Frequency (f) |C(f)| Problem 8.42 Let G T (f) and G R (f) be the frequency response of the transmitting and receiving filter. Then, the condition for zero ISI implies G T (f)C(f)G R (f)=X rc (f)=      T 0 ≤|f|≤ 1 4T T 2 [1 + cos(2πT(|f|− 1 T )] 1 4T ≤|f|≤ 3 4T 0 |f| > 3 4T Since the additive noise is white, the optimum transmitting and receiving filter characteristics are given by (see Example 8.6.1) |G T (f)| = |X rc (f)| 1 2 |C(f)| 1 2 , |G R (f)| = |X rc (f)| 1 2 |C(f)| 1 2 Thus, |G T (f)| = |G R (f)| =             T 1+0.3 cos 2πfT  1 2 0 ≤|f|≤ 1 4T  T (1+cos(2πT(|f|− 1 T ) 2(1+0.3 cos 2πfT )  1 2 1 4T ≤|f|≤ 3 4T 0 otherwise 238 Problem 8.43 A 4-PAM modulation can accommodate k = 2 bits per transmitted symbol. Thus, the symbol interval duration is T = k 9600 = 1 4800 sec Since, the channel’s bandwidth is W = 2400 = 1 2T , in order to achieve the maximum rate of transmission, R max = 1 2T , the spectrum of the signal pulse should be X(f)=T Π  f 2W  Then, the magnitude frequency response of the optimum transmitting and receiving filter is (see Section 8.6.1 and Example 8.6.1) |G T (f)| = |G R (f)| =  1+  f 2400  2  1 4 Π  f 2W  =       1+  f 2400  2  1 4 , |f| < 2400 0 otherwise Problem 8.44 1) The equivalent discrete-time impulse response of the channel is h(t)= 1  n=−1 h n δ(t − nT )=0.3δ(t + T )+0.9δ(t)+0.3δ(t −T) If by {c n } we denote the coefficients of the FIR equalizer, then the equalized signal is q m = 1  n=−1 c n h m−n which in matrix notation is written as    0.90.30. 0.30.90.3 0. 0.30.9       c −1 c 0 c 1    =    0 1 0    The coefficients of the zero-force equalizer can be found by solving the previous matrix equation. Thus,    c −1 c 0 c 1    =    −0.4762 1.4286 −0.4762    2) The values of q m for m = ±2, ±3 are given by q 2 = 1  n=−1 c n h 2−n = c 1 h 1 = −0.1429 q −2 = 1  n=−1 c n h −2−n = c −1 h −1 = −0.1429 q 3 = 1  n=−1 c n h 3−n =0 q −3 = 1  n=−1 c n h −3−n =0 239 Problem 8.45 1) The output of the zero-force equalizer is q m = 1  n=−1 c n x m n With q 0 = 1 and q m = 0 for m = 0, we obtain the system    1.00.1 −0.5 −0.21.00.1 0.05 −0.21.0       c −1 c 0 c 1    =    0 1 0    Solving the previous system in terms of the equalizer’s coefficients, we obtain    c −1 c 0 c 1    =    0.000 0.980 0.196    2) The output of the equalizer is q m =                                0 m ≤−4 c −1 x −2 =0 m = −3 c −1 x −1 + c 0 x −2 = −0.49 m = −2 0 m = −1 1 m =0 0 m =1 c 0 x 2 + x 1 c 1 =0.0098 m =2 c 1 x 2 =0.0098 m =3 0 m ≥ 4 Hence, the residual ISI sequence is residual ISI = { ,0, −0.49, 0, 0, 0, 0.0098, 0.0098, 0, } and its span is 6 symbols. Problem 8.46 The MSE performance index at the time instant k is J(c k )=E         N  n=−N c k,n y k−n − a k       62   If we define the gradient vector g k as g k = ϑJ(c k ) 2ϑc k then its l th element is g k,l = ϑJ(c k ) 2ϑc k,l = 1 2 E   2   N  n=−N c k,n y k−n − a k   y k−l   = E [−e k y k−l ]=−E [e k y k−l ] 240 Thus, the vector g k is g k =    −E[e k y k+N ] . . . −E[e k y k−N ]    = −E[e k y k ] where y k is the vector y k =[y k+N ···y k−N ] T . Since ˆ g k = −e k y k , its expected value is E[ ˆ g k ]=E[−e k y k ]=−E[e k y k ]=g k Problem 8.47 1) If {c n } denote the coefficients of the zero-force equalizer and {q m } is the sequence of the equal- izer’s output samples, then q m = 1  n=−1 c n x m−n where {x k } is the noise free response of the matched filter demodulator sampled at t = kT. With q −1 =0,q 0 = q 1 = E b , we obtain the system    E b 0.9E b 0.1E b 0.9E b E b 0.9E b 0.1E b 0.9E b E b       c −1 c 0 c 1    =    0 E b E b    The solution to the system is  c −1 c 0 c 1  =  0.2137 −0.3846 1.3248  2) The set of noise variables {ν k } at the output of the sampler is a Gaussian distributed sequence with zero-mean and autocorrelation function R ν (k)=  N 0 2 x k |k|≤2 0 otherwise Thus, the autocorrelation function of the noise at the output of the equalizer is R n (k)=R ν (k) c(k) c(−k) where c(k) denotes the discrete time impulse response of the equalizer. Therefore, the autocorrela- tion sequence of the noise at the output of the equalizer is R n (k)= N 0 E b 2                  0.9402 k =0 1.3577 k = ±1 −0.0546 k = ±2 0.1956 k = ±3 0.0283 k = ±4 0 otherwise To find an estimate of the error probability for the sequence detector, we ignore the residual interference due to the finite length of the equalizer, and we only consider paths of length two. Thus, if we start at state a 0 = 1 and the transmitted symbols are (a 1 ,a 2 )=(1, 1) an error is made by the sequence detector if the path (−1, 1) is more probable, given the received values of r 1 and r 2 . The metric for the path (a 1 ,a 2 )=(1, 1) is µ 2 (1, 1)=[ r 1 − 2E b r 2 − 2E b ]C −1  r 1 − 2E b r 2 − 2E b  241 where C = N 0 E b 2  0.9402 1.3577 1.3577 0.9402  Similarly, the metric of the path (a 1 ,a 2 )=(−1, 1) is µ 2 (−1, 1)=[ r 1 r 2 ]C −1  r 1 r 2  Hence, the probability of error is P 2 = P (µ 2 (−1, 1) <µ 2 (1, 1)) and upon substitution of r 1 =2E b + n 1 , r 2 =2E b + n 2 , we obtain P 2 = P (n 1 + n 2 < −2E b ) Since n 1 and n 2 are zero-mean Gaussian variables, their sum is also zero-mean Gaussian with variance σ 2 =(2×0.9402 + 2 × 1.3577) N 0 E b 2 =4.5958 N 0 E b 2 and therefore P 2 = Q   8E b 4.5958N 0  The bit error probability is P 2 2 . Problem 8.48 The optimum tap coefficients of the zero-force equalizer can be found by solving the system    1.00.30.0 0.21.00.3 0.00.21.0       c −1 c 0 c 1    =    0 1 0    Hence,    c −1 c 0 c 1    =    −0.3409 1.1364 −0.2273    b) The output of the equalizer is q m =                        0 m ≤−3 c −1 x −1 = −0.1023 m = −2 0 m = −1 1 m =0 0 m =1 c 1 x 1 = −0.0455 m =2 0 m ≥ 3 Hence, the residual ISI sequence is residual ISI = { ,0, −0.1023, 0, 0, 0, −0.0455, 0, } 242 Problem 8.49 1) If we assume that the signal pulse has duration T , then the output of the matched filter at the time instant t = T is y(T )=  T 0 r(τ)s(τ)dτ =  T 0 (s(τ)+αs(τ − T )+n(τ ))s(τ )dτ =  T 0 s 2 (τ)dτ +  T 0 n(τ)s(τ)dτ = E s + n where E s is the energy of the signal pulse and n is a zero-mean Gaussian random variable with variance σ 2 n = N 0 E s 2 . Similarly, the output of the matched filter at t =2T is y(2T )=α  T 0 s 2 (τ)dτ +  T 0 n(τ)s(τ)dτ = αE s + n 2) If the transmitted sequence is x(t)= ∞  n=−∞ a n s(t − nT ) with a n taking the values 1, −1 with equal probability, then the output of the demodulator at the time instant t = kT is y k = a k E s + αa k−1 E s + n k The term αa k−1 E s expresses the ISI due to the signal reflection. If a symbol by symbol detector is employed and the ISI is ignored, then the probability of error is P (e)= 1 2 P (error|a n =1,a n−1 =1)+ 1 2 P (error|a n =1,a n−1 = −1) = 1 2 P ((1 + α)E s + n k < 0) + 1 2 P ((1 − α)E s + n k < 0) = 1 2 Q    2(1 + α) 2 E s N 0   + 1 2 Q    2(1 − α) 2 E s N 0   3) To find the error rate performance of the DFE, we assume that the estimation of the parameter α is correct and that the probability of error at each time instant is the same. Since the transmitted symbols are equiprobable, we obtain P (e)=P (error at k|a k =1) = P (error at k −1)P (error at k|a k =1, error at k − 1) +P (no error at k −1)P (error at k|a k =1, no error at k − 1) = P (e)P (error at k|a k =1, error at k − 1) +(1 − P (e))P (error at k|a k =1, no error at k − 1) = P (e)p +(1− P (e))q 243 where p = P (error at k|a k =1, error at k − 1) = 1 2 P (error at k|a k =1,a k−1 =1, error at k − 1) + 1 2 P (error at k|a k =1,a k−1 = −1, error at k −1) = 1 2 P ((1+2α)E s + n k < 0) + 1 2 P ((1 − 2α)E s + n k < 0) = 1 2 Q    2(1+2α) 2 E s N 0   + 1 2 Q    2(1 − 2α) 2 E s N 0   and q = P (error at k|a k =1, no error at k − 1) = P (E s + n k < 0) = Q   2E s N 0  Solving for P (e), we obtain P (e)= q 1 − p + q = Q   2E s N 0  1 − 1 2 Q   2(1+2α) 2 E s N 0  − 1 2 Q   2(1−2α) 2 E s N 0  + Q   2E s N 0  A sketch of the detector structure is shown in the next figure. ❧ ❧ ✛ ✛ ✻ ✲✲ ✲✲✲ + Delay r k Input × − + Output ˆa k device Threshold α Estimate Problem 8.50 A discrete time transversal filter equivalent to the cascade of the transmitting filter g T (t), the channel c(t), the matched filter at the receiver g R (t) and the sampler, has tap gain coefficients {y m }, where y m =      0.9 m =0 0.3 m = ±1 0 otherwise The noise ν k , at the output of the sampler, is a zero-mean Gaussian sequence with autocorrelation function E[ν k ν l ]=σ 2 y k−l , |k −l|≤1 If the Z-transform of the sequence {y m }, Y (z), assumes the factorization Y (z)=F (z)F ∗ (z −1 ) then the filter 1/F ∗ (z −1 ) can follow the sampler to white the noise sequence ν k . In this case the output of the whitening filter, and input to the MSE equalizer, is the sequence u n =  k a k f n−k + n k 244 where n k is zero mean Gaussian with variance σ 2 . The optimum coefficients of the MSE equalizer, c k , satisfy (see (8.6.35)) 1  n=−1 c n R u (n − k)=R ua (k),k=0, ±1 where R u (n − k)=E[u l−k u l−n ]= 1  m=0 f m f m+n−k + σ 2 δ n,k =  y n−k + σ 2 δ n,k , |n − k|≤1 0 otherwise R ua (k)=E[a n u n−k ]=  f −k , −1 ≤ k ≤ 0 0 otherwise With Y (z)=0.3z +0.9+0.3z −1 =(f 0 + f 1 z −1 )(f 0 + f 1 z) we obtain the parameters f 0 and f 1 as f 0 =  ± √ 0.7854 ± √ 0.1146 ,f 1 =  ± √ 0.1146 ± √ 0.7854 The parameters f 0 and f 1 should have the same sign since f 0 f 1 =0.3. However, the sign itself does not play any role if the data are differentially encoded. To have a stable inverse system 1/F ∗ (z −1 ), we select f 0 and f 1 in such a way that the zero of the system F ∗ (z −1 )=f 0 + f 1 z is inside the unit circle. Thus, we choose f 0 = √ 0.1146 and f 1 = √ 0.7854 and therefore, the desired system for the equalizer’s coefficients is    0.9+0.10.30.0 0.30.9+0.10.3 0.00.30.9+0.1       c −1 c 0 c 1    =    √ 0.7854 √ 0.1146 0    Solving this system, we obtain c −1 =0.8596,c 0 =0.0886,c 1 = −0.0266 Problem 8.51 1) The spectrum of the band limited equalized pulse is X(f)=  1 2W  ∞ n=−∞ x( n 2W )e −j πnf W |f|≤W 0 otherwise =  1 2W  2+2cos πf W  |f|≤W 0 otherwise =  1 W  1+1cos πf W  |f|≤W 0 otherwise where W = 1 2T b 245 2) The following table lists the possible transmitted sequences of length 3 and the corresponding output of the detector. -1 -1 -1 -4 -1 -1 1 -2 -1 1 -1 0 -111 2 1-1-1 -2 1-1 1 0 11-1 2 111 4 As it is observed there are 5 possible output levels b m , with probability p(b m =0)= 1 4 , p(b m = ±2) = 1 4 and p(b m = ±4) = 1 8 . 3) The transmitting filter G T (f), the receiving filter G R (f) and the equalizer G E (f) satisfy the condition G T (f)G R (f)G E (f)=X(f) The power spectral density of the noise at the output of the equalizer is S ν (f)=S n (f)|G R (f)G E (f)| 2 = σ 2 |G R (f)G E (f)| 2 With G T (f)=G R (f)=P (f)= πT 50 2 e −πT 50 |f| the variance of the output noise is σ 2 ν = σ 2  ∞ −∞ |G R (f)G E (f)| 2 df = σ 2  ∞ −∞     X(f) G T (f)     2 df = σ 2  W −W 4 π 2 T 2 50 W 2 |1 + cos πf W | 2 e −2πT 50 |f| df = 8σ 2 π 2 T 2 50 W 2  W 0  1 + cos πf W  2 e 2πT 50 f df The value of the previous integral can be found using the formula  e ax cos n bxdx = 1 a 2 + n 2 b 2  (a cos bx + nb sin bx)e a x cos n−1 bx + n(n − 1)b 2  e ax cos n−2 bxdx  Thus, we obtain σ 2 ν = 8σ 2 π 2 T 2 50 W 2 ×   e 2πT 50 W − 1   1 2πT 50 + 2πT 50 + π 1 W 2 T 50 4π 2 T 2 50 +4 π 2 W 2  − 4πT 50 4π 2 T 2 50 + π 2 W 2  e 2πT 50 W +1   To find the probability of error using a symbol by symbol detector, we follow the same procedure as in Section 8.4.3. The results are the same with that obtained from a 3-point PAM constellation (0, ±2) used with a duobinary signal with output levels having the probability mass function given in part b). An upper bound of the symbol probability of error is 246 P (e) <P(|y m | > 1|b m =0)P (b m =0)+2P(|y m − 2| > 1|b m =2)P (b m =2) +2P (y m +4> 1|b m = −4)P (b m = −4) = P (|y m | > 1|b m =0)[P(b m =0)+2P(b m =2)+P (b m = −4)] = 7 8 P (|y m | > 1|b m =0) But P (|y m | > 1|b m =0)= 2 √ 2πσ ν  ∞ 1 e −x 2 /2σ 2 ν dx Therefore, P (e)= 14 8 Q  1 σ ν  Problem 8.52 Since the partial response signal has memory length equal to 2, the corresponding trellis has 4 states which we label as (a n−1 ,a n ). The following figure shows three frames of the trellis. The labels of the branches indicate the output of the partial response system. As it is observed the free distance between merging paths is 3, whereas the Euclidean distance is equal to d E =2 2 +4 2 +2 2 =24 ✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉     ✒ ✲✲ ❍ ❍ ❍ ❍ ❍❥ ✟ ✟ ✟ ✟ ✟✯ ✲     ✒ ✟ ✟ ✟ ✟ ✟✯ ❅ ❅ ❅ ❅ ❅❘ ❍ ❍ ❍ ❍ ❍❥ ❍ ❍ ❍ ❍ ❍❥ ✲ -2 0 -2 -4-4-4 (1,1) (1,-1) (-1,1) (-1,-1) (a n−1 ,a n ) Problem 8.53 a) The alternative expression for s(t) can be rewritten as s(t)=Re   n a  n Q(t − nT )  =Re   n a n e j2πf c nT g(t − nT ) [cos 2πf c (t − nT )+j sin 2πf c (t − nT )]  =Re   n a n g(t − nT ) [cos 2πf c nT + j sin 2πf c nT ] [cos 2πf c (t − nT )+j sin 2πf c (t − nT )]  =Re   n a n g(t − nT ) [cos 2πf c nT cos 2πf c (t − nT ) − sin 2πf c nT sin 2πf c (t − nT ) +j sin 2πf c nT cos 2πf c (t − nT )+j cos 2πf c nT sin 2πf c (t − nT )]  =Re   n a n g(t − nT ) [cos 2πf c t + j sin 2πf c t]  =Re   n a n g(t − nT )e j2πf c t  = s(t) so indeed the alternative expression for s(t) is a valid one. 247 [...]...b) e j2πfnT ani e-j2πfnT q(t) a'nr a'ni + + + anr q(t) - ^ q(t) ^ q(t) Modulator (with phase rotator) Demodulator (with phase derotator) Problem 8.54 a) The impulse response of the pulse having a square-root raised cosine characteristic, is an even function, i.e., xSQ (t) = xSQ (−t), i.e., the pulse g(t) is an... analog signal is: N /T If we sample at the Nyquist rate: 2N /T = N/T , we obtain the discrete-time sequence: 1 x(n) = x(t = nT /N ) = √ N N −1 k=0 1 Xk ej2πk(nT /N )/T = √ N N −1 Xk ej2πkn/N , n = 0, 10, , N − 1 k=0 which is simply the IDFT of the information sequence {Xk } ∗ To show that x(t) is a real-valued signal, we make use of the condition: XN −k = Xk , for k = ˜ ˜ 1.2 , N −1 By combining... pairs of complex conjugate terms, we obtain for k = 1, 2, , N −1 ∗ Xk ej2πkt/T + Xk e−j2πkt/T = 2|Xk | cos 2πkt + θk T where Xk = |Xk |ejθk We also note that X0 and XN are real Hence, x(t) is a real-valued signal ˜ 248 Problem 8.56 The filter with system function Hn (z) has the impulse response h(k) = ej2πnk/N , k = 0, 1, If we pass the sequence {Xk , k = 0, 1, , N − 1} through such a filter,... Therefore, the capacity of the cascade channel is CAB = 1 Problem 9.11 The SNR is SNR = P 2P 10 = = 104 = −9 N0 2W 2W 10 × 106 Thus the capacity of the channel is C = W log2 (1 + P ) = 106 log2 (1 + 10000) ≈ 13. 2879 × 106 bits/sec N0 W Problem 9.12 The capacity of the additive white Gaussian channel is C= P 1 log 1 + 2 N0 W For the nonwhite Gaussian noise channel, although the noise power is equal to the noise... 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 256 Problem 9 .13 1) The capacity of the binary symmetric channel with crossover probability is C = 1 − h( ) where h( ) is the binary entropy function The rate distortion function of a zero mean Gaussian source with . sequences of length 3 and the corresponding output of the detector. -1 -1 -1 -4 -1 -1 1 -2 -1 1 -1 0 -1 11 2 1-1 -1 -2 1-1 1 0 1 1-1 2 111 4 As it is observed there are 5 possible output levels b m ,. to d E =2 2 +4 2 +2 2 =24 ✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉     ✒ ✲✲ ❍ ❍ ❍ ❍ ❍❥ ✟ ✟ ✟ ✟ ✟✯ ✲     ✒ ✟ ✟ ✟ ✟ ✟✯ ❅ ❅ ❅ ❅ ❅❘ ❍ ❍ ❍ ❍ ❍❥ ❍ ❍ ❍ ❍ ❍❥ ✲ -2 0 -2 -4 - 4-4 (1,1) (1 ,-1 ) (-1 ,1) (-1 ,-1 ) (a n−1 ,a n ) Problem 8.53 a) The alternative expression for s(t) can be rewritten as s(t)=Re   n a  n Q(t − nT )  =Re   n a n e j2 πf c nT g(t. one. 247 b) + + a nr a ni a' nr a' ni q(t) q(t) ^ - e j2 π fnT q(t) q(t) ^ e - j2 π fnT Modulator (with phase rotator) Demodulator (with phase derotator) + Problem 8.54 a) The impulse response of the pulse having a square-root raised