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Proakis. Solutions for.. Digital communications, 4ed(322s)_E_.pdf |
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(c) When x(t), y (t) are uncorrelated and have zero means : φ z z (τ ) = φ x x (τ ) + φ y y (τ )...
Problem 3.9 : (a) I (xi ; yj ) = log = log
P (xi |yj ) P (xi ) P (xi ,yj ) P (xi )P (yj ) P (yj |xi ) P (yj ) 1 − log P (y1|xi ) P (yj ) j...
Problem 3.23 : (a) H (X ) = −(.05 log2 .05 + .1 log2 .1 + .1 log2 .1 + .15 log2 .15 +.05 log2 .05 + .25 log2 .25 + .3 log2 .3) = 2.5282 (b) After quantization, the new alphabet is B = {−4, 0, 4} and the corresponding symbol probabilities are given by P (−4) = P (−5) + P (−3) = .05 + .1 = .15 P (0) = P (−1) + P (0) + P (1) = .1 + .15 + .05 = .3 P (4) = P (3) + P (5) = .25 + .3 = .55 30...
(b) The following figure depicts R(D ) for λ = 0.1, .2 and .3. As it is observed from the figure, an increase of the parameter λ increases the required rate for a given distortion.
7 6 5 4 3 2 1 0 0...
The minimum MSE is : 1 = φ(0) (1 − a21 ) = 3/4 1 (b) For the second order predictor : x(n) = a21 x(n − 1) + a22 x(n − 2). Following the Levinsonˆ Durbin algorithm (Eqs 3-5-25) : a22 = φ (2) −
1
k =1 1 0− 12 a1k φ(2 − k ) 2 = = −1/3 1 3/4...
Problem 3.41 : (a) The joint probability density function is pX Y (x, y ) = y pX (x) is pX (x) = pX Y (x, y )dy . If −2 ≤ x ≤ 0,then pX (x) = If 0 ≤ x ≤ 2,then pX (x) = The next figure depicts pX (x).
x +2
1 √ (2 2)2...
The corresponding Markov chain mo del is illustrated in the following figure :
1/2
1/2 5✥ ✲ ❅ ❘❄ ❅5✥ ✠ ...
However, this is the same expression with the case of the output of the matched filter sampled at t = T . Thus, the correlator can substitute the matched filter in a demo dulation system and vice versa....
E [z (τ )] h(T − τ )dτ = 0, and the second moment : E 0 z (τ )h(T − τ )dτ T 2N0 0 h2 (T − τ )dτ = N0 T (1 − e−2 ) 89
= T
0...
Similarly we find that R2 is the set of points (r1 , r2 ) that satisfy r2 > 0, r2 > r1 and R3 is the region such that r2 < 0 and r2 < −r1 . The regions R1 , R2 and R3 are shown in the next figure.
...
For binary FSK (M = 2) the required frequency separation is 1/2T (assuming coherent receiver) and (see 5-2-86): W= M 2 W l og 2 M R→R= = W = 100 kbits/sec l og 2 M M...
Problem 5.42 : (a) The noncoherent envelope detector for the on-off keying signal is depicted in the next figure. t=T rc (· )2
❄r ❧ +✲ ✻ ✲
0 (· )d τ...
(b) Expanding the quantity (1 − 2p)n , we obtain (1 − 2p)n = 1 − n2p + Since, p n(n − 1) (2 p )2 + · · · 2...
where Q1 , Q2 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, Q1 and Q2 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 I (X ; Y ) = max I (p; Q)
p p...
and the channel capacity is
− ( ( )2 2− H − C =H H( ) H( ) ) + 2− 1 + 2−
1...
✎ ❄ ✍✌ d X ✒ ❅ DN J ❅ D2J ❅ 2 D 3 N ✲ J D2 J ❅ D J✲ ❘ ❅ ✲ o ❙ Xa Xb Xc ❙ Xa
U
DN J sing the flow graph results, we obtain the system Xc = D 3 N J X a + D N J X b Xb = D 2 J X c + D 2 J X d Xd = D N J X c + D N J X d Xa = D 2 J X b Eliminating Xb , Xc and Xd results in T (D , N , J ) =
X Xa =
a
D7N J 3 1 − DN J − D3 N J 2
(c) To find the free distance of the co de we set N = J = 1 in the transfer function, so that T1 (D ) = T (D , N , J )|N =J =1 = D7 = D7 + D8 + D9 + · · · 3 1−D−D
Hence, dfree = 7. The path, which is at a distance dfree from the all zero path, is the path X a → Xc → X b → X a . (d) The following figure shows 6 frames of the trellis diagram used by the Viterbi algorithm to deco de the sequence {111, 111, 111, 111, 111, 111}. The numbers on the no des indicate the 177
Hamming distance of the survivor paths from the received sequence. The branches that are dropped by the Viterbi algorithm have been marked with an X. In the case of a tie of two merging paths, we delete the upper path...
Beginning State 0 45 JND(2) 1 54 JND(2) 2 67 JND(2) 3 76 JND(2) 4 4 01 JND 5 10 JND 6 23 JND(2) 7 32 JND(2) Beginning State 0 61 JND(2) 1 70 JND 2 43 JND(2) 3 52 JND(2) 6 4 25 JND(2) 5 34 JND(2) 6 07 JND 7 16 JND(2)...
|GT (f )|2 = (AT )2 sinc2 (f T ) Φx (f ) = A2 T Φa (f )sinc2 (f T ) = A2 T sinc2 (f T )...
The steady-state solution is obtained when ck,(n+1) = ck , which gives : ck = or going back to matrix form : C = Λ−1 ξ ⇒ C = VC = VΛ−1 V−1ξ ⇒ C=
V...
Problem 13.14 : (a) The perio d of the maximum length shift register sequence is N = 210 − 1 = 1023 Since Tb = N Tc , then the pro cessing gain is N Tb = 1023 (30dB ) Tc...
(c) The probability of error with worst-case partial-band jamming is P2 =
e −1 (Eb /J0 )...
1−µ 1 1 1 γc ¯ ⇒ = = < 1 + γc ¯ 2 2(1 + γ c ) ¯ 2γ c ¯ 2RC γb ¯
1 . 1+γc ¯...
Problem 15.18 : (a) Since the number of arrivals in the interval T, follows a Poisson distribution with parameter λT , the average number of arrivals in the interval T, is E [k ] = λT . (b) Again, from the well-known properties of the Poisson distribution : σ 2 = (λT )2 . (c) P (k ≥ 1) = 1 − P (k = 0) = 1 − e−λT...
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