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Proakis. Solutions for.. Digital communications, 4ed(322s)_E_.pdf |
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Problem 3.37 : ˜ ˜ ˜ dW (X, X) = (X − X) W(X − X) Let W = P P. Then : ˜ ˜ ˜ dW (X, X) = (X − X) P P(X − X) P P ˜ ˜ = (X − X ) (X − X ) w Y Y 1 ˜ ˜ −Y −Y =n √ ˜√ ˜ ˜ ˜ here by definition : Y= nPX, Y = nPX . Hence : dW (X, X) = d2 (Y,Y )....
1 1 1 + = 12 12 6 whereas the resulting number of bits per (X, Y ) pair Dtotal = DX + DY = R = RX + RY = log2 4 + log2 4 = 4...
.. ✏ ... ✏✏ ❘✏ ❅ ✏✏✏ ❅ ✏ q ✠ ❅ φ = 3π /4 φ2 = 5π /4 ✒ ✑ ✒✑ 1 ✏ ✐ ✏✏ ✂ ❇ ❇ (1,1) ✍ ✂ ✍ ✏ ✂ ✂ ✏✏ ❇ ❇ ✂ ✂ ❇ ❇ ✂ ✂ ❇ ❇ ✂ ✂ ❇ (-1,-1) ❇❇ ❇ ✂ ✂ (1,1) (1,1)
(-1,-1)
❇ ❇ ✂ ❇ ✂ ✂...
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.
...
Problem 5.36 : The bandwidth required for transmission of an M -ary PAM signal is W= R Hz 2 log2 M 107...
From the results in Sec. 5-4-4 we observe that the performance of the non coherent detector metho d is about 4 dB worse than the coherent FSK detector. hence the loss is about 7 dB compared to the optimum demo dulator for the MSK signal....
Problem 6.5 : The transfer function of the RC circuit is : G (s ) =
1 R2 + C s 1 + R2 C s 1 + τ2 s = 1= R1 + R2 + C s 1 + (R1 + R2 )C s 1 + τ1 s...
I m(U ) ˆ φ 4 = − tan −1 Re(U ) Nn is a complex-valued Gaussian noise component with zero mean and variance σ 2 = N0 /2. ˆ Hence, the pdf of φ4 is given by (5-2-55) where : K√ 2 Es K Es K 2 Es γs = = = 2) 16 (2K σ 16K N0 16N0 To a first approximation, the variance of the estimate is :
2 σφ4 ≈ ˆ...
and the same is true for I (x2 ; Y ). Thus : C = 0.0817 bits/symbol sent (c) We assume that P (xi ) = 1/3, i = 1, 2, 3. Then P (y1) = P (yj ) = 1/3, j = 2, 3. Hence : I (x1 ; Y ) =
j3
=1 1 3...
✎ ❄ ✍✌ 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...
Since, each element of the vector yi can take two values, the cardinality of the set Y is 2n ...
Problem 9.6 : (a)(b) In order to calculate the frequency response based on the impulse response, we need the values of the impulse response at t = 0, ±T /2, which are not given directly by the expression of Problem 9.5. Using L’Hospital’s rule it is straightforward to show that: √ 2 (2 + π ) 12 x(0) = + , x(±T /2) = 2π 2 2π Then, the frequency response of the filters with N = 10, 15, 20 compared to the frequency response of the ideal square-ro ot raised cosine filter are depicted in the following figure...
Problem 9.13 : (a) The bandwidth of the bandpass channel is : W = 3 0 0 0 − 600 = 2400 Hz Since each symbol of the QPSK constellation conveys 2 bits of information, the symbol rate of transmission is : 2400 1 = 1200 symbols/sec R= = T 2 Thus, for spectral shaping we can use a signal pulse with a raised cosine spectrum and roll-off 1 factor β = 1, since the spectral requirements will be 21 (1 + β ) = T = 1200Hz. Hence : T
I...
with U (f ) even with respect to 0 and o dd with respect to f = 21 Since x(t) is real, V (f ) is o dd T with respect to 0 and by assumption it is even with respect to f = 21 . Then, T x(t) = F −1[X (f )] = =
2T 1 −T 2T...
Since the minimum transmission bandwidth required for bandpass signaling is R, where R is the rate of transmission, we conclude that the maximum value of the symbol rate for the given channel is Rmax = 2700. If an M -ary PAM mo dulation is used for transmission, then in order to achieve a bit-rate of 9600 bps, with maximum rate of Rmax , the minimum size of the constellation is M = 2k = 16. In this case, the symbol rate is : R= 9600 = 2400 symbols/sec k...
Thus, the capacity of the (0,1) runlength-limited co de is : √ 1± 5 C (0, 1) = log2 ( ) = 0.6942 2 The capacity of a (1, ∞) co de is found from Table 9-4-1 to be 0.6942. As it is observed, the two co des have exactly the same capacity. This result is to be expected since the (0,1) runlengthlimited co de and the (1, ∞) co de pro duce the same set of co de sequences of length n, N (n), 220...
Now we compute the metrics for the next stage : µ3 (I3 = 3, I2 = 3, I1 = 1) = µ2 (3, 1) + [−1 − 2.4 + 1.8]2 = 2.69 µ3 (3, 1, −1) = µ2 (1, −1) + [−1 − 2.4 + 0.6]2 = 9.89 µ3 (3, −1, −1) = µ2 (−1, −1) + [−1 − 2.4 − 0.6]2 = 22.53 µ3 (3, −3, −3) = µ2 (−3, −3) + [−1 − 2.4 − 1.8]2 = 42.21...
(c) Let C = Vt C, ξ = Vtξ , where V is the matrix whose columns form the eigenvectors of the covariance matrix Γ (note that Vt = V−1). Then : C(n+1) = (I − ∆Γ) C(n) + C ξ ⇒ ∆ I −1 C(n+1) = − ∆VΛV (n) + ∆ξ ⇒ I C −1 V−1 C(n+1) = V−1 − ∆VΛV−1 (n) + ∆V ξ ⇒ C(n+1) = (I − ∆Λ) C(n) + ∆ξ...
(b) If the real and imaginary parts of the information sequence {Xk } have the same average energy : E [Re(Xk )]2 = E [I m(Xk )]2 , then it is straightforward to prove that the time-domain samples {xn }, that are the output of the IDFT, have the same average energy:
k 1 N −1 (Re(Xk ) + j · I m(Xk )) exp(j 2π nk /N ), xn = √ N =0...
W /R Q = 1 0− 3 Jav /Pav Then, 500 W/Rb = =5 Jav /Pav Jav /Pav and Jav = 100 (20 d B ) Pav...
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