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Proakis. Solutions for.. Digital communications, 4ed(322s)_E_.pdf |
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As it is expected the entropy of the source is less than the average length of each co deword....
Problem 3.21 : (a) The following figure depicts the design of the Huffman co de, when enco ding a single level at a time :...
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 average probability of error is : P (e) = = = = = 1 1 P (e|A) + P (e| − A) 2 2 10 1∞ f (r |A)dr + f (r | − A)dr 2 −∞ 20 0 ∞ 1 1 λ2 e−λ|r−A| dr + λ2 e−λ|r+A| dr 2 −∞ 20 −A ∞ λ λ −λ|x| e dx + e−λ|x| dx 4 −∞ 4A 1 −λA 1 −√2A e =eσ 2 2 92...
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.
...
As it is observed there is an increase in transmitted power of approximately 3 dB per additional bit per symbol....
where we have used the identity Re [z ] = 1 (z + z ∗ ) , and the fact from Problem 5.7 (or 4.3) that 2 E [z (t)z (t + τ )] = 0, E [z ∗ (t)z ∗ (t + τ )] = 0. Similarly : E [n2c n2c ] = E [n1s n1s ] = E [n2s n2s ] = 2N0 E where for the quadrature noise term components we use the identity : I m [z ] = The covariance between the in-phase terms for the two correlators is : E [n1c n2c ] =
=
1 E 4 1 2j...
Thus, the maximum likeliho o d estimate of the carrier phase is : ˆ φc,M L = − arctan
0 0...
Then the necessary conditions for the estimates of φ and τ to be the ML estimates (6.4-8) and (6.4-9) give ˆ φM L = 0 and
n...
Problem 7.6 : (a) We assume that P (xi ) = 1/2, i = 1, 2. Then P (y1) = P (yj ) = 1/4, j = 2, 3, 4. Hence :
4
1 2...
Problem 7.13 : (a) The capacity of the channel is : C1 = max[H (Y ) − H (Y |X )]
P (x)...
Problem 7.15 : (a) Let q be the probability of the input symbol 0, and therefore (1 − q ) the probability of the input symbol 1. Then : H (Y | X ) =
x...
Problem 7.24 : Remember that the capacity of the BSC is C = p log2 2p + (1 − p) log2 (2(1 − p)) where p is the error probability (for binary antipo dal mo dulation for this particular case). Then, the plot with the comparison of the capacity vs the cutoff rate R2 for the BSC, with antipo dal signaling, is given in the following figure:
C vs R2 for BSC 1...
Using Rl (p) (with l = 4 corresponding to the last row of G,... l = 14 corresponding to the first row) for the parity matrix P we obtain :
...
Problem 8.24 : The co de of Problem 8-23 is a (3, 1) convolutional co de with K = 3. The length of the received sequence y is 15. This means that 5 symbols have been transmitted, and since we assume that the information sequence has been padded by two 0’s, the actual length of the information sequence is 3. The following figure depicts 5 frames of the trellis used by the Viterbi deco der. The numbers on the no des denote the metric (Hamming distance) of the survivor paths (the non-survivor paths are shown with an X). In the case of a tie of two merging paths at a no de, we have purged the upper path...
✎ ❄ ✍✌ 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...
Problem 9.16 : The pulse x(t) having the raised cosine spectrum given by (9-2-26/27) is : x(t) = sinc(t/T ) cos(π β t/T ) 1 − 4β 2 t2 /T 2...
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...
Problem 9.39 : The state transition matrix of the (2,7) runlength-limited co de is the 8 × 8 matrix :
...
(c) 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 |Ik = 1) = P (error at k − 1)P (error at k |Ik = 1, error at k − 1) +P (no error at k − 1)P (error at k |Ik = 1, no error at k − 1) = P (e)P (error at k |Ik = 1, error at k − 1) +(1 − P (e))P (error at k |Ik = 1, no error at k − 1) = P (e)p + (1 − P (e))q where : p = P (error at k |Ik = 1, error at k − 1) 1 P (error at k |Ik = 1, Ik−1 = 1, error at k − 1) = 2 1 + P (error at k |Ik = 1, Ik−1 = −1, error at k − 1) 2 1 1 P ((1 + 2α)Es + nk < 0) + P ((1 − 2α)Es + nk < 0) = 22 2 2 2E 1 ( 1 + 2α ) s 1 ( 1 − 2 α ) 2 E s Q = +Q 2 N0 2 N0 237...
(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...
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