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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....
The last condition shows that in order to have a real-valued lowpass signal sl (t), the positive frequency content of the corresponding bandpass signal must exhibit hermitian symmetry around the center frequency fc . In general, bandpass signals do not satisfy this property (they have Hermitian symmetry around f = 0), hence, the lowpass equivalent is generally complex-valued....
It is easy to show that n=L−1 ˆ if all ak,n , n = 0, 1, ..., L − 1 are zero (for a specific k ), and n = max {n : ak,n = 1} ˆ (2) (1 )...
(b) Each matched filter has an equivalent lowpass impulse response : hi (t) = si (T − t) . The following figure shows hi (t) : h (t) ✻1 A A h (t) ✻2...
Problem 5.25 : (a) If the power spectral density of the additive noise is Sn (f ), then the PSD of the noise at the output of the prewhitening filter is Sν (f ) = Sn (f )|Hp (f )|2 In order for Sν (f ) to be flat (white noise), Hp (f ) should be such that H p (f ) =
S...
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....
1 0 0 × 1 03 = 92.6853 Hz 0.2997 × 3600 Thus, the Doppler frequency spread is Bd = 2fD max = 185.3706 Hz. fD max =...
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 7.24 : 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 :
...
✎ ❄ ✍✌ 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 9.6 : (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...
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...
|GT (f )|2 = (AT )2 sinc2 (f T ) Φx (f ) = A2 T Φa (f )sinc2 (f T ) = A2 T sinc2 (f T )...
with a renaming of the bits from 0 to 1 and vise versa. For example, the (0,1) runlength-limited co de with a renaming of the bits, can be described as the co de with no minimum number of 1’s between 0’s in a sequence, and at most one 1 between two 0’s. In terms of 0’s, this is simply the co de with no restrictions on the number of adjacent 0’s and no consequtive 1’s, that is the (1, ∞) co de....
Problem 10.12 : (a) If {cn } denote the co efficients of the zero-force equalizer and {qm } is the sequence of the equalizer’s output samples, then : qm =
n1
=−1...
(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...
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...
Problem 11.10 : In Probl. 11.9 we found that the optimum (MSE) linear predictor for x(n), is x(n) = bx(n − 1). ˆ Since it is a first order predictor, the corresponding lattice implementation will comprise of one stage, to o, with reflection co efficient a11 . This co efficient can be found using (11-4-28) : a11 = γ x x (1) =b γ x x (0)...
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