| Home / lib / E_Engineering / | ||
Proakis. Solutions for.. Digital communications, 4ed(322s)_E_.pdf |
|
Size 1.4Mb Date Dec 13, 2006 |
Problem 3.21 : (a) The following figure depicts the design of the Huffman co de, when enco ding a single level at a time :...
1 ˆ ˆ ⇒ X (f ) = − [δ (f − f0 ) + δ (f + f0 )] = −F −1 {cos 2π ω0 t} ⇒ x(t) = − cos ω0 t 2 (e) The positive frequency content of the new signal will be : (−j )(−j )X (f ) = −X (f ), f > 0, ˆ ˆ while the negative frequency content will be : j · j X (f ) = −X (f ), f < 0. Hence, since X (f ) = ˆ −X (f ), we have : x(t) = −x(t). ˆ 44...
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....
ince the lowpass equivalent of x(t) is single-sideband, we conclude that x(t) is a single-sideband signal, to o. Suppose, for example, that s(t) has the following spectrum. Then, the spectra of the signals u(t) (shown in the figure for the case u(t) = s(t) + j s(t)) and x(t) are single-sideband ˆ...
Problem 4.25 : MFSK signal with waveforms : si (t) = sin 2πit , i = 1, 2, ..., M 0 ≤ t ≤ T T The expression for the power density spectrum is given by (4.4.60) with K = M and pi = 1/M . From Problem 4.23 we have that :
f...
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 )...
Since the signals are equally probable, the optimal detector decides in favor of s0 if PM(r, s0 ) = p(r |s0) > p(r |s1) = PM(r, s1 )...
If we want to have the filter matched to the passband signal, then the carrier phase estimate is fed into the matched filter, which should have an impulse response: ˆ h(t) = s(T − t) = g (T − t)cos(2π fc (T − t) + φ) ˆ ˆ = g (T − t)[cos(2π fc T )cos(−2π fc t + φ) + sin(2π fc T )sin(−2π fc t + φ) ˆ ˆ = g (T − t)cos(−2π fc t + φ) = g (T − t)cos(2π fc t − φ) where we have assumed that fc T is an integer so that : cos(2π fc T ) = 1, sin(2π fc T ) = 0. As we note, in this case the impulse response of the filter should change according to the carrier phase estimate, something that is difficult to implement in practise. Hence, the initial realization (shown in the figure) is preferable....
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 7.7 : We assume that P (xi ) = 1/3, i = 1, 2, 3. Then P (y1 ) = P (yj ) = 1/3, j = 2, 3. Hence :
3
P (y |x ) 1 3...
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 7.20 : Both channels can be viewed as binary symmetric channels with crossover probability the probability of deco ding a bit erroneously. Since :
...
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 9.13 : (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...
X (z ) = 0.3z + 0.9 + 0.3z −1 = (f0 + f1 z −1 )(f0 + f1 z ) √0.7854 , ± 0.1146
±√...
Problem 11.8 : (a) The gradient of the performance index J with respect to h is : time update equation becomes :
dJ dh...
(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...
Problem 12.7 : We assume binary (M = 2) orthogonal signalling with square-law detection (DPSK signals wil have the same combining loss). Using (12-1-24) and (12-1-14) we obtain the following graph for P2 (L), where SNR/bit =10log10 γb :
10
0...
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...
The simultaneous solution of these two equations yields the result v = 1/4 and a = these values, the function f (v , a) becomes 3 J0 f , 4 Ec
1 = 4 L 1...
Problem 14.14 : (a) The noise-free received waveforms {ri (t)} are given by : ri (t) = h(t) ∗ si (t), i = 1, 2, and they are shown in the following figure :
4A r1 (t)...
which shows the equivalence between the upper bounds given in (b) and (c). In order for the bound to go to zero, as k is increased to infinity we need the rest of the argument of the exponent to be negative, or (β dmin γb /ne − ln 2) > 0 ⇒ γb > ¯ ¯ ne dmin β ln 2 ⇒ γb min = ¯ 2e ln 2 β...
| © 2007 eKnigu | ||