... ,y n )= J j= 1 f X(t 1 ), ,X(t n ) (x 1 , ,x n ) |J( x j 1 , ,x j n )| where J is the number of solutions to the system y 1 = Q(x 1 ),y 2 = Q(x 2 ), ···,y n = Q(x n ) and J( x j 1 , ,x j n ) is the Jacobian ... − √ 10 · 1.256 = −3.97 18 ˆx 5 = −ˆx 12 = − √ 10 · 0.9424 = −2. 980 1 ˆx 6 = −ˆx 11 = − √ 10 · 0.65 68 = −2.0770 ˆx 7 = −ˆx 10 = − √ 10 · 0. 388 1 = −1.2273 ˆx 8...
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... then x n = 1 T T 2 − T 2 x(t)e j2 π n T t dt = 1 T T 2 − T 2 e j2 π n T t dt + 1 T T 4 − T 4 e j2 π n T t dt = j 2πn e j2 π n T t T 2 − T 2 + j 2πn e j2 π n T t T 4 − T 4 = j 2πn e j n − e j n + e j n 2 − ... frequency content. 13 SOLUTIONS MANUAL Communication Systems Engineering Second Edition John G. Proakis Masoud Salehi Prepared by Evang...
Ngày tải lên: 12/08/2014, 16:21
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 4 docx
... expansion c n = 1 T m T m 0 e jm(t) e j2 πnf m t dt = 1 T m T m 2 0 e j e j2 πnf m t dt + 1 T m T m T m 2 e j e j2 πnf m t dt = − e j T m j2 πnf m e j2 πnf m t T m 2 0 − e j T m j2 πnf m e j2 πnf m t T m T m 2 = (−1) n − 1 2πn j( e j − e j )= 0 ... 10 0J k (2) Power P f c +kf m 0 .2239 10 8 22.39 250.63 1 .5767 10 8 +10 4 57.67 1663.1 2 .35 28...
Ngày tải lên: 12/08/2014, 16:21
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 5 docx
... = 3 8 x 2 + 6 8 x + 3 8 p(X +2Y>1) = 3 7 x 2 + 6 7 x + 3 7 80 + b2= 356563 782 d+00, b3=1. 781 477937d+00, + b4 =-1 .82 1255978d+00, b5=1.330274429d+00) C- pi=4.*atan(1.) C-INPUT PRINT*, ’Enter -x-’ READ*, ... Hence, C i ,j = E[(ω 1 + ω 2 + ···+ ω min(i ,j) )(ω 1 + ω 2 + ···+ ω min(i ,j) )] = min(i ,j) k=1 min(i ,j) l=1 E[ω k ω l ]= min(i ,j) k=1 E[ω k ω k ]+ k,l=1...
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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 3 pptx
... f 0 =3. -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0 .8 1 0 0.2 0.4 0.6 0 .8 1 1.2 1.4 1.6 1 .8 2 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0 .8 1 0 0.2 0.4 0.6 0 .8 1 1.2 1.4 1.6 1 .8 2 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0 .8 0 0.2 ... +1)e j2 πft df + 1 2j 1 2 0 (−f +1)e j2 πft df = 1 2j 1 j2 πt fe j2 πft + 1 4π 2 t 2 e...
Ngày tải lên: 12/08/2014, 16:21
Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9 pdf
... Since the resulting distortion is (see Equation 6.6.17) -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0 .8 1 -1 -0 .8 -0 .6 -0 .4 -0 .2 0 0.2 0.4 0.6 0 .8 1 g(x) x D = 1 12 × 4 ν ∞ −infty [f X (x)] 1 3 dx 3 = 1 12 ... is observed the µ-law compressor is insensitive to the dynamic range of the input signal for E[ ˘ X 2 ] > 1. -5 0 0 50 100 150 200 -1...
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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 10 pot
... average transmitted power of the 4-PSK and the 8- PSK constellation is given by P 4,av = d 2 2 ,P 8, av = d 2 2 − √ 2 Thus, the additional transmitted power needed by the 8- PSK signal is P = 10 log 10 2d 2 (2 ... and s j (t)is ∞ −∞ s i (t)s j (t)dt = ∞ −∞ n k=1 c ik p(t −kT c ) n l=1 c jl p(t −lT c )dt = n k=1 n l=1 c ik c jl ∞ −∞ p(t −kT c )p(t −lT c )dt = n k=1 n l...
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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 12 pptx
... the relative difference in SNR of the error probability of 10 −6 is 2 dB. -7 -6 .5 -6 -5 .5 -5 -4 .5 -4 -3 .5 -3 -2 .5 -2 6 7 8 9 10 11 12 13 14 SNR/bit, dB log(P(e) 232 ... cases. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 -5 -4 -3 -2 -1 0 1 2 3 45 frequency f Sv(f) T=1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 .8 -5 -4 -3 -2 -1 0 1 2...
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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13 pps
... sequences of length 3 and the corresponding output of the detector. -1 -1 -1 -4 -1 -1 1 -2 -1 1 -1 0 -1 11 2 1-1 -1 -2 1-1 1 0 1 1-1 2 111 4 As it is observed there are 5 possible output levels b m , ... to d E =2 2 +4 2 +2 2 =24 ✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉ ✒ ✲✲ ❍ ❍ ❍ ❍ ❍❥ ✟ ✟ ✟ ✟ ✟✯ ✲ ✒ ✟ ✟ ✟ ✟ ✟✯ ❅ ❅ ❅ ❅ ❅❘ ❍ ❍ ❍ ❍ ❍❥ ❍ ❍ ❍ ❍ ❍❥...
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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 14 pot
... figure ♥ ❙ ❙♦ ❄ ❅ ❅ ❅ ❅ ❅❘ ✲✲ ✒ ✲ D 2 J D 2 NJ D 3 J DJ DNJ DNJ D 2 NJ X c X b X a X a Using the flow graph relations we write X c = D 2 NJX a + D 2 NJX b X b = DJX d + D 3 JX c X d = DNJX d + DNJX c X a = D 2 JX b Eliminating ... we obtain T (D, N, J) = X a X a = D 6 N 2 J 4 + D 7 NJ 3 − D 8 N 2 J 4 1 − DNJ − D 4 N 2 J 3 − D 5 NJ 2 + D 6 N 2 J 3 Thus,...
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