Proakis J (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13 pps

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 13 pps

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 1 docx

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 1 docx

... 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 Eva...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 3 pptx

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 j2 πft     ...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 4 docx

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 4 docx

... + 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 n =2l 2 π(2l+1) sin(1) n =2l +1 Hence, e jm(t) = ∞  l=−∞ 2 π(2l ... 1050950 100 2 J 4 (5) The Fourier Series expansion of e j sin(2πf m t) is c n = f m  5 4f m 1 4f m e j sin(2πf m t) e j2 πnf m...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 5 docx

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 5 docx

... j element (C i ,j ) of this matrix is C i ,j = E[(ω 1 + ω 2 + ···+ ω i )(ω 1 + ω 2 + ···+ ω j )] = E[(ω 1 + ω 2 + ···+ ω min(i ,j) )(ω 1 + ω 2 + ···+ ω min(i ,j) )] +E[(ω 1 + ω 2 + ···+ ω min(i ,j) )(ω min(i ,j) +1 + ... 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 ]+ ...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 8 docx

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 8 docx

... ,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 ... Jacobian of the transformation system evaluated at the solution {x j 1 , ,x j n }. Note that if the system has a unique solution, then J( x 1 , ,x n )=Q  (x 1...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9 pdf

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9 pdf

... g(1) = 1. 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 ... where r(t)= N  i=1 r i ψ i (t) s m (t)= N  i=1 s m,i ψ i (t) then,  ∞ −∞ r(t)s m (t)dt =  ∞ −∞ N  i=1 r i ψ i (t) N  j= 1 s m ,j ψ...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 10 pot

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 10 pot

... (f) n  k=1 c k e j2 πfkT c and therefore, the Fourier transform of the signal matched to s(t)is H(f )=S ∗ (f)e j2 πfT = S ∗ (f)e j2 πfnT c = P ∗ (f) n  k=1 c k e j2 πfkT c e j2 πfnT c = P ∗ (f) n  i=1 c n−i+1 e j2 πf(i−1)T ... 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...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 12 pptx

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 12 pptx

... 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 3 45 f...

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Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 14 pot

Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 14 pot

... figure ♥   ❙ ❙♦ ❄ ❅ ❅ ❅ ❅ ❅❘ ✲✲    ✒ ✲ D 2 J DNJ D 2 J D 2 J DNJ DNJ D 3 NJ X c X b X a  X a  Using the flow graph relations we write X c = D 3 NJX a  + DNJX b X b = D 2 JX c + D 2 JX d X d = DNJX c + DNJX d X a  = D 2 JX b Eliminating ... 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 w...

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