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     ...

Ngày tải lên: 12/08/2014, 16:21

... 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 frequency f Sv(f) T=2 3) The first spectral null ... 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...

Ngày tải lên: 12/08/2014, 16:21

... + 3) ] = sinc(f)e j2 πf3 + sinc(f)e j2 πf3 = 2sinc(f) cos(2π3f) 3) F[x(t)] = F[Λ(2t +3) +Λ(3t − 2)] = F[Λ(2(t + 3 2 )) + Λ (3( t − 2 3 )] = 1 2 sinc 2 ( f 2 )e j f3 + 1 3 sinc 2 ( f 3 )e j2 πf 2 3 4) ... = 1 T  T 3 0 3 T te j2 π n T t dt + 1 T  2T 3 T 3 e j2 π n T t dt + 1 T  T 2T 3 (− 3 T t +3) e j2 π n T t dt = 3 T 2  jT 2πn te j2 π n T t + T 2 4π 2 n...

Ngày tải lên: 12/08/2014, 16:21

... these compo- nents whose frequencies lie in the interval [10 6 −4 ×10 3 , 10 6 +4×10 3 ]. Note that J 0 (3) = −.2601, J 1 (3) =0 .33 91, J 2 (3) =0.4861, J 3 (3) =0 .30 91 and J 4 (3) =0. 132 0. ✛ ... Amplitude 10 0J k (2) Power P f c +kf m 0 .2 239 10 8 22 .39 250. 63 1 .5767 10 8 +10 4 57.67 16 63. 1 2 .35 28 10 8 +2× 10 4 35 .28 622.46 3 .1289 10 8 +3 10 4 12...

Ngày tải lên: 12/08/2014, 16:21

... ×10 3 6.214 × 10 3 3. 1 .35 × 10 3 1 .35 1 × 10 3 3.5 2 .33 ×10 −4 2 .32 8 × 10 −4 4. 3. 17 × 10 −5 3. 171 × 10 −5 4.5 3. 40 ×10 −6 3. 404 × 10 −6 5. 2.87 × 10 −7 2.874 × 10 −7 Problem 4.25 The n-dimensional ... y)dvdy =  x 0  3 8 v 2 + 3 4 v + 3 8  dv = 1 8 x 3 + 3 8 x 2 + 3 8 x Hence, f X (x|X +2Y>1) = 3 8 x 2 + 6 8 x + 3 8 p(X +2Y>1) = 3 7 x 2 + 6 7 x...

Ngày tải lên: 12/08/2014, 16:21

... ,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 ... +2 4  dx = 1 4  1 4 x 4 + 2 3 x 3      0 −2 + 1 4  − 1 4 x 4 + 2 3 x 3      2 0 = 2 3 Hence, SQNR = 3 × 4 ν × 2 3 x 2 max = 3 × 4 5 × 2...

Ngày tải lên: 12/08/2014, 16:21

... =  x+1 0 z 1 3 dz = 3 4 z 4 3     x+1 0 = 3 4 (x +1) 4 3 If x>0, then  x −∞ [f X (v)] 1 3 dv =  0 −1 (v +1) 1 3 dv +  x 0 (−v +1) 1 3 dv = 3 4 +  1 1−x z 1 3 dz = 3 4 + 3 4  1 − (1 − x) 4 3  Hence, g(x)=    g(1)  (x ... 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...

Ngày tải lên: 12/08/2014, 16:21

... is P (e)= 1 3  A 2 −∞ 1  2πσ 2 n e − (r−A) 2 2σ 2 n dr + 1 3  1 −  A 2 − A 2 1  2πσ 2 n e − r 2 2σ 2 n dr  + 1 3  ∞ − A 2 1  2πσ 2 n e − (r+A) 2 2σ 2 n dr = 1 3 Q  A 2σ n  + 1 3 2Q  A 2σ n  + 1 3 Q  A 2σ n  = 4 3 Q  A 2σ n  Problem ... (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...

Ngày tải lên: 12/08/2014, 16:21

... (e))q 2 43 2) The following table lists the possible transmitted 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 ... to d E =2 2 +4 2 +2 2 =24 ✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉     ✒ ✲✲ ❍ ❍ ❍ ❍ ❍❥ ✟ ✟ ✟ ✟ ✟✯ ✲     ✒ ✟ ✟ ✟ ✟ ✟✯ ❅ ❅ ❅ ❅ ❅❘ ❍ ❍ ❍ ❍ ❍❥ ❍ ❍...

Ngày tải lên: 12/08/2014, 16:21

... 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 ... 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 X b , X c and X d , we...

Ngày tải lên: 12/08/2014, 16:21