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

... = 1 2  1 1 Λ(t)e j nt dt = 1 2  0 1 (t +1) e j nt dt + 1 2  1 0 (−t +1) e j nt dt = 1 2  j πn te j nt + 1 π 2 n 2 e j nt      0 1 + j 2πn e j nt     0 1 − 1 2  j πn te j nt + 1 π 2 n 2 e j nt      1 0 + j 2πn e j nt     1 0 1 π 2 n 2 − 1 2π 2 n 2 (e j n + ... =  1 0 e ( j2 πn +1) t dt = 1 j2 πn +1 e ( j2 πn +1) t...

Ngày tải lên: 12/08/2014, 16:21

... Note that J 0 (1. 5) = . 511 8, J 1 (1. 5) = .5579 and J 2 (1. 5) = .23 21. 63 ✲✛ ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✻ ✛ ✲ ✻ ✻ ✻ ✻ ✻ AJ 4 (3) 2 AJ 2 (3) 2 8 10 3 10 6 f Hz 5 10 3 AJ 2 (1. 5) 2 AJ 1 (1. 5) 2 f Hz 10 6 5) ... (1C)=P (1C|1S)P (1S)+P (1C|0C)P(0C) = .8 · .7+.2 · .3=.62 where we have used the fact that P (1S)=.7, P (0C)=.3, P (1C|0C)=.2 and P (1C|1S) =1 .2= .8 2) P (1S|1C)= P (1C,...

Ngày tải lên: 12/08/2014, 16:21

... 10 −2 2.276 × 10 −2 2.5 6. 21 10 −3 6. 214 × 10 −3 3. 1. 35 × 10 −3 1. 3 51 × 10 −3 3.5 2.33 10 −4 2.328 × 10 −4 4. 3 .17 × 10 −5 3 .17 1 × 10 −5 4.5 3.40 10 −6 3.404 × 10 −6 5. 2.87 × 10 −7 2.874 × 10 −7 Problem ... figure.    ❍ ❍ ❍ . . . .      ❍ ❍ ❍ ❍ ❍ ❍ x y 1/ 3 (1, 1) x+2y =1 A 79 Thus p(X>Y,X+2Y> ;1) =  1 1 3  x 1 x 2 (x + y)dxdy =  1...

Ngày tải lên: 12/08/2014, 16:21

... R 1 = 1 2 log 2 10 00 = 5. 14 9 . . . . . . 11 1 11 11 1 10 11 10 10 0 1 0 1 0 1 0 1 0 1 0 1 2 n−2 1 2 n 1 1 2 n 1 1 4 1 2 The entropy of the source is H(X)= n 1  i =1 1 2 i log 2 2 i + 1 2 n 1 log 2 2 n 1 = n 1  i =1 1 2 i i ... = 10 , we have BW = 10 W . 15 5 Hence, the quantization levels are ˆx 1 = −ˆx 8 = −3 11 .72 − 1 2 11 .72 = − 41. 02...

Ngày tải lên: 12/08/2014, 16:21

... 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 0.4 0.6 0.8 1 1.2 1. 4 1. 6 ... Thus, y l (t)=F 1 [Y l (f)] = 1 2j  0 − 1 2 (f +1) e j2...

Ngày tải lên: 12/08/2014, 16:21

... 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 × 4 ν  3 2  3 we ... then  x −∞ [f X (v)] 1 3 dv =  x 1 (v +1) 1 3 dv =  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 −∞...

Ngày tải lên: 12/08/2014, 16:21

... adjacent nodes can be found using the previous figure. 0000 00 01 010 1 010 0 11 00 11 01 1 010 011 0 0 010 0 011 10 01 1000 10 11 011 1 11 11 111 0 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 21 2 2 1 2 1 Problem 7.46 1) ... Adjacent points are connected with solid or dashed lines. 0000 00 01 0 011 0 010 011 0 011 1 010 1 010 0 11 00 11 01 111 1 11 10 10 10 10...

Ngày tải lên: 12/08/2014, 16:21

... transmission of 1. ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✑ ✑ ✑ ✑ ✑ ✑ ✑ ✑ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗                 a m +1 1 -1 1 -1 -1 1 1 -1 11 1- 1 -1 1 -1 -1 (a m 1 ,a m ) 2 31 Problem ... is E av =24.70 × 10 −9 228 Problem 8.8 The transition probability matrix P is P = 1 2      010 1 0 011 11 00 10 10      Hence, P 2 =...

Ngày tải lên: 12/08/2014, 16:21

... -1 1 -2 -1 1 -1 0 -1 11 2 1- 1 -1 -2 1- 1 1 0 1 1- 1 2 11 1 4 As it is observed there are 5 possible output levels b m , with probability p(b m =0)= 1 4 , p(b m = ±2) = 1 4 and p(b m = ±4) = 1 8 . 3) ... to d E =2 2 +4 2 +2 2 =24 ✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉✉ ✉ ✉ ✉     ✒ ✲✲ ❍ ❍ ❍ ❍ ❍❥ ✟ ✟ ✟ ✟ ✟✯ ✲     ✒ ✟ ✟ ✟ ✟ ✟✯ ❅ ❅ ❅ ❅ ❅❘ ❍ ❍ ❍ ❍ ❍❥ ❍ ❍ ❍ ❍ ❍❥ ✲...

Ngày tải lên: 12/08/2014, 16:21

... 11 010 00 11 010 11 c 14 10 111 00 0 011 100 11 111 00 10 011 00 10 1 010 0 10 110 00 10 111 10 10 111 01 c 15 011 10 01 111 10 01 0 011 0 01 010 10 01 011 00 01 011 110 1 011 1 011 011 1000 c 16 11 111 11 011 111 1 10 111 11 110 111 1 11 1 011 1 ... 0 011 111 000 011 1 00 010 11 00 011 01 00 011 10 c 6 11 0 010 1 010 010 1 10 0 010 1 11 1 010 1 11 011 01 110...

Ngày tải lên: 12/08/2014, 16:21