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Unformatted text preview: Chapter 11 1. See Fig 1 fc = 100 MHz fc+B fcB 2B = 100 KHz Figure 1: Band of interest. B = 50 KHz, f c = 100 MHz H eq ( f ) = 1 H ( f ) = f Noise PSD = N W/Hz. Using this we get Noise Power = Z f c + B f c B N  H eq ( f )  2 df (1) = N Z f c + B f c B f 2 df (2) = N f 3 3 ( f c + B ) ( f c B ) (3) = N 3 ( f c + B ) 3 ( f c B ) 3 (4) = 10 21 N W (5) Without the equalizer, the noise power will be 2 BN = 10 5 N W. As seen from the noise power values, there is tremendous noise enhancement and so the equalizer will not improve system performance. 2. (a) For the first channel: ISI power over a bit time = A 2 T b /T b = A 2 For the 2nd channel: ISI power over a bit time = A 2 T b n =1 R ( n +1) T b nT b e t/T m dt = 2 e 1 / 2 A 2 A S(t) T m /2 1 h T m 1 (t) h 2 (t) t t t Figure 2: Problem 2a (b) No ISI: pulse interval = 11 / 2 s = 5 . 5 s Data rate = 1 / 5 . 5 s = 181 . 8 Kbps If baseband signal =100KHz: pulse width = 10 s Data rate = 2 / 10 s + 10 s = 100 Kbps 3. (a) h ( t ) = e t t o.w. (6) = 6 sec H eq ( f ) = 1 H ( f ) H ( f ) = Z e t e j 2 ft dt (7) = 1 1 + j 2 f (8) Hence, H eq ( f ) = 1 + j 2 f (b) SNR eq SNR ISI = R B B S x ( f )  H ( f )  2  H eq ( f )  2 df R B B N  H eq ( f )  2 df R B B S x ( f )  H ( f )  2 df 2 BN 1 h 10us 1 (t) t 1 X 1us (t) t h 12us 1 (t)*X(t) t 1 Figure 3: Problem 2b Assume S x ( f ) = S , B f B 2 BS N &#16; 2 B 2 + 8 2 3 B 3 &#17; S R B B  H ( f )  2 df 2 BN = 2 B &#16; 1 2 + 4 2 3 B 2 &#17; 1 . 617 10 6 = 0 . 9364 = . 28 dB (c) h [ n ] = 1 + e Ts [ n 1] + e 2 Ts [ n 2] + ... H ( z ) = 1 + e T s z 1 + e 2 T s z 2 + e 3 T s z 3 + ... = X n =0 e Ts z 1 n = z z e Ts = 1 1 e Ts z 1 H eq ( z ) = 1 H ( z )+ N . Now, we need to use some approximation to come up with the filter tap coefficient values. If we assume N u 0 (the zeroforcing assumption), we get H eq ( z ) = 1 e T s z 1 ....
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This note was uploaded on 01/11/2012 for the course EE 359 taught by Professor Goldsmith during the Fall '08 term at Stanford.
 Fall '08
 Goldsmith

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