Chapter 11

Chapter 11 - Chapter 11 1 See Fig 1 fc = 100 MHz fc B fc-B...

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Unformatted text preview: Chapter 11 1. See Fig 1 fc = 100 MHz fc+B fc-B 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 μs = 5 . 5 μs ∴ Data rate = 1 / 5 . 5 μs = 181 . 8 Kbps If baseband signal =100KHz: pulse width = 10 μs Data rate = 2 / 10 μs + 10 μs = 100 Kbps 3. (a) h ( t ) = ‰ e- t τ t ≥ o.w. (6) τ = 6 μ sec H eq ( f ) = 1 H ( f ) H ( f ) = Z ∞ e- t τ e- j 2 πft dt (7) = 1 1 τ + j 2 πf (8) Hence, H eq ( f ) = 1 τ + j 2 πf (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  2 B τ 2 + 8 π 2 3 B 3  S R B- B | H ( f ) | 2 df 2 BN = 2 B  1 τ 2 + 4 π 2 3 B 2  1 . 617 × 10- 6 = 0 . 9364 =- . 28 dB (c) h [ n ] = 1 + e- Ts τ δ [ n- 1] + e- 2 Ts τ δ [ n- 2] + ... H ( z ) = 1 + e- T s τ z- 1 + e- 2 T s τ z- 2 + e- 3 T s τ z- 3 + ... = ∞ X n =0 ‡ e- Ts τ z- 1 · n = z z- e- Ts τ = 1 1- e- Ts τ 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 zero-forcing assumption), we get H eq ( z ) = 1- e- T s τ z- 1 ....
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Chapter 11 - Chapter 11 1 See Fig 1 fc = 100 MHz fc B fc-B...

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