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Unformatted text preview: 1 ECE155A –Vlad Dorfman Class 10  Solutions Problem 1. Noise Enhancement Introduction: The problem illustrates the noise enhancement for a continuous ZF filter using the example of the Lorentzian channel. The channel s( t ), AWGN noise and the equalizer g ( t ) can be represented as shown in Figure 1a. Figure 1a. ZF Equalization Block Diagram. In the model shown in Figure 1a, 2 / 50 2 50 ) ( ω π ω PW e PW S − = . (1) and the equalizer G(w) is calculated as in (4). Note, that the inverse of S(w) is 2 / 50 50 2 ω π PW e PW , ( 2 ) and contains an exponential term that increases extremely rapidly with frequency. As the density of recording D increases, the range of frequencies w ≤π /T in (4) containing the nonzero "inverted" S ( w ) increases as well causing strong noise increase ("enhancement") illustrated in Figure 1b. At the same time, when D (density) is small and T is large, or when the AWGN noise is insignificant and ISI is dominant, ZF is an acceptable technique. + Noise n ( t ) Equalizer g ( t ) Signal + Enhanced noise Channel s ( t ) 2 a) Plot G ( e j ω ) power for D=1.0,1.6,2.0, and 2.5 105 5 10 2 4 6 8 10 12 14 w,rad Magnitude Equalizer Frequency Response, G(w) D=1.0, Noise: Analyt.=2.86 σ 2 Sim.=2.86 σ 2 D=1.6, Noise: Analyt.=7.63 σ 2 Sim.=7.64 σ 2 D=2.0, Noise: Analyt.=17.2 σ 2 Sim.=17.3 σ 2 D=2.5, Noise: Analyt.=53.1 σ 2 Sim.=53.2 σ 2 Figure 1. Noise Enhancement vs. Density of Recording "D". b) Calculate the noise power for D=1.0,1.6,2.0, and 2.5 The output noise power is calculated based on that for a linear system that has a FT magnitude H( ω ), and input random process X(t) with power spectral density S X ( ω ), the output power spectral density (not total power) will be S Y ( ω ) = H( ω ) 2 S X ( ω ). Consider that in our case the power spectral density is σ 2 , and the linear system H( ω ) = G( ω ). Then to find the total output noise power one needs to integrate ∫ ∞ ∞ − = ω ω π σ σ d G eq 2 2 2 ) ( 2 , (3) where σ 2 is the input noise power. Input noise is assumed uncorrelated. 3 Since G(w) is calculated as ⎧ T/S(w), w ≤π /T G( w )= ⎨ ( 4 ) ⎩ 0 , w> π /T Then, substituting (4) into (3), ( ) ( ) 1 50 4 1 50 4 50 4 50 2 50 2 2 ) ( 2 / 50 3 2 / 50 3 3 2 2 50 2 3 2 2 50 2 3 2 2 2 2 / 50 2 2 2 2 − ⋅ = − = = = = = = ∫ ∫ ∫ ∫ − − − T PW T PW T PW T T PW T T PW T T out e PW D e PW T d e PW T d e PW T d e PW T d S T π π π ω π π ω π π ω π π π σ π σ ω π σ ω π σ ω π π σ ω...
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This note was uploaded on 11/11/2009 for the course ECE 670377 taught by Professor Wolf during the Winter '07 term at UCSD.
 Winter '07
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