Signal Processing and Linear Systems-B.P.Lathi copy

# Show t hat t he o ft o f t his signal is fo 4 h 1 a nd

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Unformatted text preview: ar convolution of signals i [k] and h[k] using DFT. E xample 1 0.12 . i [k] Find y[k], t he output of an LTID filter with impulse response h[k] t o a n mput illustrated in Figs. 1O.14a and b. We obtain the convolution y[k] i n four steps listed above. • = 3 and Nh = 2. Therefore, we pad 1 zero to i [k] and 2 zeros to h[k], as depicted in Figs. 1O.14c and d, respectively. 2. Now, No = 4 and no = 7r / 2. The DFTs Fr and H r of the zero-padded sequences i [k] and h[k] are given by 1. In this case N t 2 Fr 2 = L i[k]e-irOok = Li[k]e-ir~k k=O = 3 + 2e-i~r 1 Hr k=O + 3e- i ".r 1 = L h[k]e-irOok = L h[k]e-irtk k=O k=O = 2 + 2e-i~r Substituting r = 0, 1, 2, 3, we obtain Fo = 8 F l = - 2j F2 = 4 Ho = 4 H l = 2 V2e-it H2 = 0 3. Multiply Fr by H r to obtain Yr. This step yields Yo = 32 4. The desired convolution is t he IDFT of Y r , given by 3 y[k] F iltering In t he Frequency Domain I f we a re given t he filter t ransfer f unction H (Q), we k now H r. We c ompute Fr, t he D FT o f i [k]. T hen follow t he s teps 3 a nd 4 t o o btain t he o utput y[k]. -2 ( c) (10.78) -4 y[kJ k...... f [kJ • We know t hat t he D FT o f a n N o-point sequence is t he s et o f u niform samples of i ts D TFT a t frequency interval Q o = 27r/No. T herefore, i f Y r is t he N o-point D FT o f y[k]' t hen ( e) JiL' l From t he w idth p roperty o f t he convolution (Sec. 9.4), t he l ength of y[k] is ( Nj + N h - 1). Let t he D TFT o f t he sequences i [k], h[k], a nd y[k] b e F (Q), H(Q), a nd Y (Q), respectively. T hen, from Eq. (1O.50a), we have Y(Q) = F (Q)H(Q) 653 10.6 Signal processing Using D FT a nd F FT 10 F ourier Analysis o f Discrete- Time Signals 3 r::;:O r =O = ~ L YreirOok = ~ LYrejr~k F3 = 2 j H3 = 2 V2eit 654 10 F ourier Analysis of Discrete- Time Signals Substitution of r = 0, 1, 2, 3 in this equation yields y[O] = 6 y[1] = 10 y[2] = 10 y[3] = 6 o C omputer E xample C IO.4 Use MATLAB to do Example 10.12. Find the answer by direct convolution as well as by using DFT. . I~ performing the convolution via DFT, we shall use the command 'fft{f,L)" which gIves .he F FT of a sequence f with sufficient zeros padded to make its length equal to L. f =[323]; h =[22]; L =length{f)+length{h)_I; k =O:I:L-l; % L inear c onvolution: D irect a pproach y l=conv(f,h); s ubplot(2,1,2) s tem(k,yl) % L inear C onvolution:via D FT F E=fft(f,L); H E=fft(h,L); y 2=ifft(FE. * HE); s ubplot(2, I , 1 ); s tem(k,y2) 0 C omputer E xample C lO.5 Use MATLAB to convolve (0.8)ku[k] and (0.5)ku[k]. Find the answer by direct convolution and using DFT. . Both the signals have infinite duration. Hence, we must truncate them beyond some SUItable value ? f k, where both functions become negligible. For this purpose k = 32 is a reasonable chOIce. R =32; m =O:l:R-l; f =(0.8). " m; h ={0.5)."m; L =length(fHlength(h)_l; k =O:l:L-l; % L inear c onvolution: D irect a pproach y l=conv(f,h); s ubplot(2,1,2) s tem(k,yl) % L inear C onvolution:via D FT F E=fft(f,L); H E=fft(h,L); y 2=ifft(FE. * HE);...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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