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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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);...
View Full Document

Ask a homework question - tutors are online