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 zeropadded sequences i [k]
and h[k] are given by
1. In this case N t 2 Fr 2 = L i[k]eirOok = Li[k]eir~k
k=O = 3 + 2ei~r 1 Hr k=O + 3e i ".r
1 = L h[k]eirOok = L h[k]eirtk
k=O k=O = 2 + 2ei~r Substituting r = 0, 1, 2, 3, we obtain Fo = 8
F l =  2j
F2 = 4
Ho = 4
H l = 2 V2eit
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 opoint 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 opoint 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:Ll;
% 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:Rl;
f =(0.8). " m;
h ={0.5)."m;
L =length(fHlength(h)_l;
k =O:l:Ll;
% 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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