Signals and Systems - Electrical Engineering

622 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

where fs= 1 /Ts is the sampling rate given in samples/second. The results are shown in

Figure 10.15.

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% Example 10.21---Improving frequency resolution of FFFT of periodic signals

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f0 = 100; f1 = 4∗f0; % frequencies in Hz of signal

Ts = 1/(3∗f1); % sampling period

t = 0:Ts:4/f0; % time for 4 periods

y = 4∗cos(2∗pi∗f0∗t) - cos(2∗pi∗f1∗t); % sampled signal (4 periods)

M = length(y);

Y = fft(y,M); Y = fftshift(Y)/4; % fft using 4 periods, shifting and normalizing

t1 = 0:Ts:12/f0; % time for 12 periods

y1 = 4∗cos(2∗pi∗f0∗t1) - cos(2∗pi∗f1∗t1); % sampled signal (12 periods)

Y1 = fft(y1);Y1 = fftshift(Y1)/12; % fft using 12 periods, shifting and normalizing

w = 2∗[0:M - 1]./M - 1;f = w/(2∗Ts); % frequency scale (4 periods)

N = length(y1);

w1 = 2∗[0:N - 1]./N - 1;f = w/(2∗Ts); % frequency scale (12 periods) n

10.4.4 Linear and Circular Convolution Sums

The most important property of the DFT is the convolution property, which permits the computation

of the linear convolution sum very efficiently by means of the FFT.

Consider the convolution sum that gives the outputy[n] of a discrete-time LTI system with impulse

responseh[n] and inputx[n]:

y[n]=

∑

m

x[m]h[n−m]

In frequency,y[n] is the inverse DTFT of the product

Y(ejω)=X(ejω)H(ejω)

Assuming thatx[n] has a finite lengthMand thath[n] has a finite lengthK, theny[n] has a finite

lengthN=M+K−1. If we choose a periodL≥Nfor the periodic extension ̃y[n] ofy[n], we would

obtain the frequency-sampled periodic sequence

Y(ejω)|ω= 2 πk/L=X(ejω)H(ejω)|ω= 2 πk/L

or the DFT ofy[n] as the product of the DFTs ofx[n] andh[n]:

Y[k]=X[k]H[k] for k=0, 1,...,L− 1

We then obtainy[n] as the inverse DFT ofY[k]. It should be noticed that theL-length DFT ofx[n] and

ofh[n] requires that we padx[n] withL−Mzeros andh[n] withL−Kzeros, so that bothX[k] and

H[k] have the same lengthLand can be multiplied at eachk.