Many of you were onto me immediately last week. I asked you to estimate the frequency of the sampled cosine signal below, and readers quickly chimed in to guess that this question was really a teaser about aliasing.
As it happened, I started with a 1 rad/second cosine, , :
t = linspace(-2.3*pi, 2.3*pi, 800); alpha = 1; x_c = cos(alpha * t); plot(t, x_c, 'k')
And I sampled that signal to get with so that .
T = 1; n = -7:7; nt = n * T; x = cos(alpha*n*T); hold on plot(nt, x, 'ok') hold off
But I could have started with a completely different frequency and still ended up with exactly the same samples. For example, let's see what happens when we try :
alpha_2 = 2*pi - 1; x2_c = cos(alpha_2 * t); plot(t, x2_c, 'k') x2 = cos(alpha_2 * n * T); hold on plot(nt, x2, 'ok') hold off
The two sets of samples are the same (within floating-point round-off error):
max(abs(x - x2))
ans = 1.1657e-015
Let's look at this phenomenon in terms of the relationship between the continuous-time Fourier transform and the discrete-time Fourier transform (DTFT). Below is the continuous-time Fourier transform of .
The DTFT is a collection of copies of the continuous-time Fourier transform, spaced apart by the sampling frequency, and with the frequency axis scaled so that the sampling frequency becomes . Here's the DTFT of .
Now has a higher frequency as you can see in the plot of its continuous-time Fourier transform:
But when you make a bunch of copies of spaced apart by the sampling frequency, you find that the DTFT of is exactly the same as the DTFT of .
Usually only a single period of the DTFT is plotted:
In other words, when you use a sampling rate of , the frequencies 1 and are indistinguishable. This is called aliasing. In general, the continuous-time frequency is indistinguishable from any other frequency of the form , where is an integer.
So far we've talked about the continuous-time Fourier transform, the discrete-time Fourier transform, their relationship, and a little bit about aliasing. Next time we'll bring the discrete Fourier transform (DFT) into the discussion. That's what the MATLAB function fft actually computes.
要发表评论，请点击 此处 登录到您的 MathWorks 帐户或创建一个新帐户。