# The DFT and the DTFT 8

Posted by **Steve Eddins**,

It's finally time to start looking at the relationship between the discrete Fourier transform (DFT) and the discrete-time Fourier transform (DTFT). Let's look at a simple rectangular pulse, for . The DTFT of is:

Let's plot for over a couple of periods:

```
M = 8;
w = linspace(-2*pi, 2*pi, 800);
X_dtft = (sin(w*M/2) ./ sin(w/2)) .* exp(-1j * w * (M-1) / 2);
plot(w, abs(X_dtft))
title('|X(\omega)|')
```

It turns out that, under certain conditions, the DFT is just equally-spaced samples of the DTFT. Suppose is the P-point DFT of . If is nonzero only over the finite domain , then equals at equally spaced intervals of :

The MATLAB function `fft` computes the DFT. Here's the 8-point DFT of our 8-point rectangular pulse:

x = ones(1, M); X = fft(x)

X = 8 0 0 0 0 0 0 0

One 8 and a bunch of zeros?? That doesn't seem anything like the DTFT plot above. But when you superimpose the output of `fft` in the right places on the DTFT plot, it all becomes clear.

P = 8; w_k = (0:P-1) * (2*pi/P); X = fft(x); plot(w, abs(X_dtft)) hold on plot(w_k, abs(X), 'o') hold off

Now you can see that the seven zeros in the output of `fft` correspond to the seven places (in each period) where the DTFT equals zero.

You can get more samples of the DTFT simply by increasing P. One way to do that is to **zero-pad**.

x16 = [x, zeros(1, 8)]

x16 = Columns 1 through 15 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 Column 16 0

P = 16; X16 = fft(x16); w_k = (0:P-1) * (2*pi/P); X = fft(x); plot(w, abs(X_dtft)) hold on plot(w_k, abs(X16), 'o') hold off

Another way to increase P is to use the `fft(x,P)` syntax of the `fft` function. This syntax computes the P-point DFT of `x` by using zero-padding. Let's try a 50-point DFT.

P = 50; Xp = fft(x, P); w_k = (0:P-1) * (2*pi/P); X = fft(x); plot(w, abs(X_dtft)) hold on plot(w_k, abs(Xp), 'o') hold off

If you've ever wondered what that whole zero-padding business was all about with Fourier transforms, now you know. When you tack on a bunch of zeros to a sequence and then compute the DFT, you're just getting more and more samples of the DTFT of the original sequence.

I think the next logical place to go in our Fourier exploration is to start considering some of the reasons why many people
find the output of `fft` so surprising or puzzling. Here's a sample:

- Why isn't the zero frequency (or "DC" frequency) in the center of the output from
`fft`? - Why isn't the output of
`fft`real when the input is symmetric?

Do you have puzzles to add? Let me know by adding your comments.

Get the MATLAB code

Published with MATLAB® 7.9

**Category:**- Fourier transforms

### Note

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*[Long code fragment edited out. -SE]*The high value output from the fft is carried forward till my end result.

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