I am trying to understand exactly how sampling the DTFT to get the DFT works. The signal I'm trying to analyze is $x(n)$ seen below.

Signal x

$$x(n) = \delta(n\pm2) + 2\delta(n\pm1) + 3\delta(n)$$

Taking the DTFT, we have \begin{align} X(\omega)&= \sum\limits_{i=-\infty}^\infty x(n)e^{-j\omega}\\ &= \left(e^{j2ω} + e^{-j2ω}\right) + 2\left(e^{jω} + e^{-jω}\right) + 3\\ &= 2\cos⁡(2ω) + 2(2\cos⁡ω) + 3\\ &= 3 + 4\cos⁡ω + 2\cos⁡(2ω) \end{align}

I next implemented this in MATLAB:

x = [1 2 3 2 1];
N = size(x,2);
w = -pi:0.01:pi;
X_DTFT_computational = freqz(x,1,w);
X_DTFT_analytical = 3 + 4*cos(w) + 2*cos(2*w);

This results in the following graphs: DTFT

Next, I compute the DFT in two ways: 1. I apply an FFT to the original signal $x(n)$. 2. I sample the DTFT.

The code that achieves this is as follows:

% FFT of x
X_DFT_computational = fftshift(fft(x))

% DFT = Sampled DTFT
X_DFT_analytical(1) = 3 + 4*cos(-4*pi/N) + 2*cos(2*(-4*pi/N));
X_DFT_analytical(2) = 3 + 4*cos(-2*pi/N) + 2*cos(2*(-2*pi/N));
X_DFT_analytical(3) = 3 + 4*cos(0)       + 2*cos(2*(0));
X_DFT_analytical(4) = 3 + 4*cos(2*pi/N)  + 2*cos(2*(2*pi/N));
X_DFT_analytical(5) = 3 + 4*cos(4*pi/N)  + 2*cos(2*(4*pi/N));

Plotting the DFT (while showing the DTFT for comparison), I have DFT

The magnitude of the FFT does give me the correct result. But just plotting the FFT alone gives me a complex signal (the graph shows the real part).

Why is this signal giving a complex FFT when $x(n)$ is both real and symmetric?


1 Answer 1


You defined the signal vector as x = [1 2 3 2 1]. Since the DFT is defined by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$

the command fft(x) computes the DFT of the signal


This signal is not symmetrical with respect to $n=0$, and it is not equal to the signal you computed the DTFT of.

If you want to compute the DFT of the signal


you have to periodically continue it in the interval $n\in [0,N-1]$ (with $N=5$):

x = [3 2 1 1 2]

This results in a real-valued and symmetrical DFT, as expected:

X = fft(x);
X =

   9.00000   2.61803   0.38197   0.38197   2.61803
  • $\begingroup$ Matt, you answered my question perfectly. Thank you friend. Now I need to figure out how to extend this to a 2D signal. $\endgroup$
    – Josh
    Oct 9, 2016 at 6:54

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