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I am trying to calculate the absolute value of a certain sinusoid to see how MATLAB calculates the Fourier Transform by using the freqz() command. Here is my Matlab code

    Np=20;
    n=0:1:Np-1;
    s1=[sin(2*pi/Np*n)];
    h1=freqz(s1);
    figure(1);
    plot(abs(h1));

And the plotted result is: enter image description here

My question is: My theoretical plot should only show one "bar". Why doesn't it do that? Is my code wrong?

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  • \$\begingroup\$ You only have a small interval, so the sine wave is truncated. Theory assumes it goes to infinity. \$\endgroup\$
    – Chu
    Feb 26 '19 at 14:20
  • \$\begingroup\$ How many sine wave cycles did your math consider? \$\endgroup\$
    – Andy aka
    Feb 26 '19 at 14:20
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    \$\begingroup\$ The transformed window does not contain an exact multiple of the desired singular frequency \$\endgroup\$
    – uglyoldbob
    Feb 26 '19 at 14:26
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    \$\begingroup\$ Mário, sure you want freqz and not just abs(fft)? \$\endgroup\$ Feb 26 '19 at 15:02
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    \$\begingroup\$ Have you read the documentation for freqz? \$\endgroup\$
    – TimWescott
    Feb 26 '19 at 15:36
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As @MarcusMüller indicated, you need to use fft instead of freqz:

Replace h1=freqz(s1); with h1=fft(s1)/Np; Also, to interpret the result, it would be better to replace plot(abs(h1)); with stem(n,abs(h1));.

In your example, the signal is exactly one period of a sine wave with amplitude 1, so you should get a spike at n=1 and n=19 with amplitude one half. The amplitude of 1 is split over two frequencies. The n=19 corresponds to a negative frequency n=-1 (a sine wave is the sum of two complex exponentials, one with a positive frequency and one with a negative frequency: \$\sin(\omega t) = \frac{1}{2 j}(e^{j \omega t} - e^{-j \omega t})\$).

If you would want to plot a single-sided FFT (this has only positive frequencies and would provide an amplitude of one at n=1 in your example), have a look at the Matlab documentation for fft, example Noisy Signal.

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