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I took even number of sample (in this case 6) from sinc, hamming window, and multiple of those two. then after fourier transform, frequency-domain variables have zero only left side, in this case its x is -pi. I'm wondering why there are no zeros in right side (where x = pi), and has positive value.

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here's my work using fourier transfrom

  • $\begingroup$ that's most likely because you have an even number of samples. You can't create perfectly symmetrical signals ( e.g windows, filter kernels etc) using an even number of points. Use an odd number instead. $\endgroup$ – dsp_user May 27 at 7:58
  • $\begingroup$ my prof asked why even number sample has none-symmetrical output. so.. i have to deal with it. $\endgroup$ – TAETAE May 27 at 8:04
  • $\begingroup$ We don't normally help with school assignments, at least not until we see what you've tried first. $\endgroup$ – dsp_user May 27 at 8:18
  • $\begingroup$ its not assignment he told just think about it but never gave a answer. i tried to solve it through fourier transform (using sigma), but i cannot got the answer of both 0 on -pi and +pi. $\endgroup$ – TAETAE May 27 at 8:25
  • $\begingroup$ i uploaded my work using fourier transform $\endgroup$ – TAETAE May 27 at 8:26

Answer: Your derivation is absolutely correct and You will see 0 for both $\omega = -\pi$ and $\omega = \pi$, if you use fvtool() function of MATLAB instead of fft(). The reason is simple : FFT does not calculate values of $H(e^{j\omega})$ at continuous $\omega \in [-\pi, \pi]$, but it calculates DFT and for that the digital frequency resolution is $\omega = \frac{2\pi k}{N}$. So, for $N=6$, it will compute values of $H(e^{j\omega})$ at $\omega = 0, \frac{2\pi}{6}, \frac{4\pi}{6}, \frac{6\pi}{6}, \frac{8\pi}{6} \ and \ \frac{10\pi}{6}$. See that $\omega = \pi$ is occurring only at $k=3$, and hence you are seeing $0$ only once.

For checking this, you can take any symmetrical h, like $h = {1,2,4,4,2,1}$. Then use the following MATLAB code to plot its DTFT for $\omega \in [-\pi, \pi]$


This will plot a Single sided DTFT from $\omega \in [0, \pi]$. You can then change from "Analysis" option to plot for frequency range $[-\pi, \pi]$.

You will get following plot for the example I have taken: SymmResp

Check that the values are $0$ at both $\omega = -\pi$ and $\omega = \pi$.

And this will happen for any symmetric values, not any particular window function.

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