1
$\begingroup$

enter image description here

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.

enter image description here

here's my work using fourier transfrom

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\omega \in [-\pi, \pi]$

fvtool(h,1);

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.