I would like to know why the spectrum of FIR filters (and maybe all DTFT spectra) start and end at the same magnitude. I guess there is something related to $H(e^{j\omega})$. Thanks


1 Answer 1


DTFT is a periodic waveform;

$$H(e^{j \omega}) = H(e^{j (\omega + 2 \pi k)}).$$

Hence for every frequency interval $[\omega_1, \omega_2]$ of length $2\pi$ it will repeat itself; i.e., beginning and ending at the same magnitude (and phase) on that interval.

  • $\begingroup$ Thanks, but it seems that you implicitly supposed that $H(e^{j\omega})$ is continuous. Is that always true? $\endgroup$
    – Ali
    Jan 3, 2020 at 13:09
  • $\begingroup$ No I didn't assume anything about continuity. The periodicity of $H(e^{j\omega})$ is true irrespective of it's being continuous or not. $\endgroup$
    – Fat32
    Jan 3, 2020 at 19:46

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