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
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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.
Fat32
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Thanks, but it seems that you implicitly supposed that $H(e^{j\omega})$ is continuous. Is that always true? – Ali Jan 03 '20 at 13:09
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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. – Fat32 Jan 03 '20 at 19:46