# Why does DTFT start and end at the same magnitude?

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

$$H(e^{j \omega}) = H(e^{j (\omega + 2 \pi k)}).$$
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.
• Thanks, but it seems that you implicitly supposed that $H(e^{j\omega})$ is continuous. Is that always true? – Ali Jan 3 '20 at 13:09
• 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 3 '20 at 19:46