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$$x(n) = \{ -1, 0, 1, 2, 1, 0, 1, 2, 1, 0, -1 \}$$

Let $X(\omega)$ be the DTFT of $x(n)$. I need to find the phase of $X (\omega)$ without computing $X(\omega)$. I notice that $x(n)$ can be a type I linear phase filter, but not sure if that is used in solving the problem.

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  • $\begingroup$ You noticed correctly, so I guess the problem is solved, isn't it? $\endgroup$
    – Matt L.
    Commented Jul 17, 2020 at 21:29

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hint: x(n) has symmetric coefficients which as a filter is a type 1 linear phase filter as the OP has stated. Such a filter will have a phase slope that is dependent on the number of taps only.

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  • $\begingroup$ So my group delay is (N-1) /2 and I integrate w.r.t w to find phase $\endgroup$
    – my_knee_Hertz
    Commented Jul 17, 2020 at 21:44
  • $\begingroup$ Yes that is true but it is even simpler than having to integrate- realize the phase for every unit sample delay goes from 0 to $-2\pi$ as the frequency goes from 0 to the sampling rate. $\endgroup$
    – Dan Boschen
    Commented Jul 17, 2020 at 22:18

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