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I study condition for fix if filter is a linear phase,but it's not clear in my mind!

I have this $h(n)$: $$h(n) = \begin{cases} \left(\frac{1}{2}\right)^{n} & 0<n<N-1 \\[2ex] 0 & \text{elsewhere} \end{cases}$$ I got the $H(z)$: $$ H(z) = \frac {\left(\frac{1}{2}\right)^{N}{z}^{-N}-1}{\frac{1}{2}{z}^{-1}-1} $$ Now, for the phase linear what I do ? I think to verify palindrome property for this $h(n)$, but how ?? or I can finally say that to the form that has definitely not linear ?

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There are four types of linear phase FIR filters, as described in this answer. They have to satisfy the following symmetry conditions:

$$h[n]=h[N-1-n]\quad \text{or}\quad h[n]=-h[N-1-n]\tag{1}$$

(assuming the impulse response $h[n]$ is zero for $n<0$ and $n\ge N$). Since your filter satisfies neither of the two conditions given by $(1)$ it can't have a linear phase response.

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  • $\begingroup$ Thank you, i understand your answer theorically, but you can you do me a practical example for this exercise ? because I think then it would be a fir filter linear phase if it had been $$h(n) = \begin{cases} \left(\frac{1}{2}\right)^{|n|} & 0<n<N-1 \\[2ex] 0 & \text{elsewhere} \end{cases}$$ $\endgroup$ – P_B Feb 11 '16 at 21:53
  • $\begingroup$ @P_B: You mean how to show that the impulse response in your question doesn't satisfy Eq. (1)? $\endgroup$ – Matt L. Feb 11 '16 at 21:56
  • $\begingroup$ Yes, exactly @MattL $\endgroup$ – P_B Feb 11 '16 at 22:03
  • $\begingroup$ @P_B: Well, your impulse response is decreasing exponentially, so it can't be symmetrical. $\endgroup$ – Matt L. Feb 11 '16 at 22:09

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