# FIR filter : $h(n)$ is palindrome?

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 ?

$$h[n]=h[N-1-n]\quad \text{or}\quad h[n]=-h[N-1-n]\tag{1}$$
(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.
• 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 $$h(n) = \begin{cases} \left(\frac{1}{2}\right)^{|n|} & 0<n<N-1 \\[2ex] 0 & \text{elsewhere} \end{cases}$$ – P_B Feb 11 '16 at 21:53