# Why do FIR systems need to be antisymmetric about M/2 to be linear phase?

Why do FIR systems need to be antisymmetric about M/2 to be linear phase? where M=delay.Also im talking about discrete time.And why are hilbert transformers antisymmetric on n=τ?

Consider the $$M$$ th order FIR system, where $$M$$ is even \begin{aligned} H(z) &= \sum_{m=0}^{M-1}b_m z^{-1} \\ &= b_0 + b_1z^{-1} + \ldots b_{M-1}z^{M-1} \\ H(e^{j\omega}) &= b_0 + b_1 e^{-j\omega} + b_2 e^{-2j\omega} + \ldots b_{M-1}e^{-j(M-1)\omega} \end{aligned} Now if our coefficients are antisymmetric, then $$b_m = -b_{M-1-m}$$, or, $$b_0 = -b_{M-1}, b_1 = -b_{M-2}, \ldots, b_{\frac{M}{2}} = -b_{\frac{M}{2}-1}$$. This gives us \begin{aligned} H(e^{j\omega}) &= b_0 - b_{M-1}e^{-j(M-1)\omega} + b_1 - b_{M-1}e^{-(j(M-1) + 1)\omega)} + \ldots \\ &= e^{-j(M-1)\omega} \left[ 2b_0 + b_1(e^{-j\omega}-e^{j\omega}) + \ldots b_{\frac{M}{2}} (e^{-j\frac{M}{2}\omega}-e^{j\frac{M}{2}\omega}) \right] \\ &= -2je^{-j(M-1)\omega}) \left[jb_0 +b_1 \sin \omega + \ldots + b_{\frac{M}{2}} \sin{\frac{M}{2}\omega} \right] \\ &= -2je^{-j(M-1)\omega})\left[jb_0 + \sum_{m=1}^{M/2} b_m \sin(m\omega)\right] \end{aligned} We see that $$H(e^{j\omega})$$ is linear phase, with its phase $$= -(M-1)\omega - \frac{\pi}{2}$$. This would have not worked out if the filter coefficients were not antisymmetric or symmetric.
• if the coefficients are antisymmetric, wouldnt $b_m = -b_{M-1-m}$ – Ben May 6 at 19:45