Why do FIR systems need to be antisymmetric about M/2 to be linear phase?
Consider the $M$ th order FIR system, where $M$ is even
$$\begin{aligned}
H(z) &amp;= \sum_{m=0}^{M-1}b_m z^{-1} \\
     &amp;= b_0 + b_1z^{-1} + \ldots b_{M-1}z^{M-1} \\
H(e^{j\omega}) &amp;= b_0 + b_1 e^{-j\omega} + b_2 e^{-2j\omega} + \ldots b_{M-1}e^{-j(M-1)\omega} \end{aligned}$$
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
$$\begin{aligned}
H(e^{j\omega}) &amp;= b_0 - b_{M-1}e^{-j(M-1)\omega} + b_1 - b_{M-1}e^{-(j(M-1) + 1)\omega)} + \ldots \\
&amp;= e^{-j(M-1)\omega} \left[ 2b_0 + b_1(e^{-j\omega}-e^{j\omega}) + \ldots b_{\frac{M}{2}} (e^{-j\frac{M}{2}\omega}-e^{j\frac{M}{2}\omega}) \right] \\
&amp;= -2je^{-j(M-1)\omega}) \left[jb_0 +b_1 \sin \omega + \ldots + b_{\frac{M}{2}} \sin{\frac{M}{2}\omega} \right] \\
&amp;= 
-2je^{-j(M-1)\omega})\left[jb_0 + \sum_{m=1}^{M/2} b_m \sin(m\omega)\right]
\end{aligned}$$
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.