# 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?

## 2 Answers

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
• Whoops my bad, in that case, only cos will change to sin. Editing my answer. – orchi_d May 6 at 19:46
• If I understand correctly, An antisymmetrics filter will have a linear phase but will have an offset of 90 degrees at 0 Hz? – Ben May 7 at 13:59
• No, at 0Hz you have H(0) = -2j.jb_0 = 2b_0 – orchi_d May 7 at 14:00
• There is a multiplicative -2j outside, since, \exp(0j) = 1. – orchi_d May 7 at 14:16

They can be either symmetric or anti symmetric to be linear phase. An intuitive view can be understood from the Fourier property that anything that is symmetric about the vertical axis in one domain will be real in the other domain. Consider that each impulse in one domain is a rotating complex phasor in the other domain, so for every pair of symmetric rotating phasors, their sum will be on the real axis always (and similarly for antisymmetric on the imaginary axis always).

The coefficients of the linear phase FIR filters represent the impulse response of the filter in time, which aren’t symmetric about t=0; rather they are symmetric about a delayed point in time (M/2 for M+1 coefficients when M+1 is odd, or at the midpoint between the two inner most samples for an even number of samples) such that the impulse response is causal. A fixed delay in time is a linear phase in frequency—- so we start with the real frequency response that has phase = 0 for all frequencies, delay that to be causal, and the result is the linear phase response given.