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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?

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2 Answers 2

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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.

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  • $\begingroup$ if the coefficients are antisymmetric, wouldnt $ b_m = -b_{M-1-m} $ $\endgroup$
    – Ben
    Commented May 6, 2021 at 19:45
  • $\begingroup$ Whoops my bad, in that case, only cos will change to sin. Editing my answer. $\endgroup$
    – orchi_d
    Commented May 6, 2021 at 19:46
  • $\begingroup$ If I understand correctly, An antisymmetrics filter will have a linear phase but will have an offset of 90 degrees at 0 Hz? $\endgroup$
    – Ben
    Commented May 7, 2021 at 13:59
  • $\begingroup$ No, at 0Hz you have H(0) = -2j.jb_0 = 2b_0 $\endgroup$
    – orchi_d
    Commented May 7, 2021 at 14:00
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    $\begingroup$ There is a multiplicative -2j outside, since, \exp(0j) = 1. $\endgroup$
    – orchi_d
    Commented May 7, 2021 at 14:16
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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.

Below is an animation I created that demonstrates this for the simple case of a filter with coefficient [1, 2, 1]. In this case specifically the filter is non-causal with an inpulse response of [1,2,1] with the first impulse appearing at $n=-1$. Each impulse in time is a spinning phasor in the frequency domain: The impulse at $n=-1$ is a phasor of magnitude $1$ and phase extending from $0$ to $2\pi$ as the frequency goes from $DC$ to the sampling rate- so on the complex plane rotates clockwise as we sweep the frequency. The impulse at $n=0$ has a magnitude of $2$ and doesn't rotate, and the impulse at $n=1$ has a magnitude of $1$ and rotates exactly opposite that for the impulse at $n=-1$. The result is a completely real response in the frequency domain due to the complex conjugate rotations. We make the filter causal by a $z^{-1}$ delay, which ends up being the phase response of the filter (linear phase as given by $z^{-1}$ alone.

one two one

If the filter was not symmetric, the result in the frequency domain cannot be real, and the phase versus frequency no longer linear. I changed the third coefficient to be 0.5 with the comparative result shown below:

one two half

The above two demonstrations are non-causal "zero-phase" filters with the center tap at $n=0$. To make them causal we simply add delay, which is linear phase: The DFT for a unit delay is $e^{-j2\pi n/N}$, so for the causal versions of the above filters the phase would go linearly from $0$ to $-2\pi$ for the frequency axis shown, in addition the phase responses shown. So the upper graphic is linear phase and the lower graphic is non-linear phase.

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