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I know that we can design a linear phase FIR filter by choosing filter coefficients symmetric or antisymmetric, what is the intuitive idea behind that?

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The reason is Euler's formula, from which you get

$$\cos(\omega)=\frac12\big(e^{j\omega}+e^{-j\omega}\big)\tag{1}$$

and

$$j\sin(x)=\frac12\big(e^{j\omega}-e^{-j\omega}\big)\tag{2}$$

If you have symmetric or anti-symmetric coefficients, the corresponding frequency response can always be decomposed in purely real-valued cosine terms $(1)$ or purely imaginary sine components $(2)$, plus a linear phase term if the filter is not centered around $n=0$.

Take as a simple example a filter of length $N=5$ with filter coefficients

$$\mathbf{h}=\big[1, 2, 3, 2, 1]\tag{3}$$

starting at index $n=0$. The corresponding frequency response is

$$\begin{align}H(e^{j\omega})&=1+2e^{-j\omega}+3e^{-2j\omega}+2e^{-3j\omega}+e^{-4j\omega}\\&=e^{-2j\omega}\big[e^{2j\omega}+2e^{j\omega}+3+2e^{-j\omega}+e^{-2j\omega}\big]\\&=2e^{-2j\omega}\big[\cos(2\omega)+2\cos(\omega)+3\big]\tag{4}\end{align}$$

Since the last term in brackets is purely real-valued, $H(e^{j\omega})$ has a linear phase response $\phi(\omega)=-2\omega$.

For even-length filters you get an additional delay of half a sample, and for anti-symmetric filters there's an additional phase shift of $\pi/2$ due to $j$ on the left side of $(2)$.

Also take a look at this answer discussing the four types of linear phase FIR filters.

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  • $\begingroup$ Thank you very much. $\endgroup$ – Ece Su Ildiz Jun 10 at 10:55

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