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Symmetric Linear-Phase Filters

As stated at the beginning of this chapter, the impulse response of every causal, linear-phase, FIR filter is symmetric:

$\displaystyle h(n) = h(N-1-n), \quad n=0,1,2,\ldots,N-1.

Assume that $ N$ is odd. Then the filter

$\displaystyle h_{\hbox{\tiny zp}}(n) = h\left(n+\frac{N-1}{2}\right), \quad n=-\frac{N-1}{2},\,\ldots\,,\frac{N-1}{2}

is a zero-phase filter. Thus, every odd-length linear-phase filter can be expressed as a delay of some zero-phase filter,

$\displaystyle h(n) = h_{\hbox{\tiny zp}}\left(n-\frac{N-1}{2}\right), \quad n=0,1,2,\ldots, N-1.

By the shift theorem for z transforms (§6.3), the transfer function of a linear-phase filter is

$\displaystyle H(z) = z^{-\frac{N-1}{2}}H_{\hbox{zp}}(z)

and the frequency response is

$\displaystyle H(e^{j\omega T}) = e^{-j\omega \frac{N-1}{2}T}H_{\hbox{zp}}(e^{j\omega T})

which is a linear phase term times $ H_{\hbox{zp}}(e^{j\omega T})$ which is real. Since $ H_{\hbox{zp}}(e^{j\omega T})$ can go negative, the phase response is

$\displaystyle \Theta(\omega) =
\displaystyle-\frac{N-1}{2}\omega T, & H_{\hbox{zp}}(e^{j\omega T})\geq 0 \\ [5pt]
\displaystyle-\frac{N-1}{2}\omega T + \pi, & H_{\hbox{zp}}(e^{j\omega T})<0 \\
\end{array} \right..

For frequencies $ \omega$ at which $ H_{\hbox{zp}}(e^{j\omega T})$ is nonnegative, the phase delay and group delay of a linear-phase filter are simply half its length:

P(\omega) &\isdef & -\displaystyle\frac{\Theta(\omega)}{\omega} &=& \displaystyle\frac{N-1}{2} T,
\qquad H_{\hbox{zp}}(e^{j\omega T})\geq0\\ [10pt]
D(\omega) &\isdef & -\displaystyle\frac{\partial}{\partial\omega}\Theta(\omega) &=& \displaystyle\frac{N-1}{2} T,
\qquad H_{\hbox{zp}}(e^{j\omega T})\geq0\\

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition).
Copyright © 2015-04-22 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University