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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) =
\left\{\begin{array}{ll}
\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:

\begin{displaymath}
\begin{array}{rclrcl}
P(\omega) &\isdef & -\displaystyle\fra...
...2} T,
\qquad H_{\hbox{zp}}(e^{j\omega T})\geq0\\
\end{array}\end{displaymath}



Subsections

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Previous: Odd Impulse Reponses
Next: Simple Linear-Phase Filter Examples

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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