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Linear Phase Terms

The reason $ e^{-j \omega_k \Delta}$ is called a linear phase term is that its phase is a linear function of frequency:

$\displaystyle \angle e^{-j \omega_k \Delta} = - \Delta \cdot \omega_k
$

Thus, the slope of the phase, viewed as a linear function of radian-frequency $ \omega_k$, is $ -\Delta$. In general, the time delay in samples equals minus the slope of the linear phase term. If we express the original spectrum in polar form as

$\displaystyle X(k) = G(k) e^{j\Theta(k)},
$

where $ G$ and $ \Theta$ are the magnitude and phase of $ X$, respectively (both real), we can see that a linear phase term only modifies the spectral phase $ \Theta(k)$:

$\displaystyle e^{-j \omega_k \Delta} X(k) \isdef
e^{-j \omega_k \Delta} G(k) e^{j\Theta(k)}
= G(k) e^{j[\Theta(k)-\omega_k\Delta]}
$

where $ \omega_k\isdeftext 2\pi k/N$. A positive time delay (waveform shift to the right) adds a negatively sloped linear phase to the original spectral phase. A negative time delay (waveform shift to the left) adds a positively sloped linear phase to the original spectral phase. If we seem to be belaboring this relationship, it is because it is one of the most useful in practice.


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[How to cite this work]  [Order a printed hardcopy]

``Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition'', by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8.
Copyright © 2008-05-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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