Spectral Phase Next  |  Prev  |  Up  |  Top  |  Index  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Spectral Phase

As for the phase of the spectrum, what do we expect? We have chosen the sinusoid phase offset to be zero. The window is causal and symmetric about its middle. Therefore, we expect a linear phase term with slope $ -(M-1)/2=-15$ samples (as discussed in connection with the shift theorem in §7.4.4). Also, the window transform has sidelobes which cause a phase of $ \pi $ radians to switch in and out. Thus, we expect to see samples of a straight line (with slope $ -15$ samples) across the main lobe of the window transform, together with a switching offset by $ \pi $ in every other sidelobe away from the main lobe, starting with the immediately adjacent sidelobes.

In Fig.8.9(a), we can see the negatively sloped line across the main lobe of the window transform, but the sidelobes are hard to follow. Even the unwrapped phase in Fig.8.9(b) is not as clear as it could be. This is because a phase jump of $ \pi $ radians and $ -\pi$ radians are equally valid, as is any odd multiple of $ \pi $ radians. Figure 8.9(c) shows what could be considered the ``canonical'' unrwapped phase for this example: We see a linear phase segment across the main lobe as before, and outside the main lobe, we have a continuation of that linear phase across all of the positive sidelobes, and only a $ \pi $-radian deviation from that linear phase across the negative sidelobes. To obtain unwrapped phase of this typoe, the unwrap function needs to alternate the sign of successive phase-jumps by $ \pi $ radians (to within some numerical tolerance). In summary, Fig.8.9(c) shows a straight linear phase at the desired slope interrupted by temporary jumps of $ \pi $ radians. In Fig.8.9(b), on the other hand, starting near frequency $ 0.3$, all phase jumps are by $ +\pi$; the more intuitive phase plot results if the 2nd, 4th, 6th, and so on are replaced by phase-jumps of $ -\pi$ radians.

To convert the expected phase slope from $ -15$ ``radians per (rad/sec)'' to ``radians per cycle-per-sample,'' we need to multiply by ``radians per cycle,'' or $ 2\pi $. Thus, in Fig.8.9(c), we expect a slope of $ -94.2$ radians per unit normalized frequency, or $ -9.42$ radians per $ 0.1$ cycles-per-sample, and this looks about right, judging from the plot.

Figure 8.9: Spectral phase and two different phase unwrappings.

Raw spectral phase and its interpolation

Unwrapped spectral phase and its interpolation

Canonically unwrapped spectral phase and its interpolation

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

``Mathematics of the Discrete Fourier Transform (DFT), with Music and Audio Applications'', by Julius O. Smith III, W3K Publishing, 2003, ISBN 0-9745607-0-7.
Copyright © 2007-02-02 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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