Spectral Phase

As for the phase of the spectrum, what do we expect? We have chosen the sinusoid phase to be zero. The window is causal and symmetric about its middle. Therefore, we expect a linear phase term with slope $-(M-1)/2$ 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 $-15$ 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, 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. 8.9b is not as clear as it could be. One could add logic to unwrap to interpret phase-jumps by $\pi$ (to within some numerical tolerance) as alternating in sign, since a phase jump of $+\pi$ is equivalent to a phase jump of $-\pi$. Doing this would give a straight line at the desired slope interrupted by temporary jumps of $\pi$ radians. In Fig. 8.9b, starting near frequency $0.3$, all phase jumps are by $+\pi$. If instead the 2nd, 4th, 6th, and so on were jumps by $-\pi$, a more intuitive phase plot would result.

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, 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: Spectral phase. a) Raw phase and its interpolation. b) Unwrapped phase and its interpolation.