Phase of a single root

For real $\omega$, a plot of real and imaginary parts of Z is the circle $(x,y) = (\cos\,\omega, \sin\,\omega)$.A smaller circle is .9Z. A right-shifted circle is 1+.9Z. Let Z₀ be a complex number, such as x₀+iy₀, or $Z{_0} = e^{i\omega_0}/\rho$,where $\rho$ and $\omega_0$ are fixed constants. Consider the complex Z plane for the two-term filter

$$\begin{eqnarray} B(Z) &=& 1 - {Z \over Z{_0}} \ B(Z(\omega )) &=& 1 -\rho e^{i(... ... &=& 1 -\rho \cos(\omega - \omega_0) -i\rho \sin(\omega -\omega_0)\end{eqnarray}$$ (34) (35) (36)

 

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Figure 18 Left, complex B plane for $\rho<1$. Right, for $\rho\gt 1$.

Real and imaginary parts of B are plotted in Figure 18. Arrows are at frequency $\omega$ intervals of $20^\circ$.Observe that for $\rho\gt 1$ the sequence of arrows has a sequence of angles that ranges over $360^\circ$,whereas for $\rho<1$ the sequence of arrows has a sequence of angles between $\pm 90^\circ$.Now let us replot equation (37) in a more conventional way, with $\omega$ as the horizontal axis. Whereas the phase is the angle of an arrow in Figure 18, in Figure 19 it is the arctangent of $\Im B/ \Re B$. Notice how different is the phase curve in Figure 19 for $\rho<1$ than for $\rho\gt 1$.

Real and imaginary parts of B are periodic functions of the frequency $\omega$,since $B(\omega ) = B(\omega +2\pi )$.We might be tempted to conclude that the phase would be periodic too. Figure 19 shows, however, that for a nonminimum-phase filter, as $\omega$ ranges from $-\pi$ to $\pi$,the phase $\phi$ increases by $2\pi$(because the circular path in Figure 18 surrounds the origin). To make Figure 19 I used the Fortran arctangent function that takes two arguments, x, and y. It returns an angle between $-\pi$ and $+\pi$.As I was plotting the nonminimum phase, the phase suddenly jumped discontinuously from a value near $\pi$ to $-\pi$, and I needed to add $2\pi$to keep the curve continuous. This is called ``phase unwinding.''

 

phase Figure 19 Left shows real and imaginary parts and phase angle of equation ((37)), for $\rho<1$. Right, for $\rho\gt 1$. Left is minimum-phase and right is nonminimum-phase.

You would use phase unwinding if you ever had to solve the following problem: given an earthquake at location (x,y), did it occur in country X? You would circumnavigate the country--compare the circle in Figure 18--and see if the phase angle from the earthquake to the country's boundary accumulated to (yes) or to $2\pi$ (no).

The word ``minimum" is used in ``minimum phase" because delaying a filter can always add more phase. For example, multiplying any polynomial by Z delays it and adds $\omega$to its phase.

For the minimum-phase filter, the group delay $d\phi / d\omega$ applied to Figure 19 is a periodic function of $\omega$.For the nonminimum-phase filter, group delay happens to be a monotonically increasing function of $\omega$.Since it is not an all-pass filter, the monotonicity is accidental.

Because group delay $d\phi / d\omega$ is the Fourier dual to instantaneous frequency $d\phi/dt$,we can now go back to Figure 5 and explain the discontinuous behavior of instantaneous frequency where the signal amplitude is near zero.