Complex roots

We have seen how a simple two-term filter can destroy the zero frequency or the Nyquist frequency. When we try to destroy any other frequency, we run into a new difficulty--we will see complex-valued signals. Let Z₀ take the complex value $Z_0 = e^{i\omega_0}$,where $\omega_0$ is real. Further, choose $\omega_0 = \pi /2$ and as a result Z₀ = i. So the filter (1-Z/Z₀)=(1+iZ) has the complex coefficients (1,i), and its output is a complex-valued signal. Naturally this is annoying, because we usually prefer a real output signal.  

The way to avoid complex-valued signals is to handle negative frequency $-\omega_0$the same way we handle $\omega_0$.To do this we use a filter with two roots, one at $\omega_0$ and one at $-\omega_0$.The filter (1+iZ)(1-iZ)= 1+Z² has real-valued time-domain coefficients, namely, (1,0,1). The factor (1+iZ) vanishes when Z=i or $\omega = \pi/2$,and (1-iZ) vanishes at $\omega =-\pi/2$.Notice what happens when the filter (1,0,1) is convolved with the time series $b_t=( \cdots 1,0,-1,0,1,0,-1, \cdots)$:the output is zero at all times. This is because b_t is a sinusoid at the half-Nyquist frequency $\pi/2$,and the filter (1,0,1) has zeros at plus and minus half-Nyquist.

Let us work out the general case for a root anywhere in the complex plane. Let the root Z₀ be decomposed into its real and imaginary parts:

$$Z_0 \eq x + i y \eq \Re Z_0 + i \Im Z_0$$ (22)

and let the root be written in a polar form:

$$Z_0 \eq { e^{i\omega_0} \over \rho}$$ (23)

where $\omega_0$ and $\rho$ are constants that can be derived from the constants $\Re Z_0$ and $\Im Z_0$and vice versa. The conjugate root is $\overline{Z}_0 =e^{-i\omega_0}/\rho$.The combined filter is

$$\begin{eqnarray} \left( 1-{Z \over Z_0} \right) \ \left( 1-{Z \over \overline{Z}... ...0} } \ &=& 1 \ - \ 2 \rho \,\cos \omega_0 \, Z \ + \ \rho^2 Z^2\end{eqnarray}$$ (24) (25)

So the convolutional coefficients of this filter are the real values $(1, -2\rho \cos \omega_0, \rho^2)$.Taking $\rho=1$, the filter completely destroys energy at frequency $\omega_0$.Other values of $\rho$ near unity suppress nearby frequencies without completely destroying them.

Recall that to keep the filter response real, any root on the positive $\omega$-axis must have a twin on the negative $\omega$-axis. In the figures I show here, the negative axis is not plotted, so we must remember the twin. Figure 10 shows a discrete approximation to the second derivative.

 

ddt2
Figure 10 Approximation to the second difference operator (1,-2,1).

It is like (1-Z)², but since both its roots are in the same place at Z=1, I pushed them a little distance apart, one going to positive frequencies and one to negative.