ECE538 Week 4: Frequency Response of LTI Systems

The purpose of this week was to understand how pole/zero placement of the impulse response can affect the frequency response of an LTI system.

All-Pass Filter

Basic Analysis

All-pass filters have the below impulse response:

$$H(e^{j\omega})=G\frac{e^{j\omega}-\frac{1}{p^*}}{e^{j\omega}-p}$$

Proof:

$$G\frac{e^{j\omega}-\frac{1}{p^*}}{e^{j\omega}-p}=G\frac{-\frac{e^{j\omega}}{p^*}(e^{-j\omega}-p^*)}{e^{j\omega}-p}=G\frac{-\frac{e^{j\omega}}{p^*}(e^{j\omega}-p)^*}{e^{j\omega}-p}$$

Therefore

$$|H(e^{j\omega})|=|G|\frac{|-\frac{e^{j\omega}}{p^*}||(e^{j\omega}-p)^*|}{|e^{j\omega}-p|}=\frac{|G|}{|p|}, \forall \omega$$

Which is an all-pass filter. Notice that $|G|=|p|$ normalizes the all-pass.

Connection to Autocorrelation

Recall that

$$DTFT\{r_{hh}[n]\}=|H(e^{j\omega})|^2=(\frac{|G|}{|p|})^2$$

Therefore

$$h[n]*h^*[-n]=(\frac{|G|}{|p|})^2\delta[n]$$

Time-Domain

$$H(z)=p\frac{z-\frac{1}{p^*}}{z-p}=p\frac{z}{z-p}-\frac{p}{p^*}z^{-1}\frac{z}{z-p}$$

Thus

$$h[n]=pp^nu[n]-e^{j2\angle p}p^{n-1}u[n-1]$$

A more common expression can be obtained with further manipulation:

$$=p\delta[n]+pp^nu[n-1]-\frac{e^{j2\angle p}}{p}p^nu[n-1]=p\delta[n]+(p-\frac{e^{j2\angle p}}{p})p^nu[n-1]=p\delta[n]+(p-\frac{e^{j2\angle p}}{p})p^{n}u[n] - (p-\frac{e^{j2\angle p}}{p})\delta[n]=\frac{e^{j2\angle p}}{p}\delta[n]+(\frac{p^2-e^{j2\angle p}}{p})p^{n}u[n]$$

Obtaining

$$h[n]=\frac{1}{p}(e^{j2\angle p}\delta[n]+(p^2-e^{j2\angle p})p^{n}u[n])$$

Notice that if $p \in \mathbb{R}$

$$\boxed{h[n]=\frac{1}{p}(\delta[n]+(p^2-1)p^{n}u[n])}$$

Notch Filter

Nothced filters have the following impulse response:

$$H(e^{j\omega})=G\frac{\Pi_{k=0}^{M}(e^{j\omega}-e^{j\omega_{k}})}{\Pi_{k=0}^{M}(e^{j\omega}-re^{j\omega_{k}})}$$

The key is that the zeros lie on the unit circle at the frequencies that are notched out. The poles create a contrast between the surrounding frequencies and the notched frequencies. The closer the poles are to the unit circle, the narrower the notching is.

Notch Filter and All-Pass Filter Connection

Two All-Pass filters in parallel can create a notch filter. Observe the following:

$$H_1(z)=\frac{z-\frac{1}{p^*}}{z-p}$$

$$H_2(z)=\frac{z+\frac{1}{p^*}}{z+p}$$

Notice that the above filters have passband gain of $\frac{1}{|p|}$

$$H_1(z)+H_2(z)=\frac{z-\frac{1}{p^*}}{z-p}+\frac{z+\frac{1}{p^*}}{z+p}=\frac{(z-\frac{1}{p^*})(z+p)+(z+\frac{1}{p^*})(z-p)}{(z-p)(z+p)}$$

$$\frac{(z-\frac{1}{p^*})(z+p)+(z+\frac{1}{p^*})(z-p)}{(z-p)(z+p)}=\frac{z^2+z(p-\frac{1}{p^*})-\frac{p}{p^*}+z^2-z(p-\frac{1}{p^*})-\frac{p}{p^*}}{(z-p)(z+p)}=2\frac{z^2-\frac{p}{p^*}}{(z-p)(z+p)}=2\frac{(z+\sqrt{\frac{p}{p^*}})(z-\sqrt{\frac{p}{p^*}})}{(z-p)(z+p)}$$

Finally, we obtain

$$H_1(z)+H_2(z)=2\frac{(z-e^{j(\angle p+\pi)})(z-e^{j\angle p})}{(z-p)(z+p)}$$

Which is a notch filter with notching frequencies $\omega = \{\angle p, \angle p + \pi\}$

Pole-Zero Cancellation: Difference Equations

Consider the following system:

$$y[n]=e^{j2\pi\frac{l}{n}}y[n-1]+x[n]-x[n-N]$$

With $l < N$

Basic Analysis

This system fits the form of IIR, but in fact is FIR.

$$H(z)=\frac{1-z^{-N}}{1-e^{j2\pi\frac{l}{n}}z^{-1}}=\frac{z^{N}-1}{z^{N}-e^{j2\pi\frac{l}{n}}z^{N-1}}=\frac{\Pi_{k=0}^{N-1}(z-e^{j\frac{2\pi k}{N}})}{z^{N-1}(z-e^{j2\pi\frac{l}{n}})}=\frac{\Pi_{k=0}^{l-1}(z-e^{j\frac{2\pi k}{N}})\Pi_{k=l+1}^{N-1}(z-e^{j\frac{2\pi k}{N}})}{z^{N-1}}$$

As can be seen, the impulse response simply consists of zeros, without any (nonzero) poles, because a pole/zero cancellation occured.

Frequency Response

We revert to the stage before the pole-zero cancellation, to make our work easier:

$$H(e^{j\omega})=\frac{e^{j\omega N}-1}{e^{j\omega (N-1)}(e^{j\omega}-e^{j2\pi\frac{l}{n}})}$$

And make use of the “half-angle trick”:

$$ H(e^{j\omega})=\frac{(e^{j\omega \frac{N}{2}}-e^{-j\omega \frac{N}{2}})e^{j\omega \frac{N}{2}}}{(e^{j\frac{\omega}{2}}e^{-j\pi\frac{l}{n}}-e^{-j\frac{\omega}{2}}e^{j\pi\frac{l}{n}})e^{j\frac{\omega}{2}}e^{j\pi\frac{l}{n}}e^{j\omega (N-1)}} =\frac{(e^{j\omega \frac{N}{2}}-e^{-j\omega \frac{N}{2}})e^{j\omega \frac{N}{2}}}{(e^{j(\frac{\omega}{2}-\pi\frac{l}{n})}-e^{-j(\frac{\omega}{2}-\pi\frac{l}{n})})e^{j(\frac{\omega}{2}+\pi\frac{l}{n}+\omega (N-1))}} = \frac{(e^{j\omega \frac{N}{2}}-e^{-j\omega \frac{N}{2}})e^{j(\omega \frac{1-N}{2}-\pi\frac{l}{n})}}{(e^{j(\frac{\omega}{2}-\pi\frac{l}{n})}-e^{-j(\frac{\omega}{2}-\pi\frac{l}{n})})} $$