0
$\begingroup$

I am studying for an exam, and a problem that my professor gave produces a transfer function that looks like the following:

$$H(z)=\frac{z^2-j}{z^2-\frac{1}{4}j} = \frac{\left(z-e^{j\frac{\pi}{4}}\right)\left(z+e^{j\frac{\pi}{4}}\right)}{\left(z-\frac{1}{2}e^{j\frac{\pi}{4}}\right)\left(z+\frac{1}{2}e^{j\frac{\pi}{4}}\right)} $$

How am I able to determine:

  1. That this is a notch filter,
  2. At which frequencies the notches occur at?

His solution says that the notches occur at $\frac{\pi}{4}$ and $-\frac{3\pi }{4}$.

$\endgroup$
1
$\begingroup$

Remember that a notch filter must have transfer function zeros on the unit circle $z=e^{j\omega}$, i.e., there are certain frequencies that are completely attenuated. But note that zeros on the unit circle are not sufficient to characterize a filter as a notch filter (even though any frequency with perfect attenuation is called a notch frequency). E.g., most frequency-selective filters will have most, if not all, of their zeros on the unit circle (to get a good stop band attenuation). However, they are no notch filters. So for IIR filters you also have to consider the pole locations. And for notch filters, the pole angles are identical to or at least very close to the angles of the zeros. So for each zero on the unit circle you get a pole at the same or a similar angle, just with a smaller radius, because for causal and stable filters the poles need to be inside the unit circle.

That's why the filter in your example qualifies as a notch filter. All its zeros are on the unit circle and its poles are at the same angle as the zeros. The notch frequencies are simply the angles at which the zeros occur:

$$z_{01}=e^{j\omega_{01}}=e^{j\pi /4},\qquad z_{02}=e^{j\omega_{02}}=-e^{j\pi/4}=e^{-j3\pi /4}$$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.