0
$\begingroup$

 

Q1:  Consider a discrete-time band-pass filter (BPF) that uses only two poles as

$$ H(z) = G \frac{1}{\big(1-r \, e^{j \omega_0} z^{-1}\big)\big(1-r \, e^{-j \omega_0} z^{-1}\big)} $$

where this filter has a center frequency of $\omega_0$. If this BPF was designed to have a very narrow bandwidth by selecting $r$ to be very close to unity, show that the 3dB bandwidth of the BPF filter is approximately given by

$$ BW_{3dB}=2(1-r) $$

(Hint:  Use a geometrical approach on the unit circle.)

 

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ If this is homework, please add the home work tag $\endgroup$ – Hilmar Jul 29 '18 at 19:18
  • 1
    $\begingroup$ this is not a band-pass filter (BPF), but is a low-pass filter (LPF). however if $$ 0 < 1-r \ll 1 $$ then it's an LPF with a resonant peak at frequency $\omega_0$. $\endgroup$ – robert bristow-johnson Jul 30 '18 at 6:47
0
$\begingroup$

Here is (I think) what you are supposed to be doing

  1. Draw a diagram of the poles and the unit circle and zoom in on the area around the resonance frequency
  2. Estimate the relative gain as the inverse of the distance between the pole and the unit circle (assuming that conjugate pole is too far away to change the gain much over a narrow frequency range)
  3. Pencil in the -3dB points by, again, using distance as a representative for gain
  4. Ignore the curvature of the circle (since r is close to 1)
  5. Do some trigonometry to calculate the distance between the two -3dB points
  6. Convert this distance to a frequency difference.
| 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.