2nd order resonance filter amplitude response(z-parameters)

  • #1
Jimmy Johnson
27
0

Homework Statement


A resonance filter is a specific second-order digital system designed to attenuate all frequencies except at and
around a given center frequency. The transfer function of a resonance filter is:

H(z)=(1 - r)(1 - rz-2) / (1 - 2rcos(ωcT)z-1+r2z-2)

where ωc = 2πfc, where fc is the center frequency (in Hz), and r is a parameter that controls the resonance
bandwidth.

Show that, for any given resonance frequency ωc and any value of the parameter r, the amplitude response
of the filter equals unity at ω = ωc, i.e. |H(ωc)| = 1.

Homework Equations



From the book Digital signal processing: applications and concepts by bernard mulgrew:

H(ω) = F[output] / F[input] = Y(ω) / X(ω) just as H(z) = Y(z)/X(z)

The Attempt at a Solution



Not entirely sure where to begin with this one. In most examples in the book going to H(ω) is done from the H(s) domain. Would I need to go change the transfer function from the z to s domain?

Also is it possible to consider it with the T = 0 inside the cosine to let the cosine value equal 1? making the equation easier to solve.

The steps given in the book to calculate output response of a digital sequence

(i) take the z-transform of the input sequence
(ii) multiply by the transfer function
(iii) take the inverse z-transform

Would I need to do these steps before converting to ω or suffice to do so after?

So far what I think could or should be included in the method is

it is not a proper fraction because highest numerator power equals highest denominator power, so divide both sides by z

use partial fractions to separate the functions (I think this is more so for difference eqn than first question)

The entire question is in the attached pdf and any hints about how to begin and if it is possible to get the answer working in z domain, making the cosine = 0 etc or if I'm on the right track at all would be much appreciated.
 

Attachments

  • q2.pdf
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Answers and Replies

  • #2
milesyoung
818
67
Recall that the frequency response of a discrete-time system is given by ##H(e^{j\omega T})##, which relates to the starred transform, if you've taken that approach in your course.

I'd suggest you convert the cosine term to its complex representation, and then evaluate ##H(e^{j\omega_c T})##.
 
  • #3
Jimmy Johnson
27
0
The complex representation of the cosine term would be (ecT+e-jωcT)/2 ? which in this case would leave
r(ecT+e-jωcT)

Do the z's of the transfer function simply get replaced by ejωT? As in would the numerator become
(1-r)(1 - re-2jωT)
 
  • #4
milesyoung
818
67
The complex representation of the cosine term would be (ejωcT+e-jωcT)/2 ? which in this case would leave
r(ejωcT+e-jωcT)
Yes (assuming you mean just the factor), don't forget the sign, though.
Do the z's of the transfer function simply get replaced by ejωT? As in would the numerator become
(1-r)(1 - re-2jωT)
Yes, you substitute in ##z = e^{j\omega T}##.
 
  • #5
Jimmy Johnson
27
0
Ok so my (assuming ωc is replaced with ω so I don't keep having to subscript it, not sure of how to use latex) H(ω) becomes

H(ω) = (1 - r)(1 - re-j2ωT) / (1 - r(ejωT+e-jωT)+r2e-j2ωT)

Show that for any given r or ω value amplitude response = unity, |H(ω)| = 1.

Do I use arbitrary values? I would think that its more likely done via algebra, partial fractions perhaps? or would I need to use the starred transform?
 
  • #6
milesyoung
818
67
When you've done the substitution, and you just have an expression with complex exponentials, try simplifying it a bit. It'll reduce nicely.

Edit: Your denominator is missing a factor of ##z^{-1}##.
 
Last edited:
  • #7
Jimmy Johnson
27
0
I've been trying but I'm just not seeing the simplification! I've tried to take a factor of (1 - r) and (1 - re-j2ωT) on the denominator but can't see how those would factor, tried expanding top and bottom then dividing the terms separately which also didn't work out nicely.. I'm guessing that this should fall pretty nicely as you say and I'm either going horribly wrong somewhere or I've just been at it so long nothing is making sense! ha

Any pointers or should it really just be jumping out at me?
 
  • #8
milesyoung
818
67
Did you remember to include the factor of ##z^{-1}## in the denominator, as I wrote in my edit in post #6?

If so, after you expand both the numerator and denominator, stop there and have a look at them. Don't they seem similar? Include your work here, if you want some help with it.
 
  • #9
Jimmy Johnson
27
0
ahhh I had actually missed that. That would make the denominator become (1 - re-j2ωT+ r2e-j2ωT).

The numerator is (1 -re-j2ωT - r +re-j2ωT)

They are pretty similar but with the stand alone r on top seems to make it a bit tricky. Is it to do with an improper fraction because the top and bottom have the same highest power or am I over thinking it?

Only noticed your reply so and the missing z-1 in the denominator so I'll be trying to simplify in the meantime
 
  • #10
milesyoung
818
67
That would make the denominator become (1 - re-j2ωT+ r2e-j2ωT)
No, that's not quite right. The cosine term is:
$$
-2r\cos(\omega T)e^{-j\omega T} = -r(e^{j\omega T} + e^{-j\omega T})e^{-j\omega T}
$$
What is it after expanding?

The numerator is (1 -re-j2ωT - r +re-j2ωT)
There should be a ##r^2## factor somewhere.
 
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  • #11
Jimmy Johnson
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sorry typo on the numerator,

numerator = (1 - re-j2ωT - r - r2e-j2ωT)

as for the expansion of

−2rcos(ωT)e−jωT=−r(ejωT+e−jωT)e−jωT

Well I see the mistake I made, I did the expansion and the e terms e-jωT*ejωT equal to 1 but forgot to keep it as 1 and dropped it instead of multiplying the bracket (1 + e-j2ωT) by -r which leaves the denominator equal to the numerator when expanded which is why I was missing the separate r term! Thanks so much for the help!
 
  • #12
milesyoung
818
67
Great, you're welcome :)
 
  • #13
Jimmy Johnson
27
0
Out of curiosity would you happen to know about the difference equation? More specifically what impact the cosine would have on the difference equation as a coefficient?

I know H(z) = Y(z)/X(z), and I separate that to Y(z)*denominator = X(z)*numerator, then take the inverse z transform, would my difference eqn become

y[n] - 2rcos(ωT)[n-1] + r2[n-2] = x[n] - r[n-2] - r[n] -r2[n-2]

or do I need to account for r2 and the cosine when performing the inverse z transform steps?

I was thinking that looked ok but now I'm thinking that since its a follow up question I should use the expansions I got from the part (a) which we just done. Is it possible to do the difference eqn from the H(ω) function?
 
  • #14
milesyoung
818
67
I know H(z) = Y(z)/X(z), and I separate that to Y(z)*denominator = X(z)*numerator, then take the inverse z transform, would my difference eqn become

y[n] - 2rcos(ωT)[n-1] + r2[n-2] = x[n] - r[n-2] - r[n] -r2[n-2]
I think you have some typo's there. Should be:
$$
y[n] - 2r\cos(\omega_c T)y[n - 1] + r^2y[n - 2] = x[n] - rx[n - 2] - rx[n] + r^2x[n - 2]
$$
But that's all there is to it. It's completely analogous to what you normally do with the Laplace transform for continuous-time systems.
 
  • #15
Jimmy Johnson
27
0
Thanks again man! You're awesome! appreciate it! And a lot of typos I forgot pretty much all the x's and y's haha
 

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