# [help] calculating the cutoff frequency

Discussion in 'Homework Help' started by cupcake, Apr 18, 2011.

1. ### cupcake Thread Starter Member

Sep 20, 2010
73
0

Ha(s).Ha(-s) where s=jΩ
I got (Ω^4)/1+Ω^6 hence this not Butterworth filter, and I managed to find all the poles, where all the poles lie on the left hand side so this filter is stable..

My problem is, I'm not quite sure how to calculate the cutoff frequency, for Butterworth filter, the cutoff frequency should be Ωc=1. However, this filter is not a Butterworth filter, so how do I calculate the cutoff frequency?

Kindly need assistance here.. thanks

Last edited: Apr 18, 2011
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
You can re-cast Ha(s) as

$H_a(s)=\frac{s^2}{(s+1)(s^2+s+1)}=\frac{s^2}{(s^3+2s^2+2s+1)}$

which looks like a band-pass function. One can find the maximum pass-band gain ....

$H_a(j\omega)=\frac{-\omega^2}{((1-2\omega^2)+j\omega(2-\omega^2))}$

From which the gain magnitude can be found as

$G=|H_a(j\omega)|=\frac{\omega^2}{\sqr{1+\omega^6}}$

Differentiating G with respect to ω leads to

$\frac{\delta G}{\delta \omega}=\frac{2\omega}{\sqr{1+\omega^6}}-\frac{3\omega^7}{(1+\omega^6)^{\frac{3}{2}}}$

Equating this to zero to find the maximum gives a maximum at

$\omega=2^{\frac{1}{6}}$

or

$\omega=1.122462$

$G_{max}=\frac{(1.122462)^2}{\sqr{1+(1.122462)^6}}$

or

$G_{max}=0.727416 \ <=>\ -2.764dB$

The -3dB points would occur when G/Gmax=0.707

One then has to solve for the value of ω when G=0.707Gmax=0.51436

Or when

$G=\frac{\omega^2}{\sqr{1+\omega^6}}=0.51436$

This can be solved to give positive roots at

ω=0.7463381 rad/sec or 0.1188Hz - the lower 3dB cut-off frequency

and

ω=1.9254192 rad/sec or 0.3064Hz - the upper 3dB cut-off frequency

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3. ### cupcake Thread Starter Member

Sep 20, 2010
73
0
Thank you very much for your help.. so, I have to find the gain first, right?
but, this afternoon my lecturer posted the answer...and here it is.. it looks quiet simple

so, omega = 1 and omega = 1.2720
I think we should worked out from H(s).H(-s) where the analog filter is stable. But, I don't quite understand where 1/2 in this equation Ω^4/(1+Ω^6) = 1/2 comes from.. maybe you could explain to me.. thanks

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
As I showed in my last post the gain magnitude for the function is given by

$G=\frac{\omega^2}{\sqr{(1+\omega^6)}}$

What your lecturer has given is based on the 3dB points occurring at a value of ω such that

$G=\frac{\omega^2}{\sqr{(1+\omega^6)}}=\frac{1}{ \sq 2}$

or

$G^2=\frac{\omega^4}{(1+\omega^6)}=\frac{1}{2}$

Incidentally this assumes a mid-band (pass band) gain of 1. In this case an incorrect assumption.

Presumably your lecturer has made a simple mistake. Dare you point it out!

The peak mid-band gain is less than unity and (as shown in my last post) is approximately 0.7274 at ω=1.1225 rad/sec.

5. ### cupcake Thread Starter Member

Sep 20, 2010
73
0
thank you very much...
I also think that he made the assumption for unity gain...as he maybe thought that this is a butterworth filter, which is actually not.

Thanks for your clear explanations earlier.