# Notch filter

## Homework Statement

Question is when

F(s) = ## \frac {(s)^2+2*ζa*Wn*s+(Wn)^2)} {(s)^2+2*ζb*Wn*s+(Wn)^2)} \ ##

prove that ## \ {|F(iw)|}^2 \ ## = 1 - ## \frac {4(ζb^2-ζa^2) \tilde w} {(1-\tilde w)^2+4ζb^2 \tilde w} \ ##

when ## \tilde w \ ## = ## (\frac {w} {Wn} )^2 \ ##

## Homework Equations

F(s) is the equation of a notch filter. |F(iw)| is the magnitude of its frequency response.

## The Attempt at a Solution

I have replaced s by iw in the F(s) equation but ended up with something weird which doesn't make sense. Can someone please help!

BvU
Homework Helper
I think you strayed somewhere. Show your calculations.

I think you strayed somewhere. Show your calculations.

F(iw) = ## \frac {(iw)^2+2ζaWn*wi+Wn^2} {(iw)^2+2ζbWn*wi+Wn^2} \ ##

= ## \frac {-w^2+2ζaWn*wi+Wn^2} {-w^2+2ζbWn*wi+Wn^2} \ ##

= ## \frac {Wn^2-w^2+2ζaWn*wi} {Wn^2-w^2+2ζbWn*wi} \ ## x ## \frac {Wn^2-w^2-2ζbWn*wi} {Wn^2-w^2-2ζbWn*wi} \ ##

= ## \frac {Wn^4-2Wn^2w^2+w^4-2ζbWn^2wi+2ζbWn*w^2i+2ζaWn^3wi-2ζaWn*w^3i+2ζaζbWn^2w^2} {Wn^4-2Wn^2w^2+w^4-2ζbWn^2wi+2ζbWn*w^2i+2ζbWn^3wi-2ζbWn*w^3i+2ζb^2Wn^2w^2} \ ##

After this I stopped doing, because it makes no sense....Can you help now?!

BvU
Homework Helper
Yep. You want ##\left |F(j\omega)\right |^2.\ ## So multiply ##F(j\omega)## with its complex conjugate !

Chestermiller
Mentor
In a complicated system like this, you need to look for tricks.

Trick #1:
$$\frac{A+Bas}{A+Bbs}=1-\frac{B(b-a)s}{A+Bbs}$$
Trick #2:
$$(A+B)(A-B)=A^2-B^2$$

Chet

Yep. You want ##\left |F(j\omega)\right |^2.\ ## So multiply ##F(j\omega)## with its complex conjugate !

If you look properly you will see I have already done that....the 3rd step incase you still can't see it...

In a complicated system like this, you need to look for tricks.

Trick #1:
$$\frac{A+Bas}{A+Bbs}=1-\frac{B(b-a)s}{A+Bbs}$$
Trick #2:
$$(A+B)(A-B)=A^2-B^2$$

Chet

I get how I am supposed to use the trick 1....but its the trick 2 I can't really figure out how to use wisely....I tried to use trick 2 to replace Wn^2-w^2 but doesn't fit there....what do you use trick 2 for?

BvU
Homework Helper
If you look properly you will see I have already done that....the 3rd step incase you still can't see it...
I see you multiplying with something that has a ##\zeta_b## in the numerator.

##
\frac {Wn^2-w^2-2ζbWn*wi} {Wn^2-w^2-2ζbWn*wi} \
##

That's not the complex conjugate, that is 1

By the way, TeX x is \times : ##\times##

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Chestermiller
Mentor
I get how I am supposed to use the trick 1....but its the trick 2 I can't really figure out how to use wisely....I tried to use trick 2 to replace Wn^2-w^2 but doesn't fit there....what do you use trick 2 for?
It's applied on the denominator.
$$(Wn^2-w^2+2ζbWn*wi)(Wn^2-w^2-2ζbWn*wi)=?$$
Chet

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I see you multiplying with something that has a ##\zeta_b## in the numerator.

##
\frac {Wn^2-w^2-2ζbWn*wi} {Wn^2-w^2-2ζbWn*wi} \
##

That's not the complex conjugate, that is 1

By the way, TeX x is \times : ##\times##

That's because the complex term "i" is with it? don't you see I changed the sign infront of the "i" term in the numerator? As far as I know for any positive complex number x+iy, its conjugate is x-yi, so just changing the sign. Don't know what type of complex number did you study

BvU
Homework Helper
That's because the complex term "i" is with it? don't you see I changed the sign infront of the "i" term in the numerator? As far as I know for any positive complex number x+iy, its conjugate is x-yi, so just changing the sign. Don't know what type of complex number did you study
Yes but you also changed ##\zeta_a## to ##\zeta_b##, effectively writing 1 instead of ##F^*(j\omega)##.

I studied the usual complex numbers. Long ago.
In those days ##|{a+bi\over c+di}|^2 = {a+bi\over c+di}\times {a-bi\over c-di}##, and not ##{a+bi\over c+di}\times {a-di\over c-di}##,

Chestermiller
Mentor
Yes but you also changed ##\zeta_a## to ##\zeta_b##, effectively writing 1 instead of ##F^*(j\omega)##.

I studied the usual complex numbers. Long ago.
In those days ##|{a+bi\over c+di}|^2 = {a+bi\over c+di}\times {a-bi\over c-di}##, and not ##{a+bi\over c+di}\times {a-di\over c-di}##,
BvU,

I had the impression that he was not trying to multiply by the complex conjugate yet. I think he was just trying to multiply numerator and denominator by the complex conjugate of the denominator to get a real number in the denominator.

Chet

BvU
Homework Helper
Dear JI, please don't worry. I've been there, done the same thing.

Dear Chet, in post #6 JI claims otherwise. And: The system isn't all that complicated. ##F^* \times F## brings us a long way towards the known ultimate expression.

Chestermiller
Mentor
Dear JI, please don't worry. I've been there, done the same thing.

Dear Chet, in post #6 JI claims otherwise. And: The system isn't all that complicated. ##F^* \times F## brings us a long way towards the known ultimate expression.
He may have claimed otherwise, but that's not what it looks like to me.

Chet

BvU
Homework Helper
JI's move now. I hope we haven't lost him/her.

Chestermiller
Mentor
JI's move now. I hope we haven't lost him/her.
By the way, I like your approach quite a bit.

Chet

BvU
Homework Helper
Hey, this is PF, not a Mutual admiration society ! ;)

Chestermiller
It's applied on the denominator.
$$(Wn^2-w^2+2ζbWn*wi)(Wn^2-w^2-2ζbWn*wi)=?$$
Chet

hmmm so that makes my denominator have A^2-B^2, but it should really be A+Bbs...also how do you get the 1 in the numerator? I mean in the denominator i multiply with denominators complex conjuguate? what about numerator? is it gonna be multiplied by numerator complex conjugate or denominator complex conjuguate?

Hey, this is PF, not a Mutual admiration society ! ;)

Hey man, do you know anything about plotting bode magnitudes or step response plots?

Chestermiller
Mentor
hmmm so that makes my denominator have A^2-B^2, but it should really be A+Bbs...also how do you get the 1 in the numerator? I mean in the denominator i multiply with denominators complex conjuguate? what about numerator? is it gonna be multiplied by numerator complex conjugate or denominator complex conjuguate?
If you're trying to find F* F, then you multiply the numerator by its complex conjugate, and the denominator by its complex conjugate. After you do that, it should be obvious (using trick #1) how to get the 1 out front.

Chet

If you're trying to find F* F, then you multiply the numerator by its complex conjugate, and the denominator by its complex conjugate. After you do that, it should be obvious (using trick #1) how to get the 1 out front.

Chet

What do you mean by F*F though? I just need to find |F(iw)|^2....

Chestermiller
Mentor
What do you mean by F*F though? I just need to find |F(iw)|^2....
Yes. That's what I meant. You are supposed to multiply F(iw) by its complex conjugate F(-iw). I used F* to represent F(-iw).

Chet

BvU
Hey man, eeeeeverything. But first we are going to work out $$\frac {Wn^2-w^2+2ζaWn*wi} {Wn^2-w^2+2ζbWn*wi}$$ times its complex conjugate, just like they taught us with the usual ordinary complex numbers to evalulate the modulus squared. And by posting the work you can show off how you deftly extract the 1 so the characteristic shape of this notch filter frequency plot is prominent, even in the expression.