# Homework Help: Notch filter

1. Jan 19, 2015

### JI567

1. The problem statement, all variables and given/known data
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 \$
2. Relevant equations

F(s) is the equation of a notch filter. |F(iw)| is the magnitude of its frequency response.
3. 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!

2. Jan 19, 2015

### BvU

I think you strayed somewhere. Show your calculations.

3. Jan 19, 2015

### JI567

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?!

4. Jan 19, 2015

### BvU

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

5. Jan 19, 2015

### Staff: 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

6. Jan 20, 2015

### JI567

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

7. Jan 20, 2015

### JI567

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?

8. Jan 20, 2015

### BvU

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 : $\times$

Last edited: Jan 20, 2015
9. Jan 20, 2015

### Staff: Mentor

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

Last edited: Jan 21, 2015
10. Jan 21, 2015

### JI567

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

11. Jan 21, 2015

### BvU

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 ${a+bi\over c+di}\times {a-di\over c-di}$,

12. Jan 21, 2015

### Staff: Mentor

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

13. Jan 21, 2015

### BvU

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.

14. Jan 21, 2015

### Staff: Mentor

He may have claimed otherwise, but that's not what it looks like to me.

Chet

15. Jan 21, 2015

### BvU

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

16. Jan 21, 2015

### Staff: Mentor

By the way, I like your approach quite a bit.

Chet

17. Jan 21, 2015

### BvU

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

18. Jan 22, 2015

### JI567

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?

19. Jan 22, 2015

### JI567

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

20. Jan 22, 2015

### Staff: Mentor

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

21. Jan 22, 2015

### JI567

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

22. Jan 22, 2015

### Staff: Mentor

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

23. Jan 22, 2015

### BvU

Are we back to post #4 now ?

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.