Notch filter

  • Thread starter JI567
  • Start date
  • #1
174
0

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!
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
14,244
3,621
I think you strayed somewhere. Show your calculations.
 
  • #3
174
0
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?!
 
  • #4
BvU
Science Advisor
Homework Helper
14,244
3,621
Yep. You want ##\left |F(j\omega)\right |^2.\ ## So multiply ##F(j\omega)## with its complex conjugate !
 
  • #5
20,970
4,603
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
174
0
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...
 
  • #7
174
0
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?
 
  • #8
BvU
Science Advisor
Homework Helper
14,244
3,621
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##
 
Last edited:
  • #9
20,970
4,603
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
 
Last edited:
  • #10
174
0
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
 
  • #11
BvU
Science Advisor
Homework Helper
14,244
3,621
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}##,
 
  • #12
20,970
4,603
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
 
  • #13
BvU
Science Advisor
Homework Helper
14,244
3,621
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
20,970
4,603
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
 
  • #15
BvU
Science Advisor
Homework Helper
14,244
3,621
JI's move now. I hope we haven't lost him/her.
 
  • #17
BvU
Science Advisor
Homework Helper
14,244
3,621
Hey, this is PF, not a Mutual admiration society ! ;)
 
  • Like
Likes Chestermiller
  • #18
174
0
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?
 
  • #19
174
0
Hey, this is PF, not a Mutual admiration society ! ;)
Hey man, do you know anything about plotting bode magnitudes or step response plots?
 
  • #20
20,970
4,603
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
 
  • #21
174
0
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....
 
  • #22
20,970
4,603
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
 
  • #23
BvU
Science Advisor
Homework Helper
14,244
3,621
Are we back to post #4 now ?

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

Related Threads on Notch filter

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
688
Replies
0
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
13
Views
760
  • Last Post
Replies
2
Views
5K
Replies
0
Views
4K
Replies
2
Views
1K
Replies
2
Views
12K
Replies
2
Views
580
Top