1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Notch filter

  1. Jan 19, 2015 #1
    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. jcsd
  3. Jan 19, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you strayed somewhere. Show your calculations.
     
  4. Jan 19, 2015 #3
    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?!
     
  5. Jan 19, 2015 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yep. You want ##\left |F(j\omega)\right |^2.\ ## So multiply ##F(j\omega)## with its complex conjugate !
     
  6. Jan 19, 2015 #5
    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
     
  7. Jan 20, 2015 #6
    If you look properly you will see I have already done that....the 3rd step incase you still can't see it...
     
  8. Jan 20, 2015 #7
    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?
     
  9. Jan 20, 2015 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
  10. Jan 20, 2015 #9
    It's applied on the denominator.
    $$(Wn^2-w^2+2ζbWn*wi)(Wn^2-w^2-2ζbWn*wi)=?$$
    Chet
     
    Last edited: Jan 21, 2015
  11. Jan 21, 2015 #10
    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
     
  12. Jan 21, 2015 #11

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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}##,
     
  13. Jan 21, 2015 #12
    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
     
  14. Jan 21, 2015 #13

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  15. Jan 21, 2015 #14
    He may have claimed otherwise, but that's not what it looks like to me.

    Chet
     
  16. Jan 21, 2015 #15

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    JI's move now. I hope we haven't lost him/her.
     
  17. Jan 21, 2015 #16
    By the way, I like your approach quite a bit.

    Chet
     
  18. Jan 21, 2015 #17

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hey, this is PF, not a Mutual admiration society ! ;)
     
  19. Jan 22, 2015 #18
    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?
     
  20. Jan 22, 2015 #19
    Hey man, do you know anything about plotting bode magnitudes or step response plots?
     
  21. Jan 22, 2015 #20
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted