Butterworth Filter Sallen Ket Cut Off Frequency?

Discussion in 'Homework Help' started by Volt, Feb 24, 2010.

  1. Volt

    Thread Starter Member

    Nov 28, 2008
    24
    0
    Hello,

    I am having a little trouble calculating the upper cut off frequency of this Butterworth Filter in the Sallen and key configuration.

    Below you can see the component values in the pic:

    http://twitpic.com/152zmr

    The upper cut off frequency here (according to my supervisors) is 28kHZ,

    however using the equation I found:

    fc=1/2pi(root(R1R2C1C2))

    ,it doesn't equal 28kHz, I am getting 44.209kHz?!

    Am I using the wrong equation?

    Any help would be much appreciated,
    Regards,
    Adam
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    2,503
    380
  3. Volt

    Thread Starter Member

    Nov 28, 2008
    24
    0
    I'm on a mac i'm afraid
     
  4. nubcookie

    New Member

    Apr 27, 2009
    4
    0
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    1. That is a filter, but it is not a Butterworth filter. You can't make a unity gain Butterworth with equal valued resistors and equal valued capacitors.
    2. Stray capacitance will cause the actual cutoff frequency to differ from the calculated value. You should use higher capacitance values.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well, your supervisors is wrong.
    Because IF R1=R2 and C1 = C2 the
    Fo=1/(2*pi*RC)=0.16/(RC)=44KHz
     
  7. msr

    Active Member

    Jul 8, 2008
    62
    1
  8. Volt

    Thread Starter Member

    Nov 28, 2008
    24
    0
    When I simulate this on multisim it comes out as 28kHz, is there an equation I could use to determine that value?

    I have tried v hard to work it out but can't!

    Thanks for all your help so far...
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    The cutoff frequency is indeed 44kHz, but that equation does not refer to the -3dB frequency. It is the natural frequency of the filter. The gain at that frequency is equal to the Q of the filter. A Butterworth filter has a Q of 1/√2. The OP's filter has a Q of 0.5. Therefore, the Butterworth is 3dB down at the natural frequency. The OP's filter is 6dB down at the natural frequency.
    A simulation of the OP's filter and a B'worth shows that the OP's filter is in fact 3dB down at 28kHz, which I'm sure can (and probably will) be calculated by another member of the forum.

    A filter with higher Q will have less loss at the natural frequency, and in fact will have peaking as the Q increases.
    See the Wikipedia entry for Sallen & Key.
     
    cgama likes this.
  10. Darren Holdstock

    Active Member

    Feb 10, 2009
    262
    11
    Just to say, all hail Ron H, this is filter gold. In one clear and concise paragraph Ron H has explained it beautifully.
    I blew out a job interview at TAG McLaren Audio many years ago by asking the technical director a similar question regarding damping and the various resonant frequencies, as I couldn't find any textbook that would explain it to my satisfaction. He didn't know either, and took his revenge for my impudence in asking. Still, I learned not to ask awkward questions at interviews, lest a thirst for knowledge be mistaken for being awkward.
     
    cgama likes this.
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Thanks for the kind words, Darren.:)
    I have to admit that I didn't know this until I looked at the simulation, which drove me to the equations in Wikipedia.
     
  12. Volt

    Thread Starter Member

    Nov 28, 2008
    24
    0

    Thankyou so much, explained it perfectly!
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    If we have the transfer function

    \frac{Vout}{Vin}=\frac{1}{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}

    And we want to know cut-of frequency for 3dB=1/√2.
    So we have

    \frac{Vout}{Vin}=\frac{1}{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}= \frac {1} {\sqrt2}

    And for R1=R2=R and C1=C2=C
    We get

    (1-(\omega*R*C)^2)^2+4*(\omega*R*C)^2=2

    And if we solve this we get:

    (1+(\omega^2*C^2*R^2))^2=2 we substitute X=(\omega^2*C^2*R^2)

    (1+X)^2=2

    X=\sqrt{2}-1

    And cut-off frequency is equal

    fg=\frac{\sqrt{\sqrt{2}-1}}{2*\Pi*R*C}

    And this is a special case only for R1=R1, C1=C2
     
  14. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    jony130,

    Shouldn't the equation for the magnitude of Vo/Vi be:

    \frac{Vout}{Vin}=\frac{1}{\sqrt{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}}

    hgmjr
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
Loading...