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2nd order RC filter with two equal poles

  1. May 15, 2016 #1
    Is it possible to make a 2nd order CR LPF or HPF where the cut off frequencies for each pole are equal?

    Here is a calculator for this system which includes the transfer function.http://sim.okawa-denshi.jp/en/CRCRkeisan.htm

    I figured that I need to try to solve the denominator of the transfer function in this way and impose the restriction that both roots of S must be equal, but I'm not sure how to do that.

    jgWR96S.jpg
     
  2. jcsd
  3. May 16, 2016 #2

    NascentOxygen

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    Staff: Mentor

    In principle, you can set the poles equal.

    Are you sure a passive filter will do the job? The source has low resistance and you won't have loading on the filter output? The filter shape doesn't matter much?
     
  4. May 16, 2016 #3
    It's going to be implemented between two discrete BJT amplifier stages. It's a filter within an amplifier basically, so yeah there's not going to be a loss of signal.
     
  5. May 16, 2016 #4

    NascentOxygen

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    The BJT input loads the filter output and changes the response.
     
  6. May 16, 2016 #5
    Yeah I know. My output and input impedances will be adjusted to account for this.
     
  7. May 16, 2016 #6

    NascentOxygen

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    How will you adjust for loading when the filter equation assumes no load?
     
  8. May 16, 2016 #7
    I'm beginning to think that two poles at the same positions isn't possible.
    My reasoning is that for them to be the same, they're going to be a complex conjugate pair.

    And for that to be true, if we look at the standard quadratic formula we see that
    4ac needs to be greater than b^2

    And using the transfer function I posted to obtain b,a, and c
    I performed the following analysis at wolfram

    http://www.wolframalpha.com/input/?i=4*(1/abcd)>((1/ac)+(1/bc)+(1/bd))^2

    and you can see that some of the resistor and capacitor values end up needing to be negative
     
  9. May 16, 2016 #8

    NascentOxygen

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    I wasn't referring to varying load. If the filter is to work into a fixed load, then that load needs to feature in the filter equation.
     
  10. May 16, 2016 #9
    okay, so even though it's not possible to have two equal poles with the transfer function I posted, maybe it's possible if the load is taken into account. I will have to formulate that transfer function and check. Also, there's a different post above to the one I initially posted if you missed it.
     
  11. May 16, 2016 #10

    LvW

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    A second-order high- or lowpass filter with a double-pole (two equal poles on the neg. real axis of the s-plane) is possible only if both filter blocks are separated by a buffer amplifier (no loading of the first stage by the second stage).
    In this case, the pole quality factor will be 0.5.
    In this case, the term R2C1 disappears in the denominator.
    Any transistor-based buffer amplifier will do the job of buffering only approximately because of finite input and output impedances, unless you include these impedances in the calculation of the filter elements.
     
    Last edited: May 16, 2016
  12. May 16, 2016 #11
    Yeah, I thought this was probably what I would have to opt towards doing, I just wanted to make sure.
     
  13. May 16, 2016 #12
    If you have gain to spare, you can put matched pads, 10dB or so, between the poles.
     
  14. May 17, 2016 #13

    jim hardy

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    Rule of thumb
    If you're cascading passive filters
    downstream one needs to have at least 10X the impedance of upstream one so it doesn't load it noticeably. More would be better, 30X or 100X.

    upload_2016-5-17_1-41-6.png

    Of course whatever is connected to "out" should be 10X higher yet.
    That's why it's best to buffer. But you can get away with this approach if you don't need precision.
     
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