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Question about design active filter

  1. May 10, 2013 #1

    i designed an active low-pass filter,but i don't know if it correct method or not

    i want fc=200 Hz and n(order)=4. i use this method:

    by let C=100 nF and use this equation "fc=1/(2*pi*C*R1)"
    i get R1= 8 Kohm,R2=R1
    this is for one stage

    to get filter of 4 order,i repeat that stage 4 times

    so is this method correct?
  2. jcsd
  3. May 11, 2013 #2


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    Staff: Mentor

    You show four stages, each of 1st order. That will work for an overdamped system (low Q), but if you are designing for a sharp Q (lower damping and with overshoot in its step response), then you'll have to use 2nd order stages. Each comprises 2 capacitors and 2 resistors (at least), and can provide a high Q response (i.e., Q greater than 0.707). In practice, the professional designer faces a bewildering choice, each offering its own set of pros and cons. But for an introduction to the topic there are a few well-tried classics. Any general second-order filter stage will do the job nicely; you'll need two stages.
  4. May 11, 2013 #3


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    Science Advisor
    Homework Helper

    No, because your first order filter has a gain of 0.5 at fc=200 Hz.

    When you put the four filters in series, the gain at 200 Hz will be ##0.5^4## = 0.0625.

    We don't know how much you already know about filter design, so it's hard to give much advice beyond "get a textbook and study it".

    You could try Google for "design a low-pass Butterworth filter" (which seems to be what you were trying to do) but you might need to study some more basic ideas first.
  5. May 11, 2013 #4
    NascentOxygen & AlephZero thank you very much
  6. May 11, 2013 #5
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