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IIR filter and FIR filter

  1. Feb 13, 2012 #1
    Hi all..

    I'm a new to filters and I need some help understanding filters (I'm in mechanical engineering)

    So far, I've found out that FIR filters use convolution and has a finite impulse response.

    On the other hand, IIR filters are recursive and have an infinite impulse response.

    But, what if I get the impulse response of an IIR filter (Butterworth, for example) and convolute the response with a signal? (given that I use only a part of the infinite signal).

    Does this still count as an IIR filtering? Can I apply an IIR filter this way?

    Any help would be appreciated.

    Thank you
     
  2. jcsd
  3. Feb 13, 2012 #2
    Well - taking the convolution sum and truncating it to some finite range is effectively taking an FIR approximation of a given IIR filter, which is actually one of the techniques to design FIR filters.
    As a practical way to model a system, it's ok, it can even be pretty accurate - eventually any IIR impulse response decreases to a certain level bellow which the quantization makes it effectively zero, but theoretically it is wrong, and off course using the recursive way to calculate the response is always more accurate than approximating with FIR.
     
  4. Feb 13, 2012 #3
    Thank you for your reply :)
     
  5. Feb 13, 2012 #4
    Can I ask one more question?

    If I have a low pass filter, whose transfer function is

    H(s) = a/(a+s),

    It it an IIR filter or FIR filter?
     
  6. Feb 13, 2012 #5
    It is neither.

    The discrete-time versions are an approximation to the continous transfer function.

    The FIR approximation will come as close as you want to analog. The IIR version is very good but not exactly the same.

    The IIR version of that filter includes a zero in the numerator at the nyquist frequency.

    It's roughly y(n+1) = b*y(n) + c*(x(n)-x(n-1)).
     
  7. Feb 13, 2012 #6
    Thank you for your reply and sorry for keep asking...

    The equation I'm dealing with contains a convolution, like A = f*B, where B is the input signal and f is the impulse response of a filter.

    Does this mean that I am forced to use FIR filtering and no Butterworth or Chebyshev for me? (man... they look so attractive)

    Also, Butterworth has a nonlinear phase, but FIR filters have a linear phase. However, if I use the impulse response of the Butterworth to do convolution (which is FIR filtering), do I still get a linear phase?

    Thank you
     
  8. Feb 13, 2012 #7
    If you must convolve then yes FIR is required. Are you sure you must convolve?

    You cannot get linear phase from an IIR.
     
  9. Feb 13, 2012 #8
    Sadly, yes. I must convolve....

    Will there be any advantages of taking FIR approximation of IIR filters?
     
  10. Feb 13, 2012 #9

    rbj

    User Avatar

    not necessarily. if the analog filter was digitized using the bilinear transform, then it's true. but not necessarily if it was converted from analog to digital by other means (like Impulse Invariant).
     
  11. Feb 14, 2012 #10
    Thank you all for your help :)
     
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