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What criteria should be used to safely decide if it is an IIR filter or FIR filter from a difference equation? FIR filter are always stable, meaning all poles are inside the unit circle AND have pole-zero cancellations i.e have equal number of poles and zeros at the same point in the z-plane. IIR filters are conditionally stable, i.e always have poles on/outside the unit circle. Is this good enough to distinguish from the difference equation? what other criteria should be considered?

PS: Please don't say IIR filters have recursive difference equations, and FIR doesn't. It is not true.

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    $\begingroup$ Why do you say it isn't true? $\endgroup$
    – Jim Clay
    Feb 12, 2014 at 21:58
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    $\begingroup$ Because FIR filters can also be realized recursively. For example, y[n] = y[n-1]+x[n]-x[n-3] is FIR even though it has a recursive equation. Please correct me if I'm wrong! $\endgroup$
    – Masked
    Feb 12, 2014 at 22:25
  • $\begingroup$ Minimum phase IIR filters do not have any poles outside the unit circle, and do not require any poles on the unit circle. $\endgroup$
    – hotpaw2
    Feb 13, 2014 at 0:37
  • $\begingroup$ @Masked I see your point. $\endgroup$
    – Jim Clay
    Feb 13, 2014 at 0:37
  • $\begingroup$ The recursive difference equation you show as an example can be converted into a non-recursive form. Not so for an IIR. $\endgroup$
    – hotpaw2
    Feb 13, 2014 at 0:45

2 Answers 2

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If the Z-transform of the feedforward section is divisible by the Z-transform of the feedback section, the filter is FIR.

Consider your example: $y[n] = y[n-1] + x[n] - x[n-3]$.

The Z-transform is $\mathrm Y(z)- z^{-1}\mathrm Y(z) = \mathrm X(z) - z^{-3}\mathrm X(z)$, and the Z-transform of the response is $\mathrm H(z) = \mathrm Y(z)/\mathrm X(z) = (1 - z^{-3})/(1 - z^{-1}) = 1 + z^{-1} + z^{-2}$. This means that the filter can be realized as $\mathrm y[n] = \mathrm x[n] + \mathrm x[n-1] + \mathrm x[n-2]$, which is clearly FIR.

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  • $\begingroup$ By the way, pole-zero cancellations are the same as the divisibility demonstrated above. $\endgroup$ Feb 12, 2014 at 23:26
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As far as i know, unless you want a deep philosophical analysis fo filters, there is absolutely no working distinction between a FIR/IIR filter or a diference equation, in the sense that a filter has "ALWAYS" an associated diference equation that you can write.

Having said that though, if you randomly pick up a piece of papaer that has a difference equation written on it, you can't automatically say "Oh they are working with filters and signal processing here", because difference equations are used in lots of other applications where you aren't even processing a signal. And the solution to the equation (giveng initial conditions or an input $X[n]$) might mean something completely different to the output of a filter.

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