I understand that FIR filters have a finite impulse response and IIR filters have an infinite impulse response. Here's the issue:

A signal is finite in time if and only if it is infinite in frequency. A signal is finite in frequency if and only if it is infinite in time.

So let's talk about the ideal LPF as an example. The ideal frequency response is 1 at low frequencies and 0 past some cutoff frequency. Therefore, the LPF is finite in frequency and its impulse response is necessarily infinite in time (because the frequency spectrum is a rectangular function, the impulse response is a sinc function). I've seen on other forums it has been said that a LPF can be made as either a FIR or an IIR filter. I can't understand how a LPF can have a finite impulse response.

  • 2
    $\begingroup$ We can't realize an ideal LPF. When we say "an LPF can be made as either a FIR or an IIR filter", we mean a realizable LPF that has a transition band between passband and stopband, and the response of its stopband is not exactly zero. $\endgroup$
    – ZR Han
    Feb 3, 2023 at 4:52
  • $\begingroup$ "a realizable LPF that has a transition band between passband and stopband". True, but if the response of the stopband was zero I would still say there is a finite bandwidth of the filter so it's an IIR. "the response of its stopband is not exactly zero." So the stopband of a FIR is not exactly zero. This may be the answer I'm looking for. However, I feel like it implies that the stopband of an IIR is exactly zero. Is this a real difference between the two? $\endgroup$
    – Levi
    Feb 3, 2023 at 5:01
  • 1
    $\begingroup$ @Levi no. it does not imply that. It's nowhere implied that a frequency response becomes zero for something to be an LPF. Whether it's an FIR or an IIR LPF makes no difference. $\endgroup$ Feb 3, 2023 at 8:24
  • 1
    $\begingroup$ (and no, the stopband of an IIR LPF does not inherently stay zero. Why should it? You said "I feel like it implies…", and now you should really explain that! Otherwise, Occam's razor applies.) $\endgroup$ Feb 3, 2023 at 8:54
  • $\begingroup$ We don't try for a perfect zero stopband (or a perfect unity passband). We aim for a filter that's good enough without wasting resources trying to get extravagant at making it "perfect". $\endgroup$
    – TimWescott
    Feb 3, 2023 at 15:13

1 Answer 1


The answer has partly been given in the comments already, but I'll add a few things for clarity and for the sake of completeness.

First of all, when DSP engineers talk about IIR filters, they usually mean implementable (i.e., stable and causal) filters, which - apart from quantization effects - can be implemented exactly. In that sense, an ideal lowpass filter is not an IIR filter (because there exists no finite filter structure for implementing it), even though it clearly has an infinitely long impulse response. An ideal lowpass filter is a concept that we sometimes try to approximate by actually implementable filters. Note that an ideal lowpass is neither causal nor stable.

Second, when DSP engineers talk about lowpass filters, they usually mean filters that pass low frequencies with negligible distortion, and that attenuate higher frequencies sufficiently well. Any filter with that property is a lowpass filter. That's why there are FIR lowpass filters as well as IIR lowpass filters. Both FIR and IIR filters can approximate the ideal lowpass well enough for a given application. Note that we don't need to suppress the stopband components of the input signal more than the noise level. Also, magnitude and phase distortions in the passbands can always be made small enough such that they are tolerable for a given application.

In sum, there are FIR as well as IIR lowpass filters, and none of them equal the ideal (non-causal and unstable) lowpass filter. Which of the two is better completely depends on the application and on the platform. Everything we said about lowpass filters is true for all other filter types as well (highpass, bandpass, Hilbert transformer, etc.).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.