# Difference between FIR and IIR filters

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.

• 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. Feb 3 at 4:52
• "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?
– Levi
Feb 3 at 5:01
• @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. Feb 3 at 8:24
• (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.) Feb 3 at 8:54
• 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". Feb 3 at 15:13