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I have doubt regarding memory of FIR and IIR, I have read that output of FIR depends upon present and previous inputs, while output of IIR depends upon, present and previous inputs as well as previous outputs. Can anyone explain me how FIR consumes more memory than IIR?

Source: https://dspguru.com/dsp/faqs/iir/basics/

Thanks very much.

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closed as unclear what you're asking by hotpaw2, jojek Apr 14 '17 at 21:39

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ "I have read" is not really a source. I'd recommend re-reading that source, and then citing it. Inherently, the claim "FIR uses more memory than IIR" is wrong. A FIR with the same transition width as an IIR might be longer, but that is usually not the design tradeoff you do in isolation. So, you're asking us to argue against a source that isn't correctly put into context, and that makes little sense $\endgroup$ – Marcus Müller Apr 13 '17 at 23:01
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    $\begingroup$ I have added source. $\endgroup$ – Ramesh Kumar Apr 14 '17 at 9:14
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The source you quote says:

IIR filters can achieve a given filtering characteristic using less memory and calculations than a similar FIR filter.

As rbj says!

What the authors mean here with "memory" are the memory elements needed to store the filter coefficients (and as discussed, these might be fewer for IIR for defined-transition-width filters), as well as the "intermediate" results (state) needed to calculate the output. "Calculations" refer to multiplications and additions needed per output sample, and that's pretty much directly a function of the number of coefficients.

So, if you define a problem that is more efficient to solve with IIRs than with FIRs (typically, a situation where you need to have a narrow transition from stop- to passband), then the IIR needs less coefficients, multiplications and state storage.

Note that what the authors didn't tell you that they're looking at a specific class of problems – which admittedly is a very common class – namely filters designed primarily to match any given stopband/passband amplitude scheme. Not all filters are only designed with that in mind!

For example, if you need a linear phase response over a significant bandwidth (and that is a relatively common thing to wish for, as it leads to constant group delay), designing an IIR that's "smaller" than a FIR might become hard (or, in fact, impossible).

Also note that whilst IIRs are often composed of structures that are highly efficient in space, multiplier and flipflop usage in hardware, the same structures aren't always inherently advantageous in software. Modern processors use long pipelines, where it can be a big disadvantage for throughput if your structure is recursive – as requiring previous output of the IIR will require the CPU to necessarily wait for instructions to complete before being able to push in the next instructions into the pipeline. (There's clever approaches to minimize that problem, but they often lead to more complex code and larger machine code memory – also a trade-off.)

Then: In many cases, where the memory usage of your filter might become critical¹, there might be aspects to the overall application of your filter that make specific kinds of IIRs easier on memory or harder.

A very intuitive problem is a narrow-band anti-aliasing filter for a digital decimator: $L$ samples go in, $1$ goes out, and you need to apply a $\frac 1L$-band low pass to avoid aliasing. That filter should of course have a transition width of $< \frac 1N$ of the Nyquist band – and might thus become pretty long!

Now, the first thing to observe is that for FIRs, there's structures that can be very efficiently implemented in hardware to save a lot of coefficients – a Nyquist-$M$ filter could be the filtering approach of choice here, and those filters consist of zeros only, aside from every $M$th coefficient. Multiplying a number with zero and adding it to the output is a non-operation and can be completely omitted in hardware!

Another rather important aspect is that for FIR filters, multirate operations (i.e. decimation, interpolation) can usually be implemented at the lower rate – e.g. our $L$-decimator could be run only every $L$ input samples – drastically reducing the number of hardware multiplications needed per output sample (by a factor of $L$, in fact), and also: often allowing the filter to actually use one and the same multiplier hardware element – when implementing DSP hardware, you usually don't actually consider the number of calculations you need per sample, but the amount of hardware you need to achieve something at a given rate, and if you have multipliers that can run at the input rate, you can let them "handle" $L$ filter coefficients, rather than needing a multiplier for each and every single coefficient.

I thus do agree with the statement

IIR filters can achieve a given filtering characteristic (defined by the transition widths and attenuations) using less […] than a (under these specific specifications) similar FIR filter.

Especially, since you'll find biquads (which are essential IIR building blocks) everywhere, but you have to realize that the statement as given is a bit of an oversimplification.


¹ not saying you shouldn't generally use the optimal filter for your problem, but you often just don't and live with a engineering effort vs optimality trade-off

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In order to accomplish a particular specified filtering task, the FIR filter order will almost certainly be much higher than the IIR filter order. For instance a low-pass filter that requires a particular "sharpness" in cutoff, the transition region between the passband (at low frequencies) and the stopband (at high frequencies). What a 4th-order IIR filter can do might correspond to what a 20th or 25th-order FIR can do. The 4th-order IIR requires 10 coefficients and 4 states. the 20th-order FIR requires 21 coefficients and 20 states.

Feedback can be useful.

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