I've come across this question in textbook. I guess it's basically asking are there any input or impulse response conditions that a FIR filter won't be able to compute. I can't think of any.
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2$\begingroup$ Welcome to SE.SP! For a start, the definition is not correct. Summing over $n$ will yield a function of $m$, which is not what is required. I believe the right-hand side of the equation should read $\displaystyle\sum_{m=-\infty}^{+\infty} h[m] x[n-m]$. $\endgroup$– Peter K. ♦Commented Dec 7, 2021 at 15:27
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2$\begingroup$ This seem mostly a question of semantics. An FIR filter can be implemented as a convolution but it can also be implemented differently. They don't define exactly "perform" means in this context, so the question is hard to answer. Given the sloppy language and sloppy equation, this may not be a great text book to work with. $\endgroup$– HilmarCommented Dec 7, 2021 at 15:35
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For a start, the definition is not correct. Summing over $n$ will yield a function of $m$, which is not what is required. I believe the right-hand side of the equation should read
$$y[n] = \sum_{m=-\infty}^{+\infty} h[m] x[n-m].$$
Now, let's impose the FIR constraint, assuming that the filter is a $M$-tap filter:
$$y[n] = \sum_{m=0}^{M-1} h[m] x[n-m].$$
One reason that this might not be convolution is if $$ h[n] = \left \{ \begin{array}{cr} H & n=0\\ 0 & n \not = 0 \end{array} \right . $$ so that the "filter" is just a gain of $H$ so no convolution is needed.
Another reason might be to do with incorrect values of $x$ or $h$: if they're NaN then the sum will be invalid.
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2$\begingroup$ Either of those "not a convolution" operations could be considered convolutions, though -- in the one case, convolution by the Kroneker delta, in the other case it's just a convolution involving
NaNthat's corrupting the result. I'm thinking it's just a semantics question, and a poorly framed one at that. $\endgroup$ Commented Dec 7, 2021 at 16:09 -
1$\begingroup$ @TimWescott Agreed! It's a really poorly framed question. I was clutching at straws. $\endgroup$– Peter K. ♦Commented Dec 7, 2021 at 16:13
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$\begingroup$ In what cases would x or h be NaN? $\endgroup$ Commented Dec 7, 2021 at 16:13
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$\begingroup$ Apologies for putting such a poorly framed question to you! It seemed weird to me as well, but I wasn't sure if there was just something I was missing. $\endgroup$ Commented Dec 7, 2021 at 16:14
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2$\begingroup$ @shakamoto we're not saying your question is poorly framed -- we're criticizing the textbook's question. $\endgroup$ Commented Dec 7, 2021 at 16:22
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A Finite Impulse Response filter will have a finite number of taps. It is usually considered to be causal as well.
Meaning that any FIR filter of N taps will not be able to produce weightings of negative time index or those beyond tap N.
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This isn't from a textbook. It's from a University examination in my department. Since it was taken in the time of COVID, candidates had to download it, answer it, and upload the solution. Looks like they got a lot of help. It's the "bit at the end" of the question which is supposed to exercise thinking in an open-book sort of way, so that the candidate has to demonstrate understanding rather than trot out maths. Only 5 marks. [5] - see? Let's hope they were practicing with past papers and not posting during the actual exam, eh?
It's a paper in a mostly practical lab-based EE course. The expected answer, as per Knut Inge's reply, is that it's truncated and shifted (to make it causal), which explains the descrepency between the ideal response and the achieved result. In this regard, the FIR as implemented (in a lab) is not doing the same thing as the convolution given in the equation. The result is delayed, and windowed (in the simplest case, with a rectangle, although in practice with a function chosen to suit the problem domain).
So, 100% to Knut. Thanks for the heads-up on the typo in the summation. Obviously it's best to post papers here for moderation, rather than send them to the external examiners.
It seems the correct variable is used in the handout.
I'd love to know how many current students looking at past papers end up here!
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$\begingroup$ Hello Nick Bailey and welcome to DSP SE. Your answer doesn't seem to provide a solution to the question being asked. I suggest you put together a simple solution and edit your answer. This will help the OP to solve their problem as well as future reference for people that may have the same (or a similar) problem. $\endgroup$– ZaellixACommented Oct 14, 2022 at 18:22
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$\begingroup$ I would have posted a comment, but, you know, I wasn't allowed. The answer is in para 2. Ignoring the typo in the convolution equation, the answer is an FIR would truncate h(n) and shift it to make it causal. $\endgroup$ Commented Oct 14, 2022 at 19:43
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$\begingroup$ I understand that you cannot post a comment as of yet (you require more reputation), but it is ill-advised to post answers that do not provide a direct solution (or at least partially) to the asked question. "Cluttering" the session with answers that do not provide solutions can make future referencing harder and impede the interested user to find what they seek. $\endgroup$– ZaellixACommented Oct 14, 2022 at 22:20
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$\begingroup$ Sure. I understand that. As you've see I've now provided the exact answer, as required by the examiner, in paragraph 2. I think that's in the spirit of Stack Exchange and to the OP's benefit rather than issue a take-down notice for copyright infringement. I know the University goes is inclined to doing that when they see bits of its exam papers get posted up, but I'm an open-source kind of guy. It's necessary to clarify the context in which the exam q was asked. The longer answer above, while being generally pertinent, wouldn't have done the candidate any good. It's about implementasiuon. $\endgroup$ Commented Oct 15, 2022 at 12:36