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Is there a way to construct FIR analog low-pass filter? (By this, I believe I am asking for no-pole analog low-pass filter, in terms of Laplace transform.)

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  • $\begingroup$ Are time delays in the continuous time allowed, or do you only want it in terms of a polynomial (in the numerator) in $s$? $\endgroup$ – fibonatic Oct 11 '16 at 4:29
  • $\begingroup$ I was curious about this myself, some time ago, but it's apparently impossible without any form of delays, since the whole transfer function only has zeroes, and even a simple differentiator has a pole in there, somewhere (real-case). I could be wrong, though. $\endgroup$ – a concerned citizen Oct 11 '16 at 5:55
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You have to clearly define what you mean by an "analog FIR filter". "No poles" is not correct because (discrete-time) FIR filters do have poles; they are just all at the origin of the $z$-plane (for causal FIR filters). Note that filters without poles do not exist. Take as an example the discrete-time transfer function

$$H(z)=1-az^{-1}\tag{1}$$

with a zero at $z=a$. Obviously, $H(z)$ has a pole at $z=0$. Similarly for

$$H(z)=z-a\tag{2}$$

which also has a zero at $z=a$, but a pole at $z=\infty$.

If you mean transversal filters implemented as a tapped delay line, then you can find analog versions of it, see e.g. this paper.

If you literally mean a filter with a finite impulse response (FIR) then, technically speaking, such a thing does not exist in the continuous-time world. But of course, any stable system has an impulse response that decays sufficiently fast such that for practical purposes it may be considered as zero (or unmeasurable) after a certain time. The discussion if such a system should be considered as FIR or not is moot.

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  • $\begingroup$ To whom it is not obvious that a causal filter must have poles: dsp.stackexchange.com/questions/13605/… But where does that pole at $z=\infty$ come from? $\endgroup$ – Olli Niemitalo Oct 11 '16 at 10:27
  • $\begingroup$ @OlliNiemitalo: If you have a non-causal FIR filter, some of its poles must be at $z=\infty$. If it's acausal (i.e., $h[n]=0$ for $n>0$, like the one in Eq. (2)) then all its poles are at $z=\infty$. What do you mean by "where does that pole come from"? $\endgroup$ – Matt L. Oct 11 '16 at 11:11
  • $\begingroup$ I mean "please explain why we have that in addition to stating it as a fact". $\endgroup$ – Olli Niemitalo Oct 11 '16 at 11:32
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    $\begingroup$ @OlliNiemitalo: If you look at Eq. (2), which corresponds to an anti-causal 2-tap filter ($h[-1]=1$ and $h[0]=-a$), we clearly have $H(z)\rightarrow\infty$ for $z\rightarrow\infty$. Is that what you mean? $\endgroup$ – Matt L. Oct 11 '16 at 12:24
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    $\begingroup$ @OlliNiemitalo: You're right, but by convention, a function $f(z)$ has a pole at $z=\infty$ if $f(1/z)$ has a pole at $z=0$. $\endgroup$ – Matt L. Oct 11 '16 at 12:44
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You can use a bucket-brigade device to introduce a delay, or a piezoelectric or other type of delay line.

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    $\begingroup$ yep. and because these are relatively complex and/or expensive devices, things that require analog memory (for oscilloscopes, for example, or for FIRs, like in this case) have, wherever possible, been exchanged for digital components $\endgroup$ – Marcus Müller Oct 11 '16 at 7:27
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You can build the analog system

$$\frac{1}{s}-\frac{e^{-sT}}{s}$$

which has a finite impulse response, no poles, and an infinite number of zeros. (The pole at the origin cancels.) The time delay can be implemented using an analog transmission line, and the integrals and subtract using an op-amp. FIR analog systems always seem to have an infinite number of zeros.

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