Continuous low-pass filter (LPF) on a periodic signal

Question

I have a continuous-time signal $x(t)$ that has the following properties:

1. It is periodic, with a periodicity of 1 (sec).
2. It has no discontinuities or infinite values at any time.

I want to come up with an LPF to pass $x(t)$ through, but with the following requirements:

1. The filter is ideal (see point #3), continuous-time, with an impulse response $h(t)$ (and frequency response $H(f)$).
2. The filter need not be a simple looking one like a brick-wall LPF. But $h(t)$ should be describable in closed form (piece-wise definition ok too). I am not looking for anything that involves stuff like dirac-delta function or infinite sums or something that can only be defined as an integral, etc.
3. Any cutoff frequency is ok, but should be a finite number. Beyond that cutoff frequency, $H(f)$ should be 0 (not slowly reaching 0 asymptotically).
4. The main requirement is that the output $y(t) = x(t)\star h(t)$, should have the same values as $x(t)$ at times $0, 1, 2, 3, \ldots$ i.e. $y(t) = x(t)$ for all integers. At other times, there is no restriction on how $y(t)$ should look like, except that it should be finite at all times.
5. The filter need not be physically realizable. Just a mathematical definition is good enough for my purpose.

Of course, since I have given very few specifics, I am not expecting an exact $h(t)$ or $H(f)$ as an answer, but looking for what generic shape such a filter will have.

Answer 1

It looks like you want a filter that is linear time invariant. I don't think it is possible to meet all your criteria with an LTI filter, but I think it you can get close with the filter $$ H(s) = \frac{1}{2} \frac{1 + \text{e}^{-L s}}{1 + \alpha ( 1 - \text{e}^{-L s})} $$ where $\alpha > 0$ and $L$ is the period of $x(t)$. There is unity gain at the harmonics, and a zero between each harmonic. It should follow the input signal due to the internal model principle.

Going through your checklist:

1. It is continuous time, and it has an impulse and frequency response.
2. The impulse response is describable in closed form, as a Fourier series, but it will be an infinite sum. The stationary response, however, is $y(t) = x(t)$.
3. This is the criterion that can not be fulfilled, as the filter magnitude alternates between 0 and 1. If $x(t)$ is not bandwidth limited, I guess a cut-off frequency can not really be defined, and you just want to attenuate frequency content that are not at the harmonics, in which case the above filter will work.
4. This was your main requirement, and the above filter will respond exactly like that using a suitable initialization function. The filter response is bounded theoretically, but it is not structurally stable, hence it will not work well in practice. Limiting the bandwidth of the zero generating functions with a unity-gain low-pass filter $Q(s)$ should help: $$ H(s) = \frac{1}{2} \frac{1 + Q(s) \text{e}^{-L s}}{1 + \alpha ( 1 - Q(s) \text{e}^{-L s})} $$
5. The filter is physically realizable if $Q(s)$ is stable and causal (e.g. not zero-phase), but it will then lose precision in the zero locations for the numerator and denominator.

Answer 2

If $x(t)$ has period $T=1$, then $y'(t)=x(t)\star g(t)$ also has period $T=1$ for any filter $g(t)$.

Assume that $y'(t)$ is not the zero signal (in other words, assume the filter $g(t)$ is such that its cutoff frequency is larger than 1 Hz).

Now, let $$h(t)=\frac{x(1)}{y'(1)}g(t).$$ Then, if $y(t)=x(t)\star h(t)$, it follows that $y(n)=x(n)$ for $n$ integer.

Note that the definition of the filter $h(t)$ depends on the values $x(1)$ and $y'(1)$. There is no low-pass filter that will meet your condition #4 for general $x(t)$.

[Thanks to RBJ for helping me see an error in my previous answer]

Answer 3

You can't make the filter general, that is, independent of $x(t)$.

For a general filter not dependent on $x(t)$ your requirement #4 means that $y(t) = x(t)$, because if the filter is general, it must work just as well for a time-shifted $x(t)$. So due to the time-invariance property of filters, in requirement #4, $y(t) = x(t)$ for all integer $t$ becomes $y(t+u) = x(t+u)$ for all integer $t$ and all real $u$, which can be written simply as $y(t) = x(t)$ for all real $t$. So the filter must pass the signal unchanged. That requires $x(t)$ to be band-limited and the filter can no longer be general: the band-limit of $x(t)$ must be known in order to decide on the filter, which must have in its magnitude frequency response a flat passband of at least the bandwidth of $x(t)$. Without knowing the band-limit of $x(t)$ we would need the filter to have infinite bandwidth which is ruled out by your #2 that prohibits using the Dirac delta function and related abominations.