# Representing a positive 0-limiting continuous time domain function, in terms of spectral samples, outside of a window

Bit of an odd question

Assume you have a known function $x(t)$ which is continuous, but approaches but is not equal to, 0, on the interval $[-\infty,\infty]$ . You know also that it approaches 0 exponentially fast on some interval $[T_a, T_b]$ e.g. the Normal or Cauchy Distribution, or some pulse with characteristics you already know with the properties as described. You also know this time domain function's closed form Fourier Transform $X(f)$, which is also continuous:

$$X(f) \triangleq \mathscr{F}\{ x(t) \} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt$$

Say now, you wanted to reconstruct an approximate $\hat{x}(t) \approx x(t)$ from only spectral samples $f_k$ coming from the Fourier domain, and a given $t$. This is perhaps in the form of an IFT of an FT. It is acceptable that this reconstruction of $\hat{x}(t)$ be equal to $0$ outside of the given time domain range $[T_a, T_b]$ at a specified error which is determined beforehand.

I have only dabbled a little in signal processing, but my current understanding suggest that one can only represent this reconstruction $\hat{x}(t)$signal by one of two ways

1. Since you know the closed form of $x(t)$, you can take a Fourier series with coefficients calculated on the given interval. This will enable one to take the Fourier Transform of this FS, and then apply the Inverse Fourier Transform to gain the desired representation. By using the FS however, it means that your reconstructed signal is inherently assumed to be periodic outside of your given interval, such that if one were to provide a $t$ such that $t<T_a$ and $t>T_b$ you would not get 0, but instead a value, in the period you're in, that corresponds to a value inside $[T_a, T_b]$
2. Since you know the close from of $X(f)$, you can directly sample in the frequency domain for a given spectral sampling rate. This however would still give the same result as 1. because the sampling impulse train is periodic, while $X(f)$ is continuous, meaning one would need an infinite number of samples, which is definitely not desirable!

The possible idea I had for a solution is to say that $x(t)$ is multiplied in the time domain with $w(t)=\operatorname{rect}(t/T)$ for some period $T$. Assume for simplicity now that our $[T_a, T_b]$ is symmetric about 0. This would mean we would require convolution with a $\operatorname{sinc}$ function in the frequency domain. But since the $\operatorname{sinc}$ function is continuous, our convolved $W(f) \star X(f)$ would then be continuous which then means we would need again infinitely many samples!

I have a feeling something is missing in my understanding but I've searched everywhere for something that hints at an answer, but everything talks about band-limited signals, not time-limited signals, and any that regard time-limited signals only applies to a solution on a specified interval.

So specifically I am curious how to use the additional information (that we know either 1. $x(t)$ is 0 outside a given interval, or 2. we have the closed form of $x(t)$). Also note that this assumed a linear shift invariance for $t$ in $x(t)$. Perhaps an additional approximate integration step between each sample in the frequency domain is required? Or some extension with Poisson Summation?

edit: - thanks Robert for the suggested clarifications

• i think we might be able to massage your question into one people might be interested in asserting some effort to solve. do you mind if i edit it and bring the math into a more clear expression? and we electrical engineers like to use "$f$" to represent frequency. mind if i make consistent changes throughout? – robert bristow-johnson May 6 '17 at 2:47
• @robertbristow-johnson Hi Robert, thanks for the reply. You're welcome to massage the question! – undercurrent May 6 '17 at 6:32
• i didn't really "clarify". but i just changed the language a little from the Empire of Mathematics to the Realm of Electrical Engineering. – robert bristow-johnson May 6 '17 at 6:39
• so $-T_a = T_b = \tfrac12 (T_b - T_a)$, right? if so, let's lose them guys and call $T_a=-\tfrac12 T$ and $T_b=+\tfrac12 T$. okay? – robert bristow-johnson May 6 '17 at 6:42
• @robertbristow-johnson yep, that's ok – undercurrent May 6 '17 at 7:28

Finite $X_k$ samples will inevitably lead to a periodic approximation. The question, as is, is unfortunately and quite interestingly, unanswerable.
Indeed, under the standard formalisms, the reconstructed signal $\hat x(t)$ will be peridic: $$F{X(f)|||(f)}=\sum^{\inf}_{i=-\inf} X(i\Delta f)e^{i2\pi i\Delta f t}\Delta f = \hat x(t)$$
If we truncate the signal $x(t)$ to be time-bounded, and using a similar argument than the Paley Wiener theorem to the direct Fourier Transform, we will obtain a exponentially converging spectrum on $X(f)$ when $f$ tends to infinity.
Based on the last, a logarithmic approximation could be sketched, in which the $f_i$ are being preferred when closest to zero, and sparse for increasing values: $$\hat x(t)=\sum^{\inf}_{f_i=-\inf} X(f_i)e^{i2\pi f_i t}\Delta f_i$$
And under such aproximation, we could let the first terms of the sum to be again, approximated with known spectrum shapes (as $sinc(f)$ as you said). But that is not a train of frequency samples. This is a different work, and perhaps a different question?.