Suppose that I have a time-limited pulse $x(t)$ and that I know the pulse's Fourier transform $X(\omega)$. Suppose further that I use the pulse to construct the repeating waveform $\sum_{k=-\infty}^{\infty} x(t-kT)$, the period $T$ being no less than the pulse's duration so that, in the repeating waveform, the repeated pulses do not overlap. In this case, I can immediately extract from the transform the coefficients of the complex-exponential Fourier series via $$ a_j = \frac{\sqrt{2\pi}}{T}X\left(\frac{2\pi j}{T}\right) $$ [Here I have styled the Fourier transform in angular frequencies as $X(\omega)=(1/\sqrt{2\pi})\int_{-\infty}^{\infty} e^{-i\omega t}x(t)\,dt$. If you prefer to answer instead using cyclic frequencies and the cyclic Fourier style, I don't mind.]

Question: I would like to evade the restriction that the pulse be time-limited. That is, I would like to let the repeating pulses overlap. Is there an elegant way to extract the series coefficients from the transform in this case?

An example would be a Gaussian pulse, which is not time-limited.

I have pages and pages of attempts here on my desk—you would not want me to post them all here—but basically what I have tried are various ways of multiplying and/or convolving $x(t)$ by a Dirac comb and/or a rectangular window.

Perhaps I cannot have what I want because that would violate the Nyquist-Shannon sampling theorem?

I can read logical, set-theoretic and maybe even topological notation if I must but am not a professional mathematician. Therefore, in your answer, if you moderated the use of notation scientists and engineers seldom recognize, this would be appreciated.

I do not require justification of convergence.

(For information, my problem is not a homework problem—though, actually, it strikes me as a pretty good problem for a senior-level homework assignment. Rather, it is a problem that has arisen during the revision of this.)


When talking about periodic pulses in the time domain, it is convenient to use the dirac comb function notation

$$\mathrm{III}(t) = \sum_{n=-\infty}^{\infty} \delta(t-n)$$

which has an impulse at every integer value of $t$. To scale the dirac comb for an arbitary period of $T$, one can use the properties of the dirac delta function to show

$$\dfrac{1}{|T|}\mathrm{III}\left(\dfrac{t}{T}\right) = \sum_{n=-\infty}^{\infty} \delta(t-nT)$$

Using this functional notation, we can now describe periodic repetition of the pulse $x(t)$ in the time domain as the following convolution

$$y(t) = x(t) * \dfrac{1}{|T|}\mathrm{III}\left(\dfrac{t}{T}\right)$$

Taking the Fourier Transform

$$\begin{align*} \mathscr{F}\left\{y(t)\right\} &= \mathscr{F}\left\{x(t) * \dfrac{1}{|T|}\mathrm{III}\left(\dfrac{t}{T}\right)\right\}\\ \\ Y(s) &= \mathscr{F}\left\{x(t)\right\} \cdot \dfrac{1}{|T|}\mathscr{F}\left\{\mathrm{III}\left(\dfrac{t}{T}\right)\right\}\\ \\ Y(s) &= X(s) \cdot \mathrm{III}\left(Ts\right) \\ \\ Y(s) &= X(s) \cdot \dfrac{1}{T} \sum_{n=-\infty}^{\infty} \delta\left(s - \dfrac{n}{T}\right) \\ \end{align*}$$

So regardless of $x(t)$ being time-limited and/or the periodic copies of $x(t)$ overlapping or not, the Fourier Transform, $Y(s)$, of the periodic pulse train is a sampled version of the Fourier Transform, $X(s)$, of a single pulse.

If you didn't notice, the dirac comb, $\mathrm{III}()$, is its own Fourier Transform. You should note that, due to the scaling property of the Fourier Transform, the closer the the $\delta()$ functions are spaced in the time domain, the farther away they will be spaced in the transform domain, and vice-versa. So, if you want a really fine sampling of the spectrum in the transform domain, you should have a very large period, $T$, between the individual pulses in the time domain.

Update to connect the above to the Fourier Series coefficients

I've worked a bit on connecting the above to the complex Fourier Series Coefficients.

To do this I will use the following notation for a unit height rectangle function:

$$\Pi(t) = \begin{cases} 1 \quad |t|<\frac{1}{2} \\ 0 \quad |t|>\frac{1}{2}\\ \end{cases}$$

which has the Fourier Transform

$$\mathscr{F}\left\{\Pi(t)\right\} = \mathrm{sinc}(s) = \dfrac{\sin(\pi s)}{\pi s}$$

So finding the complex Fourier Series coefficients of the periodic function, $y(t)$, and assuming the period, $T > 0$:

$$\begin{align*} c_k &= \dfrac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} y(t) \space e^{-i 2\pi t \frac{k}{T}} dt\\ \\ &= \dfrac{1}{T} \int_{-\infty}^{\infty} y(t) \space \Pi\left(\dfrac{t}{T}\right)\space e^{-i 2\pi t \frac{k}{T}} dt\\ \\ &= \dfrac{1}{T} \int_{-\infty}^{\infty} y(t) \space \Pi\left(\dfrac{t}{T}\right)\space e^{-i 2\pi t s} dt\\ \\ &= \dfrac{1}{T} \mathscr{F}\left\{y(t) \cdot \Pi\left(\dfrac{t}{T}\right)\right\} \\ \\ &= \mathscr{F}\left\{y(t)\right\} * \dfrac{1}{T} \mathscr{F}\left\{ \Pi\left(\dfrac{t}{T}\right)\right\} \\ \\ &= Y(s) * \mathrm{sinc}\left(Ts\right)\\ \\ &= Y\left(\dfrac{k}{T}\right) * \mathrm{sinc}\left(k\right)\\ \\ &= \left[X\left(\dfrac{k}{T}\right) \cdot \mathrm{III}\left(k\right)\right] * \mathrm{sinc}\left(k\right)\\ \\ &= \int_{-\infty}^{\infty} X\left(\dfrac{\tau}{T}\right) \space \mathrm{III}\left(\tau\right) \space \mathrm{sinc}\left(k-\tau\right) d\tau\\ \\ &= \sum_{n=-\infty}^{\infty} X\left(\dfrac{n}{T}\right) \mathrm{sinc}\left(k-n\right)\\ \\ c_k &= X\left(\dfrac{k}{T}\right) \end{align*}$$

If I haven't made a mistake.

Note that this matches your answer for when using a time limited pulse (modulo Fourier Transform conventions), but this derivation made no assumptions about the pulse, $x(t)$, being time limited, but only that it has a Fourier Transform.

The above derivation assumes that $y(t)$ is in fact periodic and that it is finite almost everywhere over the interval of one period. This is certainly the case for $x(t)$ being finite almost everywhere and time limited, even if that time limited region is much larger than $T$. For $x(t)$ not being time limited, you have to prove that the infinite sum of the shifted pulses in any interval of length $T$ is finite almost everywhere in that interval. I believe this should be the case for your example of a Gaussian pulse, where I believe the infinite sum of the ever diminishing tail segments should converge, but I haven't proven that.

  • $\begingroup$ You are right: the collated Gaussian tail indeed converges. I see what you did there. You evaluated at the nulls in the sinc function. Clever. I suspect that your answer might have lost a coefficient of $1/T$ at some point, but this is a minor detail. Otherwise, neatly done. Assuming that this checks out, I'll cite your answer, listing "Walls, Andy," in my bibliography (if you prefer your name written in a different form, let me know). Appreciated. $\endgroup$ – thb May 3 at 16:10
  • 1
    $\begingroup$ Please use "Walls, G. Andrew" for a citation. Thanks. $\endgroup$ – Andy Walls May 3 at 16:25

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