Fourier transform of two pulses of light I have laser beam path that fires two pulses of light in a gaussian distribution, so the intensity graph over time is two identical gaussians separated by a distance $t_0$. In other words, a gaussian convolved with two $\delta$ functions equidistant from the origin. When I take the Fourier transform of this, I get a graph with beats, but I'm having trouble figuring out what this graph means physically. When we feed this beam through a spectrometer, we are trying to find out if we should see fringes, and if so, what their frequency is.
Any help would be appreciated.
 A: This is a great question because you have to think carefully about what you experiment is actually doing to answer it.
As in Andreas H's answer, you may in principle (and in some practice) see the intensity comb (the "beats" as you call them) at the output of grating spectrometer. But in practice it won't always be like this and you have to analyse your kit a bit more deeply to find out whether you will see the beats or a smooth Gaussian on your image plane.
First and foremost, the laser's coherence time must be greater than the time interval spanned by the two pulses. Otherwise the two pulses' spectrums may tend to add incoherently and you may simply see the spectrum for one pulse alone but simply double in intensity. The laser's and modulation system's physics will have something to say here. If the pulses are made by pulsing the laser's supply current, or by Q-switching, then you will just get to pulses comprising mutually incoherent photons, so you do see simply a continuous Gaussian as the spectrum. If the pulses arise from mode locking, or by passive modulation of a continuous wave, long coherence time laser, then the comb spectrum will in principle be seen at the output.
Now let's think about the grating. Firstly, we need to assume our imaging system for the grating has to be of low enough aberration that the grating spectrometer's output truly is the spatial Fourier transform (Fraunhofer diffraction) of the input aperture (i.e. the grating lit by the time-pulsed field). If there is aberration, or if the output is not truly a far field image, then the output will be some more-compicated-than-simply-the-Fourier-transform diffraction integral (a Fresnel diffraction integral, for instance) of the input aperture, which may mask the comb, so that you may tend to see more like a smooth Gaussian spectrum.
Moreover, the grating's spectrometer abilities arise from the spatial frequency upconversion imposed on a spatial field distribution by multiplying this distribution by an infinitely wide comb function. This spatial frequency upconversion depends on the light's wavelength, so the grating diffracts different wavelength components at different Bragg angles and thus splits spectrum. In practice, the grating is of finite width, i.e. multiplied by some compactly supported gating function. In the Fraunhofer diffraction, this modification by the gating function corresponds to a convolution with the gating function's Fourier transform. So you theoretical Fourier transform of the form $\exp\left(-\frac{x^2}{2\,\sigma^2}\right) \cos(\alpha\,x\,\Delta t)$ get convolved with something like a $\mathrm{sinc}$ function and thus the combs can get smeared into one another, particularly if the time separation between the input pulses is very big. Here $x$ is the sideways distance from the output spectrum's centre, $\alpha$ is set by the system's dimensions and delays and $\Delta t$ is the delay between the two input pulses. The output intensity can thus look more like $\exp\left(-\frac{x^2}{2\,\sigma^2}\right)$. 
Lastly, the combs can be very very finely spaced, so, depending on the spatial resolution of your CCD array or whatever sensor you use in the spectrometer, the imaging sensor might impose a spread-out kernel that smears over the combs. In this case, you will also only tend to see a continuous Gaussian spectrum.
A: The spectrometer measures the absolute value squared of the Fourier transform of the input signal, so you see "beating". 
It is just a fact that also far "separated" pulses in time domain affect the spectrum and you see beating. But this is nothing mysterious, because if you have two pulses the frequency component corresponding to the inverse time delay is increased in the spectrum, compared to the single pulse case (because there is only a single pulse). 
I think the term "beating" is just misleading since it sounds that something needs actively to interact. But this is incorrect. Also the spectrometer, to resolve this beating, always has to measure longer than the inverse of the beat frequency (so the pulses both have to arrive during measurement). This is the uncertainty principle and quite obvious because to discriminate a frequency component from others you have to measure at least for one period length of that beat frequency. 
Essentially it lies in the definition of the Fourier transform (and definition of frequency).
If you google frequency comb (that is a laser that has a discrete line spectrum), you will notice that this in time domain are just pulses (e.g. a femtosecond laser).
A: As an aside, this effect is called spectral interferometry in optics.  But it's even seen in seismology of surface waves that go around the earth many times.  If you Fourier transform a long (say 24 hour) surface wave record, the fringes are much narrower than a short recording (say 4 hours).  Here is an example of the same effect in laser ultrasound:  http://acoustics.mines.edu/preprints/PRE46618.pdf Figure 2 shows a surface wave pulse going around a cylinder many times, and Figure 3 shows the periodogram of two pulses vs six.  In the limit of an infinite number of pulses the power spectrum becomes a Dirac comb.
