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I'm trying to take the fourier transform of a short laser pulse, represented by $E(t) = E_oe^{-(t/\Delta T)^2}\times e^{-i \omega t}$

E is the electric field of the laser pulse. $E_o$ and $\Delta T$ are both constants. Specifically I want to know if there are beats in the fourier transform, and what their frequency is. Any help would be great.

Zar
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  • The FT of a quadratic phase signal with rectangular support is another quadratic phase signal with rectangular support. Search for info on the FT of LFM (linearly frequency modulated) signals and you see examples of how to derive the FT using the principle of stationary phase. In short though, the answer to your question of whether or not there are any "beats" is no (assuming that by "beats" you mean spikes in the frequency domain). – AnonSubmitter85 Jul 26 '13 at 20:49
  • If it's a very sharp pulse, the Fourier transform of the Dirac $\delta$ is a sine – Guido Kanschat Jul 26 '13 at 20:53
  • Ignore my previous comment. I missed that the first exponential term is real. I assumed out of habit that it was a complex exponential and, thus, was a quadratic phase signal. – AnonSubmitter85 Jul 26 '13 at 20:59

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The $e^{-i\omega t}$ factor in your pulse means only a phase delay by $\omega t$, otherwise the pulse is a Gaussian pulse which means that the Fourier transform will also be Gaussian and it will be $$\large \mathcal{E}(f)=E_0\Delta T\sqrt{\pi}e^{-\Delta T^2 \pi^2\left(f+\frac{\omega}{2\pi}\right)^2}$$

So, the power will be mostly concentrated in the frequency range $\displaystyle \left[0,\frac{1}{\sqrt{2}\pi \Delta T}\right]$

Also, the frequency domain pulse has its peak at the $\frac{-\omega}{2\pi}$ frequency and then it monotonically decreases as $f$ increases.

Samrat Mukhopadhyay
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  • I thought the $e^{-i\omega t}$ added a beat to the original function, because it multiplies the Gaussian by $cos(-\omega t) + i sin(-\omega t)$ – Zar Jul 26 '13 at 21:03
  • Multiplication by $e^{-i\omega t}$ only shifts the Fourier transform to the left in frequency domain. – Samrat Mukhopadhyay Jul 26 '13 at 21:07
  • ok, in the frequency domain, i thought you meant time domain. One other quick question, what happens if its positive $i\omega t$? I tried it and it seemed to cancel, giving me a constant, which I don't think is right. – Zar Jul 26 '13 at 21:09
  • Then the frequency response will shift to right giving you a peak at $\frac{\omega}{2\pi}$ – Samrat Mukhopadhyay Jul 26 '13 at 21:12
  • @Zar Beat notes usually come about through addition and not modulation (multiplication). That is, $f(t) + \exp{j\omega t}$ will add a beat, while $f(t) \cdot \exp{j\omega t}$ will shift the spectrum by $\omega$. – AnonSubmitter85 Jul 26 '13 at 21:14