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This is an extension of a Phys.SE question I asked earlier, Fourier transform of two pulses of light.

I start with 2 Gaussians, represented by the equation $e^{-(t/a)^2}\times e^{2\pi i d t}*(\delta(t-\tau) + \delta(t+\tau))$ where a corresponds to the FWHM of the Gaussian. (FWHM $\approx 15,\ a \approx 9.$ d is the shift in the frequency domain, and $\tau$ is the separation between Gaussians. The $e^{2\pi i d t}$ is simply adding a carrier frequency. I want the Fourier transform to be centered at 900nm light, or a frequency of $3.33\times10^{14} \frac ms$.

When I take the Fourier transform of the equation, I get $2a\sqrt\pi \cos(2\pi p \tau)e^{-(a \pi p)^2} * \delta(p - d)$ I want to know which units I should use for each variable, to get the output in terms wavelength in nm, or frequency in hz. Additionally I would like to know the beat frequency of the fourier transform in terms of $\tau$. The way its written now, $\tau = $ the beat frequency, but if if i have the units wrong, it could change.

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$a$ has dimension of inverse time because $-(t/a)^2$ appears in an exponent, $\sqrt\pi$ is dimensionless much like $2$ or a cosine or exponential of anything. $\delta(y)$ has the same units as $1/y$ because it's the derivative of the dimensionless step function with respect to $y$. Convolution symbol $*$ has the same units as the variable in which the convolution takes place, from the $dt$ etc. integration measure in the convolution. The units of a product should be obvious if you know what the word "unit" means. – Luboš Motl Jul 30 '13 at 14:58
@LubošMotl I think that $a$ has dimensions of time, not inverse of time (frequency), because the exponent has to be dimensionless. – neutrino Jul 30 '13 at 21:41
Moreover I believe you're asking for the units one should use for the transform variables, not the units of the Fourier transform itself. Am I right? – WetSavannaAnimal aka Rod Vance Jul 31 '13 at 4:51
up vote 1 down vote accepted

The Fourier transform definition you are using is:

$\mathfrak{F}(h(t))(\nu) = \tilde{H}(\nu) =\int_{-\infty}^\infty e^{-2\pi\,i\,\nu\,t} \,h(t)\mathrm{d}t$

In this convention, the transform variable is in $\mathrm{Hz}$, as opposed to radians per second, when $t$ is in seconds. So, your independent variable $t$ needs to be in seconds, to get the independent frequency variable to be in hertz.

Wavelength is inversely proportional to frequency $c = \nu \lambda$, so the transformation between frequency and wavelength spectrums is nonlinear. If your pulse is narrowband, so that its frequency spread is much less than its carrier frequency, you can use $\nu = \frac{c}{\lambda} \Rightarrow \Delta \nu \approx -\frac{c}{\lambda^2} \Delta \lambda = -\frac{\nu}{\lambda} \Delta \lambda$. The scaling constant therefore is:

$\nu \approx \nu_0 - \frac{\nu_0}{\lambda_0} (\lambda - \lambda_0) = 3.33\times10^{14} \mathrm{Hz} - 3.7\times10^{11}(\lambda - 900\mathrm{nm}) = 6.67\times10^{14} - 3.7\times10^{11}\lambda $

With $\lambda$ in nanometres. So, to convert your Fourier transform with transform variable of frequency in Hz to one with transform variable of wavelength in nanometres, you plug in:

$\tilde{H}(\nu) = \tilde{H}(6.67\times10^{14} - 3.7\times10^{11}\lambda)$

Again, this formula assumes $\lambda$ is in nanomatres. Beware, though, if your source is broadband (pulse duration less than 0.1 picoseconds) on interpreting power spectrums calculated from the Fourier transform. If your power spectrum is $G(\nu) = |\tilde{H}(\nu)|^2$, and you are trying to interpret $G$ as a power per unit wavelength, you must bring in the Jacobian of the nonlinear transformation. Given $\nu = \frac{c}{\lambda}$, witness that the power in the interval spanned by $[\nu_1,\nu_2]$ is

$\int_{\nu_1}^{\nu_2} G(\nu) d\nu = -\int_{\frac{c}{\nu_1}}^{\frac{c}{\nu_2}} G(\frac{c}{\lambda}) \frac{c}{\lambda^2} d\lambda$

So to get a power spectrum in power per wavelength, you have to use $\left|\tilde{H}\left(\frac{c}{\lambda}\right)\right|^2 \frac{c}{\lambda^2} $, not simply $\left|\tilde{H}\left(\frac{c}{\lambda}\right)\right|^2$

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