I'm currently trying to calculate the Fourier transform of a top-hat function in the context of Faraday Measurement Synthesis. This is pretty straightforward, however, I cannot understand why I cannot get the desired intensity in Faraday depth space ($\phi$ space, analog to signal space).

I'm defining a top-hat function with intensity $s$ as:

$$ F(\phi) = \begin{cases} s, & \phi_0-\phi_{\text{fg}}/2 < 0 < \phi_0+\phi_{\text{fg}}/2\\ 0, & \text{elsewhere} \end{cases} $$ where $\phi_0$ and $\phi_{\text{fg}}$ are the center and the width of the top-hat function, respectively.

In the context of Faraday Measurement Synthesis, the Fourier transform is defined as: $$P(\lambda^2) = \int F(\phi) e^{2j\phi\lambda^2} d\phi$$

Replacing $F(\phi)$ we get that: $$ \begin{align} P(\lambda^2) &= s \int_{\phi_0-\phi_{\text{fg}}/2}^{\phi_0+\phi_{\text{fg}}/2} e^{2j\phi\lambda^2} d\phi \\ &= \frac{s}{2j\lambda^2}(e^{2j(\phi_0+\phi_{\text{fg}}/2)\lambda^2}-e^{2j(\phi_0-\phi_{\text{fg}}/2)\lambda^2})\\ &= \frac{s}{2j\lambda^2}e^{2j\phi_0\lambda^2}(e^{j\phi_{\text{fg}}\lambda^2} - e^{-j\phi_{\text{fg}}\lambda^2})\\ &= \frac{se^{2j\phi_0\lambda^2}}{\lambda^2} \sin(\phi_{\text{fg}}\lambda^2)\\ &= \frac{\phi_{\text{fg}}se^{2j\phi_0\lambda^2}}{\phi_{\text{fg}}\lambda^2}\sin(\phi_{\text{fg}}\lambda^2) \end{align} $$

As you see, the last expression can be also seen as a shifted sinc. Now, say I want a top-hat function with intensity 1, width 140rad/m^2 and centered at 200rad/m^2. Then the peak of my sinc in $P(\lambda^2)$ will need to be 140. If I do that I get this:

Simulated data

Inverse Fourier transform

To calculate the IFT I'm using a IDFT such that:

$$ F(\phi) = \frac{1}{N} \sum_{i=0}^{N} P(\lambda_i^2)e^{-2j\lambda_i^2\phi} $$

The first image represents the simulated data $P(\lambda^2)$, the second image represents its Fourier transform $F(\phi)$. Note that $P(\lambda^2)$ only has data $\lambda^2 > 0$, this is normal and part of the problem. My question is why the peak in Faraday depth space ($\phi$-space) is ~11 and not 1. I have missed something in the derivation?

More information about this can be found in:

Faraday Rotation Measurement Synthesis

In this paper though, they assume a slab with signal with value $s/\phi_{\text{fg}}$, therefore you end up with: $$ P(\lambda^2) = \frac{se^{2j\phi_0\lambda^2}}{\phi_{\text{fg}}\lambda^2}\sin(\phi_{\text{fg}}\lambda^2) $$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.