# Fourier transform of a top-hat function in the Faraday Measurement synthesis context

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:

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?

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