# How to use inverse Fourier transform to get the original signal

Suppose I have a signal $$u(x)$$ where $$x\in[0,2\pi]$$ and the signal has following form in Fourier space,

$$\hat{u}(\kappa)=\begin{cases} \frac{1+i\tan{(\varphi_u(\kappa))}}{\sqrt{1+\tan^2{(\varphi_u(\kappa))}}}\kappa^{\alpha/2},\ \ 0<\kappa\le\kappa_c \\ 0+i0,\ \ \ else \end{cases}$$

where $$\varphi_u(\kappa)$$ is a random number for each $$\kappa$$, and $$\alpha$$ is a constant.

How can I get $$u(x)$$ using inverse Fourier transform?

• Hi and welcome to the MathOverflow. I suspect that the sole knowledge that $\varphi_u$ is a random number is not of great help: do you know something about its probability distribution as a random variable? Aug 30, 2020 at 9:09
• @DanieleTampieri $\varphi_u(\kappa)$ is a random variable which is uniformly distributed in $[-\pi, \pi]$ for each $\kappa$. Aug 31, 2020 at 14:18

The expectation value $$E[u](x)$$ can be obtained by inverse Fourier transform of the expectation value of $$\hat{u}(k)$$, since the Fourier transform is a linear operation; If I interpret the statement in the OP that the phase $$\phi$$ is a "random number" as saying that $$\phi$$ is uniformly distributed, then the integral $$\frac{1}{\pi}\int_0^\pi \frac{1+i\tan\phi}{\sqrt{1+\tan^2\phi}}\,d\phi=\frac{2}{\pi}$$ gives $$E[u](x)$$ as the inverse Fourier transform of $$(2/\pi)\kappa^{\alpha/2}$$, which is an incomplete Gamma function.
I should note that the statement in the OP that both $$u(x)$$ and $$\hat{u}(k)$$ have compact support is inconsistent.