Fast Fourier Transform $\mathbf{\tilde{A}}=\mathbf{A}e^{i \theta(\mathbf{k})}$

I have a vector which takes form $\mathbf{\tilde{A}}=\mathbf{A}e^{i \theta(\mathbf{k})}$, where $\mathbf{k}$ is the frequency vector ($k^2=k_x^2 +k_y^2+k_z^2$), $i$ is unitary complex number, while $\mathbf{A}$ and $\theta$ are the vector amplitudes and phases, respectively. $\theta(\mathbf{k})$ froms a uniform random distribution within $0$ and $2\pi$, and $\mathbf{\tilde{A}}$ is also randomly distributed but has a probability distribution.

$P(A, \theta)= \frac{A}{2\pi \left | A_k \right |^2} \rm exp\left ( -\frac{A^2}{2\left | A_k \right |^2} \right ) dA d\theta$

$A$ and $\phi$ are amplitude and phase of $\mathbf{\tilde{A}(\mathbf{k})}$. ${\tilde{A}(\mathbf{k})}$ has the form: ${\tilde{A}(\mathbf{k})}=a k^{-n}$, where $a$ and $n$ are constant. I need to transform to real space by taking Fourier Transform of $\mathbf{\tilde{A}(\mathbf{k})}$.

The displayed equation doesn't make sense. The differentials inside the exponent should probably be outside it; if you write differentials you need to have them on both sides of the equation; you haven't introduced $A$ or $A_k$, only $\mathbf A$ and $\mathbf{\tilde{A}}$; and $A_k$ doesn't occur on the left-hand side. –  joriki May 4 '12 at 6:48