Can I simplify $x=\frac{\ln(|\mathcal F((\mathcal F^{-1}(\sqrt{(f^{-5/3})}))^{2})|)}{\ln(f)}$

Question

Given a variable in time $u_{k}$, and an other variable f which represents the frequencies in a range [a,b]:

$$|\mathcal F \left \lbrace u_{k}\right \rbrace |^{2}=f^{-5/3}\tag{1}$$

where $\mathcal F$ stands for fast Fourier transform.

Now, I want to find $x$ knowing that:

$$|\mathcal F \left \lbrace u_{k}^{2}\right \rbrace|= f^{x}\tag{2}$$

I can isolate $x$ as:

$$x=\frac{\ln(|\mathcal F \left \lbrace u_{k}^{2}\right \rbrace |)}{\ln(f)}\tag{3}$$

I can find $u$ using (1):

$$u=\mathcal F^{-1}(\sqrt{(f^{-5/3})}) \tag{4}$$

On the consequence $x$ is:

$$x=\frac{\ln(|\mathcal F \left \lbrace ( \mathcal F^{-1} \left \lbrace \sqrt{(f^{-5/3})} \right \rbrace )^{2} \right \rbrace|)}{\ln(f)} \tag{5}$$

Is there any way in which I can simplify $x$?? I am stuck with this problem for a lot of weeks...

Answer 1

Here is a potential starting point: it is considered that when a function behaves as a power law ($x(t) \sim t^\alpha, t> 0$), then its magnitude spectrum as well, with $|X(f)| \sim f^{-(1+\alpha)}$. One can see for instance:

• Computing Fourier transform of power law
• An Interesting Fourier Transform - 1/f Noise
• Sketches of a "proof" and references (since there are convergence issues) are provided in an answer to I cannot find reference (paper) of this relation $u(t)t^{α} ↔^{FT} f^{-(α+1)}$.

If I plug that in your case, since $|X(f)|^2 \sim f^{-5/3}$, so we seek $-5/6=-(1+\alpha)$. I obtain $\alpha = -\frac{1}{6}$. So you would have $x^2(t) \sim t^{2\alpha}$, with a magnitude spectrum $|\chi(f)|$ in $f^{-(1-\frac{1}{3})}$, or $f^{-\frac{2}{3}}$.

Now, is the discrete signal sufficiently well-sampled and clean to verify it in the discrete domain is another question.