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

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...

• Sure! I am working on discrete signals. There is a theory that confirm the eq. (1), and it is also confirmed in the signal in which I work on. But I have also found that , in all the signal I work with, x is a constant and it is equal to -2/3. Unfortunally I don't know how to confirm it mathematically – Mardy Owens Apr 5 at 19:10
• I think your post would be easier to read if you used $u_k$ or $u[k]$ instead of $u(t)$, and denoted the FFT as $\mathcal F \left \lbrace u_k \right \rbrace$ (\mathcal F \left \lbrace u_k \right \rbrace) Denote the IFFT as $\mathcal F^{-1} \left \lbrace u_k \right \rbrace$. Note that in strict mathematical terms, there is no FFT -- the FFT is just a fast algorithm for calculating the DFT, mapping $x_k$ onto $X_\kappa$, where $k, \kappa \in \left[0 \dots N \right )$. – TimWescott Apr 5 at 20:19

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:
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 $$f^{-\frac{2}{3}}$$.
• Indeed, they are given in I cannot find reference (paper) of this relation $u(t)t^{α} ↔^{FT} f^{-(α+1)}$ – Laurent Duval Apr 6 at 14:57
• Sorry I have a doubt about this proof. If the $|X(f)|~f^ {-5/3}$ than it means that $\alpha=-1/6$ So x(t) should be a decreasing function, isn't it? In my case the signal is a fluctuation and its time average is 0. Is this consistent with this proof? – Mardy Owens Apr 7 at 10:27
• @MardyOwens I don't think you need that your signal is $t^{\alpha}$. In my opinion, this proof uses a syllogism. 1. from $t^{\alpha}$, you find the frequency dependence of the PSD 2. from $t^{\alpha}$, you find the the frequency dependence of the magnitude spectrum of the square of the signal. So, then you can find the magnitude spectrum of the square of the signal starting from the PSD. – Luca Mirtanini May 6 at 17:42