2
$\begingroup$

I have to do a demonstration. If we do the Parseval identity of the signals $x(t)$, $y(t)$ and $z(t)$ that go from $0$ to $T$ and that are real, we have:

$\int_{0}^{T} x(t)^2dt=\int_{-\infty}^{\infty}|X(f)|^2df$

$\int_{0}^{T} y(t)^2dt=\int_{-\infty}^{\infty}|Y(f)|^2df$

$\int_{0}^{T} z(t)^2dt=\int_{-\infty}^{\infty}|Z(f)|^2df$ (1)

If we sum up the these 3 equations, and we divide both sides by the interval $T$, we obtain:

$\frac{1}{T}\int_{0}^{T} (x(t)^2+y(t)^2+z(t)^2)dt=\frac{1}{T}\int_{-\infty}^{\infty}(|X(f)|^2+|Y(f)|^2+|Z(f)|^2)df$ (2)

Can you confirm me that everything is ok in this demonstration? I am trying to demonstrate this because my advisor says that since the PSD of the signals $x$, $y$, $z$, which are $S_{X}$, $S_{Y}$, $S_{Z}$, respect this relation:

$\int_{-\infty}^{\infty}(S_{X}(f)+S_{Y}(f)+S_{Z}(f))df=\frac{1}{T}\int_{0}^{T} (x(t)^2+y(t)^2+z(t)^2)dt$ (3)

then (and this is the point that is totally wrong for me) he says that $S_{X}+ S_{Y}+S_{Z}$ is the Fourier transform of ( $x(t)^2 +y(t)^2+z(t)^2$):

$S_{X}+ S_{Y}+S_{Z} = \int_{-\infty}^{\infty}( x(t)^2 +y(t)^2+z(t)^2)\exp(-itf2\pi)dt$ (4)

this cannot be possible in my opinion. I already tried to show him the procedure that leads to the PSD of a signal etc. but he still believe (maybe because I am young, not expert and just arrived) that is true what he said. So I am trying to do the above mentioned demonstration, saying that since $|X(f)|^2$ is different from the absolute value of the Fourier transform of $x(t)^2$, it cannot be contemporary true what it is stated in the eq. (4) and the application of the Parseval identity (2).

I beg you to tell me if my procedure is good. I want to convince him with something that is unassailable. (I am almost desperate, I have been trying to convince him that is wrong since 3 month ago)

$\endgroup$
2
$\begingroup$

Your first steps seem OK to me. Even if I'd put absolute values in the time domain as well (for generality), if the first three integrals exist, the fourth is valid. Integration is linear.

The weak point resides in the reverse inference. The equality of energy integrals does not mean that the integrands are related. For one, PDSs are real and positive functions. And there is NO reason why the Fourier transform of a sum of squared functions should be real and positive in general.

To convince your advisor, I'd suggest you to build your question on a clever counter-example.

$\endgroup$
  • 1
    $\begingroup$ Thank you for your answer. When you say reverse inference, which equation do you mean? When you say "For one" are you meaning for the equation (1)? Which counter-example would you use? (I am desperate :) ) $\endgroup$ – Arkadiusz duka Jul 13 at 7:48
  • 1
    $\begingroup$ I would like to do an example, but I don't know which one. I tried to apply it to my case study. But it's difficult with one example explain why his theory is wrong. @Laurent Could you please answer to my question? $\endgroup$ – Arkadiusz duka Jul 13 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.