I am reading Introduction to quantum noise, measurement and amplification, and I need to understand the Wiener Khinchin theorem: how to derive it. I also need to understand some context around this theorem (why some object are defined the way they are).
The theorem is derived on the page 55 of this document. I will explain to you what I understand from the starting point of this theorem and where I am stuck.
I assume a stationary stochastic gaussian process. I call $R_{XX}(\tau)=\langle X(\tau) X(0) \rangle$ the autocorrelation function of a stochastic variable $X$. I assume the signals to be real.
We define the power spectral density of noise as:
$$ S_{XX}(\omega) \equiv \lim_{T \to +\infty} \langle |X_T(\omega)|^2 \rangle $$
Where $X_T(\omega)$ is the windowed Fourier transforme of $X$ on a period $T$. Basically we define the signal $X_T(t)$ that is equal to $X(t)$ for $t \in [-T/2,T/2]$ and that vanishes elsewhere. The Fourier transform of this signal can thus be defined as:
$$ X_T(\omega) = \frac{1}{\sqrt{T}} \int_{-T/2}^{+T/2} dt X_T(t) e^{i \omega t} $$ To be consistant we must define the inverse Fourier transform as:
$$ X_T(t) = \frac{\sqrt{T}}{2 \pi} \int_{-\infty}^{+\infty} d \omega X_T(\omega) e^{-i \omega t} $$
First questions:
- Why do we take such "uncommon" definition for the Fourier transform (usually it is $\frac{1}{\sqrt{2 \pi}}$ for both sides (direct+reciprocal), or $\frac{1}{2 \pi}$ in front of one of the sides. Is there a theoretical motivation for this in this context ?
- Why do we need to window the function ? Is it to avoid technical difficulties with divergent integrals ?
Now, I go to the derivation and express where I am stuck. The goal is to show that $R_{XX}(\omega)$, the Fourier transform of the autocorrelation function corresponds to the noise power spectral density $S_{XX}(\omega)$. I define $S^T_{XX}(\omega)$ the quantity before the limit of $T \rightarrow +\infty$. I know how we can derive computing the inverse Fourier transform of $S_{XX}(\omega)$ directly as suggested in a link in the comment. However I am stuck on deriving it from the other side (I have $S_{XX}(\omega)$ on the left hand side and I make appear $T.F(R_{XX})$ on the right handside). It might look as a boring question but in the document I follow they always follow this way of proving equalities, and I really don't get the manipulation they do. To understand how they do I need to understand this "other way" of deriving the theorem.
$$ S^T_{XX}(\omega)=\frac{1}{T} \int_{-T/2}^{+T/2} dt \int_{-T/2}^{+T/2} dt' \langle X_T(t) X_T(t') \rangle e^{i \omega (t-t')}$$
Now, I use the fact my process is stationnary. And I recognize the autocorrelation function:
I start from the definition (before taking the limit): $$ S^T_{XX}(\omega)=\frac{1}{T} \int_{-T/2}^{+T/2} dt \int_{-T/2}^{+T/2} dt' R_{X_T X_T}(t-t') e^{i \omega (t-t')}$$
I do the change of variables: $(t,t') \rightarrow (\tau=t, \tau'=t-t')$. It gives me:
$$ S^T_{XX}(\omega)=\frac{1}{T} \int_{-T/2}^{+T/2} d \tau \int_{t-T/2}^{t+T/2} d \tau' R_{X_T X_T}(\tau) e^{i \omega \tau}$$
At this point, I have absolutely no idea how we could proceed.
We can also try another method (which I think is the way they do in the document but I really don't understand their justification). We can do the change of variable $(t,t') \rightarrow (v=t+t',u=t-t')$. The Jacobian gives $1/2$ and the boundaries are the one I write below:
$$S^T_{XX}(\omega)=\frac{1}{T} \int_{-T}^{T} dv \int_{-B(v)}^{+B(v)} \frac{du}{2} R_{X_T X_T}(u) e^{i \omega u}$$
$B(v)=v+T$ for $-T \leq v \leq 0$ and $B(v)=t-v$ for $0 \leq v \leq T$. It gives:
$$S^T_{XX}(\omega)=\frac{1}{T} \int_{0}^{T} dv \int_{-(T-v)}^{+(T-v)} \frac{du}{2} R_{X_T X_T}(u) e^{i \omega u}+\frac{1}{T} \int_{-T}^{0} dv \int_{-(T+v)}^{+(T+v)} \frac{du}{2} R_{X_T X_T}(u) e^{i \omega u}$$
And same problem here: I really don't see how we could get to the result with this way of deriving (which I think is roughly the path taken is the document I linked)
Thus, second question: How can prove the theorem with this method ?