The periodogram is in fact the Fourier transform of the aperiodic
correlation of the windowed data sequence
image-20240907215822425image-20240907215957865
image-20240907230715637
estimating
continuous-time stationary random signal
periodogram.drawio
The sequence x[n]
is typically multiplied by a finite-duration window w[n], since the input to
the DFT must be of finite duration. This produces the
finite-length sequence v[n] = w[n]x[n]
That is, by (1)$$
\hat{P}_{ss}(\Omega) = T_s\hat{P}_{xx(\omega)} =
\frac{T_s|V(e^{j\omega})|^2}{\sum_{n=0}^{L-1}(w[n])^2}=\frac{|V(e^{j\omega})|^2}{f_{res}L\sum_{n=0}^{L-1}(w[n])^2}
$$
That is, by (2)$$
\hat{P}_{ss}(\Omega) = T_s\hat{P}_{xx(\omega)} =
\frac{T_sL|V(e^{j\omega})|^2}{\sum_{k=0}^{L-1}(W[k])^2} =
\frac{|V(e^{j\omega})|^2}{f_{res}\sum_{k=0}^{L-1}(W[k])^2}
$$
!! ENBW
Wiener-Khinchin theorem
Norbert Wiener proved this theorem for the case of a
deterministic function in 1930; Aleksandr
Khinchin later formulated an analogous result for stationary
stochastic processes and published that probabilistic
analogue in 1934. Albert Einstein explained, without proofs, the idea in
a brief two-page memo in 1914
x(t), Fourier
transform over a limited period of time [−T/2,+T/2] , XT(f) = ∫−T/2T/2x(t)e−j2πftdt
With Parseval’s theorem∫−T/2T/2|x(t)|2dt = ∫−∞∞|XT(f)|2df
So that $$
\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt =
\int_{-\infty}^{\infty}\frac{1}{T}|X_T(f)|^2df
$$
where the quantity, $\frac{1}{T}|X_T(f)|^2$ can be interpreted as
distribution of power in the frequency domain
For each f this quantity is
a random variable, since it is a function of the random process x(t)
The power spectral density (PSD) Sx(f)
is defined as the limit of the expectation of the expression above, for
large T: $$
S_x(f) = \lim _{T\to \infty}\mathrm{E}\left[ \frac{1}{T}|X_T(f)|^2
\right]
$$
The Wiener-Khinchin theorem ensures that for well-behaved
wide-sense stationary processes the limit
exists and is equal to the Fourier transform of the
autocorrelation$$\begin{align}
S_x(f) &= \int_{-\infty}^{+\infty}R_x(\tau)e^{-j2\pi f \tau}d\tau \\
R_x(\tau) &= \int_{-\infty}^{+\infty}S_x(f)e^{j2\pi f \tau}df
\end{align}$$
Note: Sx(f)
in Hz and inverse Fourier Transform in Hz ($\frac{1}{2\pi}d\omega = df$)
$$ \cos(2\pi f_0t) \overset{\mathcal{F}}{\longrightarrow} \frac{1}{2}[\delta(f -f_0)+\delta(f+f_0)] $$
