Power spectral density and auto correlation function: Frequency vs time representation

Question

I have a time correlated periodic signal $a(t)$ with $t=0,1,...,T$ and $a(t+T) = a(t)$ that has been sampled at rate $\Delta t$ and I analyse the signal using the PSD via FFT and the ACF. As far as I understand, the standard FFT treats the functions by default as periodic.

For the ACF $\langle a(t) a(0) \rangle$, the largest possible time difference is $T/2$, since the signal is periodic (everything larger would only approach the starting point from the other side), i.e., $\langle a(T/2 + t) a(0) \rangle = \langle a(T/2 - t) a(0) \rangle$, there is no new information after $t=T/2$.

However, for the PSD $\langle a_f a_{-f}\rangle$ the frequency inherited from the FFT is $f_i = [0,1,...,T]\cdot \frac{1}{T\Delta t}$, where $f_i < f_\mathrm{Ny} = \frac{1}{2\Delta t}$ (according to Nyquist). It contains information about correlations of the signal at a frequency of $f_1 = 1/(T\Delta t)$, which should in my understanding be somewhat equivalent to the longest possible time correlation $t = T\Delta t$. But that in the end is equivalent to $t=0$ and thus should be equivalent to the shortest of the frequencies. Intuitively, I would think that only frequencies larger then $f=\frac{1}{T/2}$ contain useful information.

How is that possible? What am I missing? What is the relation between frequencies in the PSD and times in the ACF?

Answer 1

there is no new information after $t=T/2$.

Correct. If the sequence is periodic, so is it's autocorrelation.

It contains information about correlations of the signal at a frequency of $f_1=1/(TΔt)$

I don't think that's a useful interpretation. What is "correlation at a specific frequency" supposed to mean?

What is the relation between frequencies in the PSD and times in the ACF?

The PSD is the Fourier Transform of the ACF.

Answer 2

The autocorrelation at the origin is simply the total mean power of a DT signal $x[n]$ i.e., $R_{x}[0]=\mu_{x}^{2}+\sigma_{x}^{2}$ so its like a DC quantity + AC quantity. Now take the DTFT of $R_{x}[0]$ what you will get is the PSD at $k=0$ so $R_{x}[0]$ can be represented as some inverse DTFT : $$ R_{x}[0]=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_{x}(e^{j\Omega})\;\text{d}\Omega $$ So what this is telling me is that I have a DT signal $x[n]$ that is contributing a portion of a power at a certain frequency $\Omega$ in the frequency domain to the total power of the signal i.e., $R_{x}[0]$.