PSD and $\lim_{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T x(t)\bar y(t)\,dt$

Question

From Wikipedia, I taken a definition of power spectral density:

For continued signals that describe, for example, stationary physical processes, it makes more sense to define a power spectral density (PSD), which describes how the power of a signal or time series is distributed over the different frequencies, as in the simple example given previously. Here, power can be the actual physical power, or more often, for convenience with abstract signals, can be defined as the squared value of the signal. For example, statisticians study the variance of a set of data, but because of the analogy with electrical signals, it is customary to refer to it as the power spectrum even when it is not, physically speaking, power. The average power P of a signal x(t) is the following time average: $$P = \lim_{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T x(t)^2\,dt.$$

I premise that I'm not an expert on the signal theory, therefore I do apologise if this question isn't much precise.

Let $x(t),y(t)$ be two complex signals and denote with $\bar y$ the complex conjugate of $y$. What represents the following formula? $$\lim_{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T x(t)\bar y(t)\,dt$$ Is it related to PSD?

Thanks in advance.

Answer 1

For power signals $x(t)$ and $y(t)$, the function

$$R_{xy}(\tau)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(t)\bar{y}(t+\tau)dt\tag{1}$$

is the cross-correlation of $x(t)$ and $y(t)$. So the expression you're asking about is the cross-correlation of $x(t)$ and $y(t)$ evaluated at lag $\tau=0$:

$$R_{xy}(0)=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}x(t)\bar{y}(t)dt\tag{2}$$

Since the Fourier transform of $(1)$ is the cross-power spectrum $S_{xy}(\omega)$ of $x(t)$ and $y(t)$, the expression in $(2)$ is related to $S_{xy}(\omega)$ by:

$$R_{xy}(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{xy}(\omega)d\omega\tag{3}$$

So the expression in your question is proportional to the integral of the cross-power spectrum of the power signals $x(t)$ and $y(t)$.

Answer 2

Usually, for power signals, we define the inner product to be

\begin{align} \left<x\,,\,y\right> &= \lim_\limits{T\rightarrow \infty} \frac 1 {2T} \int\limits_{-T}^T x(t)\bar y(t)\,dt \end{align}

which induces the vector norm

\begin{align} ||x||^2 &= \left<x\,,\,x\right>\\ &=\lim_{T\rightarrow \infty} \frac 1 {2T} \int\limits_{-T}^T x(t)\bar x(t)\,dt \end{align}

which, for real signals, is the same as the Wikipedia article claims is the "average power"; Wikipedia is a bit inexact here; for complex signals, the average power is

\begin{align} P(x) &= \left<x\,,\,x\right>\\ &= \lim_\limits{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T {|x(t)|}^2\,dt\\ &=\lim_\limits{T\rightarrow \infty} \frac 1 {2T} \int_{-T}^T x(t)\bar x(t)\,dt\text . \end{align}

We might interpret the inner product $\left<x,y\right>$ as "similarity" between $x$ and $y$.

Thanks to the corollaries of the Cauchy-Schwarz inequality ($\left<x,y\right>\le ||x||\,||y||$), we know that when varying $y$, $|\left<x(t),y(t)\right>|,\, t\in\mathbb K$ reaches a maximum iff $x = \lambda y,\, \lambda \in \mathbb K$.

Following above definition

\begin{align} \left<x\,,\,y\right> &= \left<x\,,\lambda x\right>\\ &= \lim_{T\rightarrow \infty} \frac 1 {2T} \int\limits_{-T}^T x(t)\overline{\lambda x(t)}\,dt&\lambda\in\mathbb R \Rightarrow \bar \lambda = \lambda\\ &= \lambda \lim_{T\rightarrow \infty} \frac 1 {2T} \int\limits_{-T}^T x(t)\bar x(t)\,dt \\ &= \lambda \left<x\,,\, x\right>\\ &= \lambda ||x||^2\text{ .} \end{align}

This isn't surprising, because inner products are bilinear. It's also application-wise not surprising, because no signal is more similar to $x$ than $x$ itself!

So: No, that formula is not directly related to PSD. However, this is so-called math, and you can derive a lot of relations. Just go ahead and calculate

$$\left< \mathcal F (x)\,,\, \mathcal F (y) \right>$$

to find a few of these. It's not hard to prove Parseval's Theorem this way.