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This question might be a bit stupid, anyway, i'll risk it, since i want to get better understanding of this subject.

Let's consider random signal x(t), and let's say that we know that it is ergodic by mean, and by autocorrelation (this implies that it is wide sense stationary).

This means that time averaged mean of this signal is equal to it's statistic averaged mean, same goes for autocorrelation.

So we have the following:

$\mu_X=E(x)=\frac{1}{T} \int_T x(t)dt=\langle x(t)\rangle \\ R_{xx}(\tau)=E[x(t)x(t+\tau)]= \lim_{T->\infty} \frac{1}{T} \int_T x(t)x(t+\tau)dt=\langle x(t)x(t+\tau)\rangle$

Then, as i found in one of the books i have, it says following:

$\langle x(t) \rangle$- corresponds to DC level of given signal

$\langle x(t) \rangle ^2$ - corresponds to normalized power of DC component

$\langle x^2(t) \rangle$ - corresponds to total average normalized power

$\overline{\sigma_x^2}= \langle x^2(t) \rangle - \langle x(t) \rangle^2$-corresponds to average normalized power in AC component of the signal

$\overline{\sigma_x}$-corresponds to rms of AC component

My question is, is there any way to prove this, or to gain intuitive understanding why this is true. Any help appreciated!

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Forget the stuff about random signals since your $x(t)$ is one realization (a.k.a. sample path) of the random process and this realization is an ordinary deterministic signal. What is the DC level of the signal? Well, it is the average value of the signal over all time, and is thus just $$\lim_{T\to\infty} \frac{1}{2T}\int_{-T}^T x(t) \, \mathrm dt \tag{1}$$ if you like. And if you don't like the definition of average value, make up your own and justify it to everybody else, but for heaven's sake, don't spout stuff like "To find the DC value of $x(t)$, all I need to do is add up all the values of $x(t)$ and divide by the total number of values that I added up." Now, ergodicity comes in where we look at the expected value of the random variable $X(t)$, where said expected value is the same number $\mu_X$ for all choices of $t$, and claim that $\mu_X$ equals the value computed in $(1)$ for almost all sample paths $x(t)$ (allowing for the possibility that not all sample paths are equally representative of the process, but the event that the sample path is an offbeat weirdo is an event of probability $0$).

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Dilip: I'm confident that what you said is correct but the weirdo sample path that you refer to ( the one with probability zero ) can be used to estimate the autocorrelation at different lags, namely $\hat{R}_{xx}(\tau)$. This is true even though it's only one realization and this is the case only because of the ergodicity assumption. So, to a person coming in new, my opinion is that your answer doesn't give enough significance to the one realization. It's a weirdo but a useful weirdo for sure !!!!!

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