These are Things I studied.

$ E[X]:-$ THis gives us Average value or Dc value

$E[X^2]:-$Total power

$\sigma^2=E[X^2]-E[X]^2$:-Ac power

From Autocorrelation

$R_x(0)=E[X^2]$--Total power

And $ \lim_{x\to \infty}{R_x(\tau)}=$Dc power


The Autocorrelation fuction $ R_x(\tau) $ of a Wide-sense random process shown in below figure

enter image description here

The Average Power is____

So what I did? I just Find Value At $t=\infty $ and that is Zero But the official answer is 2.

So they find $R_x(0)$.

Here is my confusion. is there any difference between DC power and average power?

I just relate from E[x], which gives the Average value or DC value.

where I am thinking wrong?


2 Answers 2


What you have studied doesn't seem to include a definition of Average Power. For deterministic signal $x(t)$, the instantaneous power delivered at time $t$ is $x^2(t)$ ${\big (}$yeh, yeh, nitpickers should choose $R=1$ in their beloved formula $\dfrac{V^2}R\big )$ which varies from instant to instant. Thus, the average power delivered by $x(t)$ can be calculated as $$\text{Average Power} =\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x^2(t)\, \mathrm dt.\tag{1}$$ If $x(t)$ has a DC value, it is given by $\displaystyle x_0 = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)\, \mathrm dt$ and so we can write $x(t) = x_0 + x_{\text{ac}}(t)$ (prove for yourself that $x_{\text{ac}}(t)$ has DC value $0$) so that $(1)$ becomes \begin{align}\require{cancel}\text{Average Power} &=\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T (x_0 + x_{\text{ac}}(t))^2\, \mathrm dt\\ &= \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x_0^2 + +2x_0x_{\text{ac}}(t) + x_{\text{ac}}^2(t)\, \mathrm dt\\ &= x_0^2 + 2x_0 \cancelto{0}{\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x_{\text{ac}}(t)\, \mathrm dt} + \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x_{\text{ac}}^2(t)\, \mathrm dt\\ &= \text{DC Power} + \text{AC Power}\\ &= \text{Total power} \end{align} On the other hand, for wide-sense-stationary random processes, the total power (which is also the same as the average power) is $R_X(0)$ by definition, and so for the figure shown, the total power is $2$. The total power is the sum of the DC Power and the AC power and since the DC Power is $0$ for the $R_x(t)$ shown, the average power asked for also has value $2$ and ot consists solely of AC power.


is there any difference between DC power and average power

Yes. DC power is the square the of the mean of the signal . Average power is the mean of the square of the signal.

It looks your teacher is using poor terminology. In this case "average power" and "total power" are the same. Neither one is a great term.


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