I know that power of a periodic signal is defined as follows:
$$P =\lim_{T\rightarrow\infty} \frac{1}{2T} \int_{-T}^{T}{{|x(t)|}^2dt}$$
or
$$P =\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}{{|x(t)|}^2dt}$$
However, I am confused by the limit term $T\rightarrow \infty$. Does that also mean that the frequency of the signal is also going to 0?
For example, let us take the function $x(t)=\cos(\omega_0t)$ where $\omega_0 = 2\pi f_0= 2\pi / T_0$.
To calculate the power of this signal, we would do the following:
$$P =\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}{\cos^2(\omega_0 t)dt}$$
$$ =\lim_{T\rightarrow\infty} \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}{\frac{1}{2}+\frac{1}{2}\cos(2\omega_0 t) }dt$$
$$ =\lim_{T\rightarrow\infty} \frac{1}{T} \left[\left.\frac{t}{2}\right|_{-\frac{T}{2}}^{\frac{T}{2}}+\left.\frac{1}{2\omega_0}\sin(2\omega_0 t)\right|_{-\frac{T}{2}}^{\frac{T}{2}}\right] $$
$$ =\lim_{T\rightarrow\infty} \frac{1}{T} \left[\frac{T/2- (-T/2)}{2}+\frac{1}{2\omega_0}\left(\sin\left(2\omega_0 \frac{T}{2}\right)-\sin\left(2\omega_0 \frac{-T}{2}\right)\right)\right] $$
$$ =\lim_{T\rightarrow\infty} \frac{1}{T} \left[\frac{T}{2}+\frac{1}{2\omega_0}\left(\sin\left(\omega_0 T\right)-\sin\left(-\omega_0 T\right)\right)\right] $$
$$ =\frac{1}{2}+\lim_{T\rightarrow\infty} \frac{1}{2\omega_0 T}\left(\sin\left(\omega_0 T\right)-\sin\left(-\omega_0 T\right)\right) \textbf{ (eq.1)}$$
Then by expanding the equation using $\omega_0 = 2\pi f_0= 2\pi / T_0$, we get:
$$ =\frac{1}{2}+\lim_{T\rightarrow\infty} \frac{T_0}{4\pi T}\left(\sin\left(\frac{2\pi T}{T_0}\right)-\sin\left(-\frac{2\pi T}{T_0}\right)\right)$$
At this point, the there are two ways that I can go forward:
I can assume that $T= T_0$. Then, the sin of an integer multiple of pi is just zero and I get the correct result $P=\frac{1}{2}$. i.e.
$$ =\frac{1}{2}+\lim_{T\rightarrow\infty} \frac{1}{4\pi}\left(\sin\left(2\pi\right)-\sin\left(-2\pi\right)\right)$$
$$ =\frac{1}{2}+ \frac{1}{4\pi}\left(0-0\right)=1/2$$
But does this mean that the meaning of the power formula assume that the period of the signal is going to infinity and is there a reason, theory or point behind that?
The other way I can go around that is that I can assume $\omega_0$ to be a constant in eq.1, and as $T$ goes to infinity, the term $\frac{1}{2\omega_0 T}\left(\sin\left(\omega_0 T\right)-\sin\left(-\omega_0 T\right)\right)$ just goes to zero.
Also, are both ways correct?
Best Answer
I am not very sure where you are getting at and what you mean by "frenquency going to $0$". At least in your example the frenquency $\omega_0$ is always a constant.
As I understand, the integral $$\frac{1}{a-b}\int_a^bf(x)dx$$ is the average of $f(x)$ over the interval $[a,b]$.
If $f(x)$ has period $T$, then one can prove that $$\frac{1}{T}\int_a^{a+T}f(x)dx\quad(*)$$ is a constant regardless of the choice of $a$. One might say in this case that $f(x)$ has a "well-defined" average.
If, unfortunately, $f(x)$ is not periodic but we still want to talk about the "average" of $f(x)$, we consider its average over some finite interval $[-T/2,T/2]$ $$\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$ and take the limit $T\to\infty$ and call the limit (if it exists) $$\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}f(x)dx$$ the average of $f(x)$ over $\mathbb{R}$.
It can also be shown that the second average is consistent with the first in the case where $f(x)$ is periodic. My opinion is that the limit definition of the average of a function is simply a generalization to deal with not strictly periodic functions. In your case, power is the average of energy over time, so you may understand the limit as the average of the energy of a signal over a very long time.
But still, I don't see how this relates to any change in frenquency.