Why do power signals have finite power and infinite energy?

[parthj09]

I have read that for
power signals : power = finite and energy = infinite
Energy signals : power = 0 and energy = finite
Can anyone give practical example for this...?
Also we can represent power signal through Fourier series only but not through Fourier trransform..unlike energy signals.. why is that..?
I googled it but didnt get any satisfactory answerReport**

 

[Jano L.]

I have read that for
power signals : power = finite and energy = infinite
Energy signals : power = 0 and energy = finite
Can anyone give practical example for this...?
Also we can represent power signal through Fourier series only but not through Fourier trransform..unlike energy signals.. why is that..?
I googled it but didnt get any satisfactory answerReport**

For example, take function

$$
f_1(t) = \sin \omega_0 t
$$

This function has "power" $|f_1|^2 \leq 1$, but "energy"
$$
\int_{-\infty}^{\infty} |f_1|^2\,dt
$$
is infinitely great.

This function does not have ordinary Fourier transform (in terms of Fourier function), because the ordinary Fourier integral does not converge. It has however Fourier series: the series contains only one term, the function itself.
(there is however generalized Fourier transform of $f_1$, which employs delta distributions).

There are also functions that have infinite "energy" and cannot be represented even by a Fourier series, for example realization of the Ornstein-Uhlenbeck process.

 

[parthj09]

Thank you..but i still have one doubt...in your explanation about Fourier transform.
As much as i know Fourier transform do exist for f_1(t) = Asin wt
It would be F_1(f) = iA/2 [(f+fm) - (f- fm)]
taking w = 2 pi fm
Now f_1(t) is a power signal then why its Fourier transform is available?

 

[Jano L.]

It would be F_1(f) = iA/2 [(f+fm) - (f- fm)]

Haven't you forgot delta distributions in the bracket? With distributions, it is true that one can write generalized Fourier transform even for sin function.

 

[parthj09]

Yeah but it still doeant answer my question.. why power signals cannot be represented in Fourier transform?

 

[Jano L.]

It is hard to prove that all "power" functions do not have Fourier representation. It probably is not true.

Instead, I will show why a function given by Fourier series (which has non-zero average power) cannot be represented via Fourier transform.

Let the function be given by

$$
f(t) = \sum_k c_k e^{ik\omega_0 t}.
$$
It has average "power", or better - average square - equal to
$$
\overline{f^2} = \sum_k |c_k|^2.
$$
The function is periodic and does not decay to zero.

However, any Fourier representation

$$
f(t) = \int_{-\infty}^{\infty} \tilde{f}(\omega)\, e^{i\omega t}\, \frac{d\omega}{2\pi}.
$$
with piecewise continuous $g(\omega)$ has to decay to 0 as $t\rightarrow \pm \infty$ (Riemann - Lebesgue effect).

Hence the Fourier series $f(t)$ cannot be represented by any piecewise continuous function $\tilde f(\omega)$.

This is not a proof, but I think it gives an idea why the Fourier transform does not work for such functions.

 

[parthj09]

Yeah thanks for this example.
I did some digging too and found out that all power signals are periodic in nature, and it is impossible to write Fourier transform of periodic signals. They can only be represented by Fourier series.
And also according to you..fourier transforms are supposed to decay..which power signals do not .. so its a win win for us both. :-)