Can somebody guide me on how I can find the energy and power of the signal $$x(t)=e^{-10t}\cos(30\pi t)u(t).$$

Whether we have to split the $\cos$ to exponential and solve or go with the usual method of solving, both seems to be challenging.

  • $\begingroup$ Do you need to calculate using formulas or with a program tool? $\endgroup$ – havakok Aug 30 '16 at 9:17
  • $\begingroup$ Energy $\endgroup$ – havakok Aug 30 '16 at 9:24
  • $\begingroup$ relation between power and energy $\endgroup$ – havakok Aug 30 '16 at 9:26
  • $\begingroup$ i have to calculate ,can you suggest me how to start? $\endgroup$ – Aneil Aug 30 '16 at 9:26
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    $\begingroup$ From zero to infinity (don't forget the step $u(t)$)! Just use an integral table if you can't solve it on your own; that's a pretty standard integral. $\endgroup$ – Matt L. Aug 30 '16 at 11:08

Let's write the signal $x(t)$ in the general form

$$x(t)=e^{-\alpha t}\cos(\omega_0t)u(t)\tag{1}$$

Note that its energy is given by

$$E_x=\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{0}^{\infty}e^{-2\alpha t}\cos^2(\omega_0)tdt\tag{2}$$

Since this is a homework type problem I won't solve it for you but I'll give you some hints:

  1. Expand the integrand of $(2)$ as $$e^{-2\alpha t}\frac12(1+\cos(2\omega_0t))=\frac12e^{-2\alpha t}+\frac12e^{-2\alpha t}\frac12(e^{j2\omega_0t}+e^{-j2\omega_0t})$$
  2. Now you have two integrals over basic exponential functions, which should be very straightforward to solve.
  3. The final result should be $$E_x=\frac{1}{4\alpha}+\frac{\alpha}{4(\alpha^2+\omega_0^2)}$$
  4. If the energy $E_x$ is finite, what can you conclude about the power of $x(t)$?
  • $\begingroup$ @robertbristow-johnson: Given the level of the question, I'm pretty sure that the OP doesn't mean some kind of short-term power as a function of $t$, but the total power of the signal. Since $x(t)$ is an energy signal, we know what the power will be ... $\endgroup$ – Matt L. Aug 30 '16 at 17:45
  • $\begingroup$ power has to be sliding, a function of $t$. half of the magnitude-square of the analytic signal is one way. or the short-term local mean of the $|x(t)|^2$ above. either way, power will be about $\frac12 e^{-2\alpha t} u(t)$. $\endgroup$ – robert bristow-johnson Aug 30 '16 at 17:52
  • $\begingroup$ Matt L ,i dont know how you arrived at the 2nd part of the result,i can easily get i/4a but the other part,how can you solve me,i am confused because we have to take real part but exponential that you splitted has imaginary too.If you can just give me idea about the real part of of e^(a+ib) also i can move ahead clear on that part.please. $\endgroup$ – Aneil Aug 31 '16 at 0:03
  • $\begingroup$ @Aneil: Just solve the two complex integrals, and of course the result of each of them will be complex. But once you add them up you'll see that the imaginary parts cancel out and the result is real-valued. Just do it carefully and you'll arrive at the correct result. $\endgroup$ – Matt L. Aug 31 '16 at 8:39

$x(t)=e^{-\alpha t}\sin(\omega_0t)u(t)$ is called damped sine wave. Its Fourier transform is $$X(j\omega)=\frac{\omega_0}{(\alpha+j\omega)^2+\omega_0^2}$$ A straightforward approach to calculate the signal energy is to use Parseval's identity to calculate the enery of this signal:

$$E=\int_{-\infty}^{\infty}\lvert x(t)\rvert^2dt=\frac 1 {2\pi}\int_{-\infty}^{\infty}\lvert X(j\omega)\rvert^2d\omega$$

If we have $\omega_0>>\alpha$, then the spectrum has a sharp peak near $\pm \omega_0$. So for instance around $\omega_0$ we can do

$$\lvert X(j\omega)\rvert\approx\frac{1/2}{\sqrt{(\omega-\omega_0)^2+\alpha^2}}$$ and by integrating we will get to the following result $$E\approx \frac {1}{8\alpha}+\frac {1}{8\alpha}=\frac {1}{4\alpha}$$

Regarding power of the signal, since this is an energy signal (i.e. a signal with bounded energy), then its power is zero. The opposite case would be a power signal (i.e. a signal with non-zero and bounded power) whose energy is infinite. An example of such power signals are periodic signals such as sine and cosine (one or two-sided).

Note that when $\alpha\to 0$, this signal tends to a power signal (a pure sine wave), and consequently $E\to\infty$

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    $\begingroup$ I don't think that using Parseval's indentity makes things much easier here. Can you show how you arrive at the (approximate) result? $\endgroup$ – Matt L. Aug 30 '16 at 12:29
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    $\begingroup$ using Parseval makes the problem harder. both for instantaneous power and for energy. $\endgroup$ – robert bristow-johnson Aug 30 '16 at 12:49
  • $\begingroup$ Maybe. However, looking at the spectrum adds a level of confidence to make sure everything is fine. The final result is generic though. Of course, one may get a similar result in time domain. $\endgroup$ – msm Aug 30 '16 at 12:54

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