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Using a plotted function, I've come up with a function consisting of step and ramp functions; say $x(t) = 50u(t+30)+20u(t-30)+5r(t-30)$, where $u(t)$ is the unit step function and $r(t)$ is the ramp function (which I believe is just $tu(t)$). I'm not sure where to begin with regards to calculating the energy and average power of this function. I know energy and average power each have their own integrals, but how do I solve such a big function, and how do I deal with the fact that there are step and ramp functions?

Thanks in advance for any assistance!

Michael Hardy
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Yuerno
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1 Answers1

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By definition, energy of the signal $s(t)$ is $$E=\int_{-\infty}^{+\infty}|s(t)|^2 dt$$ which is $\infty$ for both $u(t)$ and $r(t)$.

Also the power of signal is $$P=\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^{+T}|s(t)|^2 dt$$ which is $\infty$ for $r(t)$ and is $\frac{1}{2}$ for $u(t)$.

The integrals are simple. Example for $u(t)$:

$$E=\int_{-\infty}^{+\infty}|u(t)|^2=\int_{0}^{+\infty}1 dt=(x+c)\Biggr\vert^{+\infty}_0=\infty$$

$$P=\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^{T}|u(t)|^2 dt=\lim_{T \to \infty}\frac{1}{2T}\int_{0}^{T}1 dt=\lim_{T \to \infty}\frac{1}{2T}(x+c)\Biggr\vert^{T}_0=\frac{1}{2}$$

If there is even one term in your function that is infinity, then the result is infinity. Also since the bound of integral is infinity, shifting in time does not change anything.

msm
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  • Thank you for the reply, that definitely helps to clear things up! A couple of smaller questions as a followup: Would the fact that the u(t) and r(t) functions are shifted not affect the integrals? And do you know of any sources where I can see the integrals done for the u(t) and r(t) functions? Just for the sake of my curiosity. – Yuerno Sep 18 '16 at 02:31
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    @Yuerno see the answer in the update. – msm Sep 18 '16 at 02:50
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    This is awesome, thanks so much! One final question, in the energy integral I notice you have .1 dt, vs the power integral having 1 dt. Shouldn't the energy integral also have 1 dt, or if it is .1 dt, how does it become that? – Yuerno Sep 18 '16 at 02:54
  • you're welcome! the dot was typo! – msm Sep 18 '16 at 02:55
  • I figured as much! Thanks once again. – Yuerno Sep 18 '16 at 02:57