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I have trouble with calculating the Energy of the following simple signal
$x(t) = e^\left(j\left(t+\pi/2\right)\right)$

Shouldn't it have $A=1=|x(t)|$, So
$$E_{x}=\lim_{T\to\infty}\int_{-T/2}^{T/2}|x(t)|^2dt = \lim_{T\to\infty}\int_{-T/2}^{T/2}1^2dt$$

temp8hh666
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  • I wonder if this is on topic here - yes, it does talk about energy, but unless we know what kind of signal that is, we don't know how exactly energy relates to amplitude. I could see it being decided either way. If people think it's off topic here, we'll migrate to [dsp.SE]. – David Z Nov 07 '15 at 10:46
  • What do you mean by Energy? Mathematical functions do not have energy in the sense of physics. – Ján Lalinský Nov 07 '15 at 11:37

2 Answers2

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The signal is periodic with 2pi. Since the signal here is exponential which is euler's representation for sinusoidal signal, these sinusoidals in turn are periodic signals. Actually you are misleading here in the integration process. Since the integration will result in t with limit as specified which will give us infinite energy as the final answer; which is the case for periodic signals. If you wish you can take average power which will be 1.

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As @Deep Chandra has written, your signal has infinite energy. This is easily justified if you look at definition of the energy and (average) power: $$E_{x} = \int\limits_{-\infty}^{+\infty}|x(t)|^2dt$$ $$P_{x} = \lim\limits_{T_{0} \rightarrow 0}\frac{1}{T_{0}}\int\limits_{-T_{0}/2}^{T_{0}/2}|x(t)|^2dt$$ [For periodic signals with period $T_{0}$ you integrate from some $t_{0}$ up to $t_{0}+T_{0}$] In your case you simply have $|x(t)|^2 = 1$ so that $P_{x} = 1$. Any signal which has limited power has infinite energy (The other way: if you have finite energy the power is $0$).

WoofDoggy
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