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Earlier I have dealt with exponential functions multiplied with unit step function. But, energy and power of exponential function alone comes out to be infinite when I put limits of the integrals to from $-\infty$ to $+\infty$. How can I find its energy or power if the signal is not multiplied with unit step function?

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An ideal exponential signal $x(t)=e^{at}$ , which extends from $-\infty$ to $\infty$ has infinite energy and infinite power, as for real $a >0$ ( and similarly for real $a < 0$) you have

$$ \mathcal{E}_x = \int_{-\infty}^{\infty} e^{2at} dt = \lim_{t \to \infty} \frac{1}{2a} e^{2a t} \to \infty $$

and

$$ \mathcal{P}_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} e^{2at} dt = \lim_{T \to \infty} \frac{e^{a T}}{2aT} \to \infty $$

Therefore It's neither energy nor power signal. Note that it's not a practical signal and exists only for mathematical purposes.

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  • $\begingroup$ the context obviously implies this, but just to be explicit, it may be nice to be explicit that $a$ is real $\endgroup$ – Robert L. Sep 15 '18 at 23:33
  • $\begingroup$ thanks yes, $a$ is real, otherwise we call it a general complex exponential $e^{st}$ for complex number $s = \sigma + j \omega$ $\endgroup$ – Fat32 Sep 15 '18 at 23:35

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