# Power of a real exponential signal

I am having difficulty with the case when a is real.

But after this, I am not going anywhere.

I think the correct expression for the power is $$P = \lim_{T\rightarrow\infty}\frac{e^{-aT}-e^{aT}}{2aT}.$$

When $$T \rightarrow \infty$$, $$e^{aT}$$ tends to infinity "faster" than $$1/T$$ and the limit does not exist.

However, an interesting thing happens when $$a=0$$. The signal $$x(t) = e^{-at}$$ becomes $$e^0=1$$, which is a power signal (with power equal to 1 watt).

I'm sure that (using L'Hopital or some other trick) it can be shown that $$P=1$$ when $$a=0$$, but I have not tried it. In any case, I would say that the premise of the question is wrong, and in fact $$x(t)$$ is a power signal for at least one (real) value of $$a$$.

• Thank you for triggering another answer. You came first! Apr 1, 2021 at 17:27

[Note: this answer was triggered by @MBaz's answer and mention on the de L'Hospital rule, which can be avoided.]

Let us suppose that $$T>0$$. If you consider a complex $$a$$, the conjugate of $$x(t)=\exp^{-at}$$ is:

$$\overline{\exp^{-at}}=\exp^{\overline{-at}}=\exp^{-\overline{a}t}\,.$$

Hence the integral on the interval is:

$$P_{x,T} = \frac{1}{T}\int_{-T/2}^{T/2} e^{-2a_rt} \mathrm{d}t\,,$$ where $$a_r$$ denotes the real part of $$a$$ (which is zero when $$a$$ is purely imaginary). We can treat both cases together, with $$a_r\to 0$$ to account for the imaginary case. Then:

$$P_{x,T} = -\frac{1}{2a_rT}( e^{-a_rT}- e^{a_rT})\,.$$

We can rewrite it with leading terms as:

$$P_{x,T} = \frac{e^{|a_r|T}}{2|a_r|T}(1 - e^{-2|a_r|T})\,.$$

If $$|a_r|\ne 0$$, $$P_{x,T}$$ behaves like $$\frac{e^{|a_r|T}}{2|a_r|T}$$ and tends to infinity, because $$(1 - e^{-2|a_r|T})\to 1$$ as $$T\to\infty$$. If $$|a_r|\approx 0$$ (imaginary case), at the limit,

$$(1 - e^{-2|a_r|T})\sim 2|a_r|T$$ then the numerator and denominator cancel, and $$P_{x,T} \to 1\,.$$

Basically, when $$a_r$$ is zero (or the power is imaginary), the function $$x(t)$$ really behaves as a constant or a "cisoid". Otherwise, that is a true exponential with divergence.

• I'm glad you filled in the details. Thanks!
– MBaz
Apr 1, 2021 at 17:55