Energy or power: x(t) = e^(−at)

• Energy signal: a signal whose total energy is finite and non-zero, $0 < E < \infty$. Its average power is then zero. Example: a single pulse, $e^{-2t}u(t)$.
• Power signal: a signal whose average power is finite and non-zero, $0 < P < \infty$. Its total energy is then infinite. Example: $\cos\omega_0 t$, $u(t)$, any periodic signal.

For a continuous-time signal:

$$E = \lim_{T\to\infty}\int_{-T}^{T}|x(t)|^2\,dt, \qquad P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt$$

For a discrete-time signal:

$$E = \sum_{n=-\infty}^{\infty}|x[n]|^2, \qquad P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2$$

A signal that has neither finite energy nor finite non-zero power is neither an energy nor a power signal (e.g. $e^{t}$, $t\,u(t)$).

$x(t) = e^{-at}$, $a > 0$

The exponential is usually taken as one-sided, $x(t) = e^{-at}u(t)$, i.e. it starts at $t = 0$.

Energy:

$$\begin{aligned} E &= \int_{0}^{\infty}e^{-2at}\,dt = \left[\frac{e^{-2at}}{-2a}\right]_0^{\infty} = \frac{1}{2a} \end{aligned}$$

Finite, since $a > 0$.

Power:

$$P = \lim_{T\to\infty}\frac{1}{2T}\int_0^{T}e^{-2at}\,dt = \lim_{T\to\infty}\frac{1-e^{-2aT}}{4aT} = 0$$

Answer: $e^{-at}u(t)$ is an energy signal with $E = \frac{1}{2a}$ and $P = 0$.

Note: if $e^{-at}$ is taken for all $t$ ($-\infty < t < \infty$), it grows without bound as $t \to -\infty$: $E = \infty$ and $P = \lim_{T\to\infty} \frac{e^{2aT}}{4aT} = \infty$, so it is neither an energy nor a power signal.