Energy or power: x(t) = e^(−3t)

• 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. $e^{t}$, $t\,u(t)$).

Checking $x(t) = e^{-3t}$ (defined for all $t$)

$|x(t)|^2 = e^{-6t}$.

Energy:

$$\begin{aligned} E &= \lim_{T\to\infty}\int_{-T}^{T} e^{-6t}\,dt = \lim_{T\to\infty}\frac{e^{6T}-e^{-6T}}{6} = \infty \end{aligned}$$

Power:

$$\begin{aligned} P &= \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{e^{6T}-e^{-6T}}{6} = \lim_{T\to\infty}\frac{e^{6T}}{12T} = \infty \end{aligned}$$

(The exponential grows faster than $T$, so the limit is infinite.)

Since both $E$ and $P$ are infinite, $x(t) = e^{-3t}$ is neither an energy signal nor a power signal. The signal blows up as $t \to -\infty$.

Note: if the signal is taken as one-sided, $x(t) = e^{-3t}u(t)$, then

$$E = \int_0^{\infty} e^{-6t}\,dt = \frac{1}{6}\ \text{J}, \qquad P = 0$$

and it is an energy signal.

Answer: $e^{-3t}$ for all $t$: neither ($E = \infty$, $P = \infty$); $e^{-3t}u(t)$: energy signal with $E = 1/6$.