Energy or power: u(t) and δ(t)

• 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. $e^{t}$, $t\,u(t)$).

(a) $x(t) = u(t)$

$$E = \lim_{T\to\infty}\int_{-T}^{T}u^2(t)\,dt = \lim_{T\to\infty}\int_0^{T}1\,dt = \lim_{T\to\infty}T = \infty$$ $$P = \lim_{T\to\infty}\frac{1}{2T}\int_0^{T}1\,dt = \lim_{T\to\infty}\frac{T}{2T} = \frac{1}{2}$$

$u(t)$ is a power signal, $P = 1/2$.

(b) $x(t) = \delta(t)$

Model $\delta(t)$ as a pulse $p_\Delta(t)$ of width $\Delta$ and height $1/\Delta$ (area 1), and let $\Delta \to 0$:

$$E = \lim_{\Delta\to 0}\int_0^{\Delta}\left(\frac{1}{\Delta}\right)^2dt = \lim_{\Delta\to 0}\frac{1}{\Delta} = \infty$$

The energy is infinite, so $\delta(t)$ is not an energy signal. For any fixed pulse width, the energy is finite and the average power is

$$P = \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{1}{\Delta} = 0$$

So the power is not a finite non-zero value either. Hence $\delta(t)$ is neither an energy nor a power signal (it is not square-integrable; note that its area $\int\delta(t)dt = 1$ is not its energy).

Answer: (a) $u(t)$: power signal, $P = 0.5$, $E = \infty$. (b) $\delta(t)$: neither ($E = \infty$).