Unit step in terms of signum; energy or power

Unit step in terms of signum

The signum function is

$$\text{sgn}(t) = \begin{cases} +1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0 \end{cases}$$

and the unit step is $u(t) = 1$ for $t > 0$, $0$ for $t < 0$.

Adding 1 to sgn$(t)$ gives $2$ for $t>0$ and $0$ for $t<0$; halving it gives the unit step:

$$u(t) = \frac{1}{2}\left[1 + \text{sgn}(t)\right]$$

Check: $t>0$: $\frac12(1+1) = 1$; $t<0$: $\frac12(1-1) = 0$; $t=0$: $\frac12$ (the mid-value used at the jump).

 sgn(t)              u(t)
  1 |------          1 |------
    |                  |
 ---+------> t      ---+------> t
-1--|                  |

This relation is used to find the Fourier transform of $u(t)$: since $\mathcal{F}\{\text{sgn}(t)\} = \frac{1}{j\pi f}$ and $\mathcal{F}\{1\} = \delta(f)$,

$$U(f) = \frac12\delta(f) + \frac{1}{j2\pi f}$$

Energy or power?

Energy:

$$E = \int_{-\infty}^{\infty} u^2(t)\,dt = \int_{0}^{\infty} 1\,dt = \infty$$

So it is not an energy signal.

Power:

$$\begin{aligned} P &= \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2} u^2(t)\,dt \\ &= \lim_{T\to\infty}\frac{1}{T}\int_{0}^{T/2} 1\,dt = \lim_{T\to\infty}\frac{1}{T}\cdot\frac{T}{2} = \frac12 \end{aligned}$$

Answer: $u(t) = \frac12[1 + \text{sgn}(t)]$; the unit step has infinite energy and finite average power $P = 0.5$ W (normalised), so it is a power signal.