Sketch {x(t) + x(−t)}u(t) from figure

Energy and power signals

• Energy signal: total energy $E = \int_{-\infty}^{\infty}|x(t)|^2dt$ is finite and non-zero, and the average power is zero. Example: $x(t) = e^{-t}u(t)$ ($E = 0.5$ J); a single rectangular pulse of height $A$ and width $\tau$ ($E = A^2\tau$).
• Power signal: average power $P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt$ is finite and non-zero, so $E = \infty$. Example: $x(t) = A\sin\omega_0 t$ ($P = A^2/2$); the unit step $u(t)$ ($P = 1/2$).
• A signal cannot be both; some (e.g. $t\,u(t)$) are neither.

Sketch of $y(t) = \{x(t) + x(-t)\}u(t)$

From the figure, $x(t)$ is:

$$x(t) = \begin{cases} t + 1, & -2 < t < -1 \\ 1, & -1 < t < 0 \\ 2, & 0 < t < 1 \\ 2 - t, & 1 < t < 2 \\ 0, & |t| > 2 \end{cases}$$

Because of $u(t)$, $y(t) = 0$ for $t < 0$; only $t > 0$ is needed. For $t > 0$, $x(-t)$ takes the values of $x$ on the negative side:

• $0 < t < 1$: $-t \in (-1, 0)$, so $x(-t) = 1$.
• $1 < t < 2$: $-t \in (-2, -1)$, so $x(-t) = -t + 1 = 1 - t$.

Adding piece by piece:

So $y(t)$ is 0 for $t < 0$, jumps to 3 at $t = 0$, stays at 3 up to $t = 1$, drops to 1 at $t = 1$, then falls linearly with slope $-2$ from $1$ at $t = 1$, crossing zero at $t = 1.5$, to $-1$ at $t = 2^-$, and jumps back to 0 at $t = 2$.

 y(t)
  3 +-------+
    |       |
  2 |       |
    |       |
  1 |       +\
    |         \
  0 +-------+---\---+------> t
    |             \ |
 -1 |              \|
    0       1  1.5  2

(Note: $x(t) + x(-t) = 2x_e(t)$, so $y(t)$ is twice the even part of $x(t)$ kept for $t > 0$.)