Even/odd signals; energy or power check

Even and odd signals

• Even: $x(-t) = x(t)$, symmetric about the vertical axis. Example: $\cos t$.
• Odd: $x(-t) = -x(t)$, antisymmetric about the origin, $x(0) = 0$. Example: $\sin t$.

  Even: x(t) = |t|           Odd: x(t) = t
  \      |      /                    |     /
    \    |    /                      |   /
      \  |  /                        | /
 --------+--------> t       ---------+---------> t
         0                         / |
                                 /   |
  mirror image about        180 deg symmetry
  vertical axis             about origin

Any signal can be split as $x_e(t) = \frac{1}{2}[x(t)+x(-t)]$, $x_o(t) = \frac{1}{2}[x(t)-x(-t)]$.

(i) $x(t) = 5A\cos(\omega t + \phi)$

It is periodic with $T = 2\pi/\omega$, so compute power over one period:

$$\begin{aligned} P &= \frac{1}{T}\int_0^{T} 25A^2\cos^2(\omega t+\phi)\,dt \\ &= \frac{25A^2}{T}\int_0^{T}\frac{1+\cos(2\omega t+2\phi)}{2}\,dt = \frac{25A^2}{2} \end{aligned}$$

$E = \infty$ (infinite duration). Power signal, $P = 12.5A^2$.

(ii) $x[n] = \sin(n/4)$

Here $\omega_0 = 1/4$, $\omega_0/2\pi = 1/(8\pi)$ is irrational, so the sequence is not periodic, but it is bounded and lasts forever.

$$\begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\sin^2\frac{n}{4} \\ &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1-\cos(n/2)}{2} = \frac{1}{2} \end{aligned}$$

because the sum of $\cos(n/2)$ stays bounded while $2N+1 \to \infty$. Energy $E = \sum\sin^2(n/4) = \infty$.

Answer: (i) power signal, $P = 25A^2/2$; (ii) power signal (aperiodic), $P = 1/2$.