DT periodicity; energy/power of 5cosπt + sin5πt

Necessary condition for DT periodicity

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}$$

For $x[n+N] = x[n]$ we need

$$e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}$$ $$\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}$$

So $e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number. The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for every $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

$f(t) = 5\cos\pi t + \sin 5\pi t$

$T_1 = 2\pi/\pi = 2$ s and $T_2 = 2\pi/5\pi = 0.4$ s; the ratio is 5 (rational), so $f(t)$ is periodic with $T_0 = 2$ s. A periodic signal has infinite energy, so check its power:

$$\begin{aligned} P &= \frac{1}{T_0}\int_0^{T_0}f^2(t)\,dt \\ &= \frac{1}{2}\int_0^{2}\left[25\cos^2\pi t + \sin^2 5\pi t + 10\cos\pi t\sin 5\pi t\right]dt \\ &= \frac{1}{2}\left[25 + 1 + 0\right] = 13 \end{aligned}$$

(The cross term integrates to zero over a common period; each sinusoid of amplitude $A$ contributes $A^2/2$: $25/2 + 1/2 = 13$.)

$E = \infty$.

Answer: $f(t)$ is a power signal with $P = 13$ W (normalised).