**1 Answer**

written 7.4 years ago by |

A signal is said to be periodic signal if x(n) = x(n+N)

Where, N = Fundamental Period

i.e., if $f=\dfrac{k}{N}$ where, k & N both are integers

**$X_1 (n) = \cos(0.5 \pi n + 0.3)$**Sol: Consider Standard equation $x(n) = A \cos( \omega n+ \theta)$

Comparing given signal we get, $\omega n = 0.5 \pi n \\ \therefore 2 \pi fn = \dfrac{\pi n}{2} \\ \therefore f = \dfrac14$

**Result: Given signal, $X_1$ (n) is periodic with Fundamental Period, N = 4**

$X_2 (n) = \cos(0.3 \pi n) + \sin(0.25 \pi n)$

Sol: Consider $X_2(n)$ to be combination of $x_1(n) \& x_2(n)$

Where, $x_1(n)= \cos(0.3 \pi n_1) \& x_2(n)= \sin(0.25 \pi n_2)$

$\therefore \omega_1n_1 = 0.3 \pi n_1 \ \ \& \ \ \omega_2n_2 = 0.25 \pi n_2$

$\therefore 2 \pi f_1n_1 = \dfrac{3}{10} \pi n_1 \ \ \ \ \& \ \ 2 \pi f_2n_2 = \dfrac{1}{4} \pi n_2$

$\therefore f_1 = \dfrac{3}{20} \& f_2 =\dfrac{1}{8}$

$\therefore N_1 = 20 \ \ \ \& \ \ N_2 = 8$

$\therefore N = LCM of N_1 \ \ \ \& \ \ N_2 = 40$

**Result: Given signal, $X_2 (n)$ is periodic with Fundamental Period, N = 40**