Question

The period of the discrete-time signal \(x[n]\) described by the equation below is \(N =\) ..................... (Round off to the nearest integer). \[ x[n] = 1 + 3\sin\!\left(\frac{15\pi}{8}n + \frac{3\pi}{4}\right) - 5\sin\!\left(\frac{\pi}{3}n - \frac{\pi}{4}\right) \]

Hint:

For discrete signals, fundamental period = LCM of periods of individual components, provided each is periodic.

Answer 1

Step 1: Check periodicity of each sinusoid.
A discrete sinusoid \(\sin(\omega n + \phi)\) is periodic if \(\omega / 2\pi\) is rational. - First term: \(\omega_1 = 15\pi/8\). \[ \frac{\omega_1}{2\pi} = \frac{15/8}{2} = \frac{15}{16} \] Thus period: \[ N_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{15\pi/8} = \frac{16}{15} \] The fundamental period in integer \(n\) is denominator of fraction \(15/16\), i.e., 16. - Second term: \(\omega_2 = \pi/3\). \[ \frac{\omega_2}{2\pi} = \frac{1}{6} \] Thus period: \[ N_2 = 6 \]

Step 2: Overall period.
Overall period = LCM of individual periods: \[ N = \text{LCM}(16, 6) = 48 \]

Final Answer:
\[ \boxed{48} \]