The period of the discrete-time signal \(x[n]\) described by the equation below is \(N =\) ..................... (Round off to the nearest integer).
\[
x[n] = 1 + 3\sin\!\left(\frac{15\pi}{8}n + \frac{3\pi}{4}\right) - 5\sin\!\left(\frac{\pi}{3}n - \frac{\pi}{4}\right)
\]
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Solution and Explanation
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:
\[
\boxed{48}
\]
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