How to find the fundamental period if it is periodic? $$ x[n] = \cos((\pi/2)n)\cos((\pi/4)n) $$ I have simplified to the following but not sure what to do next: $$ \cos((\pi/2)n)\cos((\pi/4)n) = 1/2[\cos((3\pi/4)n)+\cos((\pi/4)n)] $$ Any help is appreciated!
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Please read this https://math.stackexchange.com/q/164221/399263. Then the period is a divisor of $8$. – zwim Feb 10 '21 at 20:54
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Hint:
Determine the period of $\cos\frac{3\pi}4n$ and of $\cos\frac{\pi}4n$.
The fundamental period of their sum is the l.c.m. of both periods.
Bernard
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The lcm of their period is a period of the sum, but the fundamental one may be smaller. – zwim Feb 10 '21 at 20:56
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If you simplify further expanding with the formula $\cos 3\alpha=\cos ^3(\alpha )-3 \sin ^2(\alpha ) \cos (\alpha )$ you get $$\frac{1}{2} \cos ^3\left(\frac{\pi n}{4}\right)+\frac{1}{2} \cos \left(\frac{\pi n}{4}\right)-\frac{3}{2} \sin ^2\left(\frac{\pi n}{4}\right) \cos \left(\frac{\pi n}{4}\right)$$ So it's much easier to see that the period is $8$.
Raffaele
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