Solving Periodic Signals: Determine Fund. Period

[DrunkEngineer]

Homework Statement

Determine whether or not each of the following signals is periodic if signal is periodic determine the fundamental period (note that these are discrete not continuous signals) Show your solutions
1. $$x(n) = \cos^3(\frac{\pi(n)}{8})$$
2. $$x(n) = \cos(\frac{n}{2})\cos(\frac{\pi(n)}{4})$$

Homework Equations

a. $$f = \frac{\omega}{2\pi}$$ when f is irrational it is non periodic when f is rational it is periodic

b. determining the fundamental period requires the least common multiple of all periods

The Attempt at a Solution

1. $$= \cos(\frac{n\pi}{8})\cos^2(\frac{n\pi}{8})$$
$$= \cos(\frac{n\pi}{8})(1+\cos(\frac{2n\pi}{8}))$$
$$= \cos(\frac{n\pi}{8}) + \cos(\frac{n\pi}{8})\cos(\frac{2n\pi}{8})$$
$$= \cos(\frac{n\pi}{8} + \frac{1}{2}(\cos(\frac{n\pi}{8}-\frac{n\pi}{4}) + \frac{1}{2}(\cos(\frac{n\pi}{8}+\frac{n\pi}{4})$$
$$= \cos(\frac{n\pi}{8} + \frac{1}{2}(\cos(\frac{n\pi}{8})) + \frac{1}{2}(\cos(\frac{3n\pi}{8}))$$
the first cosine: $$f = \frac{\omega}{2\pi} = \frac{\frac{\pi}{8}}{2\pi} = \frac{1}{16}$$ --- rational ---- periodic
2nd cosine $$f = \frac{\omega}{2\pi} = \frac{\frac{\pi}{8}}{2\pi} = \frac{1}{16}$$ ---- rational---- periodic
3rd cosine $$f = \frac{\omega}{2\pi} = \frac{\frac{3\pi}{8}}{2\pi}} = \frac{3}{16}$$----rational? ---- is this periodic?

now how do i get the least common multiple to get the number of samples the period has since the third cosine is 3/16?
the only problem is i need to find the Least common multiple of the three to find the total number of samples in the period or N = 1/f

N1 = 16, N2 = 16 , N3 = 16/3

k/m=N1/N2 = 16 / (16/3) = 3/1; mN1 = kN2; 1(16) = 3(16/3) ==>> is the total number of sample equal to No = 16 which is the least common multiple of N1, N2 and N3?Number 2.
$$x(n) = \cos(\frac{n}{2})\cos(\frac{\pi(n)}{4})$$
$$x(n) = \frac{1}{2} (cos(\frac{n}{2} - \frac{\pi(n)}{4}) + cos(\frac{n}{2} - \frac{\pi(n)}{4}) )$$
$$f1 = \frac{\omega}{2\pi} = \frac{\frac{2-\pi}{4}}{2\pi}$$<<<< irrational number
$$f2 = \frac{\omega}{2\pi} = \frac{\frac{2+\pi}{4}}{2\pi}$$<<<< irrational number
hence it's not periodic

Reference books
DSP by proakis
Schaum's Outline in signals and systems

 

[HallsofIvy]

For number 1, you don't really need to do all that work. $a^3= b^3$ if a= b so as long as $cos(n\pi/8)$ is periodic, $cos^3(n\pi/8)$ is also. The period of $cos^3(n\pi/8)$ is the same as the period of $cos(n\pi/8)$.

Number 2 is correct.

 

[DrunkEngineer]

Am i not going to simplify the trigonometric identity in number 1?

because i have another similar problem given in the book:
x[n] = $$\cos^2(\frac{n\pi}{8})$$

and the sol'n is
[PLAIN]http://img593.imageshack.us/img593/346/dsphomework1.png