# Signals and Systems: Determine if the signal is periodic or nonperiodic(discrete)

1. ### DrunkEngineer

18
1. The problem statement, all variables and given/known data
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. $$x(n) = \cos(\frac{n}{2})\cos(\frac{\pi(n)}{4})$$

2. Relevant 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

3. 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

Last edited: Jan 2, 2011
2. ### HallsofIvy

40,514
Staff Emeritus
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.

3. ### DrunkEngineer

18
Am i not gonna 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