Signals and Systems: Determine if the signal is periodic or nonperiodic

[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

Related example:
x(n) = cos(2n)
f = w/(2pi) = 2/(2pi) = 1/pi ----> this is an irrational number hence it is not periodic

The Attempt at a Solution

do i need to do trigonometric identities ? I read in the book that the sum of two periodic signals is "periodic" and it does not mention "product" where the given equation above is the case

 

[vela]

It wouldn't hurt to take a look at trig identities to see if they make the answer to the questions clearer. This page may help.

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula

 

[DrunkEngineer]

It wouldn't hurt to take a look at trig identities to see if they make the answer to the questions clearer. This page may help.

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula

yes i know trig identities: but the point is to find whether the two equations above are periodic or non periodic,

the first problem seems to be non periodic because when converted using trig identity to sums instead of products results on these frequency values of the three cosines being added
(1/16), (1/16) and (3/16) ---- one of the three is an irrational number which is 3/16 hence it is non periodic

same applies to the second problem the frequencies of the two cosines being added are
(2-pi)/8pi, (2+pi)/8pi --> both are irrational numbers hence it is non periodic

correct me if I am wrong sir

 

[vela]

yes i know trig identities: but the point is to find whether the two equations above are periodic or non periodic,

the first problem seems to be non periodic because when converted using trig identity to sums instead of products results on these frequency values of the three cosines being added
(1/16), (1/16) and (3/16) ---- one of the three is an irrational number which is 3/16 hence it is non periodic

I'm not sure which trig identity you used, but I pretty much pointed you to the relevant one. Rather than just describing what you did, show us your actual work. In any case, the first one is periodic.

same applies to the second problem the frequencies of the two cosines being added are
(2-pi)/8pi, (2+pi)/8pi --> both are irrational numbers hence it is non periodic

correct me if I am wrong sir

 

[DrunkEngineer]

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

 

[vela]

Here's a plot of cos x and cos 3x. Can you see the answer?

 

[DrunkEngineer]

btw the equations are not a continuous signal but discrete signal sry i didnt mention it in the 1st post
1.)
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?

btw is my solution in my third post correct in the way that it is expressed as sums?

 

[rootX]

http://www.wolframalpha.com/input/?i=cos^3%28t%29

Use wolfram alpha.

cos^3 (t) = 1/4*(3*cos(t)+cos(3t)) = x[n]

t = n*pi/8

This is periodic when x[n+K] = x[n].

Yes, you are correct that as far you can get integer least common factor it is periodic.

LCM for 16 and 16/3 is 16*3.

 

[vela]

LCM for 16 and 16/3 is 16*3.

No, it isn't. The LCM is 16. It's the smallest integer that's an integral multiple of each number.

1 \times 16 = 3 \times \frac{16}{3} = 16

btw is my solution in my third post correct in the way that it is expressed as sums?

No, you made several mistakes (plus what you had could clearly be simplified), but the basic terms are correct as you can see from what rootX posted.

The period of the signal can be no less than the period of the lowest-frequency component. In this problem, that component corresponds to the cos t term, which goes through one cycle when n=16. The cos 3t term will go through exactly 3 complete cycles in the same period — this is the insight I was hoping you'd grasp from the plot — so n=16 is the period of their sum. Hopefully, you see the general pattern. If the faster-oscillating terms have frequencies that are integral multiples of the lowest frequency, the period of the lowest-frequency component is the period of the complete signal.

 

[DrunkEngineer]

your right i do have a glitch

1. = \cos(\frac{n\pi}{8})\cos^2(\frac{n\pi}{8})
= \cos(\frac{n\pi}{8})\frac{1}{2}(1+\cos(\frac{2n\pi}{8}))
= \frac{1}{2}(\cos(\frac{n\pi}{8}) + \cos(\frac{n\pi}{8})\cos(\frac{2n\pi}{8}))
= \frac{1}{2}\cos(\frac{n\pi}{8}) + \frac{1}{2}\frac{1}{2}(\cos(\frac{n\pi}{8}-\frac{n\pi}{4})+\cos(\frac{n\pi}{8}+\frac{n\pi}{4}))
= \frac{1}{2}\cos(\frac{n\pi}{8}) + \frac{1}{4}(\cos(\frac{n\pi}{8})) + \frac{1}{4}(\cos(\frac{3n\pi}{8}))
= (\frac{1}{2}+\frac{1}{4})(\cos(\frac{n\pi}{8})) + \frac{1}{4}(\cos(\frac{3n\pi}{8}))
= (\frac{3}{4})(\cos(\frac{n\pi}{8})) + \frac{1}{4}(\cos(\frac{3n\pi}{8}))
= (\frac{1}{4})(3(\cos(\frac{n\pi}{8})) + (\cos(\frac{3n\pi}{8})))
the first cosine: f1 = \frac{\omega}{2\pi} = \frac{\frac{\pi}{8}}{2\pi} = \frac{1}{16} --- rational ---- periodic
2nd cosine: f2 = \frac{\omega}{2\pi} = \frac{\frac{3\pi}{8}}{2\pi}} = \frac{3}{16}----rational--- periodic
LCM : 16/3 * (3) = 16 * 1 = 16 = No : Ans

To double check can you check if my number 2 solution is already correct? (1st post here)
https://www.physicsforums.com/showthread.php?t=460359