You are here:

Question
QUESTION: Hello,

I am trying to find out how to test the periodicity of the sum of 2 Periodic discrete functions
If
X[n]= cos(2*pi*a (n-k))
Y[n]= cos(2*pi*b (n-c))
If X,Y are periodic meaning a,b are rational because
if a and b are rational then there exists p where X[n+p]=X[n]
I mean cos (2pi*b (n+p-k))=cos(2pi*b(n-k)+2pi*b*p)
well if b=m/n then  p=n would lead to

cos(2pi*b(n-k)+2pi*b*p)=cos(2pi*b(n-k))

Z=X+Y
anyway so how can I test if there exists p such that Z[n+p]=Z[n]
where Z[n] is the sum of 2 periodic discrete functions ??
I will have
cos(2*pi*a (n-c))+cos(2*pi*b (n-k))=cos (2pi*b (n+p-k)+cos (2pi*a (n+p-k)

I don't know how to solve that

I mean yes I can expand the cosines and all that, but is I don't even know if it would solve. I can't find this in any text book.

Country:   Al Kuwayt, Kuwait
Private:   No
Subject:   Discrete periodic functions
Question:   Hello,

I am trying to find out how to test the periodicity of the sum of 2 Periodic discrete functions
If
X[n]= cos(2*pi*a (n-k))
Y[n]= cos(2*pi*b (n-c))
If X,Y are periodic meaning a,b are rational because
if a and b are rational then there exists p where X[n+p]=X[n]
I mean cos (2pi*b (n+p-k))=cos(2pi*b(n-k)+2pi*b*p)
well if b=m/n then  p=n would lead to

cos(2pi*b(n-k)+2pi*b*p)=cos(2pi*b(n-k))

Z=X+Y
anyway so how can I test if there exists p such that Z[n+p]=Z[n]
where Z[n] is the sum of 2 periodic discrete functions ??
I will have
cos(2*pi*a (n-c))+cos(2*pi*b (n-k))=cos (2pi*b (n+p-k)+cos (2pi*a (n+p-k)

I don't know how to solve that

I mean yes I can expand the cosines and all that, but is I don't even know if it would solve. I can't find this in any text book.
............................................
I am not totally sure, but this may help:

Suppose  X[n] is periodic at period p1.  That means:

X[n + p1] = X[n]

But it also means

X[n + 2 p1] = X[n]
X[n + 3 p1] = X[n]

etc

Now if Y[n] has period p2,  choose K = any common multiple of p1 and p2.

Then both X,Y are periodic at period K, and so would be their sum.

I think you can extend this to your case where a,b are rational.  If p1 and p2 are rational, they have a common denominator.  Use that to get your p1 and p2.

---------- FOLLOW-UP ----------

QUESTION: so does that mean if a =sqrt 2 , b = 2
Z is not periodic?

ANSWER: I thought you said 'a and b are rational'.

---------- FOLLOW-UP ----------

QUESTION: yea sorry! I forgot and got carried away...

hmm so it seems if a and b are rational there sum will always be periodic?  because you can always just use p1 * p2 as the period of the sum?
then Z=cos (2pi*b (n+p1*p2-k)+cos (2pi*a (n+p1*p2-k)
and since a has period p1 and p2 is just an integer it would still be periodic right.

QUESTION: yea sorry! I forgot and got carried away...

hmm so it seems if a and b are rational there sum will always be periodic?  because you can always just use p1 * p2 as the period of the sum?
then Z=cos (2pi*b (n+p1*p2-k)+cos (2pi*a (n+p1*p2-k)
and since a has period p1 and p2 is just an integer it would still be periodic right.

Well, mostly --  there are some interesting aspects to this:

If X[n]= cos(2*pi*a (n-k))

If p1 is the period, then:

cos(2*pi*a (n-k))

2 pi a ((n+p1) - k) - 2 pi a (n-k) = 2 pi

a ((n+p1) - k) - a (n-k) = 1
an + a p1 - ak - an + ak = 1

a p1 = 1

p1 = 1/a,  which has to be an integer, but is not necessarily.

Assuming that a = p/q,  then  p1 = q/p.  Then our period has to be  p1 = q.]

That may be tricky to understand;  I'll pick some numbers:

Suppose  a = 3/5, say, then 1/a = 5/3, but n cannot be equal to 5/3.  However, any multiple of 5/3 would work.  The smallest multiple of 5/3 that is an integer is 5.  That would have to be our p1.

So you are correct -- p1 p2 is the 'common period' of X and Y, provided you find your p1 and p2 that way.

Does this make any sense?

Volunteer

#### Paul Klarreich

##### Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

##### Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
-----------