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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.

ANSWER: Questioner:   hamad
Country:   Al Kuwayt, Kuwait
Category:   Advanced Math
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.

Answer
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?

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