Periodicity of the summation of two functions

[Jncik]

Homework Statement

show whether or not x(t) = cos(t) + sin(\sqrt{2}*t)

is periodic

The Attempt at a Solution

the first period

is

T₁ = 2\pi

the second period is

T₂ = \frac{2\pi}{\sqrt{2}}

how can I determine whether it is periodic or not?

in my book(about signals and systems) it doesn't explain how to check this out

 

[micromass]

Hi Jncik!

If a function is periodic, then

f(t)=f(t+T)

But if a function is periodic, then so is it's first and second derivative!, So

f^{\prime\prime}(t)=f^{\prime\prime}(t+T)

What is the second derivative of our function here?? And do we get something nice?

 

[Jncik]

I get -cos(t) - 2*sin(sqrt(2)t)

I don't see how this helps me...

can we say that because T2/T1 is rational, then it's periodic?

 

[micromass]

I get -cos(t) - 2*sin(sqrt(2)t)

I don't see how this helps me...

can we say that because T2/T1 is rational, then it's periodic?

So we have

\cos(t)+\sin(\sqrt{2}t)=\cos(t+T)+\sin(\sqrt{2}(t+T))

and

-\cos(t)-2\sin(\sqrt{2}t)=-\cos(t+T)-\sin(\sqrt{2}(t+T))

What happens if you add both equations?

 

[Jncik]

they become -sin(sqrt(2)t) = 0

and I get t = pi*n/sqrt(2) where n is integer

 

[Ecthelion]

The way I've always approached this problem is by finding the respective Periods, checking the ratio, and if it yields a rational number than the signal is periodic.

Micromass - is your method a more elegant way (or simply a calculus based approach given where this question was posted...) to show that the 'n' found in these last steps, in order to make the period rational, must be irrational?

 

[micromass]

they become -sin(sqrt(2)t) = 0

and I get t = pi*n/sqrt(2) where n is integer

No, you end up with

\sin(\sqrt{2}t)=\sin(\sqrt{2}(t+T))

and this must hold for all possible t! What are the only values of T that satisfy this?

The way I've always approached this problem is by finding the respective Periods, checking the ratio, and if it yields a rational number than the signal is periodic.

Micromass - is your method a more elegant way (or simply a calculus based approach given where this question was posted...) to show that the 'n' found in these last steps, in order to make the period rational, must be irrational?

Yes, you are correct that you just need to check if the ratio of the periods is rational. So I'm trying to prove that it isn't periodic. I'm just trying to explain here where it comes from that you can just check the ratio of the periods...

 

[Jncik]

for sqrt(2) * T = k*pi ?

hence T = k*pi/(sqrt(2))?

as for the ratio, is it T1/T2 or T2/T1?

 

[SammyS]

So we have

\cos(t)+\sin(\sqrt{2}t)=\cos(t+T)+\sin(\sqrt{2}(t+T))

and

-\cos(t)-2\sin(\sqrt{2}t)=-\cos(t+T)-\sin(\sqrt{2}(t+T))

What happens if you add both equations?

Small but important typo in the second equation (the one for the second derivative.)

Should be: -\cos(t)-2\sin(\sqrt{2}t)=-\cos(t+T)-\underset{\uparrow}{2}\sin(\sqrt{2}(t+T))