Stuck on proving that this function\signal is not periodic

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SUMMARY

The function x(t) = sin(πt)cos(10t) is proven to be non-periodic through analysis of its components. The user attempted to find a common period by decomposing the function into sinusoids but mistakenly calculated the periods as T = 2π for non-integer values of m and n. The discussion emphasizes the importance of integer values in determining periodicity, leading to the conclusion that the function does not repeat at regular intervals.

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Roughmar
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First of all, HI! This is my first post and my first day in this forum. =)

I am having quite a problem demystifying this function. It's on a book I have and it clearly states that it isn't periodical. I can't reach that conclusion and was hoping you could help me out.

So, the function in question is [tex]x(t)=sin(\pi t)cos(10t)[/tex]

What I did was to deconstruct this into
[tex]\frac{1}{2}\left[ sin(\pi t+10t)-sin(\pi t-10t)\right][/tex]

Now, I think that the period of the first sin is [tex]\frac{2\pi}{\pi +10}[/tex] and the one from the second sin is [tex]\frac{2\pi}{\pi -10}[/tex].

I then try to find the fundamental period of the whole function:

[tex]n\frac{2\pi}{\pi +10}=m\frac{2\pi}{\pi -10}\Rightarrow m=\pi -10[/tex] and [tex]n=\pi+10 \Rightarrow T=2\pi[/tex]


I know I have to be doing something wrong and possibly it's also really basic, but I got stuck.
Can anyone help me out? =)
 
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But your m and n aren't integers. Try an indirect argument. Suppose your x(t) has period P, so x(t+P) = x(t). What does that give you? What happens if you put t = 0 in it?
 
LCKurtz said:
But your m and n aren't integers. Try an indirect argument. Suppose your x(t) has period P, so x(t+P) = x(t). What does that give you? What happens if you put t = 0 in it?


I had done the substitution before, and noticed it didn't work, but couldn't understand why.
As soon as you said "integer" however, I just facepalmed myself.

Thank you so much.
 

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