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let us consider following sinusoidal components

$\sin(2\pi 13.5t)+\sin(2\pi 13.99t)+\sin(2\pi 25.3t)+\sin(2\pi 26t)$,

clearly this is not periodic in total,because frequencies or periods are not related each other by rational numbers,but clearly we may be able to measure some number ,which could represent as a largest period of this sinusoidal components right?for example for first component

$T_0=1/13.5=0.0740740740740741$

$T_1=1/13.99=0.0714796283059328$

$T_2=1/25.3=0.0395256916996047$

$T_3=1/26=0.0384615384615385$

clearly from there $T_0$ is largest,can we somehow consider $T_0$ as a largest period of this signal?thanks for help

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2 Answers 2

up vote 4 down vote accepted

This function is periodic because the ratios of the periods are all rational.

The periods are $$ \begin{align} & \left( \frac{1}{13.5}, \frac{1}{13.99}, \frac{1}{25.3}, \frac{1}{26} \right) \\[6pt] = & 100\left(\frac{1}{1350}, \frac{1}{1399}, \frac{1}{2530}, \frac{1}{2600}\right) = 100\left( \frac 1 a, \frac 1 b,\frac 1 c,\frac 1 d\right) \\[6pt] = & \frac{100}{abcd} (bcd,acd,abd, abc). \end{align} $$

To period is $\dfrac{100}{abcd}$ times the smallest common multiple of those last four components.

To find that, it will be useful to know the prime factorizations of some numbers: \begin{align} 1350 & = 2\cdot3\cdot3\cdot3\cdot5\cdot5 \\ 1399 & = 1399 \text{ (This one is prime.)} \\ 2530 & = 2\cdot5\cdot11\cdot23 \\ 2600 & = 2\cdot2\cdot5\cdot5\cdot13 \end{align}

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I think my latest edit should answer that question. –  Michael Hardy Mar 30 at 15:15
    
yes yes,i see now,does it means that we can somehow find period even if irrational numbers are presented? –  dato datuashvili Mar 30 at 15:16
    
The question is whether the ratios of the numbers are rational; not whether the numbers themselves are rational. –  Michael Hardy Mar 30 at 15:21
    
i see,thanks in advance –  dato datuashvili Mar 30 at 15:27

Let the period of $\displaystyle \sin(2n\pi t)$ is $T$

So we have $\displaystyle \sin[2n\pi(t+T)]=\sin(2n\pi t)$

$\displaystyle\implies 2n\pi(t+T)=m\pi+(-1)^m2n\pi t $ where $m$ is any integer

Setting $\displaystyle m=2r, 2n\pi(t+T)=2r\pi+2n\pi t\iff T=\frac{r\pi}n $

Setting $\displaystyle m=2r+1, T $ is a function of $t$ and hence not constant

So, the period is $\displaystyle T=\dfrac{\pi}n $ setting $r=1$

Now use Sum of two periodic functions is periodic? or Period of the sum/product of two functions

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thanks for reply –  dato datuashvili Mar 30 at 15:13

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