Period of a continuous signal

So I have $$x_1=2 \cos(.6\sqrt\pi x+\pi/6)$$ and $$x_2= \sin(1.2\sqrt\pi x-\pi/3)$$ and need to find the period of $$(x_1+x_2)^2$$.

Let $$a=.6\sqrt\pi x+\pi/6 ~~~\text{and} ~~~ b=1.2\sqrt\pi x-\pi/3$$ so that $$x_1=2 \cos(a) ~~ \text{and} ~~ x_2=\sin(b).$$

And finally we get: $$(x_1+x_2)^2=1.5+2\cos(2a)+2\sin(a+b)+2\sin(a-b)-.5\cos(2b)$$

Now, the how do I calculate the time period of it?

Edit:

$$\text{period of the individul sinusoids in the square}~~T_1=5\sqrt\pi/3 ~~ T_2=10\sqrt\pi/9~~T_3=10\sqrt\pi/3 ~~ T_4=5\sqrt\pi/6$$

$$\text{The ratios are}~~T_4/T_3=1/4~~T_4/T_2=3/4~~T_4/T_1=1/2$$

$$\text{LCM of denominators is 4 }$$

$$\text{Hence period }~~T=4T_4=10\sqrt\pi/3$$ is that correct?

• This a homework type problem so you would need to show us where exactly you're stuck. You've computed the square, but haven't shown any attempt to figure out the period. – Matt L. Dec 21 '19 at 15:36

Let $$T_0$$ be the period of the signal $$s(t) = x_1(t) + x_2(t)$$. $$s(t) = s(t+T_0) \tag{1}$$

Then the period for $$s^2(t) = ( x_1 + x_2)^2$$ will be either $$T_0$$ or $$T_0/2$$ :

First, it's obvious that $$T_0$$ is also a period for $$s(t)^2$$ as it satisfies $$s^2(t) = s^2(t+T_0) \tag{2}$$

However, $$T_0 / 2$$ may also be period for $$s^2(t)$$. Since we are looking for the smallest number to satisfy eq(2).

Hence do the following:

1. Find the period $$T_0$$ of $$s(t) = x_1(t)+x_2(t)$$.
2. Check if $$T_0/2$$ satisfies $$s^2(t) = s^2(t+T_0)$$.
3. If yes; then period of $$s^2(t)$$ is $$T_0/2$$; otherwise $$T_0$$.
• @John Hi! Have you tried the approach I have outlined ? Will you accept an answer ? or leave the question floating ? Don't forget to upvote as well :-)) – Fat32 Dec 25 '19 at 13:49

This is just a hint since your question is a classic homework problem. Note that a sum of sinusoids is periodic if you can write it in the following form:

$$x(t)=a_0+\sum_{k=1}^{\infty} a_k \cos(2\pi kf_0t+\phi_k)+\sum_{k=1}^{\infty} b_k \sin(2\pi kf_0t+\varphi_k)$$

where $$f_0$$ is the fundamental frequency, and $$T=1/f_0$$ is the corresponding period. Note that even if $$a_1=b_1=0$$, $$f_0$$ can still be the fundamental frequency. I trust that you can take it from here.