## SHOW: if x1 & x2 have a period T then x3 = a*x1 + b*x2 also has period T

1. The problem statement, all variables and given/known data

Show that if $x_1(t)$ and $x_2(t)$ have period T, then $x_3(t)\,=\,ax_1(t)\,+\,bx_2(t)$ (a, b constant) has the same period T.

2. Relevant equations

$$x_1\left(t\,+\,T\right)\,=\,x_1(t)$$

$$x_2\left(t\,+\,T\right)\,=\,x_2(t)$$

3. The attempt at a solution

$$x_3(t)\,=\,a\,x_1(t)\,+\,b\,x_2(t)$$

$$x_3(t\,+\,T)\,=\,a\,x_1(t\,+\,T)\,+\,b\,x_2(t\,+\,T)$$

Since the relevant equations (above) are true...

$$x_3\left(t\,+\,T\right)\,=\,a\,x_1(t)\,+\,b\,x_2(t)$$

$$x_3(t)\,=\,a\,x_1(t)\,+\,b\,x_2(t)$$

$$\therefore\,x_3\left(t\,+\,T\right)\,=\,x_3(t)\,\forall\,t\,\in\,\mathb b{R}$$

Does this look right?
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 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, that's fine. That shows that if T is a period of both x1 and x2 then it is a period of ax1+ bx2 for any a and b. It is NOT always true that if T is the period (i.e. smallest period) of both x1 and x2 then it is the period of ax1+ bx2. Example: x1= sin(x)+ sin(2x), x2= -sin(x)+ sin(2x). Then $2\pi$ is the period of both x1 and x2 but x1-x2= 2sin(2x) which has fundamental period $\pi$. (But $2\pi$ is still a period.)