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

VinnyCee

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

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



2. Relevant equations

[tex]x_1\left(t\,+\,T\right)\,=\,x_1(t)[/tex]

[tex]x_2\left(t\,+\,T\right)\,=\,x_2(t)[/tex]



3. The attempt at a solution

[tex]x_3(t)\,=\,a\,x_1(t)\,+\,b\,x_2(t)[/tex]

[tex]x_3(t\,+\,T)\,=\,a\,x_1(t\,+\,T)\,+\,b\,x_2(t\,+\,T)[/tex]

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

[tex]x_3\left(t\,+\,T\right)\,=\,a\,x_1(t)\,+\,b\,x_2(t)[/tex]

[tex]x_3(t)\,=\,a\,x_1(t)\,+\,b\,x_2(t)[/tex]

[tex]\therefore\,x_3\left(t\,+\,T\right)\,=\,x_3(t)\,\forall\,t\,\in\,\mathb b{R}[/tex]



Does this look right?

HallsofIvy

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 [itex]2\pi[/itex] is the period of both x₁ and x₂ but x₁-x₂= 2sin(2x) which has fundamental period [itex]\pi[/itex]. (But [itex]2\pi[/itex] is still a period.)