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

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If x1(t) and x2(t) have a period T, then the function x3(t) = a*x1(t) + b*x2(t), where a and b are constants, also has period T. This conclusion is derived from the periodic properties of x1 and x2, as demonstrated through the equations x1(t + T) = x1(t) and x2(t + T) = x2(t). However, it is important to note that while T is a period of both x1 and x2, it is not necessarily the smallest period of x3, as illustrated by the example of x1 = sin(x) + sin(2x) and x2 = -sin(x) + sin(2x), where the fundamental period of their difference is π.

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Homework Statement



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.



Homework Equations



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

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



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\,\mathbb{R}[/tex]



Does this look right?
 
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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.)
 

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