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

  1. Sep 6, 2007 #1
    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\,\mathbb{R}[/tex]



    Does this look right?
     
  2. jcsd
  3. Sep 6, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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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