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Combining periodic function

  1. Mar 25, 2007 #1
    Given:
    x(t) =
    cos(2*pi*f1*t) 0 <= t < 4
    cos(2*pi*f2*t) 4 <= t < 8
    cos(2*pi*f3*t) 8 <= t < 12

    f1, f2, and f3 are given as well

    Would it be possible to combine all three of these conditions into one convenient equation or am I just dreaming? I tried using Matlab to concatenate the three graphs, but it won't give me an equation of it.
     
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 25, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm not sure what you mean by a "convenient" equation. What you have is a perfectly good function, You could use the "Heaviside step function" which is defined by H(x)= 0 if x<= 0, H(x)= 1 if x> 0.

    Then f(t)= cos(2pi f1 t)+ ((cos(2pi f2 t)- cos(2pi f1 t))H(t-4)+ (cos(2pi f3 t)- cos(2pi f2 t))H(t-8).

    for 0<= t<= 4, both t-4< 0 and t-8< 0 so both H(t-4) and H(t-8) are 0 and only the first term, cos(2pi f1 t), is non zero. For 4< t<= 8, t-4> 0 so H(t-4)= 1 but H(t-8) is still 0: f(t)= cos(2pi f1 t)+ cos(2pi f2 t)- cos(2pi f1 t)= cos(2pi f2 t). If 8< t<= 12, both t-4> 0 and t-8>0 so both H(t-4) and H(t-8) are 1. f(t)= cos(2pi f1 t)+ cos(2pi f2 t)- cos(2pi f1 t)+ cos(2pi f3 t)- cos(2pi f2 t)= cos(2pi f3 t).

    But is cos(2pi f1 t)+ ((cos(2pi f2 t)- cos(2pi f1 t))H(t-4)+ (cos(2pi f3 t)- cos(2pi f2 t))H(t-8) better than what you have?
     
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