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Combining Cos(t)'s from (sin2t + sin3t)/2*sint

  1. Sep 8, 2012 #1
    The question was given: x(t) = (sin2t + sin3t)/2*sint

    determine the period. So I converted numerator and denominator into the exponential forms and ended up getting (after flipping, timesing and cancelling)
    (exp(j3t)-exp(-j)-exp(j)+exp(-3j)+exp(j4)-exp(-2j)-exp(j2)+exp(-4j)) / 2

    which in turn turns into

    x(t) = cos(3t) - cos(t) - cos(2t) + cos(4t) if I'm not mistaken

    so is there a way of simplifying this into maybe one Cos or something so I can find the period?

    Or have I done this question the wrong way?


    Thanks heaps!
     
  2. jcsd
  3. Sep 9, 2012 #2

    samalkhaiat

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    use sin(2t)=2sin(t)cos(t). then write sin(3t)=sin(2t+t) and expand it. next use
    [itex]\cos(2t) = 2 \cos^{2}t - 1[/itex]. you then end up with
    [tex]x(t) = \cos t + 2 \cos^{2}t - \frac{1}{2}[/tex]
     
  4. Sep 10, 2012 #3
    That is very clever.

    So I got to:

    (2.sin(t).cos(t) + sin(2t).cos(t)+cos(2t).sin(2t)) / 2sin(t)

    [2.sin(t).cos(t)+2sin(t).cos(t).cos(t)+(cos^2(t)-sin^2(t))*(2.sin(t).cos(t)) ]
    / 2sint

    (then) although I have ended up with cost+cos^2 t + cos^3 t - sin^2(t).cost

    Have I done something wrong or am I not finished simplifying it? (I can't think of what else to do).


    Thanks
     
    Last edited: Sep 10, 2012
  5. Sep 10, 2012 #4
    Once you have a sum of cosines, the period is the lowest common multiple of the periods of the terms.
     
  6. Sep 10, 2012 #5

    samalkhaiat

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    sin(t+2t) = sin(t)cos(2t)+sin(2t)cos(t)
     
  7. Sep 10, 2012 #6
    What about if the Cosine is negative? Anyway so that would mean from the answer in my original post the period was 't'?

    Thanks
     
  8. Sep 10, 2012 #7
    Isn't that what I did?
     
  9. Sep 11, 2012 #8
    -cos(x)=cos(x+π)
    That component will have a phase of π, in respect to the other components. Does not change the period, though.
     
  10. Sep 11, 2012 #9
    Ah, I see. So it was '1'. When I plotted the sum of: x(t) = cos(3t) - cos(t) - cos(2t) + cos(4t) in matlab it generated a period of 6.275 or something, how can this be?

    Observe the picture attached
     

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  11. Sep 11, 2012 #10
    What would you expect? What is the period of cos(t)?
     
  12. Sep 11, 2012 #11
    Please enlighten me, 360 deg?
     
  13. Sep 11, 2012 #12
    In radians is 2π or about 6.28.
     
  14. Sep 11, 2012 #13
    so from the result I got, how was 2π≈6.28?
     
  15. Sep 11, 2012 #14
    I am not sure what are you asking. The value of 2π is not a consequence of your result but the other way around.
    π or "PI" is 3.1415...
    so 2π= 6.28....

    Now the period of a function like sin(x) or cos (x) is 2π. This means that sin(x+2π) = sin(x).
    In these expressions x is a non-dimensional variable.

    You can also write sin(x+360) = sin(x) but then your x should be in hexadecimal degrees too.
    Matlab uses radians, I suppose.
     
  16. Sep 11, 2012 #15
    AH! Because I was plotting it like it was in Time (seconds)

    Playing Devil's advocate, would it be fair to make 't' as seconds and plot the thing and say it took 6.28 seconds for a period?
     
  17. Sep 11, 2012 #16
    In physics we usually make the arguments of the functions non-dimensional.
    A physical quantity periodic in time will be represented by something like
    sin(2∏t/T) so t is in seconds and T is the "physical" period, in seconds.
    This has the property

    sin[2∏(t+T)/T)]=sin(2∏t/T) so the period is T.
     
  18. Sep 11, 2012 #17
    what do you mean by 'non-dimensional' exactly?

    Could I plead the case that since that T is 2π, it is 6.28 seconds?

    How did the 'sin[2∏(t+T)/T)]' come about?
     
  19. Sep 11, 2012 #18

    samalkhaiat

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    No, look again at your calculations. you put

    [tex]\sin (2t) = \sin (2t) \cos (t) + \cos (2t) \sin (2t)[/tex]

    which is wrong.
     
  20. Sep 11, 2012 #19
    You're right, thanks, would you be happy with:

    (2.sin(t).cos(t) + sin(2t).cos(t)+cos(2t).sin(t)) / 2sin(t)

    [2.sin(t).cos(t)+2sin(t).cos(t).cos(t)+(cos^2(t)-sin^2(t))*sin(t)) ]
    / 2sint

    ∴ 2cost + 2cos^2(t) +Cos^2(t)-(1-cos^2(t)) / 2

    ∴ 2Cos^2(t) + Cos(t) - 1/2

    which is about as simple as you can get, with a period 2π?

    Thanks
     
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