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Foorier question.

  1. Jul 14, 2009 #1
    [tex]\dot{v_c(0)}=2[/tex]
    [tex]v_c(0)=0[/tex]

    [tex]\ddot{v_c}+2\dot{v_c}+2v_c=\cos (t)[/tex]

    [tex]-\omega ^2V_c-2+2j\omega V_c +2V_c=cos(t)[/tex]

    in the solution i have +2 instead of the -2 that i got here(because we do -f'(0) when we find v_c double dot)

    and why cos t turns to 1
    ??
     
  2. jcsd
  3. Jul 14, 2009 #2

    HallsofIvy

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    The Fourier series for a function is a sine and cosine sum:
    [tex]f(x)= A_0 + A_1 cos(x)+ B_1sin(x)+ A_2cos(2x)+ B_2sin(2x)+ A_3cos(3x)+ \cdot\cdot\cdot[/tex]

    If the function is simply a sine or cosine (or finite combination) the Fourier series is just that combination. In particular, if f(x)= cos(x)
    [tex]f(x)= cos(x)= 0+ 1 cos(x)+ 0 sin(x)+ 0 cos(2x)+ 0 sin(2x)+ 0 cos(3x)+ \cdot\cdot\cdot[/tex]
    The only non-zero coefficient is [itex]A_1= 1[/itex].
     
  4. Jul 14, 2009 #3
    whattt nooooooooooooooo

    the forier transform of cos t is e^(i*f)
    f-phase angle

    and because its cos (t)
    the phase angle is 0

    so i get 1

    !!!
     
  5. Jul 14, 2009 #4
    is my formula correct
    ??
     
  6. Jul 14, 2009 #5

    Redbelly98

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    No, it is the sum of two delta functions located at f=(1/2π) Hz.

    What is 1?
    And what was the question asked in the problem? Is it to solve the differential equation you wrote in post #1? Fourier transforms are not necessary for that. It is just not clear what is the problem that is to be solved here.
     
  7. Jul 14, 2009 #6
    [tex]\ddot{v_c}+2\dot{v_c}+2v_c=\cos (t)[/tex]

    [tex]-\omega ^2V_c-2+2j\omega V_c +2V_c=1[/tex]


    i was told that because the formula is e^(i*f)
    f-is the phase ange
    and we have cos t - phase angle =0
    so e^(i*f)=1

    ring a bell to you?
     
  8. Jul 14, 2009 #7

    Redbelly98

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    Okay, the phase angle f=0.

    So e^(i*f) = e^0 = ____?
     
  9. Jul 14, 2009 #8

    turin

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    Nor appropriate. Since this is an initial value problem, use Laplace Transform. That way, your initial values will appear in the transformed equation. In that case, cos(t) certainly does not transform to 1.

    Alternatively, assume an exponetial solution, and then solve for the parameters of the exponential. You can use superposition to get the cos(t) from two cleverly chosen exponentials. Then, match to the boundary (initial) conditions. (I would prefer the Laplace transform method - it is automatic.)
     
    Last edited: Jul 14, 2009
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