another average power question

Discussion in 'Homework Help' started by notoriusjt2, Sep 20, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]

    textbook states...
    [​IMG]

    so
    v(t) = 2 + 5 sin(2pi60t) - 3 cos(4pi60t + 45deg)
    i(t) = 1.5 + 2 cos(2pi60t + 20deg) + 1.1cos(4pi60t)

    I assume I cannot just multiply these together then take the integral.
    question is do I have to simplify them first?
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    If the equation says to multiply them out and take the integral then you can multiply them out and take the integral ;)

    I can't think of a simpler method than just grinding away at that... the simplification can come when you get to the integral step.

    Just remember that the average of a cosine or sine over one period, no matter the phase or anything else, is zero.
    A sinusoid needs to be multiplied by another sinusoid to contribute to the average.
    So, cos(x)^2 or sin(x)^2 or cos(x)sin(x + a) will contribute, sin(x) will not.
     
  3. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    actually posted a correction to the problem.. here is the new one...

    [​IMG]
    so let me get this straight. if a sinusoid needs to be multiplied by another sinusoid to contribute to the average then none of these sinusoids are eligible to contribute?

    so were left with v(t) = 2 and i(t) = 1.5?
    and with the period = 60
    1/60[(2*1.5)*60] = 3w

    this cant be that easy
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Hm?

    You take the integral over a period after you multiply it out.

    (a + b)(c + d) = ac + ad + bc + bd

    There are plenty of terms that can contribute.

    Edit:
    Hm, I don't remember whether sinusoids of different frequencies multiplied together have a non-zero average...
     
    Last edited: Sep 21, 2010
  5. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    0
    v(t) = 2 + 5cos(2pi60t) - 3cos(4pi60t + 45deg)

    2 = voltage peak to peak
    2pi60t = frequency
    4pi60t = frequency
    45deg = phase shift

    are these statements correct?
     
  6. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Not entirely...

    2pi60t and 4pi60t are both phase... 45 degrees is phase shift.

    For cos(wt + x) you have w being angular frequency, wt being phase and x being phase shift.

    w is in radians, and w = 2pi*f, where f is in hertz.

    So for 2pi60t the frequency is 60 Hz.

    2 is the voltage DC component.
     
  7. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
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    (a + b)(c + d) = ac + ad + bc + bd

    for this statement... i understand distribution properties but I am still confused

    which values should I multiply??? just the ones outside the parenthesis? or the phases and phase shifts within the parenthesis?
     
  8. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    (a + b)(c + d) = ac + ad + bc + bd

    v(t) = 2 + 5 cos(2pi60t) - 3 cos(4pi60t + 45deg)
    i(t) = 1.5 + 2 cos(2pi60t + 20deg) + 1.1cos(4pi60t - 20deg)

    so your saying it should be set up as follows?

    [(2) + (5 cos(2pi60t)) - (3 cos(4pi60t + 45deg))]*[(1.5) + (2 cos(2pi60t + 20deg)) + (1.1cos(4pi60t - 20deg))]
     
  9. Ghar

    Active Member

    Mar 8, 2010
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  10. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    based on that formula...

    2*1.5=3
    2*(2 cos(2pi60t + 20deg))=4 cos(2pi60t + 20deg)
    2*(1.1cos(4pi60t - 20deg))=2.2cos(4pi60t - 20deg)

    (5 cos(2pi60t))*1.5=7.5 cos(2pi60t)
    (5 cos(2pi60t))*(2 cos(2pi60t + 20deg))=10....???



    [​IMG]
    (cos(2pi60t))*(cos(2pi60t+20deg)) = [cos(2pi60t-2pi60t+20deg)+cos(2pi60t+2pi60t+20deg)]/2

    =[cos(0+20deg)+cos(4pi60t+20deg)]/2
    =i dont know how to do this with the + 20deg in there and im not sure how to add the 2pi60t's together
     
  11. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    allright I am determined to figure this problem out. I have come across another equation that I think will be easier once I plug in the values

    [​IMG]
    Note that the average power is the sum of the powers at the frequencies in the Fourier series.

    so now back to the original problem
    [​IMG]
    I understand that with each frequency there is going to be a separate average power value and at the end, I will add them all together.

    So for this problem there are three groups of frequencies
    v(t) @ 2 and i(t) @ 1.5
    V(t) @ 5cos(2pi60t) and i(t) @ 2cos(2pi60t+20deg)
    v(t) @ -3cos(4pi60t+45deg) and i(t) @ 1.1cos(4pi60t-20deg)

    I also understand the concept and evaluation of Sigma or the summation, however I am really having trouble trying to plug these values into the right spot in this formula
     
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