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Time-harmonic functions (Complex Vectors)

  1. Jul 20, 2009 #1
    Statement:
    [tex]<v(t)> = \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0cos(\omega t + \phi)dt \equiv 0.[/tex] (#1)


    Relevant Question:
    If we suppose [tex]v(t)[/tex] is a complex vector, is the second equality above still true?


    Reasoning:
    If [tex]v(t)[/tex] is a complex vector, then [tex]v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y}.[/tex]

    But we also know (by sum to angle identity), [tex]cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta) \Leftrightarrow V_0cos(\omega t + \theta) = V_0[cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)= V_0[cos(\omega t)cos(\theta)][/tex]

    But is the following true:
    [tex]V_0[cos(\omega t)cos(\theta)] = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} ?[/tex]
    Because in the original equation (#1), if v(t) is a complex vector, I cannot see why [tex]v(t) = V_0cos(\omega t + \phi)?[/tex]


    thanks,



    Jeff
     
  2. jcsd
  3. Jul 20, 2009 #2

    cepheid

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    I don't really see how v(t) ( as you have defined it ) is a complex vector in the first place.
     
  4. Jul 20, 2009 #3
    What I wrote was really,
    [tex]
    v(t) = cos(\omega t)\hat{x} + sin(\omega t)\hat{y} = A\hat{x} + jB\hat{y}
    [/tex]

    the left equality is really "instantaneous form".
     
    Last edited: Jul 20, 2009
  5. Jul 20, 2009 #4

    cepheid

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    The fact remains that v(t) is not a complex vector. Sure, as we've talked about before, you can express the cosine function or the sine function as phasors if you like. I just don't see what that has to do with anything. Perhaps you could clarify your question? If your question is:

    What is [itex] \langle \textbf{v}(t) \rangle [/itex] if [itex] \textbf{v}(t) [/itex] is defined as:

    [tex] \textbf{v}(t) \equiv \cos(\omega t) \hat{\textbf{x}} + \sin(\omega t) \hat{\textbf{y}} [/tex] ​

    Then I think that the answer is:

    [tex] \langle \textbf{v}(t) \rangle \equiv \frac{1}{T} \left( \hat{\textbf{x}} \int_0^T \cos(\omega t)\, dt + \hat{\textbf{y}} \int_0^T \sin(\omega t)\, dt \right) [/tex]

    [tex] = 0\hat{\textbf{x}} + 0\hat{\textbf{y}} [/tex]​

    which kind of makes sense when you think about it, since the vector goes around in a circle and ends up back where it started.
     
  6. Jul 20, 2009 #5
    I am just curious because in my notes it says "The time-average of any time-harmonic function is always zero." That part makes sense, I think you answered that last time for me. But I don't understand why [tex]v(t) = V_0cos(\omega t + \phi)?[/tex] In previous pages of my notes, we've defined v(t) as both complex vectors, and complex vectors in the instantaneous form (I wrote above).
     
  7. Jul 20, 2009 #6

    cepheid

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    Without seeing your notes, it is difficult for me to comment further. The fact is,

    [tex] \cos(\omega t) \hat{\textbf{x}} + \sin(\omega t) \hat{\textbf{y}} \neq V_0\cos(\omega t + \phi) [/tex]​

    The thing on the left hand side is a vector and the thing on the right hand side is a scalar. EDIT: As far as I can see, two different things are being talked about and are being referred to as "v(t)."
     
  8. Jul 20, 2009 #7
    Since cosine and sine can be written as one another- by shifting the phase,

    [tex]
    \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0cos(\omega t + \phi)dt.
    [/tex]

    Suppose [tex]\phi = 90\circ[/tex], then it follows

    [tex]
    \frac{1}{T} \int^{T}_{0}v(t)dt = \frac{1}{T} \int^{T}_{0}V_0sin(\omega t)dt = - \frac{\omega V_0cos(\omega t)}{T}|^{T}_{0} \neq 0
    \Leftrightarrow T \neq 0[/tex]

    Doesn't that mean that a time-harmonic function is not always zero?

    Thanks for reading my posts,


    JL
     
    Last edited: Jul 20, 2009
  9. Jul 20, 2009 #8

    cepheid

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    [tex] \cos(\omega T) = \cos \left(\frac{2 \pi}{T}T\right) = \cos(2 \pi) = 1 [/tex]

    [tex] \cos(0) = 1 [/tex]

    [tex] \Rightarrow \cos(\omega T) - \cos(0) = 1 - 1 = 0 [/tex] ​

    Without doing any math, you can see that the integral of a sinusoidal function over any whole number of periods will be zero, because you will always be adding together an equal amount of "positive area" and "negative area" under the curve.
     
  10. Jul 20, 2009 #9
  11. Jul 20, 2009 #10

    Hootenanny

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    As I said in the associate thread that I was helping you with, this only applies if T is in integer multiple of the period.
     
  12. Jul 20, 2009 #11
    For some reason in this example, I thought v(t) was some complex vector. Instead it shows that any sinusoid for harmonic functions have an time-average of zero, which is exactly what you said.

    Thanks,


    JL
     
  13. Jul 20, 2009 #12

    cepheid

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    Oh yeah. I assumed T WAS the period. But I guess T is the time interval over which you are averaging. In any case, I already stated as well that the integral would be zero over some whole number of periods, so we both got the disclaimer in there.
     
  14. Jul 20, 2009 #13

    Hootenanny

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    No problem, I just thought that I would clarify since in the OP's previous question he stated that he didn't know what T was and it wasn't defined in his notes.
     
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