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Harmonic Function (time averages) II

  1. Jul 19, 2009 #1
    Statement:
    [tex]A(t) = |A|cos(\omega t + \alpha), B(t) = |B|cos(\omega t + \beta)[/tex]

    The time average of the product is given by:
    [tex]<A(t)B(t)> = \frac{1}{T} \int^{T}_{0}A(t)B(t)dt = \frac{|A||B|}{2T}\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt + \int^{T}_{0}cos(\alpha - \beta)dt] = \frac{1}{2}|A||B|cos(\alpha - \beta)[/tex]


    Relevant equations:
    Note: using trigonometric identities:
    [tex] A(t)B(t) = \frac{1}{2} |A||B|[cos(2\omega t + \alpha + \beta) + cos(\alpha - \beta)][/tex]


    Questions:
    How is [tex]\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = 0? [/tex]
    When I worked it out I got [tex]\int^{T}_{0}cos(2\omega t + \alpha + \beta)dt = \frac{1}{(2w)} sin(u) |^{T}_{0} = \frac{1}{2w}sin(2\omega t + \alpha + \beta)^{T}_{0} \neq 0[/tex]


    Is there a real-world application for taking the product of two harmonic functions (even an application for phasors)??

    thanks
     
    Last edited: Jul 19, 2009
  2. jcsd
  3. Jul 20, 2009 #2

    cepheid

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    HOW is it not zero? By definition, ωT = 2π



    [tex] \sin(2 \omega t + \alpha + \beta) |_0^T [/tex]

    [tex] = \sin(4 \pi + \alpha + \beta) - \sin(\alpha + \beta) = 0 [/tex]


    because,

    [tex] \sin(\theta + 2n\pi) = \sin(\theta) [/tex]​

    The integral of a cosine function over any whole number of periods is zero (that much is obvious just by looking at it).
     
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