1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Harmonic Function (time averages)

  1. Jul 18, 2009 #1
    1. The problem statement, all variables and given/known data
    The time-average of any time-harmonic function is always zero.


    2. Relevant equations
    [tex]<v(t)> = \frac{1}{T} \int_{0}^{T}v(t)dt = \frac{1}{T} \int_{0}^{T}V_{0}cos(\omega t + \phi)dt \equiv 0[/tex]


    3. The attempt at a solution
    A harmonic function I think is defined as a sinusoidal function with some sort of fixed period, and a given phase. Here, I am just trying to understand notes I am reading. I don't understand why there is fraction of [tex]\frac{1}{T}[/tex] in the equation above. Could someone also
    explain to me why the time averages is initially defined as [tex]<v(t)> = \frac{1}{T} \int_{0}^{T}v(t)dt[/tex], and why it is set to zero.


    Thanks,


    Jeffrey
     
  2. jcsd
  3. Jul 18, 2009 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In general, we define the average value of a function, f, on some interval [a,b] by,

    [tex]\bar{f} = \frac{1}{b-a}\int^{a}_{b}f\left(x\right)\; dx[/tex]

    Effectively what we are doing is summing the value of the function, f, at a set of equally spaced points along the interval [a,b], then dividing by the number of points. We then take the limit as the number of points approaches infinity to obtain the formula quoted above. See http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/average.html" for more information.

    Now in terms of your integral, T represents the time interval and therefore we set T = b - a and integrate along the segment [0,T].

    Now, let's take a closer look at your example. Your integrand is most definitely a time-harmonic function, so let's compute the average explicitly:

    [tex]\bar{v} = \frac{V_0}{T}\int^T_0 \cos\left(\omega t + \phi\right)\; dt[/tex]

    [tex]\bar{v} = \frac{V_0}{T\omega}\left.\sin\left(\omega t + \phi\right)\right|_0^T\; dt[/tex]

    Hence, [itex]\bar{v}[/itex] is zero iff,

    [tex]\sin\left(\omega T + \phi\right) = \sin\left(\phi\right)[/tex]

    which reduces to

    [tex]\omega T = 2n\pi\;\;\;,\;\;\; n\in\mathbb{Z}[/itex]

    Hence, the time-average of a time-harmonic function is zero iff,

    [tex]T = \frac{n}{f}[/tex]

    i.e. iff the time interval of integration is an integer multiple of the period.
     
    Last edited by a moderator: Apr 24, 2017
  4. Jul 18, 2009 #3
    That makes sense, but why does my notes say it is always zero?
     
  5. Jul 18, 2009 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Without seeing your notes, I couldn't possibly say. Do your notes give any conditions or definitions on T?
     
  6. Jul 18, 2009 #5
    It just says, The time-average of any time-harmonic function is always zero.
     
  7. Jul 18, 2009 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well that simply isn't true, as you can plainly see from explicitly computing the time-average of a general time-harmonic function. If T was the period, then I would agree with the statement, but in general it isn't true.

    Are you using a textbook, or are these class notes?
     
  8. Jul 18, 2009 #7
    PDF notes from the lecturer. In this particular section of notes, we are discussing complex vectors and phasor notation to discuss sinusoidal varying vectors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Harmonic Function (time averages)
Loading...