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Average of sine function

  1. Jan 24, 2016 #1
    • Member warned about posting with no effort
    How to find the average value of sin2wt?

    Mod note: Thread locked due to lack of homework template and no effort shown.
     
    Last edited by a moderator: Jan 24, 2016
  2. jcsd
  3. Jan 24, 2016 #2

    Krylov

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    How do you find the average of any periodic, integrable function?
     
  4. Jan 24, 2016 #3
    That's what I am asking
     
  5. Jan 24, 2016 #4

    Krylov

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    Do you have any ideas yourself?
     
  6. Jan 24, 2016 #5
    I know that average of any function is equal to sum of all the values of the function divided by number of values.
    So we can integrate sin2wt is the interval 0 to 2π and divided it by the total number of values.
    But what is the total number of values.How to find it?
     
  7. Jan 24, 2016 #6

    PeroK

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    First, note that ##sin^2(wt)## repeats every ##\pi## units.

    What if you could find the average value of this function, call it ##a##, and then you drew a rectangle of height ##a## from ##0## to ##\pi##. What could you say about the area of that rectangle?
     
  8. Jan 24, 2016 #7

    Krylov

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    Indeed.
    You correctly recalled the definition of the average of finitely many values ##a_1,\ldots,a_n## as
    $$
    \frac{1}{n}\sum_{i=1}^n{a_i} \qquad (1)
    $$
    Now, your problem is that for your function there is an infinitude of values. For such a case you need a new definition of "average". It is obtained by replacing the sum in (1) by an integral and dividing by the length of the interval. So, if ##f : \mathbb{R} \to \mathbb{R}## is a function, you can define its average over any interval ##[a,b]## as
    $$
    \overline{f} := \frac{1}{b-a}\int_a^b{f(x)\,dx}
    $$
    When the interval is unbounded, you have to use a limit. So, for your function ##f(t) := \sin^2{\omega t}## you get for its average over ##\mathbb{R}##,
    $$
    \overline{f} = \lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T{f(t)\,dt}
    $$
    However, because your function is periodic with period ##\tau := \tfrac{\pi}{\omega}##, all you need to do to calculate the above is to integrate from ##0## to ##\tau## and divide by ##\tau## to obtain ##\overline{f}##. (Hint: the answer does not depend on ##\omega##.)
     
    Last edited: Jan 24, 2016
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