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Math Help - Intgration of a function.

  1. #1
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    Intgration of a function.

    actually this is not a homework, I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:


    if:



    then why integration over whole period is:




    I have problem with the power of omega, my solution returns w with power 2, while the power of omega in answer is one, Can someone help me for the reason?
    Last edited by mr fantastic; February 9th 2011 at 11:29 AM. Reason: Deleted begging for help in title.
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  2. #2
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    What is T?

    Isn't T given by: \displaystyle T={{2\pi}\over{h\omega}}\ , where h=1 ?



    and shouldn't that derivative be: \displaystyle {{dV(t)}\over{dt}} not dx ?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lordcx View Post
    actually this is not a homework, I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:


    if:



    then why integration over whole period is:




    I have problem with the power of omega, my solution returns w with power 2, while the power of omega in answer is one, Can someone help me for the reason?
    Frankly I wasn't able to do the summation, but I can address the problem with the \omega. The integral becomes:
    \frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt

    The integral is trivial and can easily be seen to be proportional to \frac{1}{ \omega } leaving only one \omega.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Frankly I wasn't able to do the summation, but I can address the problem with the \omega. The integral becomes:
    \frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt

    The integral is trivial and can easily be seen to be proportional to \frac{1}{ \omega } leaving only one \omega.

    -Dan

    Thank you, May I know please which theory did you use to get \frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt =  \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' =  1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega  t)~dt
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lordcx View Post
    Thank you, May I know please which theory did you use to get \frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt =  \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' =  1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega  t)~dt
    \frac{dV}{dt} = \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right )

    Thus
    \left ( \frac{dV}{dt} \right )^2 = \left [ \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right ) \right ] \cdot \left [ \sqrt{2}~\displaystyle\sum_{h' = 1}^{H} h' \omega V_{h'}~cos \left (h' \omega t + \frac{ \pi }{2} \right ) \right ]

    Simplifying
    \left ( \frac{dV}{dt} \right )^2 = 2 \omega ^2 \displaystyle\sum_{h = 1}^{H} \displaystyle\sum_{h' = 1}^{H}~hh'V_h V_{h'}~cos \left (h \omega t + \frac{ \pi }{2} \right )~cos \left (h' \omega t + \frac{ \pi }{2} \right )

    Now expand the cosines out in the following format:
    cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)

    -Dan

    PS: For the rest of the problem there is probably a way to write the double summation as a single sum. I'm not good at that kind of thing.
    Last edited by topsquark; February 9th 2011 at 08:56 PM.
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    Quote Originally Posted by topsquark View Post
    \frac{dV}{dt} = \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right )

    Thus
    \left ( \frac{dV}{dt} \right )^2 = \left [ \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right ) \right ] \cdot \left [ \sqrt{2}~\displaystyle\sum_{h' = 1}^{H} h' \omega V_{h'}~cos \left (h' \omega t + \frac{ \pi }{2} \right ) \right ]

    Simplifying
    \left ( \frac{dV}{dt} \right )^2 = 2 \omega ^2 \displaystyle\sum_{h = 1}^{H} \displaystyle\sum_{h' = 1}^{H}~hh'V_h V_{h'}~cos \left (h \omega t + \frac{ \pi }{2} \right )~cos \left (h' \omega t + \frac{ \pi }{2} \right )

    Now expand the cosines out in the following format:
    cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)

    -Dan

    PS: For the rest of the problem there is probably a way to write the double summation as a single sum. I'm not good at that kind of thing.
    sorry if I take your time but h'=h, and V=V' then we have:



    and over whole period:



    then we will have

    not

    am I wrong??
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lordcx View Post
    sorry if I take your time but h'=h, and V=V' then we have:



    and over whole period:



    then we will have

    not

    am I wrong??
    Let's take a simpler example.
    \left ( \displaystyle\sum_{h = 1}^5 h \right ) ^2

    I say that this is
    \displaystyle\sum_{h = 1}^5 \displaystyle\sum_{h'=1}^5 h h'

    = 1 \cdot (1 + 2 + 3 + 4 + 5) + 2 \cdot (1 + 2 + 3 + 4 + 5) + ~...

    = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 +1 \cdot 5

    + 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 +2 \cdot 5 +~...

    This is the same sequence as (1 + 2 + 3 + 4 + 5)^2 = (1 + 2 + 3 + 4 + 5)(1 + 2 + 3 + 4 + 5)

     = 1 \cdot (1 + 2 + 3 + 4 + 5) + 2 \cdot (1 + 2 + 3 + 4 + 5) + ~...

    So the h and h' cannot be equal. If they were then we would have

    \left ( \displaystyle\sum_{h = 1}^5 h \right ) ^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2

    which it is not.

    -Dan
    Last edited by topsquark; February 9th 2011 at 10:52 PM.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lordcx View Post
    and over whole period:

    Also

    \frac{1}{T}~ \int_0^T sin^2(h \omega t)~dt = \frac{1}{2} - \frac{1}{4h \omega T}~sin(2h \omega T)

    -Dan
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