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Math Help - Please help derivation of Bessel Function

  1. #1
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    Please help derivation of this integration

    I am studying Bessel Function in my antenna theory book, it said:
    \pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi


    I understand:
    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta
    Can you show me how do I get to
    \pi j^m J_m(z)=\int_0^{\pi} \cos(m\theta)e^{+jz\cos\theta}d\theta

    I tried e^{jm\theta}=\cos m\theta +j\sin m \theta but it is not easy. Please help.
    Last edited by Alan0354; July 19th 2013 at 04:52 PM.
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    Re: Please help derivation of this integration

    Quote Originally Posted by Alan0354 View Post
    I am studying Bessel Function in my antenna theory book, it said:
    \pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi


    I understand:
    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta
    Can you show me how do I get to
    \pi j^m J_m(z)=\int_0^{\pi} \cos(m\theta)e^{+jz\cos\theta}d\theta

    I tried e^{jm\theta}=\cos m\theta +j\sin m \theta but it is not easy. Please help.
    You start with:

    J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin(\phi)-m\phi)} d\phi = \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos(\phi + \pi/2)-m\phi)} d\phi

    Now change the variable to \xi=\phi+\pi/2, then as before use:

    e^{jm\xi}=\cos (m\xi) +j \sin (m \xi)

    and use the symmetry properties ( and periodicity? )of the functions involved ...

    .
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    Re: Please help derivation of this integration

    Quote Originally Posted by zzephod View Post
    You start with:

    J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin(\phi)-m\phi)} d\phi = \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos(\phi + \pi/2)-m\phi)} d\phi

    Now change the variable to \xi=\phi+\pi/2, then as before use:

    e^{jm\xi}=\cos (m\xi) +j \sin (m \xi)

    and use the symmetry properties ( and periodicity? )of the functions involved ...

    .
    Thanks for you time.

    Is it supposed to be
    J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin(\phi)-m\phi)} d\phi = \frac{1}{2\pi}\int_0^{2\pi}e^{j[z\cos(\phi + \pi/2)-m(\frac{\pi}{2}+\phi)]} d\phi

    I tried something similar but with no luck, can you show more steps?

    Thanks
    Last edited by Alan0354; July 20th 2013 at 01:58 AM.
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    Re: Please help derivation of this integration

    Quote Originally Posted by Alan0354 View Post
    Thanks for you time.

    Is it supposed to be
    J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin(\phi)-m\phi)} d\phi = \frac{1}{2\pi}\int_0^{2\pi}e^{j[z\cos(\phi + \pi/2)-m(\frac{\pi}{2}+\phi)]} d\phi

    I tried something similar but with no luck, can you show more steps?

    Thanks
    No I am just using the identity \sin(\phi)=\cos(\phi-\pi/2) (note the wrong sign in the erlier post) to convert the sine to a cosine. Then the change of variable will be \xi=\phi-\pi/2

    .
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    Re: Please help derivation of this integration

    Quote Originally Posted by zzephod View Post
    No I am just using the identity \sin(\phi)=\cos(\phi-\pi/2) (note the wrong sign in the erlier post) to convert the sine to a cosine. Then the change of variable will be \xi=\phi-\pi/2
    J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos(\phi-\frac{\pi}{2})-m\phi)} d\phi
    u=\phi-\frac{\pi}{2}\;\Rightarrow\;\phi=u+\frac{\pi}{2},\  ;du=d\phi
    = \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}}e^{j[z\cos(u)-m(u+\frac{\pi}{2})]} du

    So what is the next step?

    Thanks
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    Re: Please help derivation of this integration

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi)  }[\cos (m\theta)-j\sin (m\theta)]d\theta

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi)}  \cos (m\theta)d\theta -\frac{j}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi)}\sin (m\theta)d\theta

    J_m(z) is real.

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi)}  \cos (m\theta)d\theta

    But I still don't get the \frac{1}{\pi j^m} part of it and the interval of [0, \pi].
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    Re: Please help derivation of this integration

    Quote Originally Posted by Alan0354 View Post
    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta
    One mistake that seems to be following you...there is no \phi in this equation. The function is defined as
    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta

    -Dan
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    Re: Please help derivation of this integration

    Quote Originally Posted by topsquark View Post
    One mistake that seems to be following you...there is no \phi in this equation. The function is defined as
    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta

    -Dan
    Sorry, it's a typo. But you know what I meant.
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  9. #9
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    Re: Please help derivation of Bessel Function

    I worked on a few more steps:

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u-m\frac{\pi}{2}+mu)} du

    =\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u+mu)}e^{-jm\frac{\pi}{2}} \;du=-\frac{j\sin(m\frac{\pi}{2})}{2\pi}\int_0^{2\pi}e^{  j(z\cos u+mu)}du

    With this, J_m(z)=0 for m=even. That is not going to get the answer. Also, how do you change \;\int_0^{2\pi}e^{j(z\cos u+mu)}du\; to \;2 \int_0^{\pi}e^{j(z\sin\theta-m\theta)}d\theta?
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  10. #10
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    Re: Please help derivation of Bessel Function

    Quote Originally Posted by Alan0354 View Post
    I worked on a few more steps:

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u-m\frac{\pi}{2}+mu)} du

    =\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u+mu)}e^{-jm\frac{\pi}{2}} \;du=-\frac{j\sin(m\frac{\pi}{2})}{2\pi}\int_0^{2\pi}e^{  j(z\cos u+mu)}du

    With this, J_m(z)=0 for m=even. That is not going to get the answer. Also, how do you change \;\int_0^{2\pi}e^{j(z\cos u+mu)}du\; to \;2 \int_0^{\pi}e^{j(z\sin\theta-m\theta)}d\theta?
    I was wrong in the last step, it should be how do I get to \int_0^{\pi}e^{jz\sin\theta}\cos m\theta d\theta
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  11. #11
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    Re: Please help derivation of Bessel Function

    Hi Zzephod , You are right. I since work on a few more steps

    J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta= \frac{1}{2\pi}\int_0^{2\pi} e^{j(z\cos u-m\frac{\pi}{2}+mu)} du=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos u+mu)}e^{-jm\frac{\pi}{2}} \;du

    J_m(z)=\frac{e^{-jm\frac{\pi}{2}}}{2\pi}\int_0^{2\pi}e^{jz\cos u}[\cos(m\theta)+j\sin(m\theta)]d\theta

    As for e^{-j(\frac{m\pi}{2})}=\cos\frac{m\pi}{2}-j\sin\frac{m\pi}{2}

    For m=odd, \;\cos\frac{m\pi}{2}=0\; and \;-j\sin\frac{m\pi}{2}=j^{-m}

    For m=even, \;j\sin\frac{m\pi}{2}=0

    For m=0 \Rightarrow\;\cos\frac{m\pi}{2}=1,\;\; m=2 \Rightarrow\;\cos\frac{m\pi}{2}=-1,\;\; m=4 \Rightarrow\;\cos\frac{m\pi}{2}=-1,\;\;.......Therefore \Rightarrow\;\cos\frac{m\pi}{2}=j^m,\;\; not \;j^{-m};

    What did I do wrong this time?
    Last edited by Alan0354; July 21st 2013 at 01:56 PM.
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    Re: Please help derivation of Bessel Function

    Stupidity strikes again!!! I just realize e^{j(\frac{m\pi}{2}=j^m=j{-m}!!!! So

    J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}j\sin(m\theta)d\theta
    Last edited by Alan0354; July 22nd 2013 at 01:38 AM.
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    Re: Please help derivation of Bessel Function

    Quote Originally Posted by Alan0354 View Post
    .....Therefore \Rightarrow\;\cos\frac{m\pi}{2}=j^m,\;\; not \;j^{-m};

    What did I do wrong this time?
    Nothing, \cos(x) is symmetric so \cos(x)=\cos(-x).

    Or more explicitly

    \displaystyle j^{-m}=\frac{1}{j^m}=\frac{j^m}{j^{2m}}=(-1)^m j^m=j^m for m even
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  14. #14
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    Re: Please help derivation of Bessel Function

    J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta

    How do I proof

    \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta\;\neq\;0

    AND

    \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta=0

    Thanks
    Last edited by Alan0354; July 22nd 2013 at 12:20 PM.
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    Re: Please help derivation of Bessel Function

    There is no easy answer or easy way to integrate \;\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta\; AND \;\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta

    I tried \;e^{jz\cos \theta}=\cos (z\cos \theta) +j\sin(z\cos \theta)\; Where you need to solve \;\int_0^{2\pi}[\cos (z\cos \theta) +j\sin(z\cos \theta)][\cos(m\theta)+j\sin(m\theta)] d\theta\;

    I tried \;e^{jz\cos \theta}=\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}\; Where you need to solve \;\int_0^{2\pi}\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}[\cos(m\theta)+j\sin(m\theta)]d\theta.

    Please help

    Thanks
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