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Thread: 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:
    $\displaystyle \pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi$


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

    I tried $\displaystyle e^{jm\theta}=\cos m\theta +j\sin m \theta$ but it is not easy. Please help.
    Last edited by Alan0354; Jul 19th 2013 at 04:52 PM.
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  2. #2
    Senior Member zzephod's Avatar
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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:
    $\displaystyle \pi j^n J_n(z)=\int_0^{\pi} \cos(n\phi)e^{+jz\cos\phi}d\phi$


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

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

    $\displaystyle 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 $\displaystyle \xi=\phi+\pi/2$, then as before use:

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

    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:

    $\displaystyle 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 $\displaystyle \xi=\phi+\pi/2$, then as before use:

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

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

    .
    Thanks for you time.

    Is it supposed to be
    $\displaystyle 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; Jul 20th 2013 at 01:58 AM.
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    Senior Member zzephod's Avatar
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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
    $\displaystyle 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 $\displaystyle \sin(\phi)=\cos(\phi-\pi/2)$ (note the wrong sign in the erlier post) to convert the sine to a cosine. Then the change of variable will be $\displaystyle \xi=\phi-\pi/2$

    .
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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 $\displaystyle \sin(\phi)=\cos(\phi-\pi/2)$ (note the wrong sign in the erlier post) to convert the sine to a cosine. Then the change of variable will be $\displaystyle \xi=\phi-\pi/2$
    $\displaystyle J_m(z)= \frac{1}{2\pi}\int_0^{2\pi}e^{j(z\cos(\phi-\frac{\pi}{2})-m\phi)} d\phi $
    $\displaystyle u=\phi-\frac{\pi}{2}\;\Rightarrow\;\phi=u+\frac{\pi}{2},\ ;du=d\phi$
    $\displaystyle = \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

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

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

    $\displaystyle J_m(z)$ is real.

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

    But I still don't get the $\displaystyle \frac{1}{\pi j^m} $ part of it and the interval of [0,$\displaystyle \pi$].
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Please help derivation of this integration

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

    -Dan
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  8. #8
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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 $\displaystyle \phi$ in this equation. The function is defined as
    $\displaystyle J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\theta-m\theta)} d\theta$

    -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:

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

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

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

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

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

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

    $\displaystyle 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 $

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

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

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

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

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

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

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

    $\displaystyle J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}j\sin(m\theta)d\theta$
    Last edited by Alan0354; Jul 22nd 2013 at 01:38 AM.
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  13. #13
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    Re: Please help derivation of Bessel Function

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

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

    Or more explicitly

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

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

    How do I proof

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

    AND

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

    Thanks
    Last edited by Alan0354; Jul 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 $\displaystyle \;\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta\;$ AND $\displaystyle \;\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta$

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

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

    Please help

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