• Jul 19th 2013, 04:25 PM
Alan0354
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.
• Jul 20th 2013, 12:57 AM
zzephod
Quote:

Originally Posted by Alan0354
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.

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

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 ...

.
• Jul 20th 2013, 01:50 AM
Alan0354
Quote:

Originally Posted by zzephod

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

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
• Jul 20th 2013, 08:04 AM
zzephod
Quote:

Originally Posted by Alan0354
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 $\displaystyle \xi=\phi-\pi/2$

.
• Jul 20th 2013, 11:55 AM
Alan0354
Quote:

Originally Posted by zzephod
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
• Jul 20th 2013, 03:22 PM
Alan0354
$\displaystyle J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi) }[\cos (m\theta)-j\sin (m\theta)]d\theta$

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

$\displaystyle J_m(z)$ 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,$\displaystyle \pi$].
• Jul 20th 2013, 05:33 PM
topsquark
Quote:

Originally Posted by Alan0354
$\displaystyle J_m(z)=\frac{1}{2\pi}\int_0^{2\pi}e^{j(z\sin\phi-m\theta)} d\theta$

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
• Jul 20th 2013, 06:30 PM
Alan0354
Quote:

Originally Posted by topsquark
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.
• Jul 20th 2013, 10:52 PM
Alan0354
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 $\displaystyle \;2 \int_0^{\pi}e^{j(z\sin\theta-m\theta)}d\theta$?
• Jul 21st 2013, 12:36 AM
Alan0354
Quote:

Originally Posted by Alan0354
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 $\displaystyle \int_0^{\pi}e^{jz\sin\theta}\cos m\theta d\theta$
• Jul 21st 2013, 01:45 PM
Alan0354
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?
• Jul 22nd 2013, 01:34 AM
Alan0354
Stupidity strikes again!!! I just realize $\displaystyle e^{j(\frac{m\pi}{2}=j^m=j{-m}$!!!! So

$\displaystyle J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos u}j\sin(m\theta)d\theta$
• Jul 22nd 2013, 02:23 AM
zzephod
Quote:

Originally Posted by Alan0354
.....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
• Jul 22nd 2013, 12:16 PM
Alan0354
$\displaystyle J_m(z)=\frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}\cos(m\theta)d\theta+ \frac{j^{-m}}{2\pi}\int_0^{2\pi}e^{jz\cos \theta}j\sin(m\theta)d\theta$

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
• Jul 22nd 2013, 05:45 PM
Alan0354
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 $\displaystyle \;\int_0^{2\pi}\sum_0^{\infty}\frac {(jz\cos\theta)^k}{k!}[\cos(m\theta)+j\sin(m\theta)]d\theta$.