2D Fourier of Bessel
up vote
1
down vote
favorite
I need help with the fourier transform of Bessel function of first kind on 2 dimensions.
$$G(w_1,w_2) = F[J_0(asqrt{x^2+y^2})]$$
where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.
I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.
fourier-transform bessel-functions
asked Nov 19 at 15:30
Cowboy Trader
8712
add a comment |
up vote
1
down vote
favorite
I need help with the fourier transform of Bessel function of first kind on 2 dimensions.
$$G(w_1,w_2) = F[J_0(asqrt{x^2+y^2})]$$
where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.
I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.
fourier-transform bessel-functions
asked Nov 19 at 15:30
Cowboy Trader
8712
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need help with the fourier transform of Bessel function of first kind on 2 dimensions.
$$G(w_1,w_2) = F[J_0(asqrt{x^2+y^2})]$$
where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.
I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.
fourier-transform bessel-functions
asked Nov 19 at 15:30
Cowboy Trader
8712
I need help with the fourier transform of Bessel function of first kind on 2 dimensions.
$$G(w_1,w_2) = F[J_0(asqrt{x^2+y^2})]$$
where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.
I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.
fourier-transform bessel-functions
fourier-transform bessel-functions
asked Nov 19 at 15:30
Cowboy Trader
8712
asked Nov 19 at 15:30
Cowboy Trader
8712
asked Nov 19 at 15:30
Cowboy Trader
8712
asked Nov 19 at 15:30
Cowboy Trader
8712
asked Nov 19 at 15:30
Cowboy Trader
8712
8712
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58
add a comment |
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$
Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:
$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$
So that gives you your desired transform pair for real $a$:
$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$
The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.
answered Nov 19 at 20:45
Andy Walls
1,469127
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$
Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:
$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$
So that gives you your desired transform pair for real $a$:
$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$
The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.
answered Nov 19 at 20:45
Andy Walls
1,469127
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
add a comment |
up vote
2
down vote
accepted
From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$
Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:
$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$
So that gives you your desired transform pair for real $a$:
$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$
The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.
answered Nov 19 at 20:45
Andy Walls
1,469127
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$
Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:
$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$
So that gives you your desired transform pair for real $a$:
$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$
The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.
answered Nov 19 at 20:45
Andy Walls
1,469127
From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$
Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:
$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$
So that gives you your desired transform pair for real $a$:
$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$
The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.
answered Nov 19 at 20:45
Andy Walls
1,469127
edited Nov 19 at 20:54
answered Nov 19 at 20:45
Andy Walls
1,469127
answered Nov 19 at 20:45
Andy Walls
1,469127
answered Nov 19 at 20:45
Andy Walls
1,469127
1,469127
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
add a comment |
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
You're welcome. Ugh, I'd need to look at Bracewell's manipulation and rework it for the different convention to get to a slightly different version of the Hankel transform. I can guess that the $2pi$ disappears from the argument of the Bessel function kernel, but I won't know all the details unless I rederive it.
– Andy Walls
Nov 19 at 21:40
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
The wikipedia page on the Hankel Transform has enough info for you to,convert to the other convention: en.m.wikipedia.org/wiki/Hankel_transform
– Andy Walls
Nov 19 at 21:43
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
No guarantees for complex $a$. Start from the 2D Cartesian Fourier Transform and go forward from there. Without knowing your physical problem, I'm not sure how to assign meaning to a complex radius anyway.
– Andy Walls
Nov 19 at 21:45
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
$|a|$ is always real and the result will hold for $a ne 0$. $a=0$ is a special case that needs a separate transform.
– Andy Walls
Nov 19 at 21:48
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005075%2f2d-fourier-of-bessel%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The 2D Fourier Transform in polar coordinates is equivalent to a Hankel Transform, of zero order. For $a$ real, the transform is a Dirac delta impulse ring at a radius that is a scalar multiple of $a$ from the origin.
– Andy Walls
Nov 19 at 19:58