2D Fourier of Bessel
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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
1,324126
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
1,324126
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
1,324126
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
1,324126
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
1,324126
edited Nov 19 at 20:54
answered Nov 19 at 20:45
Andy Walls
1,324126
answered Nov 19 at 20:45
Andy Walls
1,324126
answered Nov 19 at 20:45
Andy Walls
1,324126
1,324126
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 |
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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