Mixing
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 yesterday
Cowboy Trader
8712
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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 yesterday
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
yesterday
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 yesterday
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 yesterday
Cowboy Trader
8712
asked yesterday
Cowboy Trader
8712
asked yesterday
Cowboy Trader
8712
asked yesterday
Cowboy Trader
8712
asked yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
1 Answer
1
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up vote
2
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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 yesterday
Andy Walls
1,314126
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
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 yesterday
Andy Walls
1,314126
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
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 yesterday
Andy Walls
1,314126
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
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 yesterday
Andy Walls
1,314126
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 yesterday
Andy Walls
1,314126
edited yesterday
answered yesterday
Andy Walls
1,314126
answered yesterday
Andy Walls
1,314126
answered yesterday
Andy Walls
1,314126
1,314126
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
$|a|$ 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
yesterday
$|a|$ 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
yesterday
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
yesterday