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Inverting Fourier transform “on circles”
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Dear Math enthusiasts,
I am struggeling with a problem for which a solution is already given to me, but I can just not see why it is true. Here is the setting: I am given a function $f(x,y,t)$. It's well behaved let's say. Smooth and things like that. Now, this function should be reexpressed in the following form $$f(x,y,t) = int g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d k_x dk_y dw. tag{1}label{eq1}$$ If it were only this it would be very simple, $g$ is some sort of 3-D Fourier transform of $f$. However, the trouble is that the variables $k_x, k_y, w$ are not independent. They need to satisfy the relation $k_x^2 + k_y^2 = w^2$. Therefore, I would claim that the correct form of the above expression should be $$f(x,y,t) = int oint_{S(w)} g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d begin{bmatrix} k_x\ k_yend{bmatrix} dw,tag{2}label{eq2}$$ where $S(w)$ is a circle of radius $w$, so that the inner integral goes over the perimeter of the circle and the outer over circle radii.
My question is essentially: given a target function $f(x,y,t)$, how do I find $g(k_x, k_y, w)$ such that eqref{eq2} is true for every point $x,y,t$?
The reference I have for this simply redefines $g(k_y,k_y,w)$ into $h(k_x,w)$ since only two variables are independent (I'm assuming this means $h(k_x,w)=g(k_x,pm sqrt{w^2-k_x^2},w)$ though that's never written) and uses this in the first integral. This leads to $$f(x,y,t) = int int h(k_x,w) {rm e}^{-jmath (k_x x+k_y y - w t)} d k_x dw = int int tilde{g}(k_x,y,w) {rm e}^{-jmath (k_x x - w t)} d k_x dw,tag{3}label{eq3}$$ where $tilde{g}(k_x,y,w) = h(k_x,w) {rm e}^{-jmath k_y y} $ (again, omitting the argument $k_y$ for me can only mean the implicit relation $k_y = pm sqrt{w^2-k_x^2}$). From eqref{eq3}, they claim that $f$ is the 2-D Fourier transform of $tilde{g}$ along the first and third dimension so that all we need to do to find $tilde{g}$ is $$ tilde{g}(k_x,y,w) = frac{1}{4pi^2} int int f(x,y,t) {rm e}^{jmath (k_x x-wt)} dx dt tag{4}label{eq4}$$ which gives $h$ as $ h(k_x,w) = tilde{g}(k_x,y,w) {rm e}^{jmath k_y y} $.
However, I have the feeling this is oversimplifying things a bit. I'm lacking rigor. My feeling is that the original problem eqref{eq2} may not have a unique solution (due to the variable dependence) and a particular one was chosen here. Integration limits are always skipped which may be a delicate issue (after all, $k_x$ should never leave the interval $[-w,w]$, maybe this can be solved by defining $h$ zero outside this support). The fact that we cannot directly solve for $k_y$ (due to the $pm$) troubles me. Overall I have a vague feeling that this may work but I cannot quite put my finger on it and really understand what's going on.
Would anyone be able to enlighten me how to treat such problems rigorously?
edit: A concrete example I am interested in is the function $f(x,y,t)={rm e}^{-jmath left(omega_0 t - frac{omega_0}{c}sqrt{(x-x_0)^2+(y-y_0)^2}right)}$. I'm awarding a bounty to anyone who can systematically explain me how to find the (set of) function(s) $g(k_x,k_y,omega)$ that satisfy eqref{eq2} for a given $f(x,y,t)$ everywhere. A concrete example may be helpful for the understanding, it can be the one I provided in this paragraph, but I'm also happy with any other non-trivial example, as long as it aids the understanding.
integration functional-analysis fourier-analysis integral-geometry
asked Nov 7 at 15:57
Florian
1,3401720
add a comment |
up vote
5
down vote
favorite
Dear Math enthusiasts,
I am struggeling with a problem for which a solution is already given to me, but I can just not see why it is true. Here is the setting: I am given a function $f(x,y,t)$. It's well behaved let's say. Smooth and things like that. Now, this function should be reexpressed in the following form $$f(x,y,t) = int g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d k_x dk_y dw. tag{1}label{eq1}$$ If it were only this it would be very simple, $g$ is some sort of 3-D Fourier transform of $f$. However, the trouble is that the variables $k_x, k_y, w$ are not independent. They need to satisfy the relation $k_x^2 + k_y^2 = w^2$. Therefore, I would claim that the correct form of the above expression should be $$f(x,y,t) = int oint_{S(w)} g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d begin{bmatrix} k_x\ k_yend{bmatrix} dw,tag{2}label{eq2}$$ where $S(w)$ is a circle of radius $w$, so that the inner integral goes over the perimeter of the circle and the outer over circle radii.
My question is essentially: given a target function $f(x,y,t)$, how do I find $g(k_x, k_y, w)$ such that eqref{eq2} is true for every point $x,y,t$?
The reference I have for this simply redefines $g(k_y,k_y,w)$ into $h(k_x,w)$ since only two variables are independent (I'm assuming this means $h(k_x,w)=g(k_x,pm sqrt{w^2-k_x^2},w)$ though that's never written) and uses this in the first integral. This leads to $$f(x,y,t) = int int h(k_x,w) {rm e}^{-jmath (k_x x+k_y y - w t)} d k_x dw = int int tilde{g}(k_x,y,w) {rm e}^{-jmath (k_x x - w t)} d k_x dw,tag{3}label{eq3}$$ where $tilde{g}(k_x,y,w) = h(k_x,w) {rm e}^{-jmath k_y y} $ (again, omitting the argument $k_y$ for me can only mean the implicit relation $k_y = pm sqrt{w^2-k_x^2}$). From eqref{eq3}, they claim that $f$ is the 2-D Fourier transform of $tilde{g}$ along the first and third dimension so that all we need to do to find $tilde{g}$ is $$ tilde{g}(k_x,y,w) = frac{1}{4pi^2} int int f(x,y,t) {rm e}^{jmath (k_x x-wt)} dx dt tag{4}label{eq4}$$ which gives $h$ as $ h(k_x,w) = tilde{g}(k_x,y,w) {rm e}^{jmath k_y y} $.
However, I have the feeling this is oversimplifying things a bit. I'm lacking rigor. My feeling is that the original problem eqref{eq2} may not have a unique solution (due to the variable dependence) and a particular one was chosen here. Integration limits are always skipped which may be a delicate issue (after all, $k_x$ should never leave the interval $[-w,w]$, maybe this can be solved by defining $h$ zero outside this support). The fact that we cannot directly solve for $k_y$ (due to the $pm$) troubles me. Overall I have a vague feeling that this may work but I cannot quite put my finger on it and really understand what's going on.
Would anyone be able to enlighten me how to treat such problems rigorously?
edit: A concrete example I am interested in is the function $f(x,y,t)={rm e}^{-jmath left(omega_0 t - frac{omega_0}{c}sqrt{(x-x_0)^2+(y-y_0)^2}right)}$. I'm awarding a bounty to anyone who can systematically explain me how to find the (set of) function(s) $g(k_x,k_y,omega)$ that satisfy eqref{eq2} for a given $f(x,y,t)$ everywhere. A concrete example may be helpful for the understanding, it can be the one I provided in this paragraph, but I'm also happy with any other non-trivial example, as long as it aids the understanding.
integration functional-analysis fourier-analysis integral-geometry
asked Nov 7 at 15:57
Florian
1,3401720
1
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Dear Math enthusiasts,
I am struggeling with a problem for which a solution is already given to me, but I can just not see why it is true. Here is the setting: I am given a function $f(x,y,t)$. It's well behaved let's say. Smooth and things like that. Now, this function should be reexpressed in the following form $$f(x,y,t) = int g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d k_x dk_y dw. tag{1}label{eq1}$$ If it were only this it would be very simple, $g$ is some sort of 3-D Fourier transform of $f$. However, the trouble is that the variables $k_x, k_y, w$ are not independent. They need to satisfy the relation $k_x^2 + k_y^2 = w^2$. Therefore, I would claim that the correct form of the above expression should be $$f(x,y,t) = int oint_{S(w)} g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d begin{bmatrix} k_x\ k_yend{bmatrix} dw,tag{2}label{eq2}$$ where $S(w)$ is a circle of radius $w$, so that the inner integral goes over the perimeter of the circle and the outer over circle radii.
My question is essentially: given a target function $f(x,y,t)$, how do I find $g(k_x, k_y, w)$ such that eqref{eq2} is true for every point $x,y,t$?
The reference I have for this simply redefines $g(k_y,k_y,w)$ into $h(k_x,w)$ since only two variables are independent (I'm assuming this means $h(k_x,w)=g(k_x,pm sqrt{w^2-k_x^2},w)$ though that's never written) and uses this in the first integral. This leads to $$f(x,y,t) = int int h(k_x,w) {rm e}^{-jmath (k_x x+k_y y - w t)} d k_x dw = int int tilde{g}(k_x,y,w) {rm e}^{-jmath (k_x x - w t)} d k_x dw,tag{3}label{eq3}$$ where $tilde{g}(k_x,y,w) = h(k_x,w) {rm e}^{-jmath k_y y} $ (again, omitting the argument $k_y$ for me can only mean the implicit relation $k_y = pm sqrt{w^2-k_x^2}$). From eqref{eq3}, they claim that $f$ is the 2-D Fourier transform of $tilde{g}$ along the first and third dimension so that all we need to do to find $tilde{g}$ is $$ tilde{g}(k_x,y,w) = frac{1}{4pi^2} int int f(x,y,t) {rm e}^{jmath (k_x x-wt)} dx dt tag{4}label{eq4}$$ which gives $h$ as $ h(k_x,w) = tilde{g}(k_x,y,w) {rm e}^{jmath k_y y} $.
However, I have the feeling this is oversimplifying things a bit. I'm lacking rigor. My feeling is that the original problem eqref{eq2} may not have a unique solution (due to the variable dependence) and a particular one was chosen here. Integration limits are always skipped which may be a delicate issue (after all, $k_x$ should never leave the interval $[-w,w]$, maybe this can be solved by defining $h$ zero outside this support). The fact that we cannot directly solve for $k_y$ (due to the $pm$) troubles me. Overall I have a vague feeling that this may work but I cannot quite put my finger on it and really understand what's going on.
Would anyone be able to enlighten me how to treat such problems rigorously?
edit: A concrete example I am interested in is the function $f(x,y,t)={rm e}^{-jmath left(omega_0 t - frac{omega_0}{c}sqrt{(x-x_0)^2+(y-y_0)^2}right)}$. I'm awarding a bounty to anyone who can systematically explain me how to find the (set of) function(s) $g(k_x,k_y,omega)$ that satisfy eqref{eq2} for a given $f(x,y,t)$ everywhere. A concrete example may be helpful for the understanding, it can be the one I provided in this paragraph, but I'm also happy with any other non-trivial example, as long as it aids the understanding.
integration functional-analysis fourier-analysis integral-geometry
asked Nov 7 at 15:57
Florian
1,3401720
Dear Math enthusiasts,
I am struggeling with a problem for which a solution is already given to me, but I can just not see why it is true. Here is the setting: I am given a function $f(x,y,t)$. It's well behaved let's say. Smooth and things like that. Now, this function should be reexpressed in the following form $$f(x,y,t) = int g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d k_x dk_y dw. tag{1}label{eq1}$$ If it were only this it would be very simple, $g$ is some sort of 3-D Fourier transform of $f$. However, the trouble is that the variables $k_x, k_y, w$ are not independent. They need to satisfy the relation $k_x^2 + k_y^2 = w^2$. Therefore, I would claim that the correct form of the above expression should be $$f(x,y,t) = int oint_{S(w)} g(k_x,k_y,w) {rm e}^{-jmath (k_x x + k_y y - w t)} d begin{bmatrix} k_x\ k_yend{bmatrix} dw,tag{2}label{eq2}$$ where $S(w)$ is a circle of radius $w$, so that the inner integral goes over the perimeter of the circle and the outer over circle radii.
My question is essentially: given a target function $f(x,y,t)$, how do I find $g(k_x, k_y, w)$ such that eqref{eq2} is true for every point $x,y,t$?
The reference I have for this simply redefines $g(k_y,k_y,w)$ into $h(k_x,w)$ since only two variables are independent (I'm assuming this means $h(k_x,w)=g(k_x,pm sqrt{w^2-k_x^2},w)$ though that's never written) and uses this in the first integral. This leads to $$f(x,y,t) = int int h(k_x,w) {rm e}^{-jmath (k_x x+k_y y - w t)} d k_x dw = int int tilde{g}(k_x,y,w) {rm e}^{-jmath (k_x x - w t)} d k_x dw,tag{3}label{eq3}$$ where $tilde{g}(k_x,y,w) = h(k_x,w) {rm e}^{-jmath k_y y} $ (again, omitting the argument $k_y$ for me can only mean the implicit relation $k_y = pm sqrt{w^2-k_x^2}$). From eqref{eq3}, they claim that $f$ is the 2-D Fourier transform of $tilde{g}$ along the first and third dimension so that all we need to do to find $tilde{g}$ is $$ tilde{g}(k_x,y,w) = frac{1}{4pi^2} int int f(x,y,t) {rm e}^{jmath (k_x x-wt)} dx dt tag{4}label{eq4}$$ which gives $h$ as $ h(k_x,w) = tilde{g}(k_x,y,w) {rm e}^{jmath k_y y} $.
However, I have the feeling this is oversimplifying things a bit. I'm lacking rigor. My feeling is that the original problem eqref{eq2} may not have a unique solution (due to the variable dependence) and a particular one was chosen here. Integration limits are always skipped which may be a delicate issue (after all, $k_x$ should never leave the interval $[-w,w]$, maybe this can be solved by defining $h$ zero outside this support). The fact that we cannot directly solve for $k_y$ (due to the $pm$) troubles me. Overall I have a vague feeling that this may work but I cannot quite put my finger on it and really understand what's going on.
Would anyone be able to enlighten me how to treat such problems rigorously?
edit: A concrete example I am interested in is the function $f(x,y,t)={rm e}^{-jmath left(omega_0 t - frac{omega_0}{c}sqrt{(x-x_0)^2+(y-y_0)^2}right)}$. I'm awarding a bounty to anyone who can systematically explain me how to find the (set of) function(s) $g(k_x,k_y,omega)$ that satisfy eqref{eq2} for a given $f(x,y,t)$ everywhere. A concrete example may be helpful for the understanding, it can be the one I provided in this paragraph, but I'm also happy with any other non-trivial example, as long as it aids the understanding.
integration functional-analysis fourier-analysis integral-geometry
integration functional-analysis fourier-analysis integral-geometry
asked Nov 7 at 15:57
Florian
1,3401720
asked Nov 7 at 15:57
Florian
1,3401720
edited 4 hours ago
asked Nov 7 at 15:57
Florian
1,3401720
asked Nov 7 at 15:57
Florian
1,3401720
asked Nov 7 at 15:57
Florian
1,3401720
1,3401720
1
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23
add a comment |
1
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23
1
1
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
For this problem, it is much more convenient to introduce the polar coordinates ($phi in [0,2pi)$) $$k_x = w cos(phi), qquad k_y=w sin(phi)$$ which fulfill the constraint explicitly.
We then define
$$f(x,y,t)
= int_0^infty int_{0}^{2pi} g(w cosphi, w sinphi,w)e^{i w (t- x cosphi - ysinphi)} w dphi dw . tag{1} $$
Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.
We will apply, the 2D Fourier inversion theorem (in polar coordinates)
$$ H(rho, theta) = frac{1}{2pi} int_0^{infty}int_0^{2pi} h(r,phi) e^{-i rho r cos(phi-theta)} r,dphi ,dr tag{2}$$
$$h(r,phi) = frac{1}{2pi} int_0^{infty}int_0^{2pi} H(rho,theta) e^{i rho r cos(phi-theta)} rho,dtheta ,drho tag{3}$$
We introduce the function (that implicitly depends on $t$)
$$h(w,phi;t) = 2pi g(w cosphi, w sinphi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with
$$f(rho costheta,rho sintheta ,t)= H(rho, theta;t).$$ The inverse expression is thus given by (3). In particular, we have that
$$2pi g(w cosphi, w sinphi, w) e^{i w t} =frac{1}{2pi} int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho $$
or equivalently
$$ g(w cosphi, w sinphi, w) =frac{1}{(2pi)^2} e^{-i w t}int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho. $$
We note that $g$ is only specified on the points that fulfill the constraint!
answered yesterday
Fabian
19.1k3674
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For this problem, it is much more convenient to introduce the polar coordinates ($phi in [0,2pi)$) $$k_x = w cos(phi), qquad k_y=w sin(phi)$$ which fulfill the constraint explicitly.
We then define
$$f(x,y,t)
= int_0^infty int_{0}^{2pi} g(w cosphi, w sinphi,w)e^{i w (t- x cosphi - ysinphi)} w dphi dw . tag{1} $$
Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.
We will apply, the 2D Fourier inversion theorem (in polar coordinates)
$$ H(rho, theta) = frac{1}{2pi} int_0^{infty}int_0^{2pi} h(r,phi) e^{-i rho r cos(phi-theta)} r,dphi ,dr tag{2}$$
$$h(r,phi) = frac{1}{2pi} int_0^{infty}int_0^{2pi} H(rho,theta) e^{i rho r cos(phi-theta)} rho,dtheta ,drho tag{3}$$
We introduce the function (that implicitly depends on $t$)
$$h(w,phi;t) = 2pi g(w cosphi, w sinphi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with
$$f(rho costheta,rho sintheta ,t)= H(rho, theta;t).$$ The inverse expression is thus given by (3). In particular, we have that
$$2pi g(w cosphi, w sinphi, w) e^{i w t} =frac{1}{2pi} int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho $$
or equivalently
$$ g(w cosphi, w sinphi, w) =frac{1}{(2pi)^2} e^{-i w t}int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho. $$
We note that $g$ is only specified on the points that fulfill the constraint!
answered yesterday
Fabian
19.1k3674
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
add a comment |
up vote
3
down vote
accepted
For this problem, it is much more convenient to introduce the polar coordinates ($phi in [0,2pi)$) $$k_x = w cos(phi), qquad k_y=w sin(phi)$$ which fulfill the constraint explicitly.
We then define
$$f(x,y,t)
= int_0^infty int_{0}^{2pi} g(w cosphi, w sinphi,w)e^{i w (t- x cosphi - ysinphi)} w dphi dw . tag{1} $$
Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.
We will apply, the 2D Fourier inversion theorem (in polar coordinates)
$$ H(rho, theta) = frac{1}{2pi} int_0^{infty}int_0^{2pi} h(r,phi) e^{-i rho r cos(phi-theta)} r,dphi ,dr tag{2}$$
$$h(r,phi) = frac{1}{2pi} int_0^{infty}int_0^{2pi} H(rho,theta) e^{i rho r cos(phi-theta)} rho,dtheta ,drho tag{3}$$
We introduce the function (that implicitly depends on $t$)
$$h(w,phi;t) = 2pi g(w cosphi, w sinphi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with
$$f(rho costheta,rho sintheta ,t)= H(rho, theta;t).$$ The inverse expression is thus given by (3). In particular, we have that
$$2pi g(w cosphi, w sinphi, w) e^{i w t} =frac{1}{2pi} int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho $$
or equivalently
$$ g(w cosphi, w sinphi, w) =frac{1}{(2pi)^2} e^{-i w t}int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho. $$
We note that $g$ is only specified on the points that fulfill the constraint!
answered yesterday
Fabian
19.1k3674
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For this problem, it is much more convenient to introduce the polar coordinates ($phi in [0,2pi)$) $$k_x = w cos(phi), qquad k_y=w sin(phi)$$ which fulfill the constraint explicitly.
We then define
$$f(x,y,t)
= int_0^infty int_{0}^{2pi} g(w cosphi, w sinphi,w)e^{i w (t- x cosphi - ysinphi)} w dphi dw . tag{1} $$
Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.
We will apply, the 2D Fourier inversion theorem (in polar coordinates)
$$ H(rho, theta) = frac{1}{2pi} int_0^{infty}int_0^{2pi} h(r,phi) e^{-i rho r cos(phi-theta)} r,dphi ,dr tag{2}$$
$$h(r,phi) = frac{1}{2pi} int_0^{infty}int_0^{2pi} H(rho,theta) e^{i rho r cos(phi-theta)} rho,dtheta ,drho tag{3}$$
We introduce the function (that implicitly depends on $t$)
$$h(w,phi;t) = 2pi g(w cosphi, w sinphi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with
$$f(rho costheta,rho sintheta ,t)= H(rho, theta;t).$$ The inverse expression is thus given by (3). In particular, we have that
$$2pi g(w cosphi, w sinphi, w) e^{i w t} =frac{1}{2pi} int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho $$
or equivalently
$$ g(w cosphi, w sinphi, w) =frac{1}{(2pi)^2} e^{-i w t}int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho. $$
We note that $g$ is only specified on the points that fulfill the constraint!
answered yesterday
Fabian
19.1k3674
For this problem, it is much more convenient to introduce the polar coordinates ($phi in [0,2pi)$) $$k_x = w cos(phi), qquad k_y=w sin(phi)$$ which fulfill the constraint explicitly.
We then define
$$f(x,y,t)
= int_0^infty int_{0}^{2pi} g(w cosphi, w sinphi,w)e^{i w (t- x cosphi - ysinphi)} w dphi dw . tag{1} $$
Note that $w>0$ is required as otherwise the constraint $w^2 = k_x^2 + k_y^2$ does not have a solution.
We will apply, the 2D Fourier inversion theorem (in polar coordinates)
$$ H(rho, theta) = frac{1}{2pi} int_0^{infty}int_0^{2pi} h(r,phi) e^{-i rho r cos(phi-theta)} r,dphi ,dr tag{2}$$
$$h(r,phi) = frac{1}{2pi} int_0^{infty}int_0^{2pi} H(rho,theta) e^{i rho r cos(phi-theta)} rho,dtheta ,drho tag{3}$$
We introduce the function (that implicitly depends on $t$)
$$h(w,phi;t) = 2pi g(w cosphi, w sinphi, w) e^{i w t} .$$ Then, we see that (1) is equivalent to (2) with
$$f(rho costheta,rho sintheta ,t)= H(rho, theta;t).$$ The inverse expression is thus given by (3). In particular, we have that
$$2pi g(w cosphi, w sinphi, w) e^{i w t} =frac{1}{2pi} int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho $$
or equivalently
$$ g(w cosphi, w sinphi, w) =frac{1}{(2pi)^2} e^{-i w t}int_0^{infty}int_0^{2pi} f(rho costheta,rho sintheta ,t) e^{i rho r cos(phi-theta)} rho,dtheta ,drho. $$
We note that $g$ is only specified on the points that fulfill the constraint!
answered yesterday
Fabian
19.1k3674
answered yesterday
Fabian
19.1k3674
answered yesterday
Fabian
19.1k3674
answered yesterday
Fabian
19.1k3674
19.1k3674
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
add a comment |
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
@Florian: my response uses the 2D Fourier transform to obtain an inverse (I give explicit limits). Note that $g$ can only be obtained for the points that fulfill the constraints (parameterized by $w$ and $phi$). This is of course to be expected. If you agree with the initial definition (1) (which I hope is the same as your equation (2)), then my post should give you the inverse. If you want, I can apply the transform to your example...
– Fabian
yesterday
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
Wow, thanks, this looks very helpful! Polar coordinates, why didn't I think of that. This solution disagrees with what I was given in (3) and (4), right? It would be nice to see how this compares with an example. Doesn't have to be the one I provided necessarily.
– Florian
3 hours ago
add a comment |
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1
$f$ seems some kind of Radon transform of $g$: these problems are studied within the realm of integral geometry.
– Daniele Tampieri
Nov 7 at 16:04
Thanks for the suggestion. I added a tag. I know Radon transform only as integrals over lines, but you are right, this might be strongly related.
– Florian
Nov 7 at 16:23