What is the Fourier transform of the 2 dimensional airy function?
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What is the Fourier transform for the given two dimensional airy function,
$$f(x,y) = frac{J_1(r)}{r},.$$
Where $J_1$ is the Bessel function of the first kind, order one. And $r=sqrt{x^2+y^2}$.
Written explicitly,
$$mathcal{F}left[f(x,y)right] = int^infty_{-infty}int^infty_{-infty}f(x,y)expleft[ -i2pileft( f_xx+f_yyright)right], dx,dy, .$$
fourier-analysis fourier-transform bessel-functions airy-functions
asked Jan 1 at 16:52
TianTian
77112
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add a comment |
$begingroup$
What is the Fourier transform for the given two dimensional airy function,
$$f(x,y) = frac{J_1(r)}{r},.$$
Where $J_1$ is the Bessel function of the first kind, order one. And $r=sqrt{x^2+y^2}$.
Written explicitly,
$$mathcal{F}left[f(x,y)right] = int^infty_{-infty}int^infty_{-infty}f(x,y)expleft[ -i2pileft( f_xx+f_yyright)right], dx,dy, .$$
fourier-analysis fourier-transform bessel-functions airy-functions
asked Jan 1 at 16:52
TianTian
77112
$endgroup$
$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
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– reuns
Jan 1 at 17:11
add a comment |
$begingroup$
What is the Fourier transform for the given two dimensional airy function,
$$f(x,y) = frac{J_1(r)}{r},.$$
Where $J_1$ is the Bessel function of the first kind, order one. And $r=sqrt{x^2+y^2}$.
Written explicitly,
$$mathcal{F}left[f(x,y)right] = int^infty_{-infty}int^infty_{-infty}f(x,y)expleft[ -i2pileft( f_xx+f_yyright)right], dx,dy, .$$
fourier-analysis fourier-transform bessel-functions airy-functions
asked Jan 1 at 16:52
TianTian
77112
$endgroup$
What is the Fourier transform for the given two dimensional airy function,
$$f(x,y) = frac{J_1(r)}{r},.$$
Where $J_1$ is the Bessel function of the first kind, order one. And $r=sqrt{x^2+y^2}$.
Written explicitly,
$$mathcal{F}left[f(x,y)right] = int^infty_{-infty}int^infty_{-infty}f(x,y)expleft[ -i2pileft( f_xx+f_yyright)right], dx,dy, .$$
fourier-analysis fourier-transform bessel-functions airy-functions
fourier-analysis fourier-transform bessel-functions airy-functions
asked Jan 1 at 16:52
TianTian
77112
asked Jan 1 at 16:52
TianTian
77112
asked Jan 1 at 16:52
TianTian
77112
asked Jan 1 at 16:52
TianTian
77112
asked Jan 1 at 16:52
TianTian
77112
77112
$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
$endgroup$
– reuns
Jan 1 at 17:11
add a comment |
$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
$endgroup$
– reuns
Jan 1 at 17:11
$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
$endgroup$
– reuns
Jan 1 at 17:11
$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
$endgroup$
– reuns
Jan 1 at 17:11
add a comment |
1 Answer
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A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$int_{0}^{+infty}J_1(r)int_{0}^{2pi} e^{-2pi i r cos(t) w},dt,dr $$
or
$$ 2pi int_{0}^{+infty} J_1(r) J_0(2pi r w),dr. $$
Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2pi w > 1$, $pi$ if $2pi w=1$ and $2pi$ if $0leq 2pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
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1 Answer
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1 Answer
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$begingroup$
A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$int_{0}^{+infty}J_1(r)int_{0}^{2pi} e^{-2pi i r cos(t) w},dt,dr $$
or
$$ 2pi int_{0}^{+infty} J_1(r) J_0(2pi r w),dr. $$
Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2pi w > 1$, $pi$ if $2pi w=1$ and $2pi$ if $0leq 2pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
add a comment |
$begingroup$
A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$int_{0}^{+infty}J_1(r)int_{0}^{2pi} e^{-2pi i r cos(t) w},dt,dr $$
or
$$ 2pi int_{0}^{+infty} J_1(r) J_0(2pi r w),dr. $$
Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2pi w > 1$, $pi$ if $2pi w=1$ and $2pi$ if $0leq 2pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
add a comment |
$begingroup$
A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$int_{0}^{+infty}J_1(r)int_{0}^{2pi} e^{-2pi i r cos(t) w},dt,dr $$
or
$$ 2pi int_{0}^{+infty} J_1(r) J_0(2pi r w),dr. $$
Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2pi w > 1$, $pi$ if $2pi w=1$ and $2pi$ if $0leq 2pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
$endgroup$
A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$int_{0}^{+infty}J_1(r)int_{0}^{2pi} e^{-2pi i r cos(t) w},dt,dr $$
or
$$ 2pi int_{0}^{+infty} J_1(r) J_0(2pi r w),dr. $$
Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2pi w > 1$, $pi$ if $2pi w=1$ and $2pi$ if $0leq 2pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
answered Jan 1 at 19:47
Jack D'AurizioJack D'Aurizio
292k33284674
292k33284674
add a comment |
add a comment |
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$begingroup$
$mathcal{F}[f](u,v) = G(sqrt{u^2+v^2})$ where $G(w) = int_0^infty J_1(r)int_0^{2pi } e^{-2i pi rcos(t) w} dt dr$
$endgroup$
– reuns
Jan 1 at 17:11