Mixing
Convolution with unusual limits
$begingroup$
I would like to find the Fourier transform of the following function:
$$
mathcal{F}(x) = int_{vert xvert}^infty !
{f}^astleft((r-x)/2right){f}left({(r+x)}/{2}right)operatorname{d}!r
$$
where $f(t)= e^{iA t}operatorname{sinc}(B t)$, with $A, Binmathtt{R}$.
To do the Fourier transform, I thought that I may be able to manipulate the integrand so that I could apply the convolution theorem.
My attempt to do so was to do a change of limits $r = 2y+vert xvert$, such that
$$
mathcal{F}(x) = 2H(x)int_{0}^infty !
{f}^astleft(yright){f}left(y-xright)operatorname{d}!y
+ 2H(-x)int_{0}^infty !
{f}^astleft(y+xright){f}left(yright)operatorname{d}!y
$$
where $H(x)$ is the Heaviside function. Applying the Fourier transform then we obtain:
$$
F(omega) =int_{-infty}^infty d x mathcal{F}(x) e^{iomega x}= operatorname{Re}left[int_0^infty dx int_0^infty dy e^{iomega x}{f}^astleft(yright){f}left(y-xright)
right]
$$
and now I'm stuck. Is the convolution theorem still valid for these limits of integration?
fourier-analysis convolution
asked Jan 17 at 13:59
JilesJiles
112
$endgroup$
add a comment |
$begingroup$
I would like to find the Fourier transform of the following function:
$$
mathcal{F}(x) = int_{vert xvert}^infty !
{f}^astleft((r-x)/2right){f}left({(r+x)}/{2}right)operatorname{d}!r
$$
where $f(t)= e^{iA t}operatorname{sinc}(B t)$, with $A, Binmathtt{R}$.
To do the Fourier transform, I thought that I may be able to manipulate the integrand so that I could apply the convolution theorem.
My attempt to do so was to do a change of limits $r = 2y+vert xvert$, such that
$$
mathcal{F}(x) = 2H(x)int_{0}^infty !
{f}^astleft(yright){f}left(y-xright)operatorname{d}!y
+ 2H(-x)int_{0}^infty !
{f}^astleft(y+xright){f}left(yright)operatorname{d}!y
$$
where $H(x)$ is the Heaviside function. Applying the Fourier transform then we obtain:
$$
F(omega) =int_{-infty}^infty d x mathcal{F}(x) e^{iomega x}= operatorname{Re}left[int_0^infty dx int_0^infty dy e^{iomega x}{f}^astleft(yright){f}left(y-xright)
right]
$$
and now I'm stuck. Is the convolution theorem still valid for these limits of integration?
fourier-analysis convolution
asked Jan 17 at 13:59
JilesJiles
112
$endgroup$
add a comment |
$begingroup$
I would like to find the Fourier transform of the following function:
$$
mathcal{F}(x) = int_{vert xvert}^infty !
{f}^astleft((r-x)/2right){f}left({(r+x)}/{2}right)operatorname{d}!r
$$
where $f(t)= e^{iA t}operatorname{sinc}(B t)$, with $A, Binmathtt{R}$.
To do the Fourier transform, I thought that I may be able to manipulate the integrand so that I could apply the convolution theorem.
My attempt to do so was to do a change of limits $r = 2y+vert xvert$, such that
$$
mathcal{F}(x) = 2H(x)int_{0}^infty !
{f}^astleft(yright){f}left(y-xright)operatorname{d}!y
+ 2H(-x)int_{0}^infty !
{f}^astleft(y+xright){f}left(yright)operatorname{d}!y
$$
where $H(x)$ is the Heaviside function. Applying the Fourier transform then we obtain:
$$
F(omega) =int_{-infty}^infty d x mathcal{F}(x) e^{iomega x}= operatorname{Re}left[int_0^infty dx int_0^infty dy e^{iomega x}{f}^astleft(yright){f}left(y-xright)
right]
$$
and now I'm stuck. Is the convolution theorem still valid for these limits of integration?
fourier-analysis convolution
asked Jan 17 at 13:59
JilesJiles
112
$endgroup$
I would like to find the Fourier transform of the following function:
$$
mathcal{F}(x) = int_{vert xvert}^infty !
{f}^astleft((r-x)/2right){f}left({(r+x)}/{2}right)operatorname{d}!r
$$
where $f(t)= e^{iA t}operatorname{sinc}(B t)$, with $A, Binmathtt{R}$.
To do the Fourier transform, I thought that I may be able to manipulate the integrand so that I could apply the convolution theorem.
My attempt to do so was to do a change of limits $r = 2y+vert xvert$, such that
$$
mathcal{F}(x) = 2H(x)int_{0}^infty !
{f}^astleft(yright){f}left(y-xright)operatorname{d}!y
+ 2H(-x)int_{0}^infty !
{f}^astleft(y+xright){f}left(yright)operatorname{d}!y
$$
where $H(x)$ is the Heaviside function. Applying the Fourier transform then we obtain:
$$
F(omega) =int_{-infty}^infty d x mathcal{F}(x) e^{iomega x}= operatorname{Re}left[int_0^infty dx int_0^infty dy e^{iomega x}{f}^astleft(yright){f}left(y-xright)
right]
$$
and now I'm stuck. Is the convolution theorem still valid for these limits of integration?
fourier-analysis convolution
fourier-analysis convolution
asked Jan 17 at 13:59
JilesJiles
112
asked Jan 17 at 13:59
JilesJiles
112
asked Jan 17 at 13:59
JilesJiles
112
asked Jan 17 at 13:59
JilesJiles
112
asked Jan 17 at 13:59
JilesJiles
112
112
add a comment |
add a comment |
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