Mixing
If the Fourier transform of f(x) is G(w), then what is the inverse Fourier transform of f(w)?
$begingroup$
I have the Fourier Transform of $e^{-ax^2}$, which is $sqrt{frac{{pi}}{a}}$$e^{frac{-w^2}{4a}}$.
I now want to find the inverse Fourier Transform of $e^{-t(w-ib)^2}$.
I can see that this is similar to my original function, and I can see where the first shift theorem would be used (for the shift $-ib$).
However, I don't fully understand the relationship between a function and it's Fourier transform, and the reverse. I noticed that choosing $a$ as $t$, using the first shift theorem and swapping variables from $w$ to $x$ was a factor of $frac{1}{2{pi}}$ out.
The answer should be $frac{1}{2sqrt{{pi}t}}$$e^{frac{-x^2}{4t}}$.
Any help would be greatly appreciated! Exam is on Monday!
fourier-transform
asked Jan 12 at 21:18
A NA N
61
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$begingroup$
I have the Fourier Transform of $e^{-ax^2}$, which is $sqrt{frac{{pi}}{a}}$$e^{frac{-w^2}{4a}}$.
I now want to find the inverse Fourier Transform of $e^{-t(w-ib)^2}$.
I can see that this is similar to my original function, and I can see where the first shift theorem would be used (for the shift $-ib$).
However, I don't fully understand the relationship between a function and it's Fourier transform, and the reverse. I noticed that choosing $a$ as $t$, using the first shift theorem and swapping variables from $w$ to $x$ was a factor of $frac{1}{2{pi}}$ out.
The answer should be $frac{1}{2sqrt{{pi}t}}$$e^{frac{-x^2}{4t}}$.
Any help would be greatly appreciated! Exam is on Monday!
fourier-transform
asked Jan 12 at 21:18
A NA N
61
$endgroup$
add a comment |
$begingroup$
I have the Fourier Transform of $e^{-ax^2}$, which is $sqrt{frac{{pi}}{a}}$$e^{frac{-w^2}{4a}}$.
I now want to find the inverse Fourier Transform of $e^{-t(w-ib)^2}$.
I can see that this is similar to my original function, and I can see where the first shift theorem would be used (for the shift $-ib$).
However, I don't fully understand the relationship between a function and it's Fourier transform, and the reverse. I noticed that choosing $a$ as $t$, using the first shift theorem and swapping variables from $w$ to $x$ was a factor of $frac{1}{2{pi}}$ out.
The answer should be $frac{1}{2sqrt{{pi}t}}$$e^{frac{-x^2}{4t}}$.
Any help would be greatly appreciated! Exam is on Monday!
fourier-transform
asked Jan 12 at 21:18
A NA N
61
$endgroup$
I have the Fourier Transform of $e^{-ax^2}$, which is $sqrt{frac{{pi}}{a}}$$e^{frac{-w^2}{4a}}$.
I now want to find the inverse Fourier Transform of $e^{-t(w-ib)^2}$.
I can see that this is similar to my original function, and I can see where the first shift theorem would be used (for the shift $-ib$).
However, I don't fully understand the relationship between a function and it's Fourier transform, and the reverse. I noticed that choosing $a$ as $t$, using the first shift theorem and swapping variables from $w$ to $x$ was a factor of $frac{1}{2{pi}}$ out.
The answer should be $frac{1}{2sqrt{{pi}t}}$$e^{frac{-x^2}{4t}}$.
Any help would be greatly appreciated! Exam is on Monday!
fourier-transform
fourier-transform
asked Jan 12 at 21:18
A NA N
61
asked Jan 12 at 21:18
A NA N
61
asked Jan 12 at 21:18
A NA N
61
asked Jan 12 at 21:18
A NA N
61
asked Jan 12 at 21:18
A NA N
61
61
add a comment |
add a comment |
1 Answer
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$begingroup$
Here's what I get, working from the definition of the Inverse Fourier Transform
$$begin{align*}mathscr{F}^{-1}left{e^{-t(omega-ib)^2}right} &= dfrac{1}{2pi} int_{-infty}^{infty}e^{-t(omega-ib)^2}e^{iomega x}space domega\
\
& Omega = omega -ib \
\
&= dfrac{1}{2pi} int_{-infty-ib}^{infty-ib}e^{-tOmega^2}e^{i(Omega+ib) x}space dOmega\
\
&= e^{-bx}sqrt{dfrac{1/4t}{pi} } dfrac{1}{2pi}int_{-infty-ib}^{infty-ib}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&mbox{using an infinite rectangular contour in the complex plane,}\
&mbox{which has the real axis and the line parallel to the real axis}\
&mbox{at $-ib$ as two of the four segments, we can then show}\
\
&= e^{-bx}sqrt{dfrac{1}{4pi t} } dfrac{1}{2pi}int_{-infty}^{infty}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&=e^{-bx}sqrt{dfrac{1}{4pi t} } mathscr{F}^{-1}left{sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}right}\
\
&=dfrac{e^{-bx}}{2sqrt{pi t} } e^{-frac{x^2}{4t}}\
\
end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
$endgroup$
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here's what I get, working from the definition of the Inverse Fourier Transform
$$begin{align*}mathscr{F}^{-1}left{e^{-t(omega-ib)^2}right} &= dfrac{1}{2pi} int_{-infty}^{infty}e^{-t(omega-ib)^2}e^{iomega x}space domega\
\
& Omega = omega -ib \
\
&= dfrac{1}{2pi} int_{-infty-ib}^{infty-ib}e^{-tOmega^2}e^{i(Omega+ib) x}space dOmega\
\
&= e^{-bx}sqrt{dfrac{1/4t}{pi} } dfrac{1}{2pi}int_{-infty-ib}^{infty-ib}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&mbox{using an infinite rectangular contour in the complex plane,}\
&mbox{which has the real axis and the line parallel to the real axis}\
&mbox{at $-ib$ as two of the four segments, we can then show}\
\
&= e^{-bx}sqrt{dfrac{1}{4pi t} } dfrac{1}{2pi}int_{-infty}^{infty}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&=e^{-bx}sqrt{dfrac{1}{4pi t} } mathscr{F}^{-1}left{sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}right}\
\
&=dfrac{e^{-bx}}{2sqrt{pi t} } e^{-frac{x^2}{4t}}\
\
end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
$endgroup$
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
add a comment |
$begingroup$
Here's what I get, working from the definition of the Inverse Fourier Transform
$$begin{align*}mathscr{F}^{-1}left{e^{-t(omega-ib)^2}right} &= dfrac{1}{2pi} int_{-infty}^{infty}e^{-t(omega-ib)^2}e^{iomega x}space domega\
\
& Omega = omega -ib \
\
&= dfrac{1}{2pi} int_{-infty-ib}^{infty-ib}e^{-tOmega^2}e^{i(Omega+ib) x}space dOmega\
\
&= e^{-bx}sqrt{dfrac{1/4t}{pi} } dfrac{1}{2pi}int_{-infty-ib}^{infty-ib}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&mbox{using an infinite rectangular contour in the complex plane,}\
&mbox{which has the real axis and the line parallel to the real axis}\
&mbox{at $-ib$ as two of the four segments, we can then show}\
\
&= e^{-bx}sqrt{dfrac{1}{4pi t} } dfrac{1}{2pi}int_{-infty}^{infty}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&=e^{-bx}sqrt{dfrac{1}{4pi t} } mathscr{F}^{-1}left{sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}right}\
\
&=dfrac{e^{-bx}}{2sqrt{pi t} } e^{-frac{x^2}{4t}}\
\
end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
$endgroup$
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
add a comment |
$begingroup$
Here's what I get, working from the definition of the Inverse Fourier Transform
$$begin{align*}mathscr{F}^{-1}left{e^{-t(omega-ib)^2}right} &= dfrac{1}{2pi} int_{-infty}^{infty}e^{-t(omega-ib)^2}e^{iomega x}space domega\
\
& Omega = omega -ib \
\
&= dfrac{1}{2pi} int_{-infty-ib}^{infty-ib}e^{-tOmega^2}e^{i(Omega+ib) x}space dOmega\
\
&= e^{-bx}sqrt{dfrac{1/4t}{pi} } dfrac{1}{2pi}int_{-infty-ib}^{infty-ib}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&mbox{using an infinite rectangular contour in the complex plane,}\
&mbox{which has the real axis and the line parallel to the real axis}\
&mbox{at $-ib$ as two of the four segments, we can then show}\
\
&= e^{-bx}sqrt{dfrac{1}{4pi t} } dfrac{1}{2pi}int_{-infty}^{infty}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&=e^{-bx}sqrt{dfrac{1}{4pi t} } mathscr{F}^{-1}left{sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}right}\
\
&=dfrac{e^{-bx}}{2sqrt{pi t} } e^{-frac{x^2}{4t}}\
\
end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
$endgroup$
Here's what I get, working from the definition of the Inverse Fourier Transform
$$begin{align*}mathscr{F}^{-1}left{e^{-t(omega-ib)^2}right} &= dfrac{1}{2pi} int_{-infty}^{infty}e^{-t(omega-ib)^2}e^{iomega x}space domega\
\
& Omega = omega -ib \
\
&= dfrac{1}{2pi} int_{-infty-ib}^{infty-ib}e^{-tOmega^2}e^{i(Omega+ib) x}space dOmega\
\
&= e^{-bx}sqrt{dfrac{1/4t}{pi} } dfrac{1}{2pi}int_{-infty-ib}^{infty-ib}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&mbox{using an infinite rectangular contour in the complex plane,}\
&mbox{which has the real axis and the line parallel to the real axis}\
&mbox{at $-ib$ as two of the four segments, we can then show}\
\
&= e^{-bx}sqrt{dfrac{1}{4pi t} } dfrac{1}{2pi}int_{-infty}^{infty}sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}e^{iOmega x}space dOmega\
\
&=e^{-bx}sqrt{dfrac{1}{4pi t} } mathscr{F}^{-1}left{sqrt{dfrac{pi}{1/4t} }e^{-tOmega^2}right}\
\
&=dfrac{e^{-bx}}{2sqrt{pi t} } e^{-frac{x^2}{4t}}\
\
end{align*}$$
This is slightly different from what you state should be the answer (which suspiciously excludes $b$ entirely).
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
answered Jan 13 at 15:25
Andy WallsAndy Walls
1,724139
1,724139
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
add a comment |
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
$begingroup$
I realised I phrased the question wrong, but in any case it didn't come up in my exam. Thanks very much for your solution anyway!
$endgroup$
– A N
Jan 14 at 15:30
add a comment |
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