Inverse fourier transform for function with three variables
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I'm following a textbook in which they introduce
$V_{p,q}(omega,tau,z)= frac{1}{2pi} int e^{-ih(tau-(p+q)z/c)}U_{p,q}(omega,h,z)dh$
where $p$ and $q$ are integers. How do I find an expression for $U_{p,q}$?
fourier-transform
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I'm following a textbook in which they introduce
$V_{p,q}(omega,tau,z)= frac{1}{2pi} int e^{-ih(tau-(p+q)z/c)}U_{p,q}(omega,h,z)dh$
where $p$ and $q$ are integers. How do I find an expression for $U_{p,q}$?
fourier-transform
New contributor
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm following a textbook in which they introduce
$V_{p,q}(omega,tau,z)= frac{1}{2pi} int e^{-ih(tau-(p+q)z/c)}U_{p,q}(omega,h,z)dh$
where $p$ and $q$ are integers. How do I find an expression for $U_{p,q}$?
fourier-transform
New contributor
I'm following a textbook in which they introduce
$V_{p,q}(omega,tau,z)= frac{1}{2pi} int e^{-ih(tau-(p+q)z/c)}U_{p,q}(omega,h,z)dh$
where $p$ and $q$ are integers. How do I find an expression for $U_{p,q}$?
fourier-transform
fourier-transform
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New contributor
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asked Nov 15 at 14:18
kroneckerdel69
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1 Answer
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Using the Fourier Transform convention
$$mathscr{F}left{f(tau)right} = int_{-infty}^{infty} f(tau)e^{ihtau} dtau = hat{f}(h)$$
$$mathscr{F}^{-1}left{hat{f}(h)right} = dfrac{1}{2pi}int_{-infty}^{infty} hat{f}(h)e^{-ihtau} dh = f(tau)$$
Then
$$begin{align*}V_{p,q}left(omega,tau,zright)&= frac{1}{2pi} int_{-infty}^{infty} e^{-ih(tau - (p+q)z/c)}U_{p,q}left(omega,h,zright)dh\
\
&= frac{1}{2pi} int_{-infty}^{infty} e^{-ihtau}e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)dh\
\
&= mathscr{F}^{-1}left{e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)right}\
\
mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}int_{-infty}^{infty}e^{ihtau}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
\int_{-infty}^{infty}e^{ihleft(tau-frac{(p+q)z}{c}right)}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
end{align*}$$
which shouldn't be a surprise.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Using the Fourier Transform convention
$$mathscr{F}left{f(tau)right} = int_{-infty}^{infty} f(tau)e^{ihtau} dtau = hat{f}(h)$$
$$mathscr{F}^{-1}left{hat{f}(h)right} = dfrac{1}{2pi}int_{-infty}^{infty} hat{f}(h)e^{-ihtau} dh = f(tau)$$
Then
$$begin{align*}V_{p,q}left(omega,tau,zright)&= frac{1}{2pi} int_{-infty}^{infty} e^{-ih(tau - (p+q)z/c)}U_{p,q}left(omega,h,zright)dh\
\
&= frac{1}{2pi} int_{-infty}^{infty} e^{-ihtau}e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)dh\
\
&= mathscr{F}^{-1}left{e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)right}\
\
mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}int_{-infty}^{infty}e^{ihtau}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
\int_{-infty}^{infty}e^{ihleft(tau-frac{(p+q)z}{c}right)}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
end{align*}$$
which shouldn't be a surprise.
add a comment |
up vote
0
down vote
Using the Fourier Transform convention
$$mathscr{F}left{f(tau)right} = int_{-infty}^{infty} f(tau)e^{ihtau} dtau = hat{f}(h)$$
$$mathscr{F}^{-1}left{hat{f}(h)right} = dfrac{1}{2pi}int_{-infty}^{infty} hat{f}(h)e^{-ihtau} dh = f(tau)$$
Then
$$begin{align*}V_{p,q}left(omega,tau,zright)&= frac{1}{2pi} int_{-infty}^{infty} e^{-ih(tau - (p+q)z/c)}U_{p,q}left(omega,h,zright)dh\
\
&= frac{1}{2pi} int_{-infty}^{infty} e^{-ihtau}e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)dh\
\
&= mathscr{F}^{-1}left{e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)right}\
\
mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}int_{-infty}^{infty}e^{ihtau}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
\int_{-infty}^{infty}e^{ihleft(tau-frac{(p+q)z}{c}right)}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
end{align*}$$
which shouldn't be a surprise.
add a comment |
up vote
0
down vote
up vote
0
down vote
Using the Fourier Transform convention
$$mathscr{F}left{f(tau)right} = int_{-infty}^{infty} f(tau)e^{ihtau} dtau = hat{f}(h)$$
$$mathscr{F}^{-1}left{hat{f}(h)right} = dfrac{1}{2pi}int_{-infty}^{infty} hat{f}(h)e^{-ihtau} dh = f(tau)$$
Then
$$begin{align*}V_{p,q}left(omega,tau,zright)&= frac{1}{2pi} int_{-infty}^{infty} e^{-ih(tau - (p+q)z/c)}U_{p,q}left(omega,h,zright)dh\
\
&= frac{1}{2pi} int_{-infty}^{infty} e^{-ihtau}e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)dh\
\
&= mathscr{F}^{-1}left{e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)right}\
\
mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}int_{-infty}^{infty}e^{ihtau}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
\int_{-infty}^{infty}e^{ihleft(tau-frac{(p+q)z}{c}right)}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
end{align*}$$
which shouldn't be a surprise.
Using the Fourier Transform convention
$$mathscr{F}left{f(tau)right} = int_{-infty}^{infty} f(tau)e^{ihtau} dtau = hat{f}(h)$$
$$mathscr{F}^{-1}left{hat{f}(h)right} = dfrac{1}{2pi}int_{-infty}^{infty} hat{f}(h)e^{-ihtau} dh = f(tau)$$
Then
$$begin{align*}V_{p,q}left(omega,tau,zright)&= frac{1}{2pi} int_{-infty}^{infty} e^{-ih(tau - (p+q)z/c)}U_{p,q}left(omega,h,zright)dh\
\
&= frac{1}{2pi} int_{-infty}^{infty} e^{-ihtau}e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)dh\
\
&= mathscr{F}^{-1}left{e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright)right}\
\
mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=e^{ihfrac{(p+q)z}{c}}U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}mathscr{F}left{V_{p,q}left(omega,tau,zright)right} &=U_{p,q}left(omega,h,zright) \
\
e^{-ihfrac{(p+q)z}{c}}int_{-infty}^{infty}e^{ihtau}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
\int_{-infty}^{infty}e^{ihleft(tau-frac{(p+q)z}{c}right)}V_{p,q}left(omega,tau,zright)dtau &=U_{p,q}left(omega,h,zright) \
end{align*}$$
which shouldn't be a surprise.
answered Nov 16 at 14:19
Andy Walls
1,289126
1,289126
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