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 $U_{p,q}$?










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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 $U_{p,q}$?










    share|cite|improve this question
























      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 $U_{p,q}$?










      share|cite|improve this question













      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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      asked Nov 15 at 14:18









      kroneckerdel69

      85




      85






















          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.






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
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            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.






            share|cite|improve this answer

























              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.






              share|cite|improve this answer























                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 14:19









                Andy Walls

                1,324126




                1,324126






























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