Forward and Backward Projections












0












$begingroup$


I have the transform functions (forward and backward projections) such as:



$$FP{f(x,y)} = int_{-infty}^{infty}f(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz$$
$$BP{g_{theta}(r)} = int_{0}^{pi}g_{theta}(xcos(theta) + ysin(theta))dtheta$$



What i want to solve is that first forward projection, which is: $FP{delta(x,y)} = ?$, then i want to solve the backward projection of the result, which actually is: $BP{FP{delta(x,y)}}=?$.



I know that the answers must be $BP{FP{delta(x,y)}}= frac{1}{sqrt{x^2 + y^2}}$.



This problem occurs during study of the Radon transform.






fourier-analysis linear-transformations transformation projection







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  • $begingroup$
    I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
    $endgroup$
    – Marcus Müller
    Jan 6 at 19:41
















0












$begingroup$


I have the transform functions (forward and backward projections) such as:



$$FP{f(x,y)} = int_{-infty}^{infty}f(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz$$
$$BP{g_{theta}(r)} = int_{0}^{pi}g_{theta}(xcos(theta) + ysin(theta))dtheta$$



What i want to solve is that first forward projection, which is: $FP{delta(x,y)} = ?$, then i want to solve the backward projection of the result, which actually is: $BP{FP{delta(x,y)}}=?$.



I know that the answers must be $BP{FP{delta(x,y)}}= frac{1}{sqrt{x^2 + y^2}}$.



This problem occurs during study of the Radon transform.






fourier-analysis linear-transformations transformation projection







share|cite|improve this question











$endgroup$












  • $begingroup$
    I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
    $endgroup$
    – Marcus Müller
    Jan 6 at 19:41














0












0








0





$begingroup$


I have the transform functions (forward and backward projections) such as:



$$FP{f(x,y)} = int_{-infty}^{infty}f(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz$$
$$BP{g_{theta}(r)} = int_{0}^{pi}g_{theta}(xcos(theta) + ysin(theta))dtheta$$



What i want to solve is that first forward projection, which is: $FP{delta(x,y)} = ?$, then i want to solve the backward projection of the result, which actually is: $BP{FP{delta(x,y)}}=?$.



I know that the answers must be $BP{FP{delta(x,y)}}= frac{1}{sqrt{x^2 + y^2}}$.



This problem occurs during study of the Radon transform.






fourier-analysis linear-transformations transformation projection







share|cite|improve this question











$endgroup$




I have the transform functions (forward and backward projections) such as:



$$FP{f(x,y)} = int_{-infty}^{infty}f(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz$$
$$BP{g_{theta}(r)} = int_{0}^{pi}g_{theta}(xcos(theta) + ysin(theta))dtheta$$



What i want to solve is that first forward projection, which is: $FP{delta(x,y)} = ?$, then i want to solve the backward projection of the result, which actually is: $BP{FP{delta(x,y)}}=?$.



I know that the answers must be $BP{FP{delta(x,y)}}= frac{1}{sqrt{x^2 + y^2}}$.



This problem occurs during study of the Radon transform.






fourier-analysis linear-transformations transformation projection




fourier-analysis linear-transformations transformation projection






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 19:52









Marcus Müller

18418




18418










asked Jan 6 at 13:51









GuernicaGuernica

63




63












  • $begingroup$
    I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
    $endgroup$
    – Marcus Müller
    Jan 6 at 19:41


















  • $begingroup$
    I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
    $endgroup$
    – Marcus Müller
    Jan 6 at 19:41
















$begingroup$
I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
$endgroup$
– Marcus Müller
Jan 6 at 19:41




$begingroup$
I didn't like that you cross-posted here and then deleted your question including my answer over at DSP.SE without any warning :(
$endgroup$
– Marcus Müller
Jan 6 at 19:41










1 Answer
1






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oldest

votes


















0












$begingroup$

Let's –boringly– just insert the functional in question and see where it takes us:



$$newcommand{FP}[1]{mathrm{FP}{#1}(r,theta)}
$$



begin{align}
FP{delta(x,y)} &= int_{-infty}^{infty}delta(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz\
&=#left{z|rcos(theta) - zsin(theta) = 0 wedge rsin(theta) + zcos(theta)=0right}\
&= #left{z|rcos(theta) = zsin(theta) wedge rsin(theta) = - zcos(theta)right}tag{*}\[2em]
&underline{text{1. case: $costhetane0,,rne0,,ninmathbb Z$}}\
&= #left{z| r= ztan(theta) wedge rtan(theta) = - zright}\
&= #left{zBig| frac rz=tan(theta) wedge tan(theta) = - frac zrright}\
&text{$tantheta$ takes any arbitrary but fixed value $inmathbb R$:}\
&= #left{zBig| frac rz = - frac zrright}\
&= #left{z| r^2 = - z^2right}\
&= 0 text{ (because $r^2>0ge-z^2$)}\[2em]
&underline{text{2. case: $r=0$}}\
(*) &= #left{z| 0 = zsin(theta) wedge 0 = - zcos(theta)right}\
&= #left{z|0 = zsin(theta) wedge 0 = zcos(theta)right}\
&=1 text{ (namely, $z=0$, since $sinthetanecostheta$)}\[2em]
&underline{text{3. case: $costheta=0,,rne0,,ninmathbb Z$}}\
(*) &= #left{z| r0 = z1 wedge r1 = - z0right}\
&= #left{z|z=0 wedge r = 0right}\
&= 0
end{align}



Thus, in total:



$$FP{delta}(r,theta) = begin{cases}
1 &text{when} &r=0 \
0 &text{else.}
end{cases}$$






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Let's –boringly– just insert the functional in question and see where it takes us:



    $$newcommand{FP}[1]{mathrm{FP}{#1}(r,theta)}
    $$



    begin{align}
    FP{delta(x,y)} &= int_{-infty}^{infty}delta(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz\
    &=#left{z|rcos(theta) - zsin(theta) = 0 wedge rsin(theta) + zcos(theta)=0right}\
    &= #left{z|rcos(theta) = zsin(theta) wedge rsin(theta) = - zcos(theta)right}tag{*}\[2em]
    &underline{text{1. case: $costhetane0,,rne0,,ninmathbb Z$}}\
    &= #left{z| r= ztan(theta) wedge rtan(theta) = - zright}\
    &= #left{zBig| frac rz=tan(theta) wedge tan(theta) = - frac zrright}\
    &text{$tantheta$ takes any arbitrary but fixed value $inmathbb R$:}\
    &= #left{zBig| frac rz = - frac zrright}\
    &= #left{z| r^2 = - z^2right}\
    &= 0 text{ (because $r^2>0ge-z^2$)}\[2em]
    &underline{text{2. case: $r=0$}}\
    (*) &= #left{z| 0 = zsin(theta) wedge 0 = - zcos(theta)right}\
    &= #left{z|0 = zsin(theta) wedge 0 = zcos(theta)right}\
    &=1 text{ (namely, $z=0$, since $sinthetanecostheta$)}\[2em]
    &underline{text{3. case: $costheta=0,,rne0,,ninmathbb Z$}}\
    (*) &= #left{z| r0 = z1 wedge r1 = - z0right}\
    &= #left{z|z=0 wedge r = 0right}\
    &= 0
    end{align}



    Thus, in total:



    $$FP{delta}(r,theta) = begin{cases}
    1 &text{when} &r=0 \
    0 &text{else.}
    end{cases}$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let's –boringly– just insert the functional in question and see where it takes us:



      $$newcommand{FP}[1]{mathrm{FP}{#1}(r,theta)}
      $$



      begin{align}
      FP{delta(x,y)} &= int_{-infty}^{infty}delta(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz\
      &=#left{z|rcos(theta) - zsin(theta) = 0 wedge rsin(theta) + zcos(theta)=0right}\
      &= #left{z|rcos(theta) = zsin(theta) wedge rsin(theta) = - zcos(theta)right}tag{*}\[2em]
      &underline{text{1. case: $costhetane0,,rne0,,ninmathbb Z$}}\
      &= #left{z| r= ztan(theta) wedge rtan(theta) = - zright}\
      &= #left{zBig| frac rz=tan(theta) wedge tan(theta) = - frac zrright}\
      &text{$tantheta$ takes any arbitrary but fixed value $inmathbb R$:}\
      &= #left{zBig| frac rz = - frac zrright}\
      &= #left{z| r^2 = - z^2right}\
      &= 0 text{ (because $r^2>0ge-z^2$)}\[2em]
      &underline{text{2. case: $r=0$}}\
      (*) &= #left{z| 0 = zsin(theta) wedge 0 = - zcos(theta)right}\
      &= #left{z|0 = zsin(theta) wedge 0 = zcos(theta)right}\
      &=1 text{ (namely, $z=0$, since $sinthetanecostheta$)}\[2em]
      &underline{text{3. case: $costheta=0,,rne0,,ninmathbb Z$}}\
      (*) &= #left{z| r0 = z1 wedge r1 = - z0right}\
      &= #left{z|z=0 wedge r = 0right}\
      &= 0
      end{align}



      Thus, in total:



      $$FP{delta}(r,theta) = begin{cases}
      1 &text{when} &r=0 \
      0 &text{else.}
      end{cases}$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let's –boringly– just insert the functional in question and see where it takes us:



        $$newcommand{FP}[1]{mathrm{FP}{#1}(r,theta)}
        $$



        begin{align}
        FP{delta(x,y)} &= int_{-infty}^{infty}delta(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz\
        &=#left{z|rcos(theta) - zsin(theta) = 0 wedge rsin(theta) + zcos(theta)=0right}\
        &= #left{z|rcos(theta) = zsin(theta) wedge rsin(theta) = - zcos(theta)right}tag{*}\[2em]
        &underline{text{1. case: $costhetane0,,rne0,,ninmathbb Z$}}\
        &= #left{z| r= ztan(theta) wedge rtan(theta) = - zright}\
        &= #left{zBig| frac rz=tan(theta) wedge tan(theta) = - frac zrright}\
        &text{$tantheta$ takes any arbitrary but fixed value $inmathbb R$:}\
        &= #left{zBig| frac rz = - frac zrright}\
        &= #left{z| r^2 = - z^2right}\
        &= 0 text{ (because $r^2>0ge-z^2$)}\[2em]
        &underline{text{2. case: $r=0$}}\
        (*) &= #left{z| 0 = zsin(theta) wedge 0 = - zcos(theta)right}\
        &= #left{z|0 = zsin(theta) wedge 0 = zcos(theta)right}\
        &=1 text{ (namely, $z=0$, since $sinthetanecostheta$)}\[2em]
        &underline{text{3. case: $costheta=0,,rne0,,ninmathbb Z$}}\
        (*) &= #left{z| r0 = z1 wedge r1 = - z0right}\
        &= #left{z|z=0 wedge r = 0right}\
        &= 0
        end{align}



        Thus, in total:



        $$FP{delta}(r,theta) = begin{cases}
        1 &text{when} &r=0 \
        0 &text{else.}
        end{cases}$$






        share|cite|improve this answer











        $endgroup$



        Let's –boringly– just insert the functional in question and see where it takes us:



        $$newcommand{FP}[1]{mathrm{FP}{#1}(r,theta)}
        $$



        begin{align}
        FP{delta(x,y)} &= int_{-infty}^{infty}delta(rcos(theta) - zsin(theta), rsin(theta) + zcos(theta))dz\
        &=#left{z|rcos(theta) - zsin(theta) = 0 wedge rsin(theta) + zcos(theta)=0right}\
        &= #left{z|rcos(theta) = zsin(theta) wedge rsin(theta) = - zcos(theta)right}tag{*}\[2em]
        &underline{text{1. case: $costhetane0,,rne0,,ninmathbb Z$}}\
        &= #left{z| r= ztan(theta) wedge rtan(theta) = - zright}\
        &= #left{zBig| frac rz=tan(theta) wedge tan(theta) = - frac zrright}\
        &text{$tantheta$ takes any arbitrary but fixed value $inmathbb R$:}\
        &= #left{zBig| frac rz = - frac zrright}\
        &= #left{z| r^2 = - z^2right}\
        &= 0 text{ (because $r^2>0ge-z^2$)}\[2em]
        &underline{text{2. case: $r=0$}}\
        (*) &= #left{z| 0 = zsin(theta) wedge 0 = - zcos(theta)right}\
        &= #left{z|0 = zsin(theta) wedge 0 = zcos(theta)right}\
        &=1 text{ (namely, $z=0$, since $sinthetanecostheta$)}\[2em]
        &underline{text{3. case: $costheta=0,,rne0,,ninmathbb Z$}}\
        (*) &= #left{z| r0 = z1 wedge r1 = - z0right}\
        &= #left{z|z=0 wedge r = 0right}\
        &= 0
        end{align}



        Thus, in total:



        $$FP{delta}(r,theta) = begin{cases}
        1 &text{when} &r=0 \
        0 &text{else.}
        end{cases}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 19:50

























        answered Jan 6 at 19:41









        Marcus MüllerMarcus Müller

        18418




        18418






























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