Coordinates of a vector FFT












1












$begingroup$


I was reading a paper about DFT, at the end he got the relations $Y=frac{1}{sqrt N}W_n cdot y$ and $y=frac{1}{sqrt N}W_n cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
Later, the author shows how to reduce the following into smaller matrices but I don't really understand what is he doing nor how is he doing it.



begin{align*}
W_4 begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix} &= begin{bmatrix}
begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
begin{bmatrix} y_0\y_2 end{bmatrix}
+begin{bmatrix} 1&1\omega&omega^3 end{bmatrix}
begin{bmatrix} y_1\y_3 end{bmatrix}
\
begin{bmatrix} 1&omega^4\1&omega^6 end{bmatrix}
begin{bmatrix} y_0\y_2 end{bmatrix}
+begin{bmatrix} omega^2&omega^6\omega^3&omega^9 end{bmatrix}
begin{bmatrix} y_1\y_3 end{bmatrix}
end{bmatrix}
\ &=
begin{bmatrix}
begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
begin{bmatrix} y_0\y_2 end{bmatrix}
+begin{bmatrix} 1\&omega end{bmatrix}
begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
begin{bmatrix} y_1\y_3 end{bmatrix}
\
begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
begin{bmatrix} y_0\y_2 end{bmatrix}
-begin{bmatrix} 1\&omega end{bmatrix}
begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
begin{bmatrix} y_1\y_3 end{bmatrix}
end{bmatrix}
end{align*}






fast-fourier-transform







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    1












    $begingroup$


    I was reading a paper about DFT, at the end he got the relations $Y=frac{1}{sqrt N}W_n cdot y$ and $y=frac{1}{sqrt N}W_n cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
    Later, the author shows how to reduce the following into smaller matrices but I don't really understand what is he doing nor how is he doing it.



    begin{align*}
    W_4 begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix} &= begin{bmatrix}
    begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
    begin{bmatrix} y_0\y_2 end{bmatrix}
    +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix}
    begin{bmatrix} y_1\y_3 end{bmatrix}
    \
    begin{bmatrix} 1&omega^4\1&omega^6 end{bmatrix}
    begin{bmatrix} y_0\y_2 end{bmatrix}
    +begin{bmatrix} omega^2&omega^6\omega^3&omega^9 end{bmatrix}
    begin{bmatrix} y_1\y_3 end{bmatrix}
    end{bmatrix}
    \ &=
    begin{bmatrix}
    begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
    begin{bmatrix} y_0\y_2 end{bmatrix}
    +begin{bmatrix} 1\&omega end{bmatrix}
    begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
    begin{bmatrix} y_1\y_3 end{bmatrix}
    \
    begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
    begin{bmatrix} y_0\y_2 end{bmatrix}
    -begin{bmatrix} 1\&omega end{bmatrix}
    begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
    begin{bmatrix} y_1\y_3 end{bmatrix}
    end{bmatrix}
    end{align*}






    fast-fourier-transform







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I was reading a paper about DFT, at the end he got the relations $Y=frac{1}{sqrt N}W_n cdot y$ and $y=frac{1}{sqrt N}W_n cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
      Later, the author shows how to reduce the following into smaller matrices but I don't really understand what is he doing nor how is he doing it.



      begin{align*}
      W_4 begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix} &= begin{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      \
      begin{bmatrix} 1&omega^4\1&omega^6 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} omega^2&omega^6\omega^3&omega^9 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      end{bmatrix}
      \ &=
      begin{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} 1\&omega end{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      \
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      -begin{bmatrix} 1\&omega end{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      end{bmatrix}
      end{align*}






      fast-fourier-transform







      share|cite|improve this question











      $endgroup$




      I was reading a paper about DFT, at the end he got the relations $Y=frac{1}{sqrt N}W_n cdot y$ and $y=frac{1}{sqrt N}W_n cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
      Later, the author shows how to reduce the following into smaller matrices but I don't really understand what is he doing nor how is he doing it.



      begin{align*}
      W_4 begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix} &= begin{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      \
      begin{bmatrix} 1&omega^4\1&omega^6 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} omega^2&omega^6\omega^3&omega^9 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      end{bmatrix}
      \ &=
      begin{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      +begin{bmatrix} 1\&omega end{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      \
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_0\y_2 end{bmatrix}
      -begin{bmatrix} 1\&omega end{bmatrix}
      begin{bmatrix} 1&1\1&omega^2 end{bmatrix}
      begin{bmatrix} y_1\y_3 end{bmatrix}
      end{bmatrix}
      end{align*}






      fast-fourier-transform




      fast-fourier-transform






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 26 '18 at 1:02









      Rócherz

      2,9112821




      2,9112821










      asked Dec 25 '18 at 20:38









      Alejandro DuqueAlejandro Duque

      62




      62






















          1 Answer
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          $begingroup$

          In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
          begin{align*}
          begin{bmatrix} 1&1&1&1\ 1&omega&omega^2&omega^3 end{bmatrix} begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix}
          &=
          begin{bmatrix} y_0+y_1+y_2+y_3\ y_0+omega y_1+omega^2y_2+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} (y_0+y_2)+(y_1+y_3)\ (y_0+omega^2y_2)+(omega y_1+omega^3y_3) end{bmatrix}
          \ &=
          begin{bmatrix} y_0+y_2\ y_0+omega^2y_2 end{bmatrix}
          +
          begin{bmatrix} y_1+y_3\ omega y_1+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} 1&1\1&omega^2 end{bmatrix} begin{bmatrix} y_0\y_2 end{bmatrix} +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix} begin{bmatrix} y_1\y_3 end{bmatrix}.
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh ok, now i get it thanks. i will continue by myself from here
            $endgroup$
            – Alejandro Duque
            Dec 25 '18 at 21:30











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          1 Answer
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          1 Answer
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          active

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          active

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          1












          $begingroup$

          In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
          begin{align*}
          begin{bmatrix} 1&1&1&1\ 1&omega&omega^2&omega^3 end{bmatrix} begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix}
          &=
          begin{bmatrix} y_0+y_1+y_2+y_3\ y_0+omega y_1+omega^2y_2+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} (y_0+y_2)+(y_1+y_3)\ (y_0+omega^2y_2)+(omega y_1+omega^3y_3) end{bmatrix}
          \ &=
          begin{bmatrix} y_0+y_2\ y_0+omega^2y_2 end{bmatrix}
          +
          begin{bmatrix} y_1+y_3\ omega y_1+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} 1&1\1&omega^2 end{bmatrix} begin{bmatrix} y_0\y_2 end{bmatrix} +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix} begin{bmatrix} y_1\y_3 end{bmatrix}.
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh ok, now i get it thanks. i will continue by myself from here
            $endgroup$
            – Alejandro Duque
            Dec 25 '18 at 21:30
















          1












          $begingroup$

          In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
          begin{align*}
          begin{bmatrix} 1&1&1&1\ 1&omega&omega^2&omega^3 end{bmatrix} begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix}
          &=
          begin{bmatrix} y_0+y_1+y_2+y_3\ y_0+omega y_1+omega^2y_2+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} (y_0+y_2)+(y_1+y_3)\ (y_0+omega^2y_2)+(omega y_1+omega^3y_3) end{bmatrix}
          \ &=
          begin{bmatrix} y_0+y_2\ y_0+omega^2y_2 end{bmatrix}
          +
          begin{bmatrix} y_1+y_3\ omega y_1+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} 1&1\1&omega^2 end{bmatrix} begin{bmatrix} y_0\y_2 end{bmatrix} +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix} begin{bmatrix} y_1\y_3 end{bmatrix}.
          end{align*}






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh ok, now i get it thanks. i will continue by myself from here
            $endgroup$
            – Alejandro Duque
            Dec 25 '18 at 21:30














          1












          1








          1





          $begingroup$

          In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
          begin{align*}
          begin{bmatrix} 1&1&1&1\ 1&omega&omega^2&omega^3 end{bmatrix} begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix}
          &=
          begin{bmatrix} y_0+y_1+y_2+y_3\ y_0+omega y_1+omega^2y_2+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} (y_0+y_2)+(y_1+y_3)\ (y_0+omega^2y_2)+(omega y_1+omega^3y_3) end{bmatrix}
          \ &=
          begin{bmatrix} y_0+y_2\ y_0+omega^2y_2 end{bmatrix}
          +
          begin{bmatrix} y_1+y_3\ omega y_1+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} 1&1\1&omega^2 end{bmatrix} begin{bmatrix} y_0\y_2 end{bmatrix} +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix} begin{bmatrix} y_1\y_3 end{bmatrix}.
          end{align*}






          share|cite|improve this answer











          $endgroup$



          In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
          begin{align*}
          begin{bmatrix} 1&1&1&1\ 1&omega&omega^2&omega^3 end{bmatrix} begin{bmatrix} y_0\y_1\y_2\y_3 end{bmatrix}
          &=
          begin{bmatrix} y_0+y_1+y_2+y_3\ y_0+omega y_1+omega^2y_2+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} (y_0+y_2)+(y_1+y_3)\ (y_0+omega^2y_2)+(omega y_1+omega^3y_3) end{bmatrix}
          \ &=
          begin{bmatrix} y_0+y_2\ y_0+omega^2y_2 end{bmatrix}
          +
          begin{bmatrix} y_1+y_3\ omega y_1+omega^3y_3 end{bmatrix}
          \ &=
          begin{bmatrix} 1&1\1&omega^2 end{bmatrix} begin{bmatrix} y_0\y_2 end{bmatrix} +begin{bmatrix} 1&1\omega&omega^3 end{bmatrix} begin{bmatrix} y_1\y_3 end{bmatrix}.
          end{align*}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 21:32

























          answered Dec 25 '18 at 21:01









          RócherzRócherz

          2,9112821




          2,9112821












          • $begingroup$
            Oh ok, now i get it thanks. i will continue by myself from here
            $endgroup$
            – Alejandro Duque
            Dec 25 '18 at 21:30


















          • $begingroup$
            Oh ok, now i get it thanks. i will continue by myself from here
            $endgroup$
            – Alejandro Duque
            Dec 25 '18 at 21:30
















          $begingroup$
          Oh ok, now i get it thanks. i will continue by myself from here
          $endgroup$
          – Alejandro Duque
          Dec 25 '18 at 21:30




          $begingroup$
          Oh ok, now i get it thanks. i will continue by myself from here
          $endgroup$
          – Alejandro Duque
          Dec 25 '18 at 21:30


















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