How to find a transformation matrix which will make the system a chain of integrators?











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Consider a system of the form
$$dot{x}(t)=Ax(t)+Bu(t)+phi(t)+D(t)$$
I have
$$dot{x}(t)=begin{bmatrix}
-p_1 &G_b & 0 & 0 &0 \
0& -p_2 & p_3 & 0 & 0\
0& 0 & -p_4 & p_5 &0 \
0& 0 & 0 & -p_6 &p_6 \
0& 0& 0 & 0 & -p_6
end{bmatrix}x(t)+begin{bmatrix}
0\
0\
0\
0\
1\
end{bmatrix}u(t)+begin{bmatrix}
-x_1(t)x_2(t)\
0\
0\
0\
0\
end{bmatrix}+begin{bmatrix}
1\
0\
0\
0\
0\
end{bmatrix}D(t)$$

Where $phi(t)$ is a lumped nonlinearity of the system and $D(t)$ is a disturbance acting from outside. I want to convert the system of the form
$$dot{Z}_{i}=Z_{i+1}+text{maybe nonlinearities and disturbances}, i=1,2,...,r-1 \dot{Z}_{r}=u+text{maybe some function oif states}$$
i.e
$$dot{Z_1}=Z_2 \ dot{Z_2}=Z_3 \ cdots \dot{Z_r}=f(Z_1,...,Z_r,t,)+u$$
How to find a transformation matrix to do this?










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    up vote
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    down vote

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    Consider a system of the form
    $$dot{x}(t)=Ax(t)+Bu(t)+phi(t)+D(t)$$
    I have
    $$dot{x}(t)=begin{bmatrix}
    -p_1 &G_b & 0 & 0 &0 \
    0& -p_2 & p_3 & 0 & 0\
    0& 0 & -p_4 & p_5 &0 \
    0& 0 & 0 & -p_6 &p_6 \
    0& 0& 0 & 0 & -p_6
    end{bmatrix}x(t)+begin{bmatrix}
    0\
    0\
    0\
    0\
    1\
    end{bmatrix}u(t)+begin{bmatrix}
    -x_1(t)x_2(t)\
    0\
    0\
    0\
    0\
    end{bmatrix}+begin{bmatrix}
    1\
    0\
    0\
    0\
    0\
    end{bmatrix}D(t)$$

    Where $phi(t)$ is a lumped nonlinearity of the system and $D(t)$ is a disturbance acting from outside. I want to convert the system of the form
    $$dot{Z}_{i}=Z_{i+1}+text{maybe nonlinearities and disturbances}, i=1,2,...,r-1 \dot{Z}_{r}=u+text{maybe some function oif states}$$
    i.e
    $$dot{Z_1}=Z_2 \ dot{Z_2}=Z_3 \ cdots \dot{Z_r}=f(Z_1,...,Z_r,t,)+u$$
    How to find a transformation matrix to do this?










    share|cite|improve this question
























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      1





      Consider a system of the form
      $$dot{x}(t)=Ax(t)+Bu(t)+phi(t)+D(t)$$
      I have
      $$dot{x}(t)=begin{bmatrix}
      -p_1 &G_b & 0 & 0 &0 \
      0& -p_2 & p_3 & 0 & 0\
      0& 0 & -p_4 & p_5 &0 \
      0& 0 & 0 & -p_6 &p_6 \
      0& 0& 0 & 0 & -p_6
      end{bmatrix}x(t)+begin{bmatrix}
      0\
      0\
      0\
      0\
      1\
      end{bmatrix}u(t)+begin{bmatrix}
      -x_1(t)x_2(t)\
      0\
      0\
      0\
      0\
      end{bmatrix}+begin{bmatrix}
      1\
      0\
      0\
      0\
      0\
      end{bmatrix}D(t)$$

      Where $phi(t)$ is a lumped nonlinearity of the system and $D(t)$ is a disturbance acting from outside. I want to convert the system of the form
      $$dot{Z}_{i}=Z_{i+1}+text{maybe nonlinearities and disturbances}, i=1,2,...,r-1 \dot{Z}_{r}=u+text{maybe some function oif states}$$
      i.e
      $$dot{Z_1}=Z_2 \ dot{Z_2}=Z_3 \ cdots \dot{Z_r}=f(Z_1,...,Z_r,t,)+u$$
      How to find a transformation matrix to do this?










      share|cite|improve this question













      Consider a system of the form
      $$dot{x}(t)=Ax(t)+Bu(t)+phi(t)+D(t)$$
      I have
      $$dot{x}(t)=begin{bmatrix}
      -p_1 &G_b & 0 & 0 &0 \
      0& -p_2 & p_3 & 0 & 0\
      0& 0 & -p_4 & p_5 &0 \
      0& 0 & 0 & -p_6 &p_6 \
      0& 0& 0 & 0 & -p_6
      end{bmatrix}x(t)+begin{bmatrix}
      0\
      0\
      0\
      0\
      1\
      end{bmatrix}u(t)+begin{bmatrix}
      -x_1(t)x_2(t)\
      0\
      0\
      0\
      0\
      end{bmatrix}+begin{bmatrix}
      1\
      0\
      0\
      0\
      0\
      end{bmatrix}D(t)$$

      Where $phi(t)$ is a lumped nonlinearity of the system and $D(t)$ is a disturbance acting from outside. I want to convert the system of the form
      $$dot{Z}_{i}=Z_{i+1}+text{maybe nonlinearities and disturbances}, i=1,2,...,r-1 \dot{Z}_{r}=u+text{maybe some function oif states}$$
      i.e
      $$dot{Z_1}=Z_2 \ dot{Z_2}=Z_3 \ cdots \dot{Z_r}=f(Z_1,...,Z_r,t,)+u$$
      How to find a transformation matrix to do this?







      differential-equations dynamical-systems control-theory nonlinear-system optimal-control






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      asked Nov 13 at 14:47









      Darthsid1995

      1,0871418




      1,0871418






















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          For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by



          $$
          vec{v}^top = begin{bmatrix}0 & cdots & 0 & 1end{bmatrix}
          begin{bmatrix}
          B & B, A & B, A^2 & cdots & B, A^{n-1}
          end{bmatrix}^{-1}, tag{1}
          $$



          $$
          T = begin{bmatrix}
          vec{v}^top \
          vec{v}^top A \
          vec{v}^top A^2 \
          vdots \
          vec{v}^top A^{n-1}
          end{bmatrix}. tag{2}
          $$



          So using the transformation $z(t) = T,x(t)$ gives



          $$
          dot{z} = T,A,T^{-1} z(t) + T,B,u(t) + T,phi(t) + T,D(t), tag{3}
          $$



          where $(T,A,T^{-1}, T,B)$ will be in the controllable canonical form.



          If you would like to know more about how to derive this then you can look at this related question.






          share|cite|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            up vote
            1
            down vote



            accepted










            For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by



            $$
            vec{v}^top = begin{bmatrix}0 & cdots & 0 & 1end{bmatrix}
            begin{bmatrix}
            B & B, A & B, A^2 & cdots & B, A^{n-1}
            end{bmatrix}^{-1}, tag{1}
            $$



            $$
            T = begin{bmatrix}
            vec{v}^top \
            vec{v}^top A \
            vec{v}^top A^2 \
            vdots \
            vec{v}^top A^{n-1}
            end{bmatrix}. tag{2}
            $$



            So using the transformation $z(t) = T,x(t)$ gives



            $$
            dot{z} = T,A,T^{-1} z(t) + T,B,u(t) + T,phi(t) + T,D(t), tag{3}
            $$



            where $(T,A,T^{-1}, T,B)$ will be in the controllable canonical form.



            If you would like to know more about how to derive this then you can look at this related question.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by



              $$
              vec{v}^top = begin{bmatrix}0 & cdots & 0 & 1end{bmatrix}
              begin{bmatrix}
              B & B, A & B, A^2 & cdots & B, A^{n-1}
              end{bmatrix}^{-1}, tag{1}
              $$



              $$
              T = begin{bmatrix}
              vec{v}^top \
              vec{v}^top A \
              vec{v}^top A^2 \
              vdots \
              vec{v}^top A^{n-1}
              end{bmatrix}. tag{2}
              $$



              So using the transformation $z(t) = T,x(t)$ gives



              $$
              dot{z} = T,A,T^{-1} z(t) + T,B,u(t) + T,phi(t) + T,D(t), tag{3}
              $$



              where $(T,A,T^{-1}, T,B)$ will be in the controllable canonical form.



              If you would like to know more about how to derive this then you can look at this related question.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by



                $$
                vec{v}^top = begin{bmatrix}0 & cdots & 0 & 1end{bmatrix}
                begin{bmatrix}
                B & B, A & B, A^2 & cdots & B, A^{n-1}
                end{bmatrix}^{-1}, tag{1}
                $$



                $$
                T = begin{bmatrix}
                vec{v}^top \
                vec{v}^top A \
                vec{v}^top A^2 \
                vdots \
                vec{v}^top A^{n-1}
                end{bmatrix}. tag{2}
                $$



                So using the transformation $z(t) = T,x(t)$ gives



                $$
                dot{z} = T,A,T^{-1} z(t) + T,B,u(t) + T,phi(t) + T,D(t), tag{3}
                $$



                where $(T,A,T^{-1}, T,B)$ will be in the controllable canonical form.



                If you would like to know more about how to derive this then you can look at this related question.






                share|cite|improve this answer












                For a single input system the similarity transformation which transforms the system into its controllable canonical form is given by



                $$
                vec{v}^top = begin{bmatrix}0 & cdots & 0 & 1end{bmatrix}
                begin{bmatrix}
                B & B, A & B, A^2 & cdots & B, A^{n-1}
                end{bmatrix}^{-1}, tag{1}
                $$



                $$
                T = begin{bmatrix}
                vec{v}^top \
                vec{v}^top A \
                vec{v}^top A^2 \
                vdots \
                vec{v}^top A^{n-1}
                end{bmatrix}. tag{2}
                $$



                So using the transformation $z(t) = T,x(t)$ gives



                $$
                dot{z} = T,A,T^{-1} z(t) + T,B,u(t) + T,phi(t) + T,D(t), tag{3}
                $$



                where $(T,A,T^{-1}, T,B)$ will be in the controllable canonical form.



                If you would like to know more about how to derive this then you can look at this related question.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Kwin van der Veen

                5,0852826




                5,0852826






























                     

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