MIMO transformation to controller canonical form
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I am unable to prove a result concerning MIMO linear dynamical systems. Let
$$ dotX = Acdot X + Bcdot U$$ be a linear time invariant dynamical system, with $A in mathbbR^ntimes n$, $B in mathbbR^ntimes m$ and $U in mathbbR^m$ with
$$ Co(A,B) = beginbmatrixB &... &A^n-1 cdot B endbmatrix$$ having full rank $n$. Let
$$M = beginbmatrix B_1 &... &A^r_1-1cdot B_1 & ... &B_m &... &A^r_m-1cdot B_mendbmatrix$$ be a square nonsingular matrix formed with the columns of $Co(A,B)$, where $r_1 + ... + r_m = n$ are called controllability indices.
I want to obtain the controller canonical form for this system. For this I take the transformation matrix
$$T_B = beginbmatrix C_1^T\ vdots \ C_1^Tcdot A^r_1-1\ vdots \ C_m^T\ vdots \ C_m^Tcdot A^r_m-1endbmatrix$$ Take then $Y = T_Bcdot X$ and obtain:
$$ dotY = T_Bcdot dotX = T_Bcdot Acdot X + T_Bcdot Bcdot U$$ We find $C_i$s such that
$$ M^T cdot beginbmatrix C_1 &... &C_mendbmatrix = beginbmatrixe_r_1 &e_r_1+r_2 &... &e_r_1 + ... + r_m endbmatrix$$ where $r_x$ is the $x$'th column of $1_ntimes n$. I am unable to prove that
$$ T_Bcdot B = beginbmatrix0 &0 &... &0\ 0 &0 &... &0\ vdots &vdots &ddots &vdots\ 1 &* &... &*\ vdots &vdots &ddots &vdotsendbmatrix$$ More precisely is I am unable to prove that $T_Bcdot B$ has $n-m$ zero lines. can somebody give me a hint?
linear-algebra control-theory linear-control
asked Aug 27 at 14:39
C Marius
533210
add a comment |Â
up vote
0
down vote
favorite
I am unable to prove a result concerning MIMO linear dynamical systems. Let
$$ dotX = Acdot X + Bcdot U$$ be a linear time invariant dynamical system, with $A in mathbbR^ntimes n$, $B in mathbbR^ntimes m$ and $U in mathbbR^m$ with
$$ Co(A,B) = beginbmatrixB &... &A^n-1 cdot B endbmatrix$$ having full rank $n$. Let
$$M = beginbmatrix B_1 &... &A^r_1-1cdot B_1 & ... &B_m &... &A^r_m-1cdot B_mendbmatrix$$ be a square nonsingular matrix formed with the columns of $Co(A,B)$, where $r_1 + ... + r_m = n$ are called controllability indices.
I want to obtain the controller canonical form for this system. For this I take the transformation matrix
$$T_B = beginbmatrix C_1^T\ vdots \ C_1^Tcdot A^r_1-1\ vdots \ C_m^T\ vdots \ C_m^Tcdot A^r_m-1endbmatrix$$ Take then $Y = T_Bcdot X$ and obtain:
$$ dotY = T_Bcdot dotX = T_Bcdot Acdot X + T_Bcdot Bcdot U$$ We find $C_i$s such that
$$ M^T cdot beginbmatrix C_1 &... &C_mendbmatrix = beginbmatrixe_r_1 &e_r_1+r_2 &... &e_r_1 + ... + r_m endbmatrix$$ where $r_x$ is the $x$'th column of $1_ntimes n$. I am unable to prove that
$$ T_Bcdot B = beginbmatrix0 &0 &... &0\ 0 &0 &... &0\ vdots &vdots &ddots &vdots\ 1 &* &... &*\ vdots &vdots &ddots &vdotsendbmatrix$$ More precisely is I am unable to prove that $T_Bcdot B$ has $n-m$ zero lines. can somebody give me a hint?
linear-algebra control-theory linear-control
asked Aug 27 at 14:39
C Marius
533210
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am unable to prove a result concerning MIMO linear dynamical systems. Let
$$ dotX = Acdot X + Bcdot U$$ be a linear time invariant dynamical system, with $A in mathbbR^ntimes n$, $B in mathbbR^ntimes m$ and $U in mathbbR^m$ with
$$ Co(A,B) = beginbmatrixB &... &A^n-1 cdot B endbmatrix$$ having full rank $n$. Let
$$M = beginbmatrix B_1 &... &A^r_1-1cdot B_1 & ... &B_m &... &A^r_m-1cdot B_mendbmatrix$$ be a square nonsingular matrix formed with the columns of $Co(A,B)$, where $r_1 + ... + r_m = n$ are called controllability indices.
I want to obtain the controller canonical form for this system. For this I take the transformation matrix
$$T_B = beginbmatrix C_1^T\ vdots \ C_1^Tcdot A^r_1-1\ vdots \ C_m^T\ vdots \ C_m^Tcdot A^r_m-1endbmatrix$$ Take then $Y = T_Bcdot X$ and obtain:
$$ dotY = T_Bcdot dotX = T_Bcdot Acdot X + T_Bcdot Bcdot U$$ We find $C_i$s such that
$$ M^T cdot beginbmatrix C_1 &... &C_mendbmatrix = beginbmatrixe_r_1 &e_r_1+r_2 &... &e_r_1 + ... + r_m endbmatrix$$ where $r_x$ is the $x$'th column of $1_ntimes n$. I am unable to prove that
$$ T_Bcdot B = beginbmatrix0 &0 &... &0\ 0 &0 &... &0\ vdots &vdots &ddots &vdots\ 1 &* &... &*\ vdots &vdots &ddots &vdotsendbmatrix$$ More precisely is I am unable to prove that $T_Bcdot B$ has $n-m$ zero lines. can somebody give me a hint?
linear-algebra control-theory linear-control
asked Aug 27 at 14:39
C Marius
533210
I am unable to prove a result concerning MIMO linear dynamical systems. Let
$$ dotX = Acdot X + Bcdot U$$ be a linear time invariant dynamical system, with $A in mathbbR^ntimes n$, $B in mathbbR^ntimes m$ and $U in mathbbR^m$ with
$$ Co(A,B) = beginbmatrixB &... &A^n-1 cdot B endbmatrix$$ having full rank $n$. Let
$$M = beginbmatrix B_1 &... &A^r_1-1cdot B_1 & ... &B_m &... &A^r_m-1cdot B_mendbmatrix$$ be a square nonsingular matrix formed with the columns of $Co(A,B)$, where $r_1 + ... + r_m = n$ are called controllability indices.
I want to obtain the controller canonical form for this system. For this I take the transformation matrix
$$T_B = beginbmatrix C_1^T\ vdots \ C_1^Tcdot A^r_1-1\ vdots \ C_m^T\ vdots \ C_m^Tcdot A^r_m-1endbmatrix$$ Take then $Y = T_Bcdot X$ and obtain:
$$ dotY = T_Bcdot dotX = T_Bcdot Acdot X + T_Bcdot Bcdot U$$ We find $C_i$s such that
$$ M^T cdot beginbmatrix C_1 &... &C_mendbmatrix = beginbmatrixe_r_1 &e_r_1+r_2 &... &e_r_1 + ... + r_m endbmatrix$$ where $r_x$ is the $x$'th column of $1_ntimes n$. I am unable to prove that
$$ T_Bcdot B = beginbmatrix0 &0 &... &0\ 0 &0 &... &0\ vdots &vdots &ddots &vdots\ 1 &* &... &*\ vdots &vdots &ddots &vdotsendbmatrix$$ More precisely is I am unable to prove that $T_Bcdot B$ has $n-m$ zero lines. can somebody give me a hint?
linear-algebra control-theory linear-control
asked Aug 27 at 14:39
C Marius
533210
edited Aug 28 at 13:37
asked Aug 27 at 14:39
C Marius
533210
asked Aug 27 at 14:39
C Marius
533210
asked Aug 27 at 14:39
C Marius
533210
533210
add a comment |Â
add a comment |Â
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