algebraic-degeneration

State-space model, algebraic degeneration example

tableau model, equations

(%i6)

e1 : i1 + i2 = 0 ;
e2 : u1 − v1 = 0 ;
e3 : u2 − v1 = 0 ;
e4 : u2 = eg ;
e5 : C· Du2 − i2 = 0 ;
T :[ e1, e2, e3, e4, e5] ;

\begin{matrix} (e1) & i2 + i1 = 0 \end{matrix}

\begin{matrix} (e2) & u1 - v1 = 0 \end{matrix}

\begin{matrix} (e3) & u2 - v1 = 0 \end{matrix}

\begin{matrix} (e4) & u2 = eg \end{matrix}

\begin{matrix} (e5) & C Du2 - i2 = 0 \end{matrix}

\begin{matrix} (T) & [i2 + i1 = 0, u1 - v1 = 0, u2 - v1 = 0, u2 = eg, C Du2 - i2 = 0] \end{matrix}

special vectors

(%i8)

Dx :[ Du2] ;
y :[ i1, i2, u1, v1] ;

\begin{matrix} (Dx) & [Du2] \end{matrix}

\begin{matrix} (y) & [i1, i2, u1, v1] \end{matrix}

state equations

(%i9)

se : eliminate( T, y) ;

\begin{matrix} (se) & [u2 - eg] \end{matrix}

algebraic degeneration!

solve for the degenerate variable

(%i10)

se : linsolve( se, u2)[ 1] ;

\begin{matrix} (se) & u2 = eg \end{matrix}

algebraic relation!

subsitute for the degenerated state variable derivative

(%i11)

de : Du2 = Deg ;

\begin{matrix} (de) & Du2 = Deg \end{matrix}

(%i12)

Tmod : ev( T, de) ;

\begin{matrix} (Tmod) & [i2 + i1 = 0, u1 - v1 = 0, u2 - v1 = 0, u2 = eg, C Deg - i2 = 0] \end{matrix}

there are no state equations; only output equations remain

u2 is now an output variable, let's change the output vector

(%i13)

ynew :[ i1, i2, u1, u2, v1] ;

\begin{matrix} (ynew) & [i1, i2, u1, u2, v1] \end{matrix}

(%i14)

oe : linsolve( Tmod, ynew) ;

\begin{matrix} (oe) & [i1 = - C Deg, i2 = C Deg, u1 = eg, u2 = eg, v1 = eg] \end{matrix}

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