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I have two state space systems. Now I want to compute the state space system where the first state space system is the input of the other...

$$M_1 = \begin{cases}\dot{x}_1 = A_1 x_1 + B_1 u_1 \\ y_1 = C_1 x_1 + D_1 u_1 \end{cases}$$ $$M_2 = \begin{cases}\dot{x}_2 = A_2 x_2 + B_2 u_2 \\ y_2 = C_2 x_2 + D_2 u_2 \end{cases}$$

 + u1 r --->O---->[ M1 ]----+---> y1 - | | y2 +-----[ M2 ]<---+ u2 

Where $u_1 = r - y_2$ and $u_2 = y_1$.

How do you compute the new $A,B,C,D$ matrices?

When I have $A_1 = B_1 = C_1 = D_1 = 1$ and $A_2 = B_2 = C_2 = D_2 = 2$ I should obtain

$A = \begin{bmatrix} 1/3 & -2/3 \\ 2/3 & 2/3 \end{bmatrix}, B = \begin{bmatrix} 1/3 \\ 2/3 \end{bmatrix}, C = \begin{bmatrix} 1/3 & -2/3 \end{bmatrix}, D = 1/3$

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1 Answer 1

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Write $u_1=r-y_2$ and $u_2 = y_1$, then

\begin{align*} y_1 &= C_1 x_1 + D_1 (r - y_2)\\ &= C_1 x_1 - D_1 C_2 x_2 - D_1 D_2 u_2 + D_1 r \\ (I + D_1 D_2) y_1 &= \begin{bmatrix} C_1 & -D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + D_1 r \\ y_1 &= \begin{bmatrix} R C_1 & -R D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + R D_1 r \\ \end{align*}

where $R = (I + D_1 D_2)^{-1}$ provided that it exists (the system is admissible).

\begin{align*} \dot{x}_1 &= A_1 x_1 + B_1 (r - y_2) \\ &= A_1 x_1 - B_1 C_2 x_2 - B_1 D_2 u_2 + B_1 r \\ &= \begin{bmatrix} A_1 - B_1 D_2 R C_1 & -B_1 C_2 + B_1 D_2 R D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + (- B_1 D_2 R D_1 + B_1) r \\ \dot{x}_2 &= A_2 x_2 + B_2 y_1 \\ &= \begin{bmatrix} B_2 R C_1 & A_2 - B_2 R D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + B_2 R D_1 r \end{align*}

Therefore, the overall system can be written as

\begin{align*} \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} &= \begin{bmatrix} A_1 - B_1 D_2 R C_1 & B_1 (D_2 R D_1 - I) C_2 \\ B_2 R C_1 & A_2 - B_2 R D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} B_1(I - D_2 R D_1) \\ B_2 R D_1 \end{bmatrix} r \\ y_1 &= \begin{bmatrix} R C_1 & -R D_1 C_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + R D_1 r \end{align*}

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  • $\begingroup$ question btw... what does "the system is admissible" mean? $\endgroup$ Commented Sep 24, 2014 at 19:32
  • $\begingroup$ That there is no algebraic loop in the system. Some terminology is also "well-posed" $\endgroup$ Commented Feb 16, 2021 at 15:35

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