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Transfer matrix and stability of MIMO state model

Short answer

The transfer matrix G(s)=C(sI-A)^{-1}B simplifies to: row1: 1/(s+1), (2s+3)/((s+1)(s+2)), (3s+5)/((s+1)(s+2)); row2: 1/(s+1), 1/((s+1)(s+2)), (s+3)/((s+1)(s+2)); row3: 1/(s+1), 1/(s+1), 2/(s+1). The eigenvalues of A are -1 and -2, so the system is asymptotically stable.

A system is described by the following equations: ẋ(t) = [−1 1; 0 −2]x(t) + [1 0 1; 0 1 1]u(t), y(t) = [1 2; 1 0; 1 1]x(t). Find the transfer function of the system and identify if the system is stable.

For a multi-input multi-output (MIMO) system the transfer function is a transfer matrix G(s)=C(sI−A)−1B+DG(s) = C(sI - A)^{-1}B + D. Here there are 3 inputs and 3 outputs, so G(s)G(s) is 3×33\times 3.

Given

A=[−110−2],  B=[101011],  C=[121011],  D=0A = \begin{bmatrix}-1 & 1\\ 0 & -2\end{bmatrix},\; B = \begin{bmatrix}1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix},\; C = \begin{bmatrix}1 & 2\\ 1 & 0\\ 1 & 1\end{bmatrix},\; D = 0

Step 1: (sI−A)−1(sI - A)^{-1}

sI−A=[s+1−10s+2],∣sI−A∣=(s+1)(s+2)sI - A = \begin{bmatrix}s+1 & -1\\ 0 & s+2\end{bmatrix}, \qquad |sI - A| = (s+1)(s+2) (sI−A)−1=1(s+1)(s+2)[s+210s+1](sI - A)^{-1} = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 1\\ 0 & s+1\end{bmatrix}

Step 2: (sI−A)−1B(sI - A)^{-1}B

(sI−A)−1B=1(s+1)(s+2)[s+21s+30s+1s+1](sI - A)^{-1}B = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 1 & s+3\\ 0 & s+1 & s+1\end{bmatrix}

Step 3: Multiply by CC

Row by row (common factor 1(s+1)(s+2)\frac{1}{(s+1)(s+2)}):

  • Row 1, C1=[1    2]C_1 = [1\;\;2]: [ s+2,    1+2(s+1),    (s+3)+2(s+1) ]=[ s+2,    2s+3,    3s+5 ][\,s+2,\;\; 1 + 2(s+1),\;\; (s+3) + 2(s+1)\,] = [\,s+2,\;\; 2s+3,\;\; 3s+5\,]
  • Row 2, C2=[1    0]C_2 = [1\;\;0]: [ s+2,    1,    s+3 ][\,s+2,\;\; 1,\;\; s+3\,]
  • Row 3, C3=[1    1]C_3 = [1\;\;1]: [ s+2,    s+2,    2s+4 ][\,s+2,\;\; s+2,\;\; 2s+4\,]

So

G(s)=1(s+1)(s+2)[s+22s+33s+5s+21s+3s+2s+22(s+2)]G(s) = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 2s+3 & 3s+5\\ s+2 & 1 & s+3\\ s+2 & s+2 & 2(s+2)\end{bmatrix}

After cancelling common factors:

G(s)=[1s+12s+3(s+1)(s+2)3s+5(s+1)(s+2)1s+11(s+1)(s+2)s+3(s+1)(s+2)1s+11s+12s+1]G(s) = \begin{bmatrix} \frac{1}{s+1} & \frac{2s+3}{(s+1)(s+2)} & \frac{3s+5}{(s+1)(s+2)}\\[4pt] \frac{1}{s+1} & \frac{1}{(s+1)(s+2)} & \frac{s+3}{(s+1)(s+2)}\\[4pt] \frac{1}{s+1} & \frac{1}{s+1} & \frac{2}{s+1} \end{bmatrix}

Here Gij(s)=Yi(s)/Uj(s)G_{ij}(s) = Y_i(s)/U_j(s) with the other inputs set to zero.

Step 4: Stability

Characteristic equation:

∣sI−A∣=(s+1)(s+2)=s2+3s+2=0  ⇒  s=−1,  −2|sI - A| = (s+1)(s+2) = s^2 + 3s + 2 = 0 \;\Rightarrow\; s = -1,\; -2

Since AA is upper triangular, its eigenvalues are simply the diagonal entries −1-1 and −2-2.

CheckResult
Eigenvalues of AA−1-1, −2-2
Real partsBoth negative
Poles of every Gij(s)G_{ij}(s)Subset of {−1,−2}\{-1, -2\}
Routh array (1,3,21, 3, 2)No sign change

Answer: The transfer matrix is as given above. All eigenvalues of AA lie in the left half s-plane, so the system is asymptotically stable (and also BIBO stable).

✓ Answer checked by Claude Opus 5.5 (high effort)

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