---
title: "Transfer matrix and stability of MIMO state model"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-mimo-tf-stability"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-mimo-tf-stability
site: BCT Engineering
---

# Transfer matrix and stability of MIMO state model

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2078 Bhadra, 8 marks
- Video: [State Space Analysis for MIMO Systems in Control Systems](https://www.youtube.com/watch?v=c1IEWCckLgM) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-mimo-tf-stability
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

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.

## 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.

## Answer

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

#### Given
$$
A = \begin{bmatrix}-1 & 1\\ 0 & -2\end{bmatrix},\;
B = \begin{bmatrix}1 & 0 & 1\\ 0 & 1 & 1\end{bmatrix},\;
C = \begin{bmatrix}1 & 2\\ 1 & 0\\ 1 & 1\end{bmatrix},\; D = 0
$$

#### Step 1: $(sI - A)^{-1}$
$$
sI - A = \begin{bmatrix}s+1 & -1\\ 0 & s+2\end{bmatrix}, \qquad |sI - A| = (s+1)(s+2)
$$
$$
(sI - A)^{-1} = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 1\\ 0 & s+1\end{bmatrix}
$$

#### Step 2: $(sI - A)^{-1}B$
$$
(sI - A)^{-1}B = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 1 & s+3\\ 0 & s+1 & s+1\end{bmatrix}
$$

#### Step 3: Multiply by $C$
Row by row (common factor $\frac{1}{(s+1)(s+2)}$):
- Row 1, $C_1 = [1\;\;2]$: $[\,s+2,\;\; 1 + 2(s+1),\;\; (s+3) + 2(s+1)\,] = [\,s+2,\;\; 2s+3,\;\; 3s+5\,]$
- Row 2, $C_2 = [1\;\;0]$: $[\,s+2,\;\; 1,\;\; s+3\,]$
- Row 3, $C_3 = [1\;\;1]$: $[\,s+2,\;\; s+2,\;\; 2s+4\,]$

So
$$
G(s) = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s+2 & 2s+3 & 3s+5\\ s+2 & 1 & s+3\\ s+2 & s+2 & 2(s+2)\end{bmatrix}
$$

After cancelling common factors:
$$
G(s) = \begin{bmatrix}
\frac{1}{s+1} & \frac{2s+3}{(s+1)(s+2)} & \frac{3s+5}{(s+1)(s+2)}\\[4pt]
\frac{1}{s+1} & \frac{1}{(s+1)(s+2)} & \frac{s+3}{(s+1)(s+2)}\\[4pt]
\frac{1}{s+1} & \frac{1}{s+1} & \frac{2}{s+1}
\end{bmatrix}
$$
Here $G_{ij}(s) = Y_i(s)/U_j(s)$ with the other inputs set to zero.

#### Step 4: Stability
Characteristic equation:
$$
|sI - A| = (s+1)(s+2) = s^2 + 3s + 2 = 0 \;\Rightarrow\; s = -1,\; -2
$$
Since $A$ is upper triangular, its eigenvalues are simply the diagonal entries $-1$ and $-2$.

| Check | Result |
|---|---|
| Eigenvalues of $A$ | $-1$, $-2$ |
| Real parts | Both negative |
| Poles of every $G_{ij}(s)$ | Subset of $\{-1, -2\}$ |
| Routh array ($1, 3, 2$) | No sign change |

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the size of G(s) for this system with B (2x3) and C (3x2)?
   A. 2 x 2
   B. 3 x 3
   C. 2 x 3
   D. 1 x 1
   Answer: B. There are 3 inputs and 3 outputs, so G(s) = C(sI-A)^-1 B is a 3 x 3 matrix with Gij = Yi/Uj. (Key point: Transfer matrix size)

2. What is (sI - A)^-1 for A = [-1 1; 0 -2]?
   A. 1/((s+1)(s+2)) [s+2 1; 0 s+1]
   B. 1/((s+1)(s+2)) [s+1 1; 0 s+2]
   C. 1/((s+1)(s+2)) [s+2 -1; 0 s+1]
   D. [1/(s+1) 0; 0 1/(s+2)]
   Answer: A. sI-A = [s+1 -1; 0 s+2], determinant (s+1)(s+2); the adjugate gives [s+2 1; 0 s+1]. (Key point: Inverse of sI - A)

3. How are the poles of the system found, and what are they?
   A. From |sI - A| = 0; s = -1, -2
   B. From |sI - A| = 0; s = 1, 2
   C. From the entries of B; s = 0, 1
   D. From C only; s = -1, -3
   Answer: A. A is upper triangular, so its eigenvalues are the diagonal entries -1 and -2. (Key point: Stability from eigenvalues)

4. What is the stability conclusion?
   A. Unstable, because B has a column of ones
   B. Marginally stable, because G(s) has simple poles
   C. Asymptotically stable, because both eigenvalues of A are in the left half plane
   D. Cannot be decided for a MIMO system
   Answer: C. All eigenvalues of A have negative real parts, so the system is asymptotically stable and also BIBO stable. (Key point: Stability conclusion)

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