---
title: "State space analysis; TF of second-order model"
subject: "Control Systems"
chapter: 8
key: "semester_3/control_systems:ss-adv-limit-tf-3-1"
source: https://www.bctengineer.com/q/semester_3/control_systems/ss-adv-limit-tf-3-1
site: BCT Engineering
---

# State space analysis; TF of second-order model

Control Systems · Chapter 8: State Space Analysis

- Asked in: 2076 Chaitra, 2+6 marks
- Video: [State Space Analysis to Determine Transfer Function](https://www.youtube.com/watch?v=GuIvzps_4RI) — 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-adv-limit-tf-3-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8

## Question

Discuss the advantages and limitations of state space analysis of control systems. Find the transfer function for the system represented by following state space model: ẋ = [−3 1; −2 0]x + [0; 1]u, y = [1 0]x.

## Short answer

Advantages: handles MIMO, nonlinear, time-varying systems, includes initial conditions, reveals internal behaviour, supports modern design. Limitations: less intuitive, state model not unique, frequency-domain specs not directly visible. For the given model, G(s)=1/(s^2+3s+2)=1/[(s+1)(s+2)], poles at -1, -2, stable.

## Answer

#### Advantages of state space analysis
State space analysis describes a system by a set of first-order differential equations in terms of state variables, $\dot{x} = Ax + Bu$, $y = Cx + Du$.

1. Works for **MIMO** systems as easily as for single-input single-output systems.
2. Applies to **non-linear** and **time-varying** systems; the transfer function method does not.
3. Takes **non-zero initial conditions** into account.
4. Gives the **internal behaviour** (all states), not only the input-output relation; controllability and observability can be tested.
5. Suited to **computer solution** and to modern design methods (pole placement, optimal control, observers).
6. Works in the time domain directly.

#### Limitations
1. Needs more mathematics (matrices, eigenvalues); less physical insight for simple systems.
2. The state model is **not unique**: different choices of states give different $A$, $B$, $C$.
3. Frequency-domain specifications (gain margin, phase margin, bandwidth) are not seen directly.
4. Some states may not be measurable, so observers are needed for feedback.

#### Transfer function of the given model
$$
A = \begin{bmatrix}-3 & 1\\ -2 & 0\end{bmatrix},\; B = \begin{bmatrix}0\\ 1\end{bmatrix},\; C = \begin{bmatrix}1 & 0\end{bmatrix},\; D = 0
$$
$$
G(s) = C(sI - A)^{-1}B + D
$$

Step 1:
$$
sI - A = \begin{bmatrix}s+3 & -1\\ 2 & s\end{bmatrix}, \qquad |sI - A| = s(s+3) + 2 = s^2 + 3s + 2
$$

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

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

**Answer:**
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 3s + 2} = \frac{1}{(s+1)(s+2)}
$$
The poles are at $s = -1$ and $s = -2$, so the system is stable and overdamped.

## Check yourself

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

1. Which is an advantage of state space analysis?
   A. It takes non-zero initial conditions into account
   B. It requires less mathematics than transfer functions
   C. It shows gain and phase margin directly
   D. It only applies to linear time-invariant SISO systems
   Answer: A. State space includes initial conditions and handles MIMO, non-linear and time-varying systems. (Key point: Advantages of state space)

2. Which is a limitation of state space analysis?
   A. The state model is unique for a given system
   B. The model is not unique and frequency-domain specs are not seen directly
   C. It cannot handle MIMO systems
   D. It cannot be solved by computer
   Answer: B. Different state choices give different A, B, C, and gain margin or bandwidth are not directly visible. (Key point: Limitations)

3. What is |sI - A| for A = [-3 1; -2 0]?
   A. s^2 + 3s - 2
   B. s^2 + 3s + 2
   C. s^2 - 3s + 2
   D. s^2 + 2s + 3
   Answer: B. |sI-A| = (s+3)s - (-1)(2) = s^2 + 3s + 2. (Key point: Resolvent matrix)

4. Given C = [1 0] and B = [0; 1], what is G(s)?
   A. (s+3)/((s+1)(s+2))
   B. s/((s+1)(s+2))
   C. 1/((s+1)(s+2))
   D. 1/(s^2 + 3s - 2)
   Answer: C. (sI-A)^-1 B = [1; s+3]/(s^2+3s+2); C = [1 0] picks the first entry, giving 1/(s^2+3s+2). (Key point: Transfer function with C = [1 0])

## Close variants set in other exams

- [State space analysis; TF and stability (−4, 10)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-adv-limit-tf-4-1) — Control Systems, chapter 8, asked 2×

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