--- title: "Transfer function from state equations, output x2" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:tf-from-de-3x1-x2-y" source: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-de-3x1-x2-y site: BCT Engineering --- # Transfer function from state equations, output x2 Control Systems · Chapter 8: State Space Analysis - Asked in: 2080 Bhadra, 8 marks - Video: [Derivation of Transfer Function from State Model](https://www.youtube.com/watch?v=vY2Byo2SJ48) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-from-de-3x1-x2-y - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question The differential equations related to the system are dx1/dt = −3x1 + x2 and dx2/dt = −2x1 + u for t > 0. Its output is given by y = x2. Derive the transfer function of the system with these differential equations and check stability. ## Short answer The transfer function is G(s) = (s+3)/(s^2+3s+2) = (s+3)/[(s+1)(s+2)], obtained via G(s)=C(sI-A)^{-1}B+D. The poles are -1 and -2, both real and negative, so the system is asymptotically stable (overdamped). ## Answer The transfer function of a state model is $G(s) = C(sI - A)^{-1}B + D$, and the system is stable if all roots of $|sI - A| = 0$ (the eigenvalues of $A$) lie in the left half of the s-plane. #### Step 1: Write the state model $$ \dot{x}_1 = -3x_1 + x_2, \qquad \dot{x}_2 = -2x_1 + u, \qquad y = x_2 $$ $$ A = \begin{bmatrix}-3 & 1\\ -2 & 0\end{bmatrix},\; B = \begin{bmatrix}0\\ 1\end{bmatrix},\; C = \begin{bmatrix}0 & 1\end{bmatrix},\; D = 0 $$ #### Step 2: Find $(sI - A)$ and its determinant $$ sI - A = \begin{bmatrix}s+3 & -1\\ 2 & s\end{bmatrix} $$ $$ |sI - A| = s(s+3) - (-1)(2) = s^2 + 3s + 2 = (s+1)(s+2) $$ #### Step 3: Inverse $$ (sI - A)^{-1} = \frac{1}{(s+1)(s+2)}\begin{bmatrix}s & 1\\ -2 & s+3\end{bmatrix} $$ #### Step 4: Transfer function $$ \begin{aligned} (sI - A)^{-1}B &= \frac{1}{(s+1)(s+2)}\begin{bmatrix}1\\ s+3\end{bmatrix} \\ G(s) = C(sI - A)^{-1}B &= \begin{bmatrix}0 & 1\end{bmatrix}\frac{1}{(s+1)(s+2)}\begin{bmatrix}1\\ s+3\end{bmatrix} = \frac{s+3}{(s+1)(s+2)} \end{aligned} $$ **Transfer function:** $$ \frac{Y(s)}{U(s)} = \frac{s+3}{s^2 + 3s + 2} $$ #### Cross-check by Laplace transform With zero initial conditions: $sX_1 = -3X_1 + X_2$ gives $X_1 = X_2/(s+3)$. Then $sX_2 = -2X_1 + U = -\frac{2X_2}{s+3} + U$, so $$ X_2\left(s + \frac{2}{s+3}\right) = U \;\Rightarrow\; \frac{X_2}{U} = \frac{s+3}{s^2 + 3s + 2} $$ Same result. #### Step 5: Stability check Characteristic equation: $$ |sI - A| = s^2 + 3s + 2 = 0 \;\Rightarrow\; s = -1,\; -2 $$ | Item | Value | |---|---| | Eigenvalues (poles) | $-1$, $-2$ | | Zero | $-3$ | | Location of poles | Left half s-plane | | Damping | Overdamped (real, distinct poles) | Routh array for $s^2 + 3s + 2$: first column $1, 3, 2$, all positive, no sign change. **Answer:** $G(s) = \dfrac{s+3}{(s+1)(s+2)}$. Both poles ($-1$ and $-2$) are real and negative, so the system is **stable** (asymptotically stable, overdamped response). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the transfer function from the state model? A. G(s) = (sI - A)^-1 B C B. G(s) = C(sI - A)^-1 B + D C. G(s) = C(sI + A)B + D D. G(s) = B(sI - A)^-1 C Answer: B. Taking the Laplace transform with zero initial conditions gives G(s) = C(sI - A)^-1 B + D. (Key point: Transfer function formula) 2. From x1' = -3x1 + x2, x2' = -2x1 + u, y = x2, what are A and C? A. A = [-3 1; -2 0], C = [0 1] B. A = [-3 -2; 1 0], C = [0 1] C. A = [-3 1; -2 0], C = [1 0] D. A = [-3 1; 0 -2], C = [0 1] Answer: A. Rows of A come from the coefficients in each state equation, and y = x2 gives C = [0 1]. (Key point: Matrices from equations) 3. What is the resulting transfer function? A. 1/((s+1)(s+2)) B. (s+3)/((s+1)(s+2)) C. s/((s+1)(s+2)) D. (s+3)/(s^2 + 3s - 2) Answer: B. (sI-A)^-1 B = [1; s+3]/((s+1)(s+2)); multiplying by C = [0 1] picks s+3. (Key point: Transfer function result) 4. What is the stability conclusion and why? A. Unstable, because there is a zero at s = -3 B. Stable, because the poles s = -1 and s = -2 lie in the left half plane C. Marginally stable, because the poles are real D. Unstable, because the poles are real and distinct Answer: B. |sI - A| = s^2+3s+2 gives eigenvalues -1 and -2, both with negative real parts, so the system is stable (overdamped). (Key point: Stability by eigenvalues) ## Close variants set in other exams - [Transfer function from state equations, output x1](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-de-3x1-x2) — Control Systems, chapter 8, asked 2× ## Related in Control Systems - [Two-mass system equations, X2/F, F-I and F-V analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ade-tf-fi-fv-m1-m2) — Control Systems, chapter 2 - [Transfer function and stability from [1 0; 1 1]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-tf-1-0-1-1) — Control Systems, chapter 8 - [F-V and F-I analogy of two-mass system, X2 output](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Afv-fi-tf-m1-m2-k) — Control Systems, chapter 2, asked 3× - [Transfer function and stability from [0 −1; −2 −3]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-tf-0-1-2-3) — Control Systems, chapter 8 - [State model of two-mass spring system, output x2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-m1-m2-k1-k2-x2) — Control Systems, chapter 8 - [Two-mass system transfer function and F-V analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-two-mass-k1k2b1b2) — Control Systems, chapter 2 ## Related in other subjects - [Transfer function of DT low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-lowpass-transfer-function) — Signals and Systems, chapter 6