--- title: "Pulse transfer function matrix (G = [−4 1; −2 −1], D = 0)" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:pulse-tf-matrix-g-4-1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-g-4-1 site: BCT Engineering --- # Pulse transfer function matrix (G = [−4 1; −2 −1], D = 0) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2082 Chaitra, 6 marks; 2076 Bhadra, 8 marks (asked 2 times) - Video: [Transfer Function Derivation from State Model Statement Problem No 4 - Control Systems](https://www.youtube.com/watch?v=n_spUERHzP8) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-g-4-1 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain pulse transfer function matrix of the following state space representation: x(k+1)=Gx(k)+Hu(k), y(k)=Cx(k)+Du(k), where G=[-4 1; -2 -1], H=[1; 1], C=[1 1; 2 1] and D=0. ## Short answer For G=[-4,1;-2,-1], H=[1;1], C=[1,1;2,1], D=0, compute F(z)=C(zI-G)⁻¹H. (zI-G)⁻¹ = [z+1,1;-2,z+4]/[(z+2)(z+3)]. Multiplying gives F(z)=[2/(z+3); 3/(z+3)]. The pole at z=-2 cancels, indicating an uncontrollable mode; both poles are outside unit circle (unstable). ## Answer The pulse transfer function matrix is $\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}$. Here there is one input and two outputs, so $\mathbf{F}(z)$ is 2×1. #### Step 1: $(z\mathbf{I}-\mathbf{G})^{-1}$ $$ z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+4 & -1 \\ 2 & z+1 \end{bmatrix} $$ $$ |z\mathbf{I}-\mathbf{G}| = (z+4)(z+1) + 2 = z^2 + 5z + 6 = (z+2)(z+3) $$ $$ (z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+1 & 1 \\ -2 & z+4 \end{bmatrix} $$ #### Step 2: $(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$ $$ (z\mathbf{I}-\mathbf{G})^{-1}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+2 \\ z+2 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix} $$ #### Step 3: Multiply by C (D = 0) $$ \mathbf{F}(z) = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{z+3} \\[2mm] \dfrac{3}{z+3} \end{bmatrix} $$ #### Remarks - $\dfrac{Y_1(z)}{U(z)} = \dfrac{2}{z+3}$ and $\dfrac{Y_2(z)}{U(z)} = \dfrac{3}{z+3}$. - The eigenvalue $z = -2$ is cancelled by a zero: the mode at $-2$ is not excited by $u$ (it is uncontrollable). This is why the transfer functions are first order although the system is second order. - Both eigenvalues ($-2$, $-3$) lie outside the unit circle, so the system is unstable. **Answer: $\mathbf{F}(z) = \begin{bmatrix} 2/(z+3) \\ 3/(z+3) \end{bmatrix}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With one input and two outputs, F(z) = C(zI-G)^-1 H + D has what size? A. 2 x 1 B. 1 x 2 C. 2 x 2 D. 1 x 1 Answer: A. Rows equal the number of outputs and columns the number of inputs. (Key point: Formula and size) 2. What is |zI-G| for G = [-4 1; -2 -1]? A. (z-2)(z-3) = z^2-5z+6 B. (z+2)(z+3) = z^2+5z+6 C. z^2+5z+2 D. z^2+3z+6 Answer: B. (z+4)(z+1) - (-1)(2) = z^2+5z+6, whose roots are -2 and -3. (Key point: Characteristic polynomial) 3. What is F(z)? A. [2/(z+2); 3/(z+2)] B. [2/((z+2)(z+3)); 3/((z+2)(z+3))] C. [2/(z+3); 3/(z+3)] D. [1/(z+3); 1/(z+3)] Answer: C. (zI-G)^-1 H = [1/(z+3); 1/(z+3)], and C times that gives 2/(z+3) and 3/(z+3). (Key point: Result) 4. Why are the transfer functions first order although the system is second order? A. The mode at z = -3 is cancelled B. D is zero C. H has two equal entries so the system is stable D. The mode at z = -2 is cancelled (uncontrollable) Answer: D. The (z+2) factor cancels because the input does not excite that mode. (Key point: Cancellation meaning) ## Close variants set in other exams - [Pulse transfer function matrix of a general companion form](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-general-companion) — Digital Control System, chapter 5, asked 7× - [Pulse transfer function matrix (G = [−4 1; −2 −1], D = [1; 0])](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-d-nonzero) — Digital Control System, chapter 5, asked 7× ## Related in Digital Control System - [Pulse transfer function from state space form](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aderive-pulse-tf-state-space) — Digital Control System, chapter 5, asked 3× - [Pulse transfer function from a 3×3 state model](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-third-order-numeric) — Digital Control System, chapter 5 - [Pulse TF matrix of a continuous state model](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-continuous-model) — Digital Control System, chapter 5 - [Discretize a continuous state model; pulse TF matrix](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adiscretize-double-integrator-pulse-matrix) — Digital Control System, chapter 5, asked 2× - [Pulse TF matrix (G = [3 1; −2 1], D = [1; 0])](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-g-3-1) — Digital Control System, chapter 5, asked 2× - [Pulse transfer function of digital PID controller](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aderive-pid-pulse-tf) — Digital Control System, chapter 3, asked 2× ## Related in other subjects - [Space partitioning representation](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aspace-partitioning-representation) — Computer Graphics, chapter 6