--- title: "Discrete state equations and pulse TF (T = 1)" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:discrete-state-equations-t-1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/discrete-state-equations-t-1 site: BCT Engineering --- # Discrete state equations and pulse TF (T = 1) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2072 Magh, 8 marks - Video: [Introduction to State Variable Analysis of Discrete Time Control Systems.](https://www.youtube.com/watch?v=oNQ38diHYnQ) — Korra Chinna - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/discrete-state-equations-t-1 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain the discrete-time state and output equations and pulse transfer function (when the sampling period T = 1) of the following continuous-time system: G(s)=Y(s)/U(s)=1/s(s+2), which may be represented in state space by the equations [ẋ₁; ẋ₂]=[0 1; 0 -2][x₁; x₂]+[0; 1]u, y=[1 0][x₁; x₂]. ## Short answer For T=1 s, the discrete state matrices are G=[1 0.4323; 0 0.1353], H=[0.2838; 0.4323], C=[1 0]. The pulse transfer function is Y(z)/U(z)=(0.2838z+0.1485)/(z^2−1.1353z+0.1353). ## Answer With a zero-order hold, the discrete model is $\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}u(k)$, $y(k) = \mathbf{C}\mathbf{x}(k)$, where $\mathbf{G} = e^{\mathbf{A}T}$ and $\mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$. #### Step 1: State transition matrix of the continuous system $$ (s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 0 & s+2 \end{bmatrix}^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix} $$ $$ e^{\mathbf{A}t} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix} $$ #### Step 2: G and H for T = 1 s $e^{-2} = 0.1353$: $$ \mathbf{G} = e^{\mathbf{A}} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2}) \\ 0 & e^{-2} \end{bmatrix} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix} $$ $$ \begin{aligned} \mathbf{H} &= \int_0^1\begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{2}\left[1 - \tfrac{1}{2}(1-e^{-2})\right] \\ \tfrac{1}{2}(1-e^{-2}) \end{bmatrix} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix} \end{aligned} $$ #### Discrete state and output equations $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} 1 & 0 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} $$ #### Step 3: Pulse transfer function $$ (z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-0.1353)}\begin{bmatrix} z-0.1353 & 0.4323 \\ 0 & z-1 \end{bmatrix} $$ $$ \begin{aligned} F(z) &= \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{0.2838(z-0.1353) + 0.4323\times0.4323}{(z-1)(z-0.1353)}\\ &= \frac{0.2838z + 0.1485}{(z-1)(z-0.1353)} = \frac{0.2838z + 0.1485}{z^2 - 1.1353z + 0.1353} \end{aligned} $$ #### Check by direct transformation $$ (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} = \frac{1.1353z + 0.5940}{4(z-1)(z-0.1353)} $$ This is the same result. **Answer: $F(z) = \dfrac{0.2838z + 0.1485}{(z-1)(z-0.1353)} = \dfrac{0.2838z^{-1} + 0.1485z^{-2}}{1 - 1.1353z^{-1} + 0.1353z^{-2}}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is C = [1 0]? A. The output is x2 B. C comes from H C. C is computed from e^(AT) D. The output y is x1 Answer: D. C and D are unchanged by ZOH discretisation. (Key point: Output matrix) 2. What is G? A. [1 0.4323; 0 0.1353] B. [1 0.1353; 0 0.4323] C. [0.1353 0; 0 1] D. [1 0.2838; 0 0.1353] Answer: A. e^-2 = 0.1353 and (1-e^-2)/2 = 0.4323. (Key point: G at T = 1) 3. What is H? A. [0.4323; 0.2838] B. [0.2838; 0.4323] C. [0; 1] D. [0.5; 0.5] Answer: B. Integrate the second column of e^(A lambda) from 0 to 1. (Key point: H at T = 1) 4. What is F(z)? A. 0.4323/(z-0.1353) B. (0.2838z-0.1485)/(z^2+1.1353z+0.1353) C. (0.2838z+0.1485)/((z-1)(z-0.1353)) D. 1/(z-1) Answer: C. No cancellation occurs since y = x1. 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