--- title: "Discretize a continuous state model; pulse TF matrix" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:discretize-double-integrator-pulse-matrix" source: https://www.bctengineer.com/q/semester_6/digital_control_system/discretize-double-integrator-pulse-matrix site: BCT Engineering --- # Discretize a continuous state model; pulse TF matrix Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2080 Chaitra, 8 marks; 2073 Bhadra, 8 marks (asked 2 times) - Video: [Discretization of State-Space Models](https://www.youtube.com/watch?v=ME8h5zEQE20) — NPTEL-NOC IITM - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/discretize-double-integrator-pulse-matrix - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question State space representation of a system is given by [ẋ₁; ẋ₂]=[0 1; 0 -2][x₁; x₂]+[0; 1]u and y=[0 1][x₁; x₂]. Discretize the above system and also find the pulse transfer function matrix. ## Short answer The system is discretized using zero-order hold: G = e^(AT) and H = ∫e^(Aλ)B dλ. For T=1s, G = [[1, 0.4323], [0, 0.1353]], H = [[0.2838], [0.4323]]. Pulse transfer function is F(z) = 0.5(1-e^{-2T})/(z-e^{-2T}) = 0.4323/(z-0.1353) for T=1s. ## Answer Discretization with a zero-order hold gives $\mathbf{G} = e^{\mathbf{A}T}$ and $\mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$, with $\mathbf{C}$ and $\mathbf{D}$ unchanged. The sampling period is not given, so the result is found for a general T and then evaluated for **T = 1 s**. #### Step 1: State transition matrix $e^{\mathbf{A}t}$ $$ \mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},\quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix} $$ $$ (s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 0 & s+2 \end{bmatrix}^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix} $$ $$ e^{\mathbf{A}t} = \mathcal{L}^{-1}\left[(s\mathbf{I}-\mathbf{A})^{-1}\right] = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix} $$ #### Step 2: G and H $$ \mathbf{G} = e^{\mathbf{A}T} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix} $$ $$ \mathbf{H} = \int_0^T \begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{4}\left(2T - 1 + e^{-2T}\right) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix} $$ #### Discretized model $$ \mathbf{x}(k+1) = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} \tfrac{1}{4}(2T-1+e^{-2T}) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k) $$ For T = 1 s ($e^{-2} = 0.1353$): $$ \mathbf{G} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix},\qquad \mathbf{H} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix} $$ #### Step 3: Pulse transfer function $$ (z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-e^{-2T})}\begin{bmatrix} z-e^{-2T} & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & z-1 \end{bmatrix} $$ With $\mathbf{C} = [0\ \ 1]$ only the second row is needed: $$ \begin{aligned} F(z) &= \frac{[0\ \ z-1]\,\mathbf{H}}{(z-1)(z-e^{-2T})} = \frac{(z-1)\tfrac{1}{2}(1-e^{-2T})}{(z-1)(z-e^{-2T})}\\ &= \frac{\tfrac{1}{2}(1-e^{-2T})}{z-e^{-2T}} \end{aligned} $$ For T = 1 s: $$ F(z) = \frac{0.4323}{z - 0.1353} $$ #### Remark The output $y = x_2$ is the velocity, so the integrator pole at $z = 1$ cancels. The result matches the direct ZOH transform of $1/(s+2)$: $(1-z^{-1})\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = \frac{0.5(1-e^{-2T})}{z-e^{-2T}}$. **Answer: $F(z) = \dfrac{0.5(1-e^{-2T})}{z-e^{-2T}}$, which is $\dfrac{0.4323}{z-0.1353}$ for T = 1 s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a ZOH discretisation, how are G and H obtained? A. G = e^(AT), H = (integral of e^(A lambda) over 0..T) B B. G = A T, H = B T C. G = e^(AT), H = B D. G = (sI-A)^-1, H = B T Answer: A. Holding the input constant over a period gives exactly these expressions. (Key point: ZOH formulas) 2. What is e^(At) for A = [0 1; 0 -2]? A. [1 t; 0 e^-2t] B. [1 (1-e^-2t)/2; 0 e^-2t] C. [e^-2t 0; 0 1] D. [1 e^-2t; 0 (1-e^-2t)/2] Answer: B. Inverse Laplace of (sI-A)^-1 gives the 1/(s(s+2)) entry as (1-e^-2t)/2. (Key point: State transition matrix) 3. What are G and H for T = 1 s? A. G = [1 0.1353; 0 0.4323], H = [0.4323; 0.2838] B. G = [1 1; 0 0.1353], H = [0; 1] C. G = [1 0.4323; 0 0.1353], H = [0.2838; 0.4323] D. G = [1 0.4323; 0 0.1353], H = [0; 1] Answer: C. With e^-2 = 0.1353: (1-0.1353)/2 = 0.4323 and (2-1+0.1353)/4 = 0.2838. (Key point: Numbers at T = 1) 4. Why is F(z) = 0.4323/(z-0.1353) first order when y = x2? A. The pole at z = 0.1353 cancels B. C is zero C. H is zero D. The pole at z = 1 cancels with the zero (z-1) Answer: D. With C = [0 1], the numerator contains (z-1), which cancels the integrator pole. 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