--- title: "Pulse transfer function from a 3×3 state model" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:pulse-tf-third-order-numeric" source: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-third-order-numeric site: BCT Engineering --- # Pulse transfer function from a 3×3 state model Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2081 Chaitra, 8 marks - 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-third-order-numeric - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain the pulse transfer function of the system defined by: x(k+1)=Gx(k)+Hu(k), y(k)=Cx(k)+Du(k), where G=[-0.3679 1 0; -0.1353 0 1; -0.0497 0 0], H=[0.5; 0.25; 0.125], C=[1 0 0] and D=1.25. ## Short answer Using F(z)=C(zI−G)^(−1)H+D with C=[1 0 0], only the first row of adj(zI−G) is required. The pulse transfer function is (1.25z^3+0.9599z^2+0.4191z+0.1871)/(z^3+0.3679z^2+0.1353z+0.0497). ## Answer The pulse transfer function is $F(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D$. Because $\mathbf{C} = [1\ 0\ 0]$, only the first row of the inverse is needed. #### Step 1: Characteristic polynomial $$ z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+0.3679 & -1 & 0 \\ 0.1353 & z & -1 \\ 0.0497 & 0 & z \end{bmatrix} $$ $$ \begin{aligned} |z\mathbf{I}-\mathbf{G}| &= (z+0.3679)(z^2) + 1\,(0.1353z + 0.0497)\\ &= z^3 + 0.3679z^2 + 0.1353z + 0.0497 \end{aligned} $$ #### Step 2: First row of adj$(z\mathbf{I}-\mathbf{G})$ $$ \begin{bmatrix} z^2 & z & 1 \end{bmatrix} $$ (Cofactors of the first column: $z\cdot z - 0$, $-(-z - 0)$, $(+1) - 0$.) #### Step 3: $\mathbf{C}\,\text{adj}\,\mathbf{H}$ $$ \begin{bmatrix} z^2 & z & 1 \end{bmatrix}\begin{bmatrix} 0.5 \\ 0.25 \\ 0.125 \end{bmatrix} = 0.5z^2 + 0.25z + 0.125 $$ #### Step 4: Add D $$ \begin{aligned} F(z) &= 1.25 + \frac{0.5z^2 + 0.25z + 0.125}{z^3 + 0.3679z^2 + 0.1353z + 0.0497}\\ &= \frac{1.25z^3 + (0.5 + 1.25\times0.3679)z^2 + (0.25 + 1.25\times0.1353)z + (0.125 + 1.25\times0.0497)}{z^3 + 0.3679z^2 + 0.1353z + 0.0497} \end{aligned} $$ #### Result $$ F(z) = \frac{Y(z)}{U(z)} = \frac{1.25z^3 + 0.9599z^2 + 0.4191z + 0.1871}{z^3 + 0.3679z^2 + 0.1353z + 0.0497} $$ In powers of $z^{-1}$: $$ F(z) = \frac{1.25 + 0.9599z^{-1} + 0.4191z^{-2} + 0.1871z^{-3}}{1 + 0.3679z^{-1} + 0.1353z^{-2} + 0.0497z^{-3}} $$ The given model is an observable-type companion form, so in general $b_i = h_i + a_ib_0$. This matches the numerator found above. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is only the first row of the inverse needed? A. H has one column B. D is scalar C. G is companion D. C = [1 0 0] picks the first row Answer: D. C times the inverse selects only its first row. (Key point: Why only first row) 2. What is |zI-G|? A. z^3 + 0.3679z^2 + 0.1353z + 0.0497 B. z^3 - 0.3679z^2 - 0.1353z - 0.0497 C. z^3 + 0.5z^2 + 0.25z + 0.125 D. 1.25z^3 + 0.9599z^2 Answer: A. The first column of G carries the negative denominator coefficients. (Key point: Characteristic polynomial) 3. What is the first row of adj(zI-G)? A. [1 z z^2] B. [z^2 z 1] C. [z z^2 1] D. [z^2 1 z] Answer: B. The cofactors give z^2, z and 1. (Key point: Adjugate row) 4. How do D = 1.25 and the denominator combine in the numerator? A. Numerator = 1.25 x (0.5z^2+...) B. Numerator is unchanged C. Numerator = 0.5z^2+0.25z+0.125 + 1.25 x denominator D. Numerator = D only Answer: C. Adding D means putting 1.25 over the common denominator. 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