--- title: "Controllable canonical form of a third-order system" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:ccf-1-3z-5z-4z" source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-1-3z-5z-4z site: BCT Engineering --- # Controllable canonical form of a third-order system Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2077 Chaitra, 8 marks - Video: [Examples On Controllable Canonical Form And Observable Canonical Form | GATE Control System](https://www.youtube.com/watch?v=5TNpxBrULzA) — Ekeeda GATE & ESE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-1-3z-5z-4z - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain the state space representation of following pulse transfer function in controllable canonical form: Y(z)/U(z)=(1+3z⁻¹+5z⁻²+4z⁻³)/(1+2z⁻¹+7z⁻²+5z⁻³). ## Short answer Rewrite as (z³+3z²+5z+4)/(z³+2z²+7z+5). The controllable canonical form matrices are G=[0 1 0; 0 0 1; −5 −7 −2], H=[0;0;1], C=[−1 −2 1], D=1. ## Answer Since the numerator and denominator have the same degree, there is a direct term $D = b_0$, and the output row uses $b_i - a_ib_0$. #### Standard result For $$ \frac{Y(z)}{U(z)} = \frac{b_0z^n + b_1z^{n-1} + \cdots + b_n}{z^n + a_1z^{n-1} + \cdots + a_n} $$ the controllable canonical form (Ogata) is $$ \mathbf{G} = \begin{bmatrix} 0 & 1 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ -a_n & -a_{n-1} & \cdots & -a_1 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} $$ $$ \mathbf{C} = \begin{bmatrix} b_n - a_nb_0 & \cdots & b_1 - a_1b_0 \end{bmatrix},\quad D = b_0 $$ #### Coefficients $$ \frac{Y(z)}{U(z)} = \frac{1+3z^{-1}+5z^{-2}+4z^{-3}}{1+2z^{-1}+7z^{-2}+5z^{-3}} = \frac{z^3+3z^2+5z+4}{z^3+2z^2+7z+5} $$ | $b_0$ | $b_1$ | $b_2$ | $b_3$ | $a_1$ | $a_2$ | $a_3$ | |---|---|---|---|---|---|---| | 1 | 3 | 5 | 4 | 2 | 7 | 5 | $$ \begin{aligned} b_3 - a_3b_0 &= 4 - 5 = -1\\ b_2 - a_2b_0 &= 5 - 7 = -2\\ b_1 - a_1b_0 &= 3 - 2 = 1 \end{aligned} $$ #### Controllable canonical form $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -7 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} -1 & -2 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + u(k) $$ #### Check $$ \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D = \frac{z^2 - 2z - 1}{z^3+2z^2+7z+5} + 1 = \frac{z^3+3z^2+5z+4}{z^3+2z^2+7z+5} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is there a direct term? A. The denominator is monic B. The poles are repeated C. a3 is non-zero D. Numerator and denominator have the same degree so D = b0 = 1 Answer: D. Equal degrees give b0 non-zero. (Key point: Direct term) 2. How is C computed? A. C = [b3-a3b0, b2-a2b0, b1-a1b0] B. C = [b1, b2, b3] C. C = [a3, a2, a1] D. C = [b0, b1, b2] Answer: A. The b0 contribution is removed from the dynamics. (Key point: Output coefficients) 3. What is C? A. [4 5 3] B. [-1 -2 1] C. [1 -2 -1] D. [-5 -7 -2] Answer: B. 4-5 = -1, 5-7 = -2, 3-2 = 1. (Key point: Values) 4. What is the last row of G? A. [-2 -7 -5] B. [5 7 2] C. [-5 -7 -2] D. [0 0 1] Answer: C. [-a3 -a2 -a1]. 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