--- title: "Controllable canonical form of (z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²)" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:ccf-z1-2z2-4z1-3z2" source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-z1-2z2-4z1-3z2 site: BCT Engineering --- # Controllable canonical form of (z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2082 Chaitra (new course), 4 marks - Video: [Controllable canonical form and its state space representation](https://www.youtube.com/watch?v=df7cNBxM5gk) — Electric Fund - 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-z1-2z2-4z1-3z2 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain a state-space representation of the following pulse-transfer-function system in the controllable canonical form: Y(z)/U(z)=(z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²). ## Short answer Controllable canonical form: G = [[0,1],[-3,-4]], H = [[0],[1]], C = [2,1]. Obtained from Y/U = (z+2)/(z^2+4z+3) with states as successive delays of intermediate variable. ## Answer In the controllable canonical form the states are successive delays of an intermediate variable. The denominator coefficients form the last row of $\mathbf{G}$. #### Write in powers of z $$ \frac{Y(z)}{U(z)} = \frac{z^{-1}+2z^{-2}}{1+4z^{-1}+3z^{-2}} = \frac{z+2}{z^2+4z+3} $$ So $a_1 = 4$, $a_2 = 3$, $b_0 = 0$, $b_1 = 1$, $b_2 = 2$. #### Derivation Introduce $Q(z)$ with $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+4z+3}$, so $Y(z) = (z+2)Q(z)$. Choose $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$: $$ \begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= q(k+2) = -3x_1(k) - 4x_2(k) + u(k)\\ y(k) &= q(k+1) + 2q(k) = 2x_1(k) + x_2(k) \end{aligned} $$ #### Controllable canonical form $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -3 & -4 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} $$ Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{[2\ \ 1]\begin{bmatrix} 1 \\ z \end{bmatrix}}{z^2+4z+3} = \dfrac{z+2}{z^2+4z+3}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Y/U in powers of z? A. (z+2)/(z^2+4z+3) B. (z+2)/(z^2+3z+4) C. (2z+1)/(z^2+4z+3) D. (z+2)/(4z^2+3z+1) Answer: A. Multiply numerator and denominator by z^2. (Key point: Convert to powers of z) 2. What is the second row of G? A. [-4 -3] B. [-3 -4] C. [3 4] D. [0 1] Answer: B. The last row is [-a2 -a1] with a1 = 4 and a2 = 3. (Key point: G matrix) 3. What is C? A. [1 2] B. [3 4] C. [2 1] D. [0 1] Answer: C. y = q(k+1) + 2q(k) = 2x1 + x2, so C = [2 1]. 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