--- title: "Controllable canonical form of (4z³+3z²+5z+4)/(2z³+5z²+2z+3)" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:ccf-4z3-3z2" source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-4z3-3z2 site: BCT Engineering --- # Controllable canonical form of (4z³+3z²+5z+4)/(2z³+5z²+2z+3) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2081 Chaitra, 4 marks; 2075 Baisakh, 6 marks (asked 2 times) - Video: [Controllable Canonical Form Numerical Of State Space Analysis ,Digital Control System IOE....](https://www.youtube.com/watch?v=K59S1ypjCgU) — Aman Grid - 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-4z3-3z2 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain the state space representation of the following pulse transfer function in controllable canonical form: Y(z)/X(z)=(4z³+3z²+5z+4)/(2z³+5z²+2z+3). ## Short answer Normalize denominator to monic: divide by 2. Controllable canonical form: G=[[0,1,0],[0,0,1],[-1.5,-1,-2.5]], H=[0;0;1], C=[-1,0.5,-3.5], D=2. Coefficients from bᵢ-aᵢb₀: b₃-a₃b₀=-1, b₂-a₂b₀=0.5, b₁-a₁b₀=-3.5. Input renamed u(k) to avoid confusion with state. ## Answer The controllable canonical form needs a monic denominator, so first divide the numerator and denominator by 2. #### Normalise $$ \frac{Y(z)}{X(z)} = \frac{2z^3 + 1.5z^2 + 2.5z + 2}{z^3 + 2.5z^2 + z + 1.5} $$ | $b_0$ | $b_1$ | $b_2$ | $b_3$ | $a_1$ | $a_2$ | $a_3$ | |---|---|---|---|---|---|---| | 2 | 1.5 | 2.5 | 2 | 2.5 | 1 | 1.5 | Since the degrees are equal, $D = b_0 = 2$ and $$ \begin{aligned} b_3 - a_3b_0 &= 2 - 3 = -1\\ b_2 - a_2b_0 &= 2.5 - 2 = 0.5\\ b_1 - a_1b_0 &= 1.5 - 5 = -3.5 \end{aligned} $$ #### Controllable canonical form The input here is called $x(k)$, so rename it $u(k)$ to avoid confusion with the state vector. Choosing the states as successive delays of an intermediate variable gives: $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1.5 & -1 & -2.5 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} -1 & 0.5 & -3.5 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + 2u(k) $$ #### Check $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{-3.5z^2 + 0.5z - 1}{z^3 + 2.5z^2 + z + 1.5}$, and adding $D = 2$ gives $\dfrac{2z^3 + 1.5z^2 + 2.5z + 2}{z^3 + 2.5z^2 + z + 1.5}$, which is the given function. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must be done first before writing the controllable canonical form of (4z^3+3z^2+5z+4)/(2z^3+5z^2+2z+3)? A. Divide numerator and denominator by 2 to make the denominator monic B. Multiply both by 2 C. Remove the z^3 terms D. Factor the denominator Answer: A. The standard form assumes the leading denominator coefficient is 1. (Key point: Normalise first) 2. What is D? A. 4 B. 2 C. 0.5 D. 1.5 Answer: B. After normalising, the leading numerator coefficient b0 = 4/2 = 2. (Key point: Direct term) 3. What is the last row of G? A. [-2.5 -1 -1.5] B. [1.5 1 2.5] C. [-1.5 -1 -2.5] D. [-3 -2 -5] Answer: C. The last row is [-a3 -a2 -a1] with a1 = 2.5, a2 = 1, a3 = 1.5. (Key point: Last row of G) 4. What is the first element of C, b3 - a3 b0? A. 2 B. -3.5 C. 0.5 D. -1 Answer: D. b3 - a3 b0 = 2 - 1.5 x 2 = -1, and C = [-1 0.5 -3.5]. 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