--- title: "Controllable and observable forms of a 0.368/0.264 system" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:ccf-ocf-0-368-0-264" source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-ocf-0-368-0-264 site: BCT Engineering --- # Controllable and observable forms of a 0.368/0.264 system Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2072 Magh, 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-ocf-0-368-0-264 - 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 (i) controllable canonical form (ii) observable canonical form: Y(z)/U(z)=(0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.368z⁻²). ## Short answer Controllable form: G=[[0,1],[-0.368,1.368]], H=[[0],[1]], C=[0.264,0.368]. Observable form: G=[[0,-0.368],[1,1.368]], H=[[0.264],[0.368]], C=[0,1]. Both yield Y/U=(0.368z+0.264)/(z^2-1.368z+0.368). Poles at z=1, 0.368. ## Answer Both forms come from the same coefficients. The controllable form puts the denominator in the last row of $\mathbf{G}$, and the observable form puts it in the last column. #### Coefficients $$ \frac{Y(z)}{U(z)} = \frac{0.368z^{-1}+0.264z^{-2}}{1-1.368z^{-1}+0.368z^{-2}} = \frac{0.368z + 0.264}{z^2 - 1.368z + 0.368} $$ $a_1 = -1.368$, $a_2 = 0.368$, $b_0 = 0$, $b_1 = 0.368$, $b_2 = 0.264$. Since $b_0 = 0$: $b_2 - a_2b_0 = 0.264$ and $b_1 - a_1b_0 = 0.368$. #### (i) Controllable canonical form Let $Q(z) = \dfrac{U(z)}{z^2-1.368z+0.368}$, $x_1 = q(k)$, $x_2 = q(k+1)$, $y = 0.264x_1 + 0.368x_2$: $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.368 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} 0.264 & 0.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} $$ #### (ii) Observable canonical form Write $Y = z^{-1}\left[0.368U + 1.368Y + z^{-1}(0.264U - 0.368Y)\right]$ and take $x_2 = y$, $x_1(k+1) = -0.368x_2(k) + 0.264u(k)$: $$ \begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.368 \\ 1 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} $$ #### Check For both forms, $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{0.368z+0.264}{z^2-1.368z+0.368}$. The observable form is the transpose of the controllable one: $\mathbf{G}_o = \mathbf{G}_c^T$, $\mathbf{H}_o = \mathbf{C}_c^T$, $\mathbf{C}_o = \mathbf{H}_c^T$. (The denominator factors as $(z-1)(z-0.368)$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where does the denominator appear in controllable versus observable form? A. Last column in controllable, last row in observable B. First row in both C. Only in C D. Last row of G in controllable, last column in observable Answer: D. The two forms are transposes of each other. (Key point: Where the denominator goes) 2. What is G in controllable form? A. [0 1; -0.368 1.368] B. [0 1; 0.368 -1.368] C. [0 -0.368; 1 1.368] D. [1 0; -1.368 0.368] Answer: A. The last row is [-a2 -a1] = [-0.368 1.368]. (Key point: Controllable G) 3. What is C in controllable form? A. [0.368 0.264] B. [0.264 0.368] C. [0 1] D. [1 0] Answer: B. y = q(k+1)*0.368 + q(k)*0.264 gives C = [b2 b1]. (Key point: Controllable C) 4. In observable form, what are H and C? A. H = [0; 1], C = [0.264 0.368] B. H = [0.368; 0.264], C = [1 0] C. H = [0.264; 0.368], C = [0 1] D. H = [1; 1], C = [0 1] Answer: C. Observable form is dual: H_o = C_c^T and C_o = H_c^T. 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