---
title: "Controllable and observable forms (denominator 0.36z⁻²)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ccf-ocf-0-36-variant"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-ocf-0-36-variant
site: BCT Engineering
---

# Controllable and observable forms (denominator 0.36z⁻²)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2075 Bhadra, 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/ccf-ocf-0-36-variant
- 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 (i) controllable canonical form (ii) observable canonical form: Y(z)/U(z)=(0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.36z⁻²).

## Short answer

Controllable form: G=[[0,1],[-0.36,1.368]], H=[[0],[1]], C=[0.264,0.368]. Observable form: G=[[0,-0.36],[1,1.368]], H=[[0.264],[0.368]], C=[0,1]. Both give Y/U=(0.368z+0.264)/(z^2-1.368z+0.36). Poles at 0.3556, 1.0124 (unstable).

## Answer

Here the constant term of the denominator is 0.36, not 0.368. The method is the same; only the entry $-a_2$ changes.

#### Coefficients
$$
\frac{Y(z)}{U(z)} = \frac{0.368z^{-1}+0.264z^{-2}}{1-1.368z^{-1}+0.36z^{-2}} = \frac{0.368z+0.264}{z^2-1.368z+0.36}
$$
| $a_1$ | $a_2$ | $b_0$ | $b_1$ | $b_2$ |
|---|---|---|---|---|
| −1.368 | 0.36 | 0 | 0.368 | 0.264 |

#### (i) Controllable canonical form
Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2-1.368z+0.36}$ with $x_1(k) = q(k)$, $x_2(k) = q(k+1)$:
$$
\begin{aligned}
x_1(k+1) &= x_2(k)\\
x_2(k+1) &= -0.36x_1(k) + 1.368x_2(k) + u(k)\\
y(k) &= 0.264x_1(k) + 0.368x_2(k)
\end{aligned}
$$
$$
\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.36 & 1.368 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0.264 & 0.368 \end{bmatrix},\quad D = 0
$$

#### (ii) Observable canonical form
From $Y = z^{-1}\left[0.368U + 1.368Y + z^{-1}(0.264U - 0.36Y)\right]$ with $x_2 = y$:
$$
\begin{aligned}
x_1(k+1) &= -0.36x_2(k) + 0.264u(k)\\
x_2(k+1) &= x_1(k) + 1.368x_2(k) + 0.368u(k)\\
y(k) &= x_2(k)
\end{aligned}
$$
$$
\mathbf{G} = \begin{bmatrix} 0 & -0.36 \\ 1 & 1.368 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix},\quad D = 0
$$

#### Check and remark
- Both give $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{0.368z+0.264}{z^2-1.368z+0.36}$.
- The poles are $z = 0.3556$ and $z = 1.0124$. The second lies just outside the unit circle, so this system is (slightly) unstable, unlike the 0.368 version, which has a pole at exactly $z = 1$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Compared with the 0.368 version, what changes in the matrices?
   A. Only the numerator
   B. Only D
   C. Nothing
   D. Only the -a2 entry (0.36 in place of 0.368)
   Answer: D. Numerator and a1 are the same; only a2 changes. (Key point: What changes)

2. What is G in controllable form?
   A. [0 1; -0.36 1.368]
   B. [0 1; -0.368 1.368]
   C. [0 -0.36; 1 1.368]
   D. [0 1; 0.36 -1.368]
   Answer: A. The last row is [-a2 -a1] = [-0.36 1.368]. (Key point: Controllable G)

3. What is G in observable form?
   A. [0 1; -0.36 1.368]
   B. [0 -0.36; 1 1.368]
   C. [0 0.36; 1 -1.368]
   D. [0 -0.368; 1 1.368]
   Answer: B. G_o = transpose of G_c. (Key point: Observable form)

4. Why is this system slightly unstable?
   A. A pole is at z = 1 exactly
   B. Both poles are zero
   C. A pole at z = 1.0124 is outside the unit circle
   D. D is non-zero
   Answer: C. The roots are 0.3556 and 1.0124. (Key point: Stability)

## Close variants set in other exams

- [Controllable and observable forms of a 0.368/0.264 system](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-ocf-0-368-0-264) — Digital Control System, chapter 5, asked 14×
- [Observable canonical form of (0.368z⁻¹+0.264z⁻²)/(1−1.368z⁻¹+0.368z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aocf-0-368-0-264) — Digital Control System, chapter 5, asked 13×
- [Controllable canonical form of (4z³+3z²+5z+4)/(2z³+5z²+2z+3)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-4z3-3z2) — Digital Control System, chapter 5, asked 14×
- [Controllable canonical form of a third-order system](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-1-3z-5z-4z) — Digital Control System, chapter 5, asked 14×
- [Observable canonical form of (0.3z⁻¹+0.6z⁻²)/(1−0.2z⁻¹+0.5z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aocf-0-3-0-6) — Digital Control System, chapter 5, asked 13×
- [Controllable canonical form of (z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-z1-2z2-4z1-3z2) — Digital Control System, chapter 5, asked 15×
- [Controllable and observable forms of (z+1)/(z²+1.3z+0.4)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-ocf-z1-1-3z) — Digital Control System, chapter 5, asked 14×
- [Diagonal canonical form of (z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adiagonal-form-0-7-0-12) — Digital Control System, chapter 5, asked 13×
- [Controllable canonical form of (z+1)/(z²+1.3z+0.4)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-z1-1-3z) — Digital Control System, chapter 5, asked 10×

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