---
title: "Controllable and observable forms of (z+1)/(z²+1.3z+0.4)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ccf-ocf-z1-1-3z"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-ocf-z1-1-3z
site: BCT Engineering
---

# Controllable and observable forms of (z+1)/(z²+1.3z+0.4)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2073 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-z1-1-3z
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Consider the following system: Y(z)/U(z)=(z+1)/(z²+1.3z+0.4). Obtain the state space representation of the above pulse transfer function in the controllable canonical form and observable canonical form.

## Short answer

For (z+1)/(z^2+1.3z+0.4), the controllable canonical form is x(k+1)=[0 1; −0.4 −1.3]x+[0;1]u, y=[1 1]x. The observable canonical form is x(k+1)=[0 −0.4; 1 −1.3]x+[1;1]u, y=[0 1]x.

## Answer

Both forms are found from the same coefficients: $a_1 = 1.3$, $a_2 = 0.4$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.
$$
\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+1.3z+0.4}
$$

#### Controllable canonical form
Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+1.3z+0.4}$, so $Y = (z+1)Q$. With $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:
$$
\begin{aligned}
x_1(k+1) &= x_2(k)\\
x_2(k+1) &= -0.4x_1(k) - 1.3x_2(k) + u(k)\\
y(k) &= x_1(k) + x_2(k)
\end{aligned}
$$
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.4 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\mathbf{x}(k)
$$

#### Observable canonical form
Divide by $z^2$ and cross-multiply:
$$
Y(z) = z^{-1}\Big[U - 1.3Y + z^{-1}\big(U - 0.4Y\big)\Big]
$$
Let $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(U - 0.4Y)$:
$$
\begin{aligned}
x_1(k+1) &= -0.4x_2(k) + u(k)\\
x_2(k+1) &= x_1(k) - 1.3x_2(k) + u(k)\\
y(k) &= x_2(k)
\end{aligned}
$$
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.4 \\ 1 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)
$$

#### Check
- Controllable form: $\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\mathbf{H} = [1\ \ 1]\begin{bmatrix} 1 \\ z \end{bmatrix} = z+1$.
- Observable form: $[0\ \ 1]\begin{bmatrix} z+1.3 & -0.4 \\ 1 & z \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 1 + z$.
- Both have denominator $z^2 + 1.3z + 0.4$.

The two forms are duals: $\mathbf{G}_o = \mathbf{G}_c^T$, $\mathbf{H}_o = \mathbf{C}_c^T$, $\mathbf{C}_o = \mathbf{H}_c^T$.

## Check yourself

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

1. What are G and C in controllable form?
   A. G = [0 -0.4; 1 -1.3], C = [0 1]
   B. G = [0 1; 0.4 1.3], C = [1 1]
   C. G = [0 1; -0.4 -1.3], C = [0 1]
   D. G = [0 1; -0.4 -1.3], C = [1 1]
   Answer: D. Denominator coefficients in last row. (Key point: Controllable form)

2. What is G in observable form?
   A. [0 -0.4; 1 -1.3]
   B. [0 1; -0.4 -1.3]
   C. [0 0.4; 1 1.3]
   D. [-1.3 -0.4; 1 0]
   Answer: A. Transpose of the controllable G. (Key point: Observable form G)

3. What are H and C in observable form?
   A. H = [0; 1], C = [1 1]
   B. H = [1; 1], C = [0 1]
   C. H = [1; 0], C = [1 1]
   D. H = [1; 1], C = [1 0]
   Answer: B. H_o = C_c^T and C_o = H_c^T. (Key point: Observable H and C)

4. How are the two forms related?
   A. They are identical
   B. G_o = -G_c
   C. G_o = G_c^T, H_o = C_c^T, C_o = H_c^T
   D. G_o = G_c^-1
   Answer: C. Transpose duality. (Key point: Duality)

## Close variants set in other exams

- [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×
- [Diagonal canonical form of (z+1)/(z²+1.3z+0.4)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adiagonal-form-z1-1-3z) — Digital Control System, chapter 5, asked 6×
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- [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×
- [Controllable and observable forms (denominator 0.36z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Accf-ocf-0-36-variant) — 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×

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