---
title: "Observable canonical form of (0.368z⁻¹+0.264z⁻²)/(1−1.368z⁻¹+0.368z⁻²)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ocf-0-368-0-264"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ocf-0-368-0-264
site: BCT Engineering
---

# Observable canonical form of (0.368z⁻¹+0.264z⁻²)/(1−1.368z⁻¹+0.368z⁻²)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2082 Chaitra, 5 marks; 2074 Bhadra, 6 marks; 2071 Bhadra, 8 marks; 2070 Magh, 6 marks (asked 4 times)
- Video: [LCS - 53b - Observable Canonical Form (OCF) and signal flow graphs](https://www.youtube.com/watch?v=Q9RmG8X4OCo) — MAFarooqi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/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 observable canonical form: Y(z)/U(z)=(0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.368z⁻²).

## Short answer

Observable canonical form: x(k+1)=[[0, -0.368],[1, 1.368]]x(k)+[0.264, 0.368]^T u(k), y(k)=[0, 1]x(k). This realises the transfer function (0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.368z⁻²).

## Answer

In the observable canonical form the denominator coefficients appear in the last column of $\mathbf{G}$, and the numerator coefficients appear in $\mathbf{H}$.

#### Standard form
For
$$
\frac{Y(z)}{U(z)} = \frac{b_0 + b_1z^{-1} + b_2z^{-2}}{1 + a_1z^{-1} + a_2z^{-2}}
$$
the observable canonical form (Ogata) is
$$
\mathbf{x}(k+1) = \begin{bmatrix} 0 & -a_2 \\ 1 & -a_1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} b_2 - a_2b_0 \\ b_1 - a_1b_0 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k) + b_0u(k)
$$

#### Derivation
Cross-multiplying gives $Y(z) = -a_1z^{-1}Y - a_2z^{-2}Y + b_1z^{-1}U + b_2z^{-2}U$ (here $b_0 = 0$), so
$$
Y(z) = z^{-1}\Big[b_1U - a_1Y + z^{-1}\big(b_2U - a_2Y\big)\Big]
$$
Define $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(b_2U - a_2Y)$. Then
$$
\begin{aligned}
x_1(k+1) &= -a_2x_2(k) + b_2u(k)\\
x_2(k+1) &= x_1(k) - a_1x_2(k) + b_1u(k)\\
y(k) &= x_2(k)
\end{aligned}
$$

#### Coefficients
| Coefficient | Value |
|---|---|
| $a_1$ | −1.368 |
| $a_2$ | 0.368 |
| $b_0$ | 0 |
| $b_1$ | 0.368 |
| $b_2$ | 0.264 |

#### Result
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.368 \\ 1 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}
$$

#### Check
$$
\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{0.368z + 0.264}{z^2 - 1.368z + 0.368}
$$
This is the given function (multiply numerator and denominator by $z^{-2}$).

## Check yourself

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

1. For Y/U = (0.368z^-1 + 0.264z^-2)/(1 - 1.368z^-1 + 0.368z^-2), the coefficients are:
   A. a1 = 1.368, a2 = 0.368, b0 = 0.368
   B. a1 = -1.368, a2 = 0.368, b0 = 0, b1 = 0.368, b2 = 0.264
   C. a1 = 0.368, a2 = 0.264, b1 = -1.368
   D. a1 = -1.368, a2 = -0.368, b1 = 0.264
   Answer: B. Compare with (b0 + b1z^-1 + b2z^-2)/(1 + a1z^-1 + a2z^-2); the numerator has no constant term so b0 = 0. (Key point: Identify coefficients)

2. In observable canonical form for a second-order system, G is:
   A. [0 1; -a2 -a1]
   B. [-a1 1; -a2 0]
   C. [0 -a2; 1 -a1]
   D. [1 0; 0 1]
   Answer: C. Denominator coefficients appear in the last column of G, with 1 on the subdiagonal. (Key point: Form of G matrix)

3. With b0 = 0, the input vector H contains:
   A. [b1; b2]
   B. [b2 - a2 b0 ; b1 - a1 b0] = [0.264; 0.368]
   C. [1; 0]
   D. [a1; a2]
   Answer: B. H = [b2 - a2 b0; b1 - a1 b0], which gives [0.264; 0.368]. (Key point: H vector)

4. The output equation of the result is:
   A. y = [1 0] x
   B. y = [0 1] x + 0.368 u
   C. y = [1 1] x
   D. y = [0 1] x, because y(k) = x2(k)
   Answer: D. State x2 was defined as Y(z), so y(k) = x2(k) and D = b0 = 0. (Key point: Output equation)

## Close variants set in other exams

- [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 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×
- [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×
- [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 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×
- [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×

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