---
title: "Observable canonical form of (0.3z⁻¹+0.6z⁻²)/(1−0.2z⁻¹+0.5z⁻²)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ocf-0-3-0-6"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ocf-0-3-0-6
site: BCT Engineering
---

# Observable canonical form of (0.3z⁻¹+0.6z⁻²)/(1−0.2z⁻¹+0.5z⁻²)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2076 Bhadra, 8 marks
- 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-0-3-0-6
- 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 observable canonical form: Y(z)/U(z)=(0.3z⁻¹+0.6z⁻²)/(1-0.2z⁻¹+0.5z⁻²).

## Short answer

Cross-multiply the pulse transfer function to get difference equations. With states defined as x2(k)=y(k) and x1(k+1)=−0.5x2(k)+0.6u(k), the observable canonical form is G=[[0,−0.5],[1,0.2]], H=[0.6;0.3], C=[0,1].

## Answer

In the observable canonical form the denominator coefficients sit in the last column of $\mathbf{G}$, the numerator terms form $\mathbf{H}$, and $\mathbf{C} = [0\ \ 1]$.

#### Coefficients
$$
\frac{Y(z)}{U(z)} = \frac{0.3z^{-1}+0.6z^{-2}}{1-0.2z^{-1}+0.5z^{-2}}
$$
$a_1 = -0.2$, $a_2 = 0.5$, $b_0 = 0$, $b_1 = 0.3$, $b_2 = 0.6$.

#### Derivation
Cross-multiplying:
$$
Y(z) = 0.2z^{-1}Y - 0.5z^{-2}Y + 0.3z^{-1}U + 0.6z^{-2}U
$$
$$
Y(z) = z^{-1}\Big[0.3U + 0.2Y + z^{-1}\big(0.6U - 0.5Y\big)\Big]
$$
Define $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(0.6U - 0.5Y)$. Then $X_2 = z^{-1}(0.3U + 0.2X_2 + X_1)$:
$$
\begin{aligned}
x_1(k+1) &= -0.5x_2(k) + 0.6u(k)\\
x_2(k+1) &= x_1(k) + 0.2x_2(k) + 0.3u(k)\\
y(k) &= x_2(k)
\end{aligned}
$$

#### Observable canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.5 \\ 1 & 0.2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.6 \\ 0.3 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}
$$

#### Check
$$
|z\mathbf{I}-\mathbf{G}| = \begin{vmatrix} z & 0.5 \\ -1 & z-0.2 \end{vmatrix} = z^2 - 0.2z + 0.5
$$
$$
\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H} = \begin{bmatrix} 1 & z \end{bmatrix}\begin{bmatrix} 0.6 \\ 0.3 \end{bmatrix} = 0.3z + 0.6
$$
So $F(z) = \dfrac{0.3z+0.6}{z^2-0.2z+0.5} = \dfrac{0.3z^{-1}+0.6z^{-2}}{1-0.2z^{-1}+0.5z^{-2}}$, as given.

```text
 u --+----------[0.3]----------+
     |                         v
     +-[0.6]->(+)->[z^-1]-x1->(+)->[z^-1]--x2--+--> y
               ^                ^              |
               +----[-0.5]------+-----[0.2]----+
```

## Check yourself

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

1. In the observable canonical form of Y/U = (0.3z^-1 + 0.6z^-2)/(1 - 0.2z^-1 + 0.5z^-2), what is the last column of G?
   A. [-0.5; 0.2]
   B. [0.5; -0.2]
   C. [0.2; -0.5]
   D. [0.6; 0.3]
   Answer: A. The last column of G holds -a2 and -a1 with a1 = -0.2 and a2 = 0.5, giving [-0.5; 0.2]. The first column is [0; 1]. (Key point: Denominator in last column of G)

2. Why is C = [0 1] in this observable canonical form?
   A. Because x1 is defined as the output
   B. Because the output is y = x2, since X2(z) = Y(z) was chosen
   C. Because the numerator has no constant term b0
   D. Because C must always equal the last row of G
   Answer: B. The derivation defines X2 = Y, so y(k) = x2(k), which means C = [0 1]. (Key point: Output matrix C)

3. What is H, and where do its entries come from?
   A. [0.3; 0.6], numerator coefficients b1 then b2
   B. [0.6; 0.3], b2 then b1 (with b0 = 0)
   C. [-0.5; 0.2], the denominator coefficients
   D. [0.6; -0.5], taken from the term 0.6U - 0.5Y
   Answer: B. With b0 = 0, H = [b2; b1] = [0.6; 0.3]. The state x1 receives 0.6u and x2 receives 0.3u. (Key point: Input matrix H)

4. In the check, what does C adj(zI - G) H = [1 z][0.6; 0.3] equal, and what does it confirm?
   A. 0.6z + 0.3, the numerator, confirming the realisation
   B. 0.3z + 0.6, the numerator of F(z), confirming the realisation
   C. z^2 - 0.2z + 0.5, the characteristic polynomial
   D. 0.9, the sum of the numerator coefficients
   Answer: B. [1 z][0.6; 0.3] = 0.6 + 0.3z. Dividing by |zI-G| = z^2 - 0.2z + 0.5 reproduces the given pulse transfer function. (Key point: Verification by C adj(zI-G) H)

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