---
title: "Observable canonical form of (4z²+3z+5)/(5z²+z−4)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ocf-4z2-3z-5"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ocf-4z2-3z-5
site: BCT Engineering
---

# Observable canonical form of (4z²+3z+5)/(5z²+z−4)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2068 Magh, 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-4z2-3z-5
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Represent system having pulse transfer function G(z)=(4z²+3z+5)/(5z²+z-4) in observable canonical form.

## Short answer

Normalise to 0.8+(0.44z+1.64)/(z²+0.2z−0.8). The observable canonical form is G=[[0,0.8],[1,−0.2]], H=[1.64;0.44], C=[0,1] with D=0.8.

## Answer

The leading coefficient of the denominator is 5, so divide the numerator and denominator by 5 first. Because the degrees are equal, there is a direct term $D = b_0$.

#### Normalise
$$
G(z) = \frac{4z^2+3z+5}{5z^2+z-4} = \frac{0.8z^2 + 0.6z + 1}{z^2 + 0.2z - 0.8}
$$
| $b_0$ | $b_1$ | $b_2$ | $a_1$ | $a_2$ |
|---|---|---|---|---|
| 0.8 | 0.6 | 1 | 0.2 | −0.8 |

$$
\begin{aligned}
b_2 - a_2b_0 &= 1 - (-0.8)(0.8) = 1.64\\
b_1 - a_1b_0 &= 0.6 - (0.2)(0.8) = 0.44
\end{aligned}
$$

#### Derivation
Separate the direct term:
$$
G(z) = 0.8 + \frac{0.44z + 1.64}{z^2 + 0.2z - 0.8}
$$
For the strictly proper part, $W = z^{-1}\left[0.44U - 0.2W + z^{-1}(1.64U + 0.8W)\right]$. With $x_2 = w$:
$$
\begin{aligned}
x_1(k+1) &= 0.8x_2(k) + 1.64u(k)\\
x_2(k+1) &= x_1(k) - 0.2x_2(k) + 0.44u(k)\\
y(k) &= x_2(k) + 0.8u(k)
\end{aligned}
$$

#### Observable canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 0.8 \\ 1 & -0.2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1.64 \\ 0.44 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + 0.8u(k)
$$

#### Check
$$
\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D = \frac{0.44z + 1.64}{z^2+0.2z-0.8} + 0.8 = \frac{0.8z^2 + 0.6z + 1}{z^2+0.2z-0.8} = \frac{4z^2+3z+5}{5z^2+z-4}
$$
(Poles: $z = 0.8$ and $z = -1$. The pole at $-1$ is on the unit circle, so the system is only marginally stable.)

## Check yourself

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

1. What must be done first with G(z) = (4z^2+3z+5)/(5z^2+z-4)?
   A. Divide numerator and denominator by 5 so the denominator is monic
   B. Divide by 4 so the numerator is monic
   C. Multiply by z^-2 only
   D. Take the inverse z-transform directly
   Answer: A. The standard forms need a leading denominator coefficient of 1, so divide by 5 to get (0.8z^2+0.6z+1)/(z^2+0.2z-0.8). (Key point: Normalising the denominator)

2. Why is there a term D, and what is its value?
   A. The numerator and denominator have equal degree; D = b0 = 0.8
   B. The system has a pole at -1; D = 1
   C. The denominator is not monic; D = 5
   D. The numerator is larger; D = 4
   Answer: A. Equal degrees make G(z) biproper, so a constant direct term D = b0 = 0.8 (after normalising) appears in y(k). (Key point: Direct feed-through term)

3. What is H, and how are its entries computed?
   A. [1; 0.6], the raw numerator constants
   B. [1.64; 0.44], from b2 - a2 b0 and b1 - a1 b0
   C. [0.44; 1.64], from b1 - a1 b0 then b2 - a2 b0
   D. [0.8; 0.6], from b0 and b1
   Answer: B. b2 - a2 b0 = 1 + 0.64 = 1.64 and b1 - a1 b0 = 0.6 - 0.16 = 0.44. In observable form H = [1.64; 0.44]. (Key point: Input matrix H)

4. What are the poles, and what is the stability conclusion?
   A. z = 0.8 and -1; marginally stable since one pole is on the unit circle
   B. z = -0.8 and 1; unstable
   C. z = 0.8 and -1; asymptotically stable
   D. z = 0.4 and -0.5; asymptotically stable
   Answer: A. z^2 + 0.2z - 0.8 = (z - 0.8)(z + 1). The pole at -1 is on the unit circle, so the system is only marginally stable. (Key point: Poles and stability)

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