---
title: "Diagonal canonical form of (z+1)/(z²+1.3z+0.4)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:diagonal-form-z1-1-3z"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/diagonal-form-z1-1-3z
site: BCT Engineering
---

# Diagonal canonical form of (z+1)/(z²+1.3z+0.4)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2071 Magh, 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/diagonal-form-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 diagonal canonical form.

## Short answer

Perform partial-fraction expansion: (z+1)/((z+0.5)(z+0.8)) = 5/3/(z+0.5) − 2/3/(z+0.8). Define states X1=U/(z+0.5), X2=U/(z+0.8). The diagonal state-space form has G=diag(−0.5,−0.8), H=[1;1], C=[5/3, −2/3].

## Answer

The diagonal canonical form uses the partial-fraction expansion. Each distinct pole becomes a decoupled first-order state.

#### Poles
$$
z^2 + 1.3z + 0.4 = (z+0.5)(z+0.8)
$$
The poles are $p_1 = -0.5$ and $p_2 = -0.8$ (distinct). $b_0 = 0$.

#### Partial fractions
$$
\frac{Y(z)}{U(z)} = \frac{z+1}{(z+0.5)(z+0.8)} = \frac{c_1}{z+0.5} + \frac{c_2}{z+0.8}
$$
$$
\begin{aligned}
c_1 &= \left[\frac{z+1}{z+0.8}\right]_{z=-0.5} = \frac{0.5}{0.3} = \frac{5}{3} = 1.6667\\
c_2 &= \left[\frac{z+1}{z+0.5}\right]_{z=-0.8} = \frac{0.2}{-0.3} = -\frac{2}{3} = -0.6667
\end{aligned}
$$

#### States
$X_1(z) = \dfrac{U(z)}{z+0.5}$, $X_2(z) = \dfrac{U(z)}{z+0.8}$:
$$
\begin{aligned}
x_1(k+1) &= -0.5x_1(k) + u(k)\\
x_2(k+1) &= -0.8x_2(k) + u(k)\\
y(k) &= \tfrac{5}{3}x_1(k) - \tfrac{2}{3}x_2(k)
\end{aligned}
$$

#### Diagonal canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} -0.5 & 0 \\ 0 & -0.8 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} \tfrac{5}{3} & -\tfrac{2}{3} \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}
$$

#### Check
$$
\frac{5/3}{z+0.5} - \frac{2/3}{z+0.8} = \frac{\tfrac{5}{3}z + \tfrac{4}{3} - \tfrac{2}{3}z - \tfrac{1}{3}}{(z+0.5)(z+0.8)} = \frac{z+1}{z^2+1.3z+0.4}
$$

```text
        +-->[1/(z+0.5)]--x1-->[ 5/3]--+
 u -----|                            (+)--> y
        +-->[1/(z+0.8)]--x2-->[-2/3]--+
```
Both eigenvalues lie inside the unit circle, so the system is stable. In this form each mode can be seen directly.

## Check yourself

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

1. Factorise z^2+1.3z+0.4.
   A. (z-0.5)(z-0.8)
   B. (z+0.4)(z+0.9)
   C. (z+0.1)(z+1.2)
   D. (z+0.5)(z+0.8)
   Answer: D. 0.5+0.8 = 1.3 and 0.5 x 0.8 = 0.4. (Key point: Poles)

2. What are c1 and c2?
   A. 5/3 and -2/3
   B. 2/3 and -5/3
   C. 1 and 1
   D. 0.5 and 0.8
   Answer: A. c1 = 0.5/0.3, c2 = 0.2/(-0.3). (Key point: Residues)

3. What are G and B?
   A. G = [0 1; -0.4 -1.3], B = [0; 1]
   B. G = diag(-0.5, -0.8), B = [1; 1]
   C. G = diag(0.5, 0.8), B = [1; 1]
   D. G = diag(-0.5, -0.8), B = [5/3; -2/3]
   Answer: B. Each state is U/(z+p). (Key point: Form)

4. Is the system stable?
   A. No, poles are negative
   B. No, poles at -0.8 are outside
   C. Yes, both poles are inside the unit circle
   D. Cannot tell
   Answer: C. |-0.5| and |-0.8| are less than 1. (Key point: Stability)

## 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×
- [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+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×

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