---
title: "Why state representation; controllable canonical form"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:state-representation-why-ccf"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/state-representation-why-ccf
site: BCT Engineering
---

# Why state representation; controllable canonical form

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2078 Chaitra, 8 marks
- Video: [Controllable Canonical Form Numerical Of State Space Analysis ,Digital Control System IOE....](https://www.youtube.com/watch?v=K59S1ypjCgU) — Aman Grid
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/state-representation-why-ccf
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

What do you mean by state representation and why is it required? Consider the following system: Y(z)/U(z)=(z+1)/(z²+1.3z+0.4). Obtain the state representation of the above pulse transfer function in the controllable canonical form.

## Short answer

State representation describes a system by first-order difference equations using a minimal set of state variables. It handles MIMO systems, initial conditions, and internal modes, enabling modern design methods. The controllable canonical form is x(k+1)=[0 1; −0.4 −1.3]x+[0;1]u, y=[1 1]x.

## Answer

#### State representation
The **state** of a system is the smallest set of variables $x_1(k), \ldots, x_n(k)$ such that knowing them at $k = k_0$, together with the input for $k \ge k_0$, completely fixes the future behaviour. A discrete-time system is then described by first-order vector difference equations:
$$
\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k),\qquad \mathbf{y}(k) = \mathbf{C}\mathbf{x}(k) + \mathbf{D}\mathbf{u}(k)
$$

#### Why it is required
- It handles **multi-input multi-output** systems as easily as single-input ones; transfer functions become clumsy for these.
- It includes **initial conditions**, whereas a transfer function assumes zero initial state.
- It applies to **time-varying and nonlinear** systems, not only LTI ones.
- It shows **internal behaviour**: hidden modes, controllability and observability. A transfer function can hide pole–zero cancellations.
- It is the basis of modern design: **pole placement, state observers, optimal (LQR) control** and Liapunov stability.
- The first-order form is ideal for **computer simulation and implementation**.

#### Coefficients of the given system
$$
\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+1.3z+0.4} = \frac{0\cdot z^2 + 1\cdot z + 1}{z^2 + 1.3z + 0.4}
$$
$a_1 = 1.3$, $a_2 = 0.4$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.

#### Derivation
Introduce $Q(z)$ such that
$$
\frac{Q(z)}{U(z)} = \frac{1}{z^2+1.3z+0.4},\qquad Y(z) = (z+1)Q(z)
$$
In the time domain, $q(k+2) + 1.3q(k+1) + 0.4q(k) = u(k)$ and $y(k) = q(k+1) + q(k)$.
Choose $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:
$$
\begin{aligned}
x_1(k+1) &= x_2(k)\\
x_2(k+1) &= -0.4x_1(k) - 1.3x_2(k) + u(k)\\
y(k) &= x_1(k) + x_2(k)
\end{aligned}
$$

#### Controllable canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.4 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}
$$

#### Check
$$
\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} z+1.3 & 1 \\ -0.4 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2+1.3z+0.4} = \frac{1 + z}{z^2+1.3z+0.4}
$$

## Check yourself

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

1. What is the state of a system?
   A. Any set of output variables
   B. The input history
   C. The transfer function
   D. Smallest set of variables that with future input fixes future behaviour
   Answer: D. State captures all memory needed. (Key point: Definition of state)

2. Which is an advantage of state space over a transfer function?
   A. Handles MIMO and initial conditions
   B. Needs no matrices
   C. Works only for LTI
   D. Hides internal modes
   Answer: A. Transfer functions assume zero initial state and are clumsy for MIMO. (Key point: Advantage 1)

3. What can state space reveal that a transfer function can hide?
   A. Only the DC gain
   B. Controllability, observability and hidden modes
   C. Only poles
   D. Nothing
   Answer: B. Pole-zero cancellations disappear from transfer functions. (Key point: Internal behaviour)

4. What are G and C for (z+1)/(z^2+1.3z+0.4)?
   A. G = [0 1; -1.3 -0.4], C = [1 1]
   B. G = [0 1; -0.4 -1.3], C = [0 1]
   C. G = [0 1; -0.4 -1.3], C = [1 1]
   D. G = [1 0; 0 1], C = [1 1]
   Answer: C. Same as the standard controllable form. (Key point: CCF matrices)

## 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×

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