---
title: "Controllable, observable and diagonal forms"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:three-canonical-forms-z-0-6"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/three-canonical-forms-z-0-6
site: BCT Engineering
---

# Controllable, observable and diagonal forms

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2072 Asoj, 10 marks
- Video: [Examples On Controllable Canonical Form And Observable Canonical Form | GATE Control System](https://www.youtube.com/watch?v=5TNpxBrULzA) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/three-canonical-forms-z-0-6
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Consider the system defined by Y(z)/U(z)=(z+1)/(z²+z+0.6). Obtain state-space representation for this system in the following three different forms: (i) controllable canonical form (ii) observable canonical form (iii) diagonal canonical form.

## Short answer

Controllable form: G=[0 1; −0.6 −1], H=[0;1], C=[1 1]. Observable form: G=[0 −0.6; 1 −1], H=[1;1], C=[0 1]. Diagonal form uses poles −0.5±j0.5916 and residues 0.5∓j0.4226, giving diagonal G with conjugate entries and H=[1;1], C=[0.5−j0.4226, 0.5+j0.4226].

## Answer

The three forms all realise
$$
\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+z+0.6}
$$
with $a_1 = 1$, $a_2 = 0.6$, $b_0 = 0$, $b_1 = 1$, $b_2 = 1$.

#### (i) Controllable canonical form
Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+z+0.6}$, so $Y = (z+1)Q$. With $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:
$$
\begin{aligned}
x_1(k+1) &= x_2(k)\\
x_2(k+1) &= -0.6x_1(k) - x_2(k) + u(k)\\
y(k) &= x_1(k) + x_2(k)
\end{aligned}
$$
$$
\mathbf{x}(k+1) = \begin{bmatrix} 0 & 1 \\ -0.6 & -1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\mathbf{x}(k)
$$

#### (ii) Observable canonical form
$Y = z^{-1}\left[U - Y + z^{-1}(U - 0.6Y)\right]$. Take $x_2 = y$ and $X_1 = z^{-1}(U - 0.6Y)$:
$$
\begin{aligned}
x_1(k+1) &= -0.6x_2(k) + u(k)\\
x_2(k+1) &= x_1(k) - x_2(k) + u(k)\\
y(k) &= x_2(k)
\end{aligned}
$$
$$
\mathbf{x}(k+1) = \begin{bmatrix} 0 & -0.6 \\ 1 & -1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)
$$

#### (iii) Diagonal canonical form
Poles:
$$
z^2 + z + 0.6 = 0 \;\Rightarrow\; p_{1,2} = \frac{-1 \pm \sqrt{1-2.4}}{2} = -0.5 \pm j0.5916
$$
They are distinct, but complex. Residues:
$$
\begin{aligned}
c_1 &= \left[\frac{z+1}{z-p_2}\right]_{z=p_1} = \frac{0.5 + j0.5916}{j1.1832} = 0.5 - j0.4226\\
c_2 &= \bar{c}_1 = 0.5 + j0.4226
\end{aligned}
$$
$$
\frac{Y(z)}{U(z)} = \frac{0.5 - j0.4226}{z + 0.5 - j0.5916} + \frac{0.5 + j0.4226}{z + 0.5 + j0.5916}
$$
With $X_i = U/(z - p_i)$:
$$
\mathbf{x}(k+1) = \begin{bmatrix} -0.5 + j0.5916 & 0 \\ 0 & -0.5 - j0.5916 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 0.5 - j0.4226 & 0.5 + j0.4226 \end{bmatrix}\mathbf{x}(k)
$$
The states are complex conjugates of each other, but the output $y(k)$ is real. Exact values: $p_{1,2} = -\tfrac12 \pm j\tfrac{\sqrt{35}}{10}$ and $c_{1,2} = \tfrac12 \mp j\tfrac{\sqrt{35}}{14}$.

#### Check
- $c_1 + c_2 = 1$, the coefficient of z in the numerator.
- $-(c_1p_2 + c_2p_1) = 1$, the constant term.

All three forms give $\dfrac{z+1}{z^2+z+0.6}$. The poles have $|p| = \sqrt{0.6} = 0.775 < 1$, so the system is stable.

## Check yourself

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

1. For (z+1)/(z^2+z+0.6), which (G, H, C) is the controllable canonical form?
   A. G = [0 1; -0.6 -1], H = [0; 1], C = [1 1]
   B. G = [0 -0.6; 1 -1], H = [1; 1], C = [0 1]
   C. G = [0 1; -1 -0.6], H = [0; 1], C = [1 1]
   D. G = [0 1; -0.6 -1], H = [1; 1], C = [0 1]
   Answer: A. The controllable form has the denominator coefficients in the last row of G, H = [0; 1], and C taken from the numerator: y = x1 + x2. (Key point: Controllable canonical form)

2. In the observable canonical form, what are H and C?
   A. H = [0; 1], C = [1 1]
   B. H = [1; 1], C = [0 1]
   C. H = [1; 1], C = [1 1]
   D. H = [0.6; 1], C = [0 1]
   Answer: B. The numerator coefficients (b2, b1) = (1, 1) go into H = [1; 1], and y = x2 gives C = [0 1]. G = [0 -0.6; 1 -1]. (Key point: Observable canonical form)

3. Why is the diagonal canonical form here complex?
   A. The numerator has complex coefficients
   B. The denominator roots are -0.5 +/- j0.5916, a complex pair, since the discriminant 1 - 2.4 is negative
   C. The poles are repeated
   D. The system is unstable
   Answer: B. z^2 + z + 0.6 has discriminant 1 - 2.4 = -1.4 < 0, so the poles are complex conjugates and G is diagonal with complex entries. (Key point: Diagonal form requires poles)

4. What does the answer conclude about stability, and why?
   A. Unstable, since the poles are complex
   B. Stable, since |p| = sqrt(0.6) = 0.775 < 1
   C. Marginally stable, since |p| = 1
   D. Stable, since the real part of the poles is negative
   Answer: B. A discrete system is stable when all poles are inside the unit circle. Both poles have magnitude sqrt(0.6) = 0.775. (Key point: Stability from the poles)

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