---
title: "Diagonal canonical form of (z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:diagonal-form-0-7-0-12"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/diagonal-form-0-7-0-12
site: BCT Engineering
---

# Diagonal canonical form of (z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2081 Chaitra, 4 marks; 2080 Chaitra, 8 marks (asked 2 times)
- Video: [LCS - 53c - Diagonal canonical form (DCF) representation of state equations](https://www.youtube.com/watch?v=u7F3L3FDbs0) — 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/diagonal-form-0-7-0-12
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Obtain the state space representation of the following pulse transfer function in the diagonal canonical form: Y(z)/U(z)=(z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²).

## Short answer

Convert to positive powers: (z+2)/[(z+0.3)(z+0.4)]. Partial fractions: 17/(z+0.3) - 16/(z+0.4). Diagonal canonical form: G=diag(-0.3,-0.4), H=[1;1], C=[17,-16], D=0. Distinct poles yield decoupled states; each state corresponds to one pole.

## Answer

The diagonal canonical form is obtained from the partial-fraction expansion when all poles are distinct. Each pole becomes one decoupled state.

#### Write in powers of z
$$
\frac{Y(z)}{U(z)} = \frac{z^{-1}+2z^{-2}}{1+0.7z^{-1}+0.12z^{-2}} = \frac{z+2}{z^2+0.7z+0.12} = \frac{z+2}{(z+0.3)(z+0.4)}
$$
Poles: $p_1 = -0.3$, $p_2 = -0.4$ (distinct). $b_0 = 0$.

#### Partial fractions
$$
\begin{aligned}
c_1 &= \left[\frac{z+2}{z+0.4}\right]_{z=-0.3} = \frac{1.7}{0.1} = 17\\
c_2 &= \left[\frac{z+2}{z+0.3}\right]_{z=-0.4} = \frac{1.6}{-0.1} = -16
\end{aligned}
$$
$$
\frac{Y(z)}{U(z)} = \frac{17}{z+0.3} - \frac{16}{z+0.4}
$$

#### States
$X_1(z) = \dfrac{U(z)}{z+0.3}$, $X_2(z) = \dfrac{U(z)}{z+0.4}$:
$$
\begin{aligned}
x_1(k+1) &= -0.3x_1(k) + u(k)\\
x_2(k+1) &= -0.4x_2(k) + u(k)\\
y(k) &= 17x_1(k) - 16x_2(k)
\end{aligned}
$$

#### Diagonal canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} -0.3 & 0 \\ 0 & -0.4 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 17 & -16 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}
$$

#### Check
$$
\frac{17}{z+0.3} - \frac{16}{z+0.4} = \frac{17z + 6.8 - 16z - 4.8}{(z+0.3)(z+0.4)} = \frac{z+2}{z^2+0.7z+0.12}
$$

```text
        +-->[1/(z+0.3)]--x1-->[17]---+
 u -----|                            (+)--> y
        +-->[1/(z+0.4)]--x2-->[-16]--+
```

## Check yourself

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

1. When can the diagonal canonical form be used?
   A. When the poles are distinct
   B. When the poles are repeated
   C. When all zeros are at the origin
   D. When D is non-zero
   Answer: A. Distinct poles give simple partial fractions, each a decoupled first-order state. (Key point: Condition for diagonal form)

2. What are the poles of (z^-1+2z^-2)/(1+0.7z^-1+0.12z^-2)?
   A. 0.3 and 0.4
   B. -0.3 and -0.4
   C. -0.3 and -0.12
   D. -0.7 and -0.12
   Answer: B. Multiplying by z^2 gives (z+2)/(z^2+0.7z+0.12) = (z+2)/((z+0.3)(z+0.4)). (Key point: Poles)

3. What are c1 and c2?
   A. 1.7 and -1.6
   B. -17 and 16
   C. 17 and -16
   D. 10 and -10
   Answer: C. c1 = 1.7/0.1 = 17 at z = -0.3 and c2 = 1.6/(-0.1) = -16 at z = -0.4. (Key point: Residues)

4. In the final model, what do B and C contain?
   A. B = [17; -16] and C = [1 1]
   B. B = [0; 1] and C = [17 -16]
   C. B = [1; 1] and C = [-0.3 -0.4]
   D. B = [1; 1] and C = [17 -16]
   Answer: D. Each state is U/(z+p), so B is all ones and C holds the residues. (Key point: Structure of B and C)

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