---
title: "Controllable canonical form of (3z²+7z+15)/(z²+2z+4)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ccf-3z2-7z-15"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ccf-3z2-7z-15
site: BCT Engineering
---

# Controllable canonical form of (3z²+7z+15)/(z²+2z+4)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2068 Jestha, 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/ccf-3z2-7z-15
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Represent X(z)=(3z²+7z+15)/(z²+2z+4) in controllable canonical form.

## Short answer

Express the transfer function as 3+(z+3)/(z²+2z+4). The controllable canonical form is G=[[0,1],[−4,−2]], H=[0;1], C=[3,1] and D=3.

## Answer

The numerator and denominator have the same degree, so first separate the direct term $D = b_0$. Then use the standard controllable canonical form.

#### Coefficients
$$
X(z) = \frac{3z^2+7z+15}{z^2+2z+4}
$$
Treat this as $Y(z)/U(z)$ with $b_0 = 3$, $b_1 = 7$, $b_2 = 15$, $a_1 = 2$, $a_2 = 4$:
$$
\begin{aligned}
b_2 - a_2b_0 &= 15 - 12 = 3\\
b_1 - a_1b_0 &= 7 - 6 = 1
\end{aligned}
$$
So
$$
\frac{Y(z)}{U(z)} = 3 + \frac{z + 3}{z^2 + 2z + 4}
$$

#### Derivation
Let $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+2z+4}$, so $q(k+2) = -4q(k) - 2q(k+1) + u(k)$. Take $x_1 = q(k)$ and $x_2 = q(k+1)$:
$$
\begin{aligned}
x_1(k+1) &= x_2(k)\\
x_2(k+1) &= -4x_1(k) - 2x_2(k) + u(k)\\
y(k) &= 3x_1(k) + x_2(k) + 3u(k)
\end{aligned}
$$

#### Controllable canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -4 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 3 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + 3u(k)
$$

#### Check
$$
[3\ \ 1]\frac{1}{z^2+2z+4}\begin{bmatrix} 1 \\ z \end{bmatrix} + 3 = \frac{z + 3 + 3z^2 + 6z + 12}{z^2+2z+4} = \frac{3z^2 + 7z + 15}{z^2+2z+4}
$$

## Check yourself

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

1. Why is a direct term D needed for X(z) = (3z^2+7z+15)/(z^2+2z+4), and what is it?
   A. Numerator and denominator have equal degree; D = b0 = 3
   B. The denominator is monic; D = 1
   C. The constant 15 is large; D = 15
   D. There is no direct term since the system is controllable
   Answer: A. Equal degrees mean the transfer function has a constant part: D = b0 = 3. (Key point: Direct term)

2. After removing D, what is the remaining strictly proper part?
   A. (z + 3)/(z^2 + 2z + 4)
   B. (7z + 15)/(z^2 + 2z + 4)
   C. (z - 3)/(z^2 + 2z + 4)
   D. (3z + 1)/(z^2 + 2z + 4)
   Answer: A. b2 - a2 b0 = 15 - 12 = 3 and b1 - a1 b0 = 7 - 6 = 1, giving (z + 3)/(z^2 + 2z + 4). (Key point: Strictly proper remainder)

3. What are G and H in the controllable canonical form?
   A. G = [0 1; -4 -2], H = [0; 1]
   B. G = [0 -4; 1 -2], H = [1; 0]
   C. G = [0 1; -2 -4], H = [0; 1]
   D. G = [0 1; 4 2], H = [0; 1]
   Answer: A. From q(k+2) = -4q(k) - 2q(k+1) + u(k), the last row of G is [-a2, -a1] = [-4, -2] and H = [0; 1]. (Key point: State matrix and input vector)

4. What is y(k) in terms of the states and input?
   A. y = [1 3] x + 3u
   B. y = [3 1] x + 3u
   C. y = [15 7] x + u
   D. y = [0 1] x + 3u
   Answer: B. C = [b2 - a2 b0, b1 - a1 b0] = [3 1] and D = 3, so y = 3x1 + x2 + 3u. (Key point: Output equation)

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