---
title: "Observable canonical form of (3z²+7z+1)/(z²+2z+4)"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:ocf-3z2-7z-1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ocf-3z2-7z-1
site: BCT Engineering
---

# Observable canonical form of (3z²+7z+1)/(z²+2z+4)

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2067 Mangsir, 8 marks
- Video: [LCS - 53b - Observable Canonical Form (OCF) and signal flow graphs](https://www.youtube.com/watch?v=Q9RmG8X4OCo) — 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/ocf-3z2-7z-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Represent X(z)=(3z²+7z+1)/(z²+2z+4) in observable canonical form.

## Short answer

The direct feedthrough term is D=3. The observable canonical form realization is x(k+1)=[0 -4; 1 -2]x(k)+[-11; 1]u(k), y(k)=[0 1]x(k)+3u(k).

## Answer

The numerator and denominator have equal degree, so there is a direct term $D = b_0 = 3$. The observable form then follows the standard pattern.

#### Coefficients
$$
\frac{Y(z)}{U(z)} = \frac{3z^2+7z+1}{z^2+2z+4}
$$
$b_0 = 3$, $b_1 = 7$, $b_2 = 1$, $a_1 = 2$, $a_2 = 4$:
$$
\begin{aligned}
b_2 - a_2b_0 &= 1 - 12 = -11\\
b_1 - a_1b_0 &= 7 - 6 = 1
\end{aligned}
$$
So
$$
\frac{Y(z)}{U(z)} = 3 + \frac{z - 11}{z^2 + 2z + 4}
$$

#### Derivation
For the strictly proper part $W(z) = \dfrac{z-11}{z^2+2z+4}U(z)$:
$$
W = z^{-1}\Big[U - 2W + z^{-1}\big(-11U - 4W\big)\Big]
$$
Let $x_2 = w$ and $X_1 = z^{-1}(-11U - 4W)$:
$$
\begin{aligned}
x_1(k+1) &= -4x_2(k) - 11u(k)\\
x_2(k+1) &= x_1(k) - 2x_2(k) + u(k)\\
y(k) &= x_2(k) + 3u(k)
\end{aligned}
$$

#### Observable canonical form
$$
\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -4 \\ 1 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} -11 \\ 1 \end{bmatrix}u(k)
$$
$$
y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + 3u(k)
$$

#### Check
$$
\mathbf{C}\,\text{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H} = \begin{bmatrix} 1 & z \end{bmatrix}\begin{bmatrix} -11 \\ 1 \end{bmatrix} = z - 11
$$
$$
\frac{z-11}{z^2+2z+4} + 3 = \frac{3z^2 + 7z + 1}{z^2+2z+4}
$$

## Check yourself

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

1. For (3z^2+7z+1)/(z^2+2z+4), what is the strictly proper remainder after taking out D = 3?
   A. (z - 11)/(z^2 + 2z + 4)
   B. (z + 3)/(z^2 + 2z + 4)
   C. (7z + 1)/(z^2 + 2z + 4)
   D. (z + 11)/(z^2 + 2z + 4)
   Answer: A. b2 - a2 b0 = 1 - 12 = -11 and b1 - a1 b0 = 7 - 6 = 1, so the remainder is (z - 11)/(z^2 + 2z + 4). (Key point: Separating the direct term)

2. What is G in the observable canonical form?
   A. [0 1; -4 -2]
   B. [0 -4; 1 -2]
   C. [0 -2; 1 -4]
   D. [0 4; 1 2]
   Answer: B. Observable form has the first column [0; 1] and the last column [-a2; -a1] = [-4; -2]. (Key point: State matrix of observable form)

3. What is H?
   A. [1; -11]
   B. [-11; 1]
   C. [1; 3]
   D. [-1; 11]
   Answer: B. H = [b2 - a2 b0; b1 - a1 b0] = [-11; 1]. (Key point: Input vector H)

4. What is the output equation?
   A. y = [1 0] x + 3u
   B. y = [0 1] x + 3u
   C. y = [0 1] x
   D. y = [0 3] x + u
   Answer: B. The output is taken from x2 (x2 = w) and the direct term is added: y = x2 + 3u. (Key point: Output equation)

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