---
title: "Pulse transfer function matrix of a general companion form"
subject: "Digital Control System"
chapter: 5
key: "semester_6/digital_control_system:pulse-tf-matrix-general-companion"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-general-companion
site: BCT Engineering
---

# Pulse transfer function matrix of a general companion form

Digital Control System · Chapter 5: Discrete Time State Equations

- Asked in: 2082 Kartik, 8 marks; 2078 Chaitra, 8 marks; 2077 Chaitra, 8 marks; 2075 Baisakh, 10 marks (asked 4 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/pulse-tf-matrix-general-companion
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5

## Question

Obtain pulse transfer function matrix of the following state space representation: x(k+1)=Gx(k)+Hu(k), y(k)=Cx(k)+Du(k), where G=[-a₁ 1 0; -a₂ 0 1; -a₃ 0 0], H=[h₁; h₂; h₃], C=[1 0 0] and D=b₀.

## Answer

The pulse transfer function is $F(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D$, obtained by taking the z-transform of the state equations with zero initial state.

#### Given
$$
\mathbf{G} = \begin{bmatrix} -a_1 & 1 & 0 \\ -a_2 & 0 & 1 \\ -a_3 & 0 & 0 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix},\quad D = b_0
$$

#### Step 1: $z\mathbf{I} - \mathbf{G}$ and its determinant
$$
z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+a_1 & -1 & 0 \\ a_2 & z & -1 \\ a_3 & 0 & z \end{bmatrix}
$$
Expanding along the first row:
$$
\begin{aligned}
|z\mathbf{I}-\mathbf{G}| &= (z+a_1)z^2 + 1\,(a_2z + a_3)\\
&= z^3 + a_1z^2 + a_2z + a_3
\end{aligned}
$$

#### Step 2: First row of the adjoint
Only the first row of $(z\mathbf{I}-\mathbf{G})^{-1}$ is needed, because $\mathbf{C} = [1\ 0\ 0]$. The first row of the adjoint (the cofactors of the first column, transposed) is
$$
\begin{bmatrix} z^2 & z & 1 \end{bmatrix}
$$
Check: cofactor $C_{11} = z\cdot z - 0 = z^2$, $C_{21} = -(-1\cdot z - 0) = z$, $C_{31} = (-1)(-1) - 0 = 1$.

#### Step 3: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$
$$
\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} z^2 & z & 1 \end{bmatrix}\begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix}}{z^3+a_1z^2+a_2z+a_3} = \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}
$$

#### Step 4: Add D
$$
\begin{aligned}
F(z) &= b_0 + \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}\\
&= \frac{b_0z^3 + (h_1 + a_1b_0)z^2 + (h_2 + a_2b_0)z + (h_3 + a_3b_0)}{z^3 + a_1z^2 + a_2z + a_3}
\end{aligned}
$$

#### Remark
This is the observable canonical form written with the state order reversed. Comparing with $\dfrac{b_0z^3+b_1z^2+b_2z+b_3}{z^3+a_1z^2+a_2z+a_3}$ gives $h_i = b_i - a_ib_0$, which is the standard result for the observable form. Since there is one input and one output, the "matrix" is the 1×1 function above.

**Answer: $F(z) = \dfrac{b_0z^3 + (h_1+a_1b_0)z^2 + (h_2+a_2b_0)z + (h_3+a_3b_0)}{z^3+a_1z^2+a_2z+a_3}$**

## Check yourself

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

1. The pulse transfer function of x(k+1) = Gx + Hu, y = Cx + Du is:
   A. C(zI - G)^-1 H + D
   B. (zI - G)^-1 H C + D
   C. C G H + D
   D. C(zI + G)^-1 H
   Answer: A. Taking z-transforms with zero initial state gives F(z) = C(zI-G)^-1 H + D. (Key point: Formula)

2. For the given G, |zI - G| equals:
   A. z^3 - a1z^2 - a2z - a3
   B. z^3 + a1z^2 + a2z + a3
   C. z^3 + a1 + a2 + a3
   D. (z + a1)(z + a2)(z + a3)
   Answer: B. Expanding along the first row gives z^3 + a1z^2 + a2z + a3. (Key point: Determinant)

3. Because C = [1 0 0], which part of the inverse is needed and what is it?
   A. Last row, [1 z z^2]
   B. Full matrix
   C. First row of the adjoint, [z^2 z 1]
   D. Diagonal only, [z z z]
   Answer: C. C picks out the first row; its cofactors give [z^2 z 1]. (Key point: First row of adjoint)

4. Adding D = b0, the z^2 coefficient of the numerator is:
   A. h1
   B. b0
   C. h1 - a1 b0
   D. h1 + a1 b0
   Answer: D. b0 + (h1z^2 + ...)/(den) gives numerator b0z^3 + (h1 + a1b0)z^2 + ... (Key point: Numerator with D)

5. Comparing with (b0z^3 + b1z^2 + b2z + b3)/(z^3 + a1z^2 + ...), h_i equals:
   A. b_i - a_i b0, the observable-form result
   B. b_i + a_i b0
   C. b_i
   D. a_i b_i
   Answer: A. From h_i + a_i b0 = b_i we get h_i = b_i - a_i b0. (Key point: Meaning of h_i)

## Close variants set in other exams

- [Pulse transfer function matrix (G = [−4 1; −2 −1], D = 0)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-g-4-1) — Digital Control System, chapter 5, asked 7×
- [Pulse transfer function matrix (G = [−4 1; −2 −1], D = [1; 0])](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Apulse-tf-matrix-d-nonzero) — Digital Control System, chapter 5, asked 7×

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