---
title: "Series programming of a third-order transfer function"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:series-programming-z-plus-1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/series-programming-z-plus-1
site: BCT Engineering
---

# Series programming of a third-order transfer function

Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System

- Asked in: 2068 Jestha, 8 marks
- Video: [Cascade Digital Controller Implementation 🌟 Third-Order System Flowchart 💡 Example 4](https://www.youtube.com/watch?v=4flqGLW8KrU) — CAN Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/series-programming-z-plus-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Realize the following transfer function by series programming: X(z)=(z+1)(z²+3z+7)/((z-2)(z²+z+9)).

## Short answer

Series programming splits X(z) into sections: G1(z)=(1+z^-1)/(1-2z^-1) and G2(z)=(1+3z^-1+7z^-2)/(1+z^-1+9z^-2). Each section uses standard programming with one and two delays respectively, total three delays.

## Answer

In **series programming**, $X(z)$ (the given pulse transfer function) is split into cascaded low-order sections. Each section is realized by standard programming.

#### Step 1: Sections in powers of $z^{-1}$
$$
X(z) = \underbrace{\frac{1 + z^{-1}}{1 - 2z^{-1}}}_{G_1(z)}\cdot\underbrace{\frac{1 + 3z^{-1} + 7z^{-2}}{1 + z^{-1} + 9z^{-2}}}_{G_2(z)}
$$

Both quadratics have complex roots, so each is kept as one second-order factor.

#### Step 2: Difference equations
Let $u(k)$ be the input and $y(k)$ the output.

**Section 1:**
- $w_1(k) = u(k) + 2w_1(k-1)$
- $y_1(k) = w_1(k) + w_1(k-1)$

**Section 2:**
- $w_2(k) = y_1(k) - w_2(k-1) - 9w_2(k-2)$
- $y(k) = w_2(k) + 3w_2(k-1) + 7w_2(k-2)$

#### Step 3: Block diagram
Section 1:
```text
u(k)-->(S)------------o--[1]-------->(S)--> y1(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w1(k-1)       |
        +<[2]---------o--[1]----------+
```
Section 2:
```text
y1(k)-->(S)------------o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-1)       |
        (S)<[-1]-------o--[3]-------->(S)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-9]--------o--[7]----------+
```

```text
u(k) --> [ G1(z) ] --y1(k)--> [ G2(z) ] --> y(k)
```

Three delay elements are used. Each section can be scaled and checked separately.

**Note:** The poles $z = 2$ and $z = -0.5 \pm j2.96$ lie outside the unit circle, so the system is unstable. The series structure is still a valid realization.

## Check yourself

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

1. The factor (z+1)/(z-2) becomes which section in powers of z^-1?
   A. (1 + z^-1)/(1 - 2z^-1)
   B. (1 - z^-1)/(1 + 2z^-1)
   C. (1 + z^-1)/(1 + 2z^-1)
   D. (z^-1 + 1)/(2 - z^-1)
   Answer: A. Dividing numerator and denominator by z gives (1+z^-1)/(1-2z^-1). (Key point: Cascade sections)

2. For G2 = (1+3z^-1+7z^-2)/(1+z^-1+9z^-2), the equations are:
   A. w2(k) = y1(k) + w2(k-1) + 9w2(k-2); y = w2(k) + 3w2(k-1) + 7w2(k-2)
   B. w2(k) = y1(k) - w2(k-1) - 9w2(k-2); y = w2(k) + 3w2(k-1) + 7w2(k-2)
   C. w2(k) = y1(k) - w2(k-1) - 9w2(k-2); y = w2(k) - 3w2(k-1) - 7w2(k-2)
   D. w2(k) = y1(k) + 3w2(k-1) + 7w2(k-2); y = w2(k) - w2(k-1) - 9w2(k-2)
   Answer: B. Denominator coefficients enter the feedback with negative sign; numerator coefficients go forward with their own sign. (Key point: Second-order section equations)

3. How are the two sections connected?
   A. Their outputs are added
   B. The output of section 1 is the input of section 2
   C. Both receive x(k) in parallel
   D. Section 2 feeds back into section 1
   Answer: B. In series programming the output y1 of G1 drives G2, so the overall response is G1 times G2. (Key point: Cascade connection)

4. Why is the system unstable, even though the realization is valid?
   A. The numerator has a zero at -1
   B. It uses three delays
   C. Poles at 2 and -0.5 +/- j2.96 lie outside the unit circle
   D. The order is odd
   Answer: C. Stability depends on pole locations, not on the realization. The poles have magnitude greater than 1. (Key point: Stability remark)

## Close variants set in other exams

- [Ladder programming of a third-order transfer function](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-z3-4z2) — Digital Control System, chapter 4, asked 2×

## Related in Digital Control System

- [Series programming of a third-order system](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aseries-programming-z-plus-3) — Digital Control System, chapter 4
- [Ladder programming of a third-order controller](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-8z3) — Digital Control System, chapter 4
- [Direct programming of a third-order controller](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adirect-programming-6z2-7z) — Digital Control System, chapter 4, asked 4×
- [Direct programming of a third-order controller](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adirect-programming-6z2-7z-16) — Digital Control System, chapter 4, asked 4×
- [Ladder programming of a third-order filter](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-third-order-128) — Digital Control System, chapter 4, asked 2×
- [Transfer function of zero order hold](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Azoh-transfer-function) — Digital Control System, chapter 1, asked 7×

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