---
title: "Ladder programming of a third-order controller"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:ladder-8z3"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-8z3
site: BCT Engineering
---

# Ladder programming of a third-order controller

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

- Asked in: 2073 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-8z3
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Realize the given digital controller by ladder programming whose transfer function is (8z³+7z²+10z+6)/(4z³+6z²+9z+10).

## Short answer

Ladder programming expands G(z) as a continued fraction. For G(z)=(8z^3+7z^2+10z+6)/(4z^3+6z^2+9z+10), the gains are A0=2, B1=-0.8, A1=12.5, B2=-0.02051, A2=-3.8604, B3=0.05031, A3=-10.04. Three delays are used.

## Answer

Ladder programming expands $G(z)$ as a continued fraction:

$$
G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}
$$

with $G(z) = \dfrac{8z^3 + 7z^2 + 10z + 6}{4z^3 + 6z^2 + 9z + 10}$.

#### Step 1: $A_0 = 8/4 = 2$
$$
G(z) = 2 + \frac{-5z^2 - 8z - 14}{4z^3 + 6z^2 + 9z + 10}
$$

#### Step 2: $B_1$
$$
\frac{4z^3 + 6z^2 + 9z + 10}{-5z^2 - 8z - 14} = -0.8z + \frac{-0.4z^2 - 2.2z + 10}{-5z^2 - 8z - 14}
$$

So $B_1 = -0.8$.

#### Step 3: $A_1$
$$
\frac{-5z^2 - 8z - 14}{-0.4z^2 - 2.2z + 10} = 12.5 + \frac{19.5z - 139}{-0.4z^2 - 2.2z + 10}
$$

So $A_1 = 12.5$.

#### Step 4: $B_2$
$$
\frac{-0.4z^2 - 2.2z + 10}{19.5z - 139} = -0.02051z + \frac{-5.0513z + 10}{19.5z - 139}
$$

So $B_2 = -0.02051$.

#### Step 5: $A_2$
$$
\frac{19.5z - 139}{-5.0513z + 10} = -3.8604 + \frac{-100.396}{-5.0513z + 10}
$$

So $A_2 = -3.8604$.

#### Step 6: $B_3$ and $A_3$
$$
\frac{-5.0513z + 10}{-100.396} = 0.05031z - 0.09961
$$

So $B_3 = 0.05031$ and $A_3 = -1/0.09961 = -10.04$.

| $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ | $B_3$ | $A_3$ |
|---|---|---|---|---|---|---|
| 2 | −0.8 | 12.5 | −0.02051 | −3.8604 | 0.05031 | −10.04 |

#### Ladder block diagram
```text
X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+
```

**Signal equations** of the ladder ($o$ = take-off point, (S) = summer):

$$
\begin{aligned}
Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\
Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\
Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\
Y &= A_0X + Y_1
\end{aligned}
$$

Numerical gains: $1/B_1 = -1.25$, $1/A_1 = 0.08$, $1/B_2 = -48.75$, $1/A_2 = -0.2590$, $1/B_3 = 19.875$, $1/A_3 = -0.0996$.

The expansion was checked by substituting back. It reproduces $G(z)$ exactly, and the ladder's impulse response matches the direct difference equation. Three delays are used.

## Check yourself

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

1. For (8z^3+7z^2+10z+6)/(4z^3+6z^2+9z+10), what is A0 and why?
   A. 8, the numerator leading coefficient
   B. 4, the denominator leading coefficient
   C. 0.6, the ratio of constants 6/10
   D. 2, the ratio of leading coefficients 8/4
   Answer: D. Equal degrees, so the quotient is 8/4 = 2, leaving remainder (-5z^2-8z-14)/(4z^3+...). (Key point: A0 from leading coefficients)

2. After A0, how does the continued-fraction expansion proceed?
   A. Invert the remainder fraction, divide to get B_i z, invert again to get A_i, and repeat
   B. Factor the denominator into poles
   C. Differentiate the numerator
   D. Evaluate at z = 1
   Answer: A. Each step inverts the remainder, divides out a term in z (B_i) or a constant (A_i), and repeats alternately. (Key point: Alternating division)

3. Here B1 = -0.8 comes out negative. What does that mean for the realization?
   A. The filter is unrealizable
   B. It is allowed; the gain 1/B1 = -1.25 is simply negative
   C. The filter is unstable
   D. The numerator must be rescaled to make it positive
   Answer: B. Negative constants are fine; the block z^-1/B1 just has gain -1.25. (Key point: Negative constants)

4. How many delays does the ladder realization of this third-order filter use?
   A. 6
   B. 3
   C. 1
   D. 4
   Answer: B. One delay for each B_i z term: B1, B2, B3, so 3 delays. (Key point: Delay count)

## Related in Digital Control System

- [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×
- [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×
- [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×
- [Series programming of a third-order transfer function](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aseries-programming-z-plus-1) — Digital Control System, chapter 4, asked 2×
- [Jury test: 2z⁴+7z³+10z²+4z+1](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ajury-2z4-7z3) — Digital Control System, chapter 3, asked 8×
- [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×

## Related in other subjects

- [Transfer function from third-order state model](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-tf-3x3-1-2-3) — Control Systems, chapter 8, asked 3×
- [3rd order Butterworth HPF from LPF transfer function](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Athird-order-hpf-from-lpf-tf) — Filter Design, chapter 5
- [Realize 3rd order Butterworth TF with R2 = 2 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aladder-third-order-r2-2-ohm) — Filter Design, chapter 5
- [PID controller block diagram, transfer function, characteristics](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Apid-bd-tf-characteristics) — Control Systems, chapter 7