---
title: "Ladder programming of a third-order transfer function"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:ladder-z3-4z2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-z3-4z2
site: BCT Engineering
---

# Ladder programming of a third-order transfer function

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

- Asked in: 2067 Mangsir, 10 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-z3-4z2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Realize the following transfer function by ladder programming: X(z)=(z³+4z²+10z+7)/(z³+3z²+11z+18).

## Short answer

Ladder programming expands X(z) as a continued fraction. For X(z)=(z^3+4z^2+10z+7)/(z^3+3z^2+11z+18), the gains are A0=1, B1=1, A1=0.25, B2=-0.6154, A2=-0.5216, B3=-2.0392, A3=-0.3395. Three delays are used.

## Answer

Ladder programming expands $X(z) = \dfrac{z^3 + 4z^2 + 10z + 7}{z^3 + 3z^2 + 11z + 18}$ as a continued fraction:

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

#### Step 1: $A_0 = 1$
$$
X(z) = 1 + \frac{z^2 - z - 11}{z^3 + 3z^2 + 11z + 18}
$$

#### Step 2: $B_1$
$$
\frac{z^3 + 3z^2 + 11z + 18}{z^2 - z - 11} = z + \frac{4z^2 + 22z + 18}{z^2 - z - 11}
$$

So $B_1 = 1$.

#### Step 3: $A_1$
$$
\frac{z^2 - z - 11}{4z^2 + 22z + 18} = 0.25 + \frac{-6.5z - 15.5}{4z^2 + 22z + 18}
$$

So $A_1 = 0.25$.

#### Step 4: $B_2$
$$
\frac{4z^2 + 22z + 18}{-6.5z - 15.5} = -0.6154z + \frac{12.4615z + 18}{-6.5z - 15.5}
$$

So $B_2 = -0.6154$.

#### Step 5: $A_2$
$$
\frac{-6.5z - 15.5}{12.4615z + 18} = -0.5216 + \frac{-6.1111}{12.4615z + 18}
$$

So $A_2 = -0.5216$.

#### Step 6: $B_3$ and $A_3$
$$
\frac{12.4615z + 18}{-6.1111} = -2.0392z - 2.9455
$$

So $B_3 = -2.0392$ and $A_3 = -1/2.9455 = -0.3395$.

| $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ | $B_3$ | $A_3$ |
|---|---|---|---|---|---|---|
| 1 | 1 | 0.25 | −0.6154 | −0.5216 | −2.0392 | −0.3395 |

Exact values: $B_2 = -8/13$, $A_2 = -169/324$, $B_3 = -1458/715$, $A_3 = -55/162$.

#### 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):

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

Gains in the diagram: $1/B_1 = 1$, $1/A_1 = 4$, $1/B_2 = -1.625$, $1/A_2 = -1.917$, $1/B_3 = -0.4904$, $1/A_3 = -2.9455$.

Substituting back reproduces $X(z)$ exactly. Three delays are used.

## Check yourself

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

1. For (z^3+4z^2+10z+7)/(z^3+3z^2+11z+18), A0 and the remainder numerator are:
   A. A0 = 4, remainder z^2 - z - 11
   B. A0 = 1, remainder z^2 + z + 11
   C. A0 = 7/18, remainder z^2 - z - 11
   D. A0 = 1, remainder z^2 - z - 11
   Answer: D. Subtracting the denominator from the numerator gives z^2 - z - 11 with quotient 1. (Key point: Step 1 A0)

2. The first constants after A0 are:
   A. B1 = 1, A1 = 0.25
   B. B1 = 0.25, A1 = 1
   C. B1 = 4, A1 = 1
   D. B1 = 1, A1 = 4
   Answer: A. (z^3+3z^2+11z+18)/(z^2-z-11) = z + ..., so B1=1; then (z^2-z-11)/(4z^2+22z+18) = 0.25 + ..., so A1=0.25. (Key point: Value of B1 and A1)

3. The diagram uses the reciprocals of constants. Which is correct for A1 = 0.25?
   A. 1/A1 = 0.25
   B. 1/A1 = -4
   C. 1/A1 = 4
   D. 1/A1 = 0.5
   Answer: C. Each A_i block has gain 1/A_i, so 1/0.25 = 4. (Key point: Gains in diagram)

## Close variants set in other exams

- [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×

## Related in Digital Control System

- [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
- [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×
- [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×
- [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
- [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×

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