---
title: "Ladder realization (denominator −0.1z⁻²)"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:ladder-realization-variant"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-realization-variant
site: BCT Engineering
---

# Ladder realization (denominator −0.1z⁻²)

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

- Asked in: 2080 Chaitra, 8 marks
- Video: [Implementation of Digital system-Ladder Structure](https://www.youtube.com/watch?v=JH65O38GVIA) — E2 Knowledge
- 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-realization-variant
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Realize the given digital filter by ladder programming: G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.1z⁻²).

## Short answer

Ladder continued-fraction expansion yields parameters: A0=2, B1=0.7143, A1=12.25, B2=0.07033, A2=-16.25. The structure cascades delays and gain blocks, using only two delay elements and achieving low coefficient sensitivity.

## Answer

In positive powers of $z$:

$$
G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.1z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.1}
$$

Ladder programming expands $G(z)$ as:

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

#### Step 1: $A_0$
$$
G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.1)}{z^2+0.4z-0.1} = 2 + \frac{1.4z + 0.4}{z^2 + 0.4z - 0.1}
$$

#### Step 2: $B_1$
$$
\frac{z^2 + 0.4z - 0.1}{1.4z + 0.4} = 0.7143z + \frac{0.1143z - 0.1}{1.4z + 0.4}
$$

The remainder is $0.4z - 0.7143(0.4)z = 0.1143z$ (exactly $4z/35$). So $B_1 = 0.7143$.

#### Step 3: $A_1$
$$
\frac{1.4z + 0.4}{0.1143z - 0.1} = 12.25 + \frac{0.4 + 12.25(0.1)}{0.1143z - 0.1} = 12.25 + \frac{1.625}{0.1143z - 0.1}
$$

So $A_1 = 12.25$.

#### Step 4: $B_2$ and $A_2$
$$
\frac{0.1143z - 0.1}{1.625} = 0.07033z - 0.06154 = B_2z + \frac{1}{A_2}
$$

So $B_2 = 0.07033$ and $A_2 = -16.25$.

$$
G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{12.25 + \cfrac{1}{0.07033z + \cfrac{1}{-16.25}}}}
$$

| $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ |
|---|---|---|---|---|
| 2 | 0.7143 | 12.25 | 0.07033 | −16.25 |

#### 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
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+
```

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

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

Numerical gains: $1/B_1 = 1.4$, $1/A_1 = 0.08163$, $1/B_2 = 14.22$, $1/A_2 = -0.06154$.

Substituting back reproduces $G(z)$ exactly, so the realization is correct. It uses two delay elements.

## Check yourself

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

1. For G(z) = (2z^2+2.2z+0.2)/(z^2+0.4z-0.1), A0 is
   A. 2
   B. 1.4
   C. 0.4
   D. 0.1
   Answer: A. The z^2 coefficients give quotient 2, leaving (1.4z + 0.4)/(z^2+0.4z-0.1). (Key point: A0)

2. B1 is found from dividing (z^2+0.4z-0.1) by (1.4z+0.4). B1 equals
   A. 1.4
   B. 0.7143
   C. 0.4
   D. 12.25
   Answer: B. The leading term is z^2/(1.4z) = 0.7143z, so B1 = 0.7143. (Key point: B1)

3. A1 for this variant is
   A. 16.333
   B. 12.25
   C. 1.625
   D. 0.08163
   Answer: B. A1 = 1.4/0.1143 = 12.25. (Key point: A1)

4. The final stage gives
   A. B2 = 0.07033, A2 = -16.25
   B. B2 = 0.0357, A2 = -20
   C. B2 = 14.22, A2 = 16.25
   D. B2 = 0.07033, A2 = 16.25
   Answer: A. (0.1143z - 0.1)/1.625 = 0.07033z - 0.06154, so A2 = -1/0.06154 = -16.25. (Key point: B2 and A2)

5. In the ladder, each B_i z term is realized as
   A. A delay z^-1/B_i
   B. A gain B_i only
   C. An integrator 1/z
   D. A gain 1/A_i
   Answer: A. 1/(B_i z) = z^-1/B_i is a delay with gain 1/B_i; the A_i terms become constant gains. (Key point: Ladder signal flow)

## Close variants set in other exams

- [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 Digital Control System

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- [Ladder realization of a second-order filter](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-realization-filter) — Digital Control System, chapter 4, asked 7×
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