---
title: "Ladder realization of a second-order filter"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:ladder-realization-filter"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-realization-filter
site: BCT Engineering
---

# Ladder realization of a second-order filter

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

- Asked in: 2082 Chaitra (new course), 5 marks; 2082 Chaitra, 5 marks; 2082 Kartik, 5 marks; 2078 Chaitra, 8 marks; 2074 Bhadra, 8 marks; 2071 Magh, 8 marks (asked 6 times)
- Video: [Ladder structure Digital signal processing b.tech (EN)](https://www.youtube.com/watch?v=yrvo_StmTeg) — SHIVAM KUMAR SINGH
- 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-filter
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Consider the digital filter defined by G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²) (equivalently GD(z)=(2z²+2.2z+0.2)/(z²+0.4z-0.12)). Realize this filter in the ladder scheme.

## Short answer

Write G(z) as a continued fraction in z: A₀=2, B₁=0.7143, A₁=16.333, B₂=0.0357, A₂=−20. The ladder structure uses two delays and realizes the filter via the equations Y₁ = (z⁻¹/B₁)(X−E₁), E₁ = (1/A₁)(Y₁−Y₂), Y₂ = (z⁻¹/B₂)(E₁−E₂), E₂ = (1/A₂)Y₂, with output Y = A₀X + Y₁.

## Answer

In **ladder programming**, $G(z)$ is written as a continued fraction in $z$:

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

Each $1/(B_iz)$ becomes a delay $z^{-1}/B_i$, and each $A_i$ becomes a constant gain.

Write the filter in positive powers of $z$:

$$
G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}
$$

#### Step 1: Constant term $A_0$
$$
G(z) = 2 + \frac{(2z^2+2.2z+0.2) - 2(z^2+0.4z-0.12)}{z^2+0.4z-0.12} = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}
$$

So $A_0 = 2$.

#### Step 2: $B_1$
$$
\frac{z^2+0.4z-0.12}{1.4z+0.44} = \frac{1}{1.4}z + \frac{0.0857z - 0.12}{1.4z+0.44}
$$

So $B_1 = 1/1.4 = 0.7143$. The remainder is $0.4z - 0.3143z = 0.0857z$ (exactly $3z/35$).

#### Step 3: $A_1$
$$
\frac{1.4z+0.44}{0.0857z-0.12} = \frac{1.4}{0.0857} + \frac{0.44 + 16.333(0.12)}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z-0.12}
$$

So $A_1 = 49/3 = 16.333$.

#### Step 4: $B_2$ and $A_2$
$$
\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05 = B_2z + \frac{1}{A_2}
$$

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$.

$$
G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z + \cfrac{1}{-20}}}}
$$

| $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ |
|---|---|---|---|---|
| 2 | 0.7143 | 16.333 | 0.0357 | −20 |

#### 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** (these define the ladder diagram; $o$ marks a take-off point):

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

With the values: $1/B_1 = 1.4$, $1/A_1 = 0.0612$, $1/B_2 = 28$, $1/A_2 = -0.05$.

Simulating these equations gives the same impulse response as the original difference equation $y(k) = -0.4y(k-1) + 0.12y(k-2) + 2x(k) + 2.2x(k-1) + 0.2x(k-2)$. The realization needs only 2 delays.

## Check yourself

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

1. In ladder programming, G(z) is written as
   A. A sum of partial fractions
   B. A product of first and second order sections
   C. A continued fraction A0 + 1/(B1 z + 1/(A1 + 1/(B2 z + 1/A2)))
   D. A single polynomial ratio in z^-1
   Answer: C. The ladder scheme uses a continued fraction expansion in z. (Key point: Ladder form)

2. For G(z) = (2z^2+2.2z+0.2)/(z^2+0.4z-0.12), the constant A0 is
   A. 0.2
   B. 2
   C. 1.4
   D. 0.12
   Answer: B. Equal degrees; the quotient is 2, leaving remainder (1.4z + 0.44)/(z^2+0.4z-0.12). (Key point: A0)

3. The coefficients B1 and A1 are
   A. B1 = 1.4, A1 = 0.0612
   B. B1 = 0.7143, A1 = 16.333
   C. B1 = 0.7143, A1 = 12.25
   D. B1 = 16.333, A1 = 0.7143
   Answer: B. B1 = 1/1.4 = 0.7143 from dividing z^2+0.4z-0.12 by 1.4z+0.44, then A1 = 1.4/0.0857 = 16.333. (Key point: B1 and A1)

4. The last coefficients B2 and A2 are
   A. B2 = 0.0357, A2 = -20
   B. B2 = 0.0357, A2 = 20
   C. B2 = 28, A2 = -0.05
   D. B2 = -20, A2 = 0.0357
   Answer: A. (0.0857z - 0.12)/2.4 = 0.0357z - 0.05, so B2 = 1/28 and A2 = -1/0.05 = -20. (Key point: B2 and A2)

5. How many delay elements does this ladder realization need?
   A. One
   B. Two
   C. Four
   D. Five
   Answer: B. A second-order filter needs only two delays; the ladder uses one delay per B_i z term. (Key point: Number of delays)

## Close variants set in other exams

- [Parallel and ladder realization of a filter](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aparallel-and-ladder-realization) — Digital Control System, chapter 4, asked 8×

## Related in Digital Control System

- [Parallel realization of a second-order filter](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aparallel-realization-filter) — Digital Control System, chapter 4, asked 2×
- [Why decompose higher-order filters; ladder realization](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adecomposition-ladder-scheme) — Digital Control System, chapter 4
- [Ladder realization (denominator −0.1z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-realization-variant) — Digital Control System, chapter 4, asked 2×
- [Diagonal canonical form of (z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adiagonal-form-0-7-0-12) — Digital Control System, chapter 5, asked 13×
- [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×
- [Standard and series realization of a filter](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Astandard-series-realization) — Digital Control System, chapter 4

## Related in other subjects

- [Saint-Venant, second-order explicit scheme](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asaint-venant-second-order-explicit) — Computational Techniques in Civil Engineering, chapter 5, asked 2×
- [Second-order scheme, linear kinematic wave](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asecond-order-linear-kinematic-scheme) — Computational Techniques in Civil Engineering, chapter 5
- [Cascade form realization of second order system](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Acascade-form-3-4-1-8) — Digital Signal Analysis and Processing, chapter 4, asked 4×
- [FDNR realization of 5th order ladder at 25,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fifth-order-25000) — Filter Design, chapter 8