---
title: "Why decompose higher-order filters; ladder realization"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:decomposition-ladder-scheme"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/decomposition-ladder-scheme
site: BCT Engineering
---

# Why decompose higher-order filters; ladder realization

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

- Asked in: 2076 Bhadra, 2+6 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/decomposition-ladder-scheme
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Higher order pulse transfer functions are generally realized by decomposing it into several lower order pulse transfer function why? Consider a digital filter shown below. Realize this filter in ladder scheme. G(z)=(2+2.2z⁻¹+0.2z⁻²)/(1+0.4z⁻¹-0.12z⁻²)

## Short answer

High-order transfer functions are decomposed to reduce sensitivity to coefficient quantization errors. Ladder expansion gives: A0=2, B1=0.7143, A1=16.333, B2=0.0357, A2=-20. The ladder uses two delay elements in a continued-fraction topology.

## Answer

#### Why higher-order pulse transfer functions are decomposed
A high-order $G(z)$ realized directly (direct or standard programming) is very **sensitive to coefficient errors**. The controller coefficients are stored with finite word length. A small rounding error in one coefficient of a high-order denominator can move all the poles a lot, even outside the unit circle, and make the filter unstable. So the transfer function is split into first- and second-order blocks (series, parallel or ladder programming):
- A coefficient error then affects only the one or two poles of that block.
- The structure is less sensitive to quantization, and the computations are better conditioned.
- Low-order blocks are easier to program, test and scale.

#### Ladder realization
Write the filter in positive powers of $z$:

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

In ladder programming, $G(z)$ is expanded as a continued fraction:

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

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

**Step 1: $A_0$.**
$$
G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}
$$
So $A_0 = 2$.

**Step 2: $B_1$.**
$$
\frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44}
$$
So $B_1 = 1/1.4 = 0.7143$.

**Step 3: $A_1$.**
$$
\frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12}
$$
So $A_1 = 49/3 = 16.333$.

**Step 4: $B_2$ and $A_2$.**
$$
\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05
$$
So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$.

$$
G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}}
$$

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

**Ladder 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):

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

With the values: $1/B_1 = 1.4$, $1/A_1 = 0.06122$, $1/B_2 = 28$, $1/A_2 = -0.05$.

The ladder needs only two delays. Because of its continued-fraction structure, it has low sensitivity to coefficient rounding.

## Check yourself

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

1. Why are higher-order pulse transfer functions decomposed into lower-order ones?
   A. Direct realization is very sensitive to coefficient errors from finite word length
   B. Lower order blocks have more delays
   C. Higher order functions cannot be realized
   D. To increase the order of the system
   Answer: A. Small rounding errors in a high-order denominator can move all poles, even out of the unit circle. (Key point: Why decompose)

2. After decomposition, an error in one coefficient affects
   A. All poles of the system
   B. Only the one or two poles of its block
   C. Only the zeros
   D. Nothing
   Answer: B. Each low-order block is independent, so errors stay local. (Key point: Effect of a coefficient error)

3. For G(z) = (2+2.2z^-1+0.2z^-2)/(1+0.4z^-1-0.12z^-2), the ladder constants A0, B1, A1, B2, A2 are
   A. 2, 0.7143, 16.333, 0.0357, -20
   B. 2, 0.7143, 12.25, 0.0703, -16.25
   C. 2, 1.4, 0.0612, 28, -0.05
   D. 1, 0.7143, 16.333, 0.0357, 20
   Answer: A. The continued fraction gives these values; the reciprocals 1.4, 0.0612, 28, -0.05 are the diagram gains. (Key point: Ladder constants)

4. The ladder realization of this second-order filter uses
   A. Two delays, with low sensitivity to rounding
   B. Four delays
   C. One delay
   D. No delays
   Answer: A. One delay per B_i z term gives two delays for this order-2 filter. (Key point: Delays in ladder)

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