---
title: "Direct form II of y(n)=−0.1y(n−1)+0.72y(n−2)…"
subject: "Digital Signal Analysis and Processing"
chapter: 4
key: "semester_7/digital_signal_analysis_and_processing:direct-form-2-0-72"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/direct-form-2-0-72
site: BCT Engineering
---

# Direct form II of y(n)=−0.1y(n−1)+0.72y(n−2)…

Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures

- Asked in: 2072 Kartik, 4 marks; 2069 Chaitra, 4 marks (asked 2 times)
- Video: [IIR realization - DIRECT FORM 1 and DIRECT FORM 2](https://www.youtube.com/watch?v=56bqoEvxEOY) — Shrenik Jain
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/direct-form-2-0-72
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4

## Question

Determine the Direct Form II realization of the following system: y(n) = −0.1y(n−1) + 0.72y(n−2) + 0.7x(n) − 0.252x(n−2).

## Short answer

H(z)=(0.7-0.252z^{-2})/(1+0.1z^{-1}-0.72z^{-2}). Direct Form II uses a shared delay line: w(n)=x(n)-0.1w(n-1)+0.72w(n-2), y(n)=0.7w(n)-0.252w(n-2). Only two delays are needed.

## Answer

#### System function
y(n) = −0.1y(n−1) + 0.72y(n−2) + 0.7x(n) − 0.252x(n−2)

Taking the z-transform: Y(z)[1 + 0.1z⁻¹ − 0.72z⁻²] = X(z)[0.7 − 0.252z⁻²]

H(z) = (0.7 − 0.252z⁻²)/(1 + 0.1z⁻¹ − 0.72z⁻²) = 0.7(1 − 0.36z⁻²)/((1 − 0.8z⁻¹)(1 + 0.9z⁻¹))

Poles: 0.8 and −0.9 (inside the unit circle, so the system is stable). Zeros: ±0.6.

#### Direct form II (canonical)
Split H(z) into an all-pole part followed by an all-zero part, sharing one delay line:

```text
w(n) = x(n) − 0.1·w(n−1) + 0.72·w(n−2)
y(n) = 0.7·w(n) − 0.252·w(n−2)
```
(Here the feedback coefficients are −a₁ = −0.1 and −a₂ = +0.72.) Only two delays are needed (order N = 2).

```text
x(n) ──►(+)───────────●──── ×0.7 ──────►(+)──► y(n)
         ▲            │ w(n)             ▲
         │          [z⁻¹]                │
         ├─ ×(−0.1) ◄─● w(n−1)           │
         │            │                  │
         │          [z⁻¹]                │
         └─ ×0.72 ◄───● w(n−2) ×(−0.252) ┘
```
Reading the diagram: the left adder forms w(n) = x(n) + (−0.1)w(n−1) + 0.72w(n−2); the right adder forms y(n) = 0.7w(n) + (−0.252)w(n−2). The feedback and feed-forward taps are both taken from the same two delays. There is no w(n−1) feed-forward tap, because the x(n−1) coefficient is 0.

| Quantity | Direct form I | Direct form II |
|---|---|---|
| Delays | 4 | 2 |
| Multipliers | 4 | 4 |
| Adders (two-input) | 3 | 3 |

Direct form II saves memory (2 delays instead of 4) and is called canonic.

## Check yourself

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

1. What is H(z)?
   A. (0.7 - 0.252z^-2)/(1 - 0.1z^-1 + 0.72z^-2)
   B. (0.7 - 0.252z^-2)/(1 + 0.1z^-1 - 0.72z^-2)
   C. (1 + 0.1z^-1 - 0.72z^-2)/(0.7 - 0.252z^-2)
   D. (0.7 + 0.252z^-2)/(1 + 0.1z^-1 + 0.72z^-2)
   Answer: B. Moving the y terms to the left gives the denominator. (Key point: System function)

2. Which equations describe Direct Form II?
   A. w(n) = x(n) + 0.1w(n-1) - 0.72w(n-2); y(n) = 0.7w(n) + 0.252w(n-2)
   B. y(n) = 0.7x(n) - 0.252x(n-2) followed by feedback in separate delay lines
   C. w(n) = x(n) - 0.1w(n-1) + 0.72w(n-2); y(n) = 0.7w(n) - 0.252w(n-2)
   D. w(n) = 0.7x(n) - 0.252x(n-2); y(n) = w(n) - 0.1y(n-1)
   Answer: C. The all-pole part comes first, then the all-zero part, sharing one delay line. (Key point: Direct form II equations)

3. How many delays does Direct Form II need?
   A. 4
   B. 1
   C. 5
   D. 2
   Answer: D. The shared delay line has order N = 2. (Key point: Number of delays)

4. Why is the system stable?
   A. The poles 0.8 and -0.9 lie inside the unit circle
   B. The zeros +/- 0.6 lie inside the unit circle
   C. It has no poles
   D. The gain 0.7 is below 1
   Answer: A. Stability depends on poles; both have magnitude below 1. (Key point: Stability)

## Close variants set in other exams

- [Direct form I and II, y(n)=−0.1y(n−1)+0.2y(n−2)+3x(n)…](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adirect-form-1-2-3x-3-6x) — Digital Signal Analysis and Processing, chapter 4, asked 5×
- [Direct form I and II of 3y[n]+y[n−1]+2y[n−4]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Adirect-form-1-2-delay4) — Digital Signal Analysis and Processing, chapter 4, asked 4×
- [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×

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