---
title: "Direct form I and II of 3y[n]+y[n−1]+2y[n−4]"
subject: "Digital Signal Analysis and Processing"
chapter: 4
key: "semester_7/digital_signal_analysis_and_processing:direct-form-1-2-delay4"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/direct-form-1-2-delay4
site: BCT Engineering
---

# Direct form I and II of 3y[n]+y[n−1]+2y[n−4]

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

- Asked in: 2074 Chaitra, 5 marks
- Video: [Draw direct form-I and direct form-II realization.](https://www.youtube.com/watch?v=mRKPGt97ytw) — Engg-Course-Made-Easy
- 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-1-2-delay4
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4

## Question

Obtain the Direct Form I and Direct Form II realization of the following system: 3y[n] + y[n−1] + 2y[n−4] = 2x[n] + x[n−3].

## Short answer

Normalize to y[n] = −(1/3)y[n−1] − (2/3)y[n−4] + (2/3)x[n] + (1/3)x[n−3]. Direct Form I uses 7 delays; Direct Form II (canonic) uses 4 delays with feedforward taps at w[n] and w[n−3] (2/3, 1/3) and feedback at w[n−1] (−1/3) and w[n−4] (−2/3).

## Answer

Divide the equation by 3 so that the coefficient of y[n] is 1:

y[n] = −(1/3)y[n−1] − (2/3)y[n−4] + (2/3)x[n] + (1/3)x[n−3]

So H(z) = (2/3 + (1/3)z⁻³) / (1 + (1/3)z⁻¹ + (2/3)z⁻⁴).

Feed-forward coefficients: b₀ = 2/3, b₃ = 1/3 (b₁ = b₂ = 0). Feedback coefficients: a₁ = 1/3, a₄ = 2/3 (a₂ = a₃ = 0), entering with a minus sign.

#### Direct Form I
Uses separate delay lines for input (3 delays) and output (4 delays): 7 delays in all.

```text
 x[n] ──────►[×2/3]──────────┐
   │z⁻¹                      │
 x[n-1]                      │
   │z⁻¹                      ▼
 x[n-2]                    ( Σ )────┬──► y[n]
   │z⁻¹                      ▲       │
 x[n-3] ────►[×1/3]──────────┤       │ (y[n] enters the
                             │       │  output delay chain)
 y[n-1] ────►[×−1/3]─────────┤       ▼
   │z⁻¹   (y[n-1] = y[n] delayed)  z⁻¹ chain
 y[n-2]                      │
   │z⁻¹                      │
 y[n-3]                      │
   │z⁻¹                      │
 y[n-4] ────►[×−2/3]─────────┘
```
- Input side: x[n] → ×2/3 into the sum. x[n] → z⁻¹ → z⁻¹ → z⁻¹ → ×1/3 into the sum.
- Output side: y[n] → z⁻¹ → ×(−1/3) into the sum; y[n] → z⁻¹ → z⁻¹ → z⁻¹ → z⁻¹ → ×(−2/3) into the sum.
- y[n] = sum of the four branches.

#### Direct Form II (canonic)
Introduce w[n] between the poles and zeros:

- w[n] = x[n] − (1/3)w[n−1] − (2/3)w[n−4]
- y[n] = (2/3)w[n] + (1/3)w[n−3]

The two delay lines share the same delays, so only max(3, 4) = **4 delays** are needed.

```text
 x[n]──►(+)──────────────┬─►[×2/3]──►(+)──► y[n]
         ▲ w[n]          │             ▲
         │               ▼ z⁻¹ (w[n-1]) │
         ├──[×−1/3]──────┤             │
         │               ▼ z⁻¹ (w[n-2]) │
         │               ▼ z⁻¹ (w[n-3])─┴─[×1/3]
         │               ▼ z⁻¹ (w[n-4])
         └──[×−2/3]──────┘
```
Taps: w[n−1] feeds back through −1/3; w[n−4] feeds back through −2/3; w[n] and w[n−3] feed forward through 2/3 and 1/3.

**Answer: DF-I needs 7 delays; DF-II needs 4 delays. Coefficients: b₀ = 2/3, b₃ = 1/3; feedback −1/3 (z⁻¹) and −2/3 (z⁻⁴).**

## Check yourself

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

1. What is the first step for 3y[n] + y[n-1] + 2y[n-4] = 2x[n] + x[n-3]?
   A. Divide by 3 so the y[n] coefficient is 1
   B. Multiply by 3
   C. Move x terms to the left
   D. Take the z-transform of x only
   Answer: A. Dividing by 3 gives y[n] = -(1/3)y[n-1] - (2/3)y[n-4] + (2/3)x[n] + (1/3)x[n-3]. (Key point: Normalise y[n] coefficient)

2. How many delays does Direct Form I need here?
   A. 4
   B. 3
   C. 7
   D. 8
   Answer: C. DF-I uses separate lines: 3 input delays and 4 output delays, 7 in all. (Key point: Delay count DF-I)

3. Why does Direct Form II need only 4 delays?
   A. It drops the x[n-3] term
   B. It shares one delay line w[n] between feedback and feed-forward, needing max(3,4) = 4
   C. It uses a delay for each coefficient
   D. It has no feedback
   Answer: B. The poles and zeros share the same delays, so only max of the two orders, 4, is required (canonic form). (Key point: Delay count DF-II)

4. With w[n] as the DF-II middle variable, which is correct?
   A. w[n] = x[n] + (1/3)w[n-1] + (2/3)w[n-4]
   B. w[n] = x[n] - (1/3)w[n-1] - (2/3)w[n-4]
   C. w[n] = x[n] - (1/3)w[n-1] - (2/3)w[n-3]
   D. w[n] = (2/3)x[n] + (1/3)x[n-3]
   Answer: B. Feedback coefficients enter with a minus sign: a1 = 1/3 on w[n-1] and a4 = 2/3 on w[n-4]. (Key point: Feedback sign)

## Close variants set in other exams

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