---
title: "Standard and series realization of a filter"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:standard-series-realization"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/standard-series-realization
site: BCT Engineering
---

# Standard and series realization of a filter

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

- Asked in: 2079 Chaitra, 8 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_6/digital_control_system/standard-series-realization
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Consider the digital filter defined by G(z)=(6z²+7z)/(10z³+6z²+7z+5). Realize this filter in the standard and series scheme.

## Short answer

Standard: define w(k)=x(k)-0.6w(k-1)-0.7w(k-2)-0.5w(k-3) and y(k)=0.6w(k-1)+0.7w(k-2) using 3 delays. Series: factor as G(z)=0.6 * [1/(1+0.6697z^-1)] * [(z^-1+1.1667z^-2)/(1-0.0697z^-1+0.7467z^-2)], giving two cascaded first- and second-order blocks.

## Answer

Divide by $10z^3$:

$$
G(z) = \frac{6z^2 + 7z}{10z^3 + 6z^2 + 7z + 5} = \frac{0.6z^{-1} + 0.7z^{-2}}{1 + 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}}
$$

#### (a) Standard programming
Introduce an intermediate variable $W(z) = X(z)/(\text{denominator})$:

$$
\begin{aligned}
w(k) &= x(k) - 0.6w(k-1) - 0.7w(k-2) - 0.5w(k-3) \\
y(k) &= 0.6w(k-1) + 0.7w(k-2)
\end{aligned}
$$

```text
x(k)-->(S)------------o
        ^             |
        |          [z^-1]
        |             | w(k-1)
       (S)<[-0.6]-----o--[0.6]------>(S)--> y(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-2)        |
       (S)<[-0.7]-----o--[0.7]--------+
        ^             |
        |          [z^-1]
        |             | w(k-3)
        +<[-0.5]------o
```

Only $n = 3$ delays are used, the minimum number.

#### (b) Series programming
Factor the denominator. Its real root, found numerically, is $z = -0.6697$:

$$
10z^3 + 6z^2 + 7z + 5 = 10(z + 0.6697)(z^2 - 0.0697z + 0.7467)
$$

The quadratic has complex roots $0.0348 \pm j0.8634$, so it is kept as one second-order block. The numerator is $6z^2 + 7z = 6z(z + 1.1667)$. So

$$
G(z) = 0.6\cdot\underbrace{\frac{1}{1 + 0.6697z^{-1}}}_{G_1}\cdot\underbrace{\frac{z^{-1} + 1.1667z^{-2}}{1 - 0.0697z^{-1} + 0.7467z^{-2}}}_{G_2}
$$

**Check:** $0.6(z^{-1} + 1.1667z^{-2}) = 0.6z^{-1} + 0.7z^{-2}$, and $(1 + 0.6697z^{-1})(1 - 0.0697z^{-1} + 0.7467z^{-2}) = 1 + 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}$. ✔

**Difference equations:**
- $w_1(k) = x(k) - 0.6697w_1(k-1)$, and $y_1(k) = 0.6w_1(k)$
- $w_2(k) = y_1(k) + 0.0697w_2(k-1) - 0.7467w_2(k-2)$
- $y(k) = w_2(k-1) + 1.1667w_2(k-2)$

Block $G_1$ (with gain 0.6):
```text
x(k)-->(S)------------o--[1]-------->(S)->[0.6]-> y1(k)
        ^             |
        |          [z^-1]
        |             | w1(k-1)
        +<[-0.6697]---o
```
Block $G_2$:
```text
y1(k)-->(S)------------o
         ^             |
         |          [z^-1]
         |             | w2(k-1)
        (S)<[0.0697]---o--[1]-------->(S)--> y(k)
         ^             |               ^
         |          [z^-1]             |
         |             | w2(k-2)       |
         +<[-0.7467]---o--[1.1667]-----+
```

The series form also uses 3 delays. Each block is first- or second-order, so it is less sensitive to coefficient rounding.

## Check yourself

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

1. In standard programming, the intermediate variable W(z) is defined as:
   A. X(z) times the denominator
   B. X(z) divided by the denominator polynomial
   C. Y(z) divided by the numerator
   D. X(z) times the numerator
   Answer: B. W = X/(1 + 0.6z^-1 + 0.7z^-2 + 0.5z^-3), so w(k) = x(k) - 0.6w(k-1) - 0.7w(k-2) - 0.5w(k-3), and y uses the numerator on w. (Key point: Standard programming variable)

2. Why does standard programming use only n = 3 delays for this third-order filter?
   A. The numerator has fewer terms
   B. The delay chain on w is shared by both the feedback and feed-forward paths
   C. Feedback is removed
   D. The poles are complex
   Answer: B. One shared delay chain of w supplies both the denominator and numerator taps, giving the minimum n delays. (Key point: Minimum delays)

3. In the series scheme, why is the complex-root quadratic kept as one second-order block?
   A. Complex roots cannot be realized
   B. Splitting it would need complex coefficients; keeping it gives real coefficients
   C. It reduces the number of delays to 2
   D. The quadratic has a repeated root
   Answer: B. The roots 0.0348 +/- j0.8634 are a conjugate pair; the quadratic z^2 - 0.0697z + 0.7467 has real coefficients and is realized as one section. (Key point: Series factorisation)

4. What is the practical advantage of the series form over the standard form?
   A. It uses fewer delays
   B. It uses no multipliers
   C. Each block is only first- or second-order, so it is less sensitive to coefficient rounding
   D. It is always stable
   Answer: C. Both use 3 delays, but low-order blocks are less sensitive to rounded coefficients. (Key point: Series advantage)

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