---
title: "Scaling a Butterworth ladder to 100 Hz"
subject: "Filter Design"
chapter: 1
key: "semester_5/filter_design:scaling-ladder-100hz-numerical"
source: https://www.bctengineer.com/q/semester_5/filter_design/scaling-ladder-100hz-numerical
site: BCT Engineering
---

# Scaling a Butterworth ladder to 100 Hz

Filter Design · Chapter 1: Introduction

- Asked in: 2079 Baisakh, 2+4 marks
- Video: [Magnitude & Frequency Scaling in Analog Filter Design 🌟 Step by Step Examples](https://www.youtube.com/watch?v=k3zaELUgwac) — CAN Education
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/scaling-ladder-100hz-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1

## Question

What is Normalization and De-Normalization in filter design? The circuit given below is a Butterworth lowpass filter with half power frequency of 1 rad/s. Convert its half power frequency to 100 Hz using capacitor of 0.01μF. [Figure: 1 Ω source resistor, series 1 H inductor, shunt 2 F capacitor, series 1 H inductor, 1 Ω load resistor]

## Short answer

Frequency scaling kf = 2π×100 = 628.32. With capacitor 0.01 μF, magnitude scaling km = 2/(628.32×0.01e-6) = 3.183e5. Scaled values: Rs = RL = 318.3 kΩ, L1 = L3 = 506.6 H, C2 = 0.01 μF, yielding 100 Hz half-power frequency.

## Answer

#### Normalization and de-normalization
**Normalization** is designing a filter for a reference frequency of 1 rad/s and a reference impedance of 1 Ω, so standard tables and simple numbers can be used. **De-normalization** is converting the normalized element values to the actual cut-off frequency and impedance level by frequency scaling ($k_f$) and magnitude scaling ($k_m$):
$$
R' = k_m R, \qquad L' = \frac{k_m L}{k_f}, \qquad C' = \frac{C}{k_m k_f}
$$

#### Numerical
Given: $R_s = R_L = 1\ \Omega$, $L_1 = L_3 = 1$ H, $C_2 = 2$ F, $\omega_o$ = 1 rad/s. Required: $f_o$ = 100 Hz, capacitor = 0.01 μF.

**Frequency scaling factor:**
$$
k_f = \frac{\omega_{new}}{\omega_{old}} = \frac{2\pi\times100}{1} = 628.32
$$
**Magnitude scaling factor** (chosen so that the 2 F capacitor becomes 0.01 μF):
$$
\begin{aligned}
C' = \frac{C}{k_m k_f} &\Rightarrow k_m = \frac{C}{k_f C'} = \frac{2}{628.32 \times 0.01\times10^{-6}} \\
&= 3.183\times10^{5}
\end{aligned}
$$
**New element values:**
$$
\begin{aligned}
R_s' = R_L' &= k_m \times 1 = 318.3\ \text{k}\Omega \\
L_1' = L_3' &= \frac{k_m L}{k_f} = \frac{3.183\times10^5 \times 1}{628.32} = 506.6\ \text{H} \\
C_2' &= 0.01\ \mu\text{F}
\end{aligned}
$$

```text
 318.3k   506.6 H      506.6 H
 o-/\/\/--UUUU----+----UUUU----+
 Vs               |            |
               0.01uF       318.3k  Vo
                  |            |
 o----------------+------------+
```

**Answer:** $R_s = R_L$ = 318.3 kΩ, $L_1 = L_3$ = 506.6 H, $C_2$ = 0.01 μF; half-power frequency 100 Hz (628.3 rad/s).

Note: the inductors are very large because the capacitor is small and the frequency is low. In practice such a filter would be built as an active RC filter or with inductors simulated by a GIC.

## Check yourself

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

1. What is normalization in filter design?
   A. Making the passband gain exactly 0 dB at all frequencies
   B. Removing all resistors from the circuit
   C. Designing for a reference of 1 rad/s and 1 ohm so standard tables and simple numbers can be used
   D. Converting the filter from analog to digital form
   Answer: C. Normalization uses wc = 1 rad/s and R = 1 ohm; scaling later gives the actual frequency and impedance. (Key point: Meaning of normalization)

2. For a new half-power frequency of 100 Hz from 1 rad/s, what is kf?
   A. 628.32 (2π × 100)
   B. 100
   C. 6.283
   D. 1.59
   Answer: A. kf = new / old frequency in rad/s = 2π×100/1 = 628.32. (Key point: Frequency scaling factor)

3. How is km found when the 2 F capacitor must become 0.01 μF?
   A. km = C' kf / C = 3.14×10^-9
   B. km = kf = 628.32
   C. km = C' / C = 5×10^-9
   D. km = C / (kf C') = 2 / (628.32 × 0.01×10^-6) = 3.183×10^5
   Answer: D. From C' = C/(km kf), solve for km. (Key point: Finding km from the capacitor)

4. With km = 3.183×10^5 and kf = 628.32, what are the resistor and inductor values?
   A. 1 Ω resistors and 506.6 H inductors
   B. 318.3 kΩ resistors and 506.6 H inductors
   C. 318.3 kΩ resistors and 1.59 mH inductors
   D. 318.3 Ω resistors and 0.506 H inductors
   Answer: B. R' = km·1 = 318.3 kΩ and L' = km·1/kf = 506.6 H. (Key point: Resulting element values)

## Related in Filter Design

- [Scaling a ladder to 1000 rad/s and 75 Ω load](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ascaling-ladder-75-ohm-numerical) — Filter Design, chapter 1
- [Frequency and magnitude scaling to 10⁵ rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ascaling-ladder-1e5-numerical) — Filter Design, chapter 1
- [Band pass (1 krad/s, 100 rad/s) from 4th order ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-1k-100-fourth-order) — Filter Design, chapter 3
- [FDNR LPF at 10⁴ rad/s from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-lpf-1e4-from-ladder) — Filter Design, chapter 8
- [FDNR LPF at 5 kHz with 0.01 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-lpf-5khz-from-ladder) — Filter Design, chapter 8, asked 2×
- [Band pass (10000 rad/s, 1000 rad/s) from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-10000-1000-from-ladder) — Filter Design, chapter 3

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