--- title: "Frequency and magnitude scaling to 10⁵ rad/s" subject: "Filter Design" chapter: 1 key: "semester_5/filter_design:scaling-ladder-1e5-numerical" source: https://www.bctengineer.com/q/semester_5/filter_design/scaling-ladder-1e5-numerical site: BCT Engineering --- # Frequency and magnitude scaling to 10⁵ rad/s Filter Design · Chapter 1: Introduction - Asked in: 2073 Chaitra, 3+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-1e5-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1 ## Question Define normalization and denormalisation. Following circuit is a lowpass filter designed at normalization frequency of ωo = 1 rad/s. Apply frequency and magnitude scaling so that ωo = 10⁵ rad/s and practically realizable elements. [Figure 1: source V1, R1 = 1 Ω, series L1 = 2.024 H, shunt C1 = 0.994 F, series L2 = 2.024 H, load R2 = 1 Ω] ## Short answer Set kf=1e5 for ωo=1e5 rad/s. Choose km=1000 for 1 kΩ terminations. Scaled: R1=R2=1 kΩ, L1=L2= (1000×2.024)/1e5 = 20.24 mH, C1=0.994/(1000×1e5)=9.94 nF (≈10 nF). Cutoff at 1e5 rad/s (≈15.9 kHz). ## Answer #### Normalization and denormalization **Normalization** is designing the filter for $\omega_o$ = 1 rad/s and 1 Ω terminations, so that standard tables and simple numbers can be used. **Denormalization** is converting the normalized values to the actual frequency and impedance level using 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: $R_1 = R_2 = 1\ \Omega$, $L_1 = L_2 = 2.024$ H, $C_1 = 0.994$ F at $\omega_o$ = 1 rad/s. Required $\omega_o = 10^5$ rad/s. **Frequency scaling factor:** $k_f = 10^5/1 = 10^5$. **Magnitude scaling factor:** with $k_m = 1$ the capacitor would be $0.994/10^5 \approx 10\ \mu$F and the resistors 1 Ω, which are not practical. Choose $k_m = 1000$ (1 kΩ terminations). $$ \begin{aligned} R_1' = R_2' &= k_m R = 1000 \times 1 = 1\ \text{k}\Omega \\ L_1' = L_2' &= \frac{k_m L}{k_f} = \frac{1000 \times 2.024}{10^5} = 20.24\ \text{mH} \\ C_1' &= \frac{C}{k_m k_f} = \frac{0.994}{1000 \times 10^5} = 9.94\ \text{nF} \end{aligned} $$ ```text 1k 20.24 mH 20.24 mH o-/\/\--UUUU----+----UUUU----+ V1 | | 9.94 nF 1k V2 | | o---------------+------------+ ``` **Answer:** $R_1 = R_2$ = 1 kΩ, $L_1 = L_2$ = 20.24 mH, $C_1$ = 9.94 nF (≈ 10 nF standard), with $\omega_o = 10^5$ rad/s (≈ 15.9 kHz). Check: $L'C'$ product scales by $1/k_f^2$: $2.024\times0.994\times10^{-10} = 20.24\times10^{-3}\times9.94\times10^{-9}$, so the response shape is unchanged and only moved to $10^5$ rad/s. Any other $k_m$ (e.g. 10 kΩ) is also correct if it gives practical values. ## 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. Why is km = 1000 chosen instead of km = 1? A. km must always equal 1000 for ωo = 10^5 rad/s B. Because km sets the cut-off frequency C. With km = 1 the capacitor would be about 10 μF and resistors 1 Ω, which are not practical D. Because km = 1 would change the response shape Answer: C. Magnitude scaling is a free choice made to get practical element values; it does not alter the response shape. (Key point: Why choose km) 3. What is kf to move ωo from 1 rad/s to 10^5 rad/s? A. 10^5 B. 1000 C. 10^-5 D. 2π × 10^5 Answer: A. kf = 10^5 / 1 = 10^5. (Key point: Frequency scaling) 4. With kf = 10^5 and km = 1000, what is C1' for C1 = 0.994 F? A. 9.94 μF B. 9.94 nF C. 99.4 pF D. 0.994 nF Answer: B. C' = 0.994/(1000 × 10^5) = 9.94×10^-9 F. 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