--- title: "Normalization and denormalization with example" subject: "Filter Design" chapter: 1 key: "semester_5/filter_design:normalization-importance-example" source: https://www.bctengineer.com/q/semester_5/filter_design/normalization-importance-example site: BCT Engineering --- # Normalization and denormalization with example Filter Design · Chapter 1: Introduction - Asked in: 2071 Shrawan, 6 marks - Video: [Butterworth Filter - 02 - Normalized Filter](https://www.youtube.com/watch?v=-EHRFrDujhc) — Adam Panagos - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/normalization-importance-example - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1 ## Question What is normalization and denormalization? Explain the importance of normalization and denormalization in filter design with example. ## Short answer Normalization designs at 1 rad/s, 1 Ω for simplicity; denormalization scales using kf=ω_actual/1 and km=R_actual/1. With R'=kmR, L'=(km/kf)L, C'=C/(km kf), a normalized 2nd-order Butterworth (L=1.414 H, C=1.414 F) becomes at 1 kHz, 600 Ω: L'≈0.135 H, C'≈0.375 μF. ## Answer **Normalization** means designing the filter with convenient reference values, usually a cut-off (or half-power) frequency of $\omega_o = 1$ rad/s and a termination of $R = 1\ \Omega$. **Denormalization (scaling)** converts the normalized element values to the actual frequency and impedance level required. Importance: - Filter tables (Butterworth, Chebyshev, Bessel poles and ladder element values) are published only in normalized form; one table serves every frequency and impedance level. - Calculations use simple numbers like 1, 1.414, 2 instead of values like $10^{-9}$ and $10^{4}$, so errors are fewer. - Designs can be compared and reused: a 1 rad/s prototype becomes a 1 kHz or 1 MHz filter by simple scaling. - Scaling lets the designer choose practical element values (e.g. capacitors in nF–μF, resistors in kΩ) without changing the shape of the response. #### Scaling relations used for denormalization With frequency scaling factor $k_f = \omega_{actual}/\omega_{normalized}$ and magnitude scaling factor $k_m = R_{actual}/R_{normalized}$: $$ R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f} $$ #### Example A normalized second-order Butterworth low-pass filter ($\omega_o$ = 1 rad/s, 1 Ω terminations) has $L = 1.414$ H and $C = 1.414$ F (doubly terminated ladder). Required: $f_o$ = 1 kHz and 600 Ω terminations. $$ \begin{aligned} k_f &= 2\pi\times1000 = 6283.2, \quad k_m = 600 \\ L' &= \frac{600 \times 1.414}{6283.2} = 0.135\ \text{H} \\ C' &= \frac{1.414}{600 \times 6283.2} = 0.375\ \mu\text{F} \end{aligned} $$ ```text 600R 0.135 H o-/\/\/--UUUU--+------+ Vs | | 0.375uF 600R Vo | | o--------------+------+ ``` The design was done once with simple numbers (1.414) and then adapted to the real requirement only by scaling. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a normalized filter design usually assume? A. Cut-off 1 kHz and termination 50 Ω B. Cut-off frequency 1 rad/s and termination 1 Ω C. Cut-off 1 rad/s and termination 1 kΩ D. Cut-off 1 MHz and termination 1 Ω Answer: B. Normalization uses convenient references, wo = 1 rad/s and R = 1 Ω, which denormalization later converts to actual values. (Key point: What is normalization) 2. Why are filter tables published only in normalized form? A. Real element values cannot be tabulated because they are negative B. Normalized filters have a different response shape that is easier to build C. Denormalization changes the response shape to the required one D. One table serves every frequency and impedance level after scaling Answer: D. Scaling keeps the shape of the response, so a single normalized table can be scaled to any requirement. (Key point: Why normalize) 3. What are the combined (km and kf) scaling formulas? A. R' = km R, L' = (km/kf) L, C' = C/(km kf) B. R' = km R, L' = (kf/km) L, C' = C/(km kf) C. R' = km R, L' = km kf L, C' = C/(km kf) D. R' = R/km, L' = (km/kf) L, C' = C km kf Answer: A. Applying kf then km multiplies L by km and divides by kf; C is divided by both. (Key point: Combined scaling) 4. A 2nd-order Butterworth ladder (L = C = 1.414) is scaled to 1 kHz with 600 Ω terminations. What is kf? A. kf = 1000 and km = 600 B. kf = 6283.2 and km = 1/600 C. kf = 2π x 1000 = 6283.2 and km = 600 D. kf = 600 and km = 6283.2 Answer: C. kf = w_actual/w_normalized = 2π x 1000, and km = 600/1. 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