--- title: "Normalization and 78 dB/90 dB attenuation numerical" subject: "Filter Design" chapter: 1 key: "semester_5/filter_design:attenuation-output-amplitude-normalization" source: https://www.bctengineer.com/q/semester_5/filter_design/attenuation-output-amplitude-normalization site: BCT Engineering --- # Normalization and 78 dB/90 dB attenuation numerical Filter Design · Chapter 1: Introduction - Asked in: 2078 Bhadra, 2+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/attenuation-output-amplitude-normalization - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1 ## Question What is the significance of Normalization and Denormalization in filter design? At frequency f = 20 KHz and f = 30 KHz a filter is designed to attenuate the input signal by 78 dB and 90 dB respectively. Find the amplitude of the output signal if the 30 KHz input signal has amplitude of 1V. ## Short answer Normalization designs filters at a convenient reference (1 rad/s, 1 Ω); denormalization scales element values to actual frequency and impedance. Attenuation α = 20 log10(Vin/Vout) gives Vout = Vin × 10^(−α/20). For 1 V input at 30 kHz and 90 dB, Vout = 10^(−90/20) = 3.16e−5 V = 31.6 μV. ## Answer #### Significance of normalization and denormalization **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. The scaling relations are $R' = k_m R$, $L' = k_m L/k_f$, $C' = C/(k_m k_f)$. #### Numerical Attenuation in dB is defined as $$ \alpha = 20\log_{10}\frac{V_{in}}{V_{out}}\ \text{dB} \;\Rightarrow\; V_{out} = V_{in}\times10^{-\alpha/20} $$ At $f$ = 30 kHz, $\alpha$ = 90 dB and $V_{in}$ = 1 V: $$ \begin{aligned} V_{out} &= 1 \times 10^{-90/20} = 10^{-4.5} \\ &= 3.162\times10^{-5}\ \text{V} \end{aligned} $$ **Answer:** output amplitude at 30 kHz ≈ 31.6 μV. (The 78 dB at 20 kHz applies only to a 20 kHz input; a 1 V signal at 20 kHz would come out as $10^{-78/20} = 126\ \mu$V.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the output amplitude found from an attenuation of α dB? A. Vout = Vin × 10^(-α/20) B. Vout = Vin × 10^(-α/10) C. Vout = Vin / α D. Vout = Vin × 10^(α/20) Answer: A. α = 20 log10(Vin/Vout), so Vout = Vin·10^(-α/20) for voltage. (Key point: Attenuation formula) 2. The 1 V input is at 30 kHz. Which attenuation figure is used? A. 78 dB, because it is the smaller value B. 90 dB, because it is specified for 30 kHz C. The average of 78 and 90 dB D. 168 dB, the sum Answer: B. Each attenuation applies only at its own frequency. (Key point: Which attenuation applies) 3. What is the output amplitude at 30 kHz? A. About 126 μV B. About 31.6 mV C. About 31.6 μV D. About 1 nV Answer: C. 10^(-4.5) = 3.162×10^-5 V. (Key point: Result) 4. Which is a reason normalization is useful? A. It increases the stopband attenuation B. Filter tables (Butterworth, Chebyshev, Bessel) are published only in normalized form C. It makes inductors unnecessary D. It fixes the filter order at 5 Answer: B. One normalized table serves all frequencies and impedance levels after scaling. 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