--- title: "αmax, αmin definitions and output amplitude numerical" subject: "Filter Design" chapter: 1 key: "semester_5/filter_design:attenuation-output-amplitude-definitions" source: https://www.bctengineer.com/q/semester_5/filter_design/attenuation-output-amplitude-definitions site: BCT Engineering --- # αmax, αmin definitions and output amplitude numerical Filter Design · Chapter 1: Introduction - Asked in: 2070 Chaitra, 3+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-definitions - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1 ## Question Define αmax, αmin and half power bandwidth with necessary diagrams. 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 αmax is the maximum allowed passband attenuation; αmin is the minimum required stopband attenuation. Half-power bandwidth is the frequency range where output power is at least half the maximum (3 dB point). Vout =1 × 10^(−90/20) = 3.16e−5 V ≈ 31.6 μV. ## Answer #### Definitions Attenuation (loss) is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB, so a large $\alpha$ means a small output. ```text alpha(dB) | ______________ amin |- - - - - - - - - - | | / | / transition amax |~~~~~~~~~~~~~~~~ / 0 +---------------+---+---------------> w passband wp ws stopband ``` - **$\alpha_{max}$ ($\alpha_p$):** the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop. - **$\alpha_{min}$ ($\alpha_s$):** the minimum attenuation required everywhere in the stopband (e.g. 40 dB). - **Half-power frequency $\omega_{3dB}$:** the frequency at which output power falls to half of its maximum, i.e. $|T(j\omega)| = |T|_{max}/\sqrt{2}$, an attenuation of 3.01 dB. - **Bandwidth:** the width of the passband. For a low-pass filter it is $0$ to $\omega_p$ (or to $\omega_{3dB}$ for the half-power bandwidth); for a band-pass filter $BW = \omega_2 - \omega_1$, where $\omega_1, \omega_2$ are the lower and upper half-power frequencies. Half-power frequency and bandwidth of a low-pass filter: ```text |T| 1.0 |------. | `. 0.707 |- - - - - -o <- half-power point (3 dB down) | `. | `-.___ 0 +----------+-----------> w |<--- BW ---> w3dB ``` #### 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. What attenuation corresponds to the half-power frequency? A. 0.5 dB B. 6 dB C. 20 dB D. About 3 dB (|T| = |T|max/√2) Answer: D. Half power means 1/√2 in amplitude, i.e. 3.01 dB. 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