--- title: "αp, αs, ωp, ωs and element scaling" subject: "Filter Design" chapter: 1 key: "semester_5/filter_design:define-attenuation-terms-scaling" source: https://www.bctengineer.com/q/semester_5/filter_design/define-attenuation-terms-scaling site: BCT Engineering --- # αp, αs, ωp, ωs and element scaling Filter Design · Chapter 1: Introduction - Asked in: 2074 Asoj, 4+4 marks - Video: [FEC26D Scaling Filter Elements and Frequency](https://www.youtube.com/watch?v=76HeghwHXxs) — EngrCS - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/define-attenuation-terms-scaling - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-1 ## Question Define and explain the following terms with necessary diagrams: αp, αs, ωp, ωs. What is scaling? Derive element scaling equations. ## Short answer αmax (αp) is maximum passband attenuation; αmin (αs) is minimum stopband attenuation; ωp is passband edge, ωs stopband edge. Scaling converts normalized (1 rad/s, 1 Ω) designs to practical values using km, kf: R'=kmR, L'=(km/kf)L, C'=C/(km kf). ## Answer #### $\alpha_p$, $\alpha_s$, $\omega_p$, $\omega_s$ 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 ``` - **$\alpha_p$ ($\alpha_{max}$):** the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop. - **$\alpha_s$ ($\alpha_{min}$):** the minimum attenuation required everywhere in the stopband (e.g. 40 dB). - **$\omega_p$:** passband edge frequency, the last frequency at which $\alpha \le \alpha_{max}$. - **$\omega_s$:** stopband edge frequency, from which $\alpha \ge \alpha_{min}$. The smaller $\alpha_p$, the larger $\alpha_s$ and the narrower the transition band $\omega_s/\omega_p$, the higher the filter order required. #### Scaling **Scaling** changes element values so that the response moves to another frequency or impedance level without changing its shape. It converts normalized designs (1 rad/s, 1 Ω) into practical filters. #### Magnitude (impedance) scaling Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change. $$ \begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned} $$ #### Frequency scaling The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$: $$ \begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned} $$ The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis. #### Combined scaling Applying both: $$ R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is attenuation defined? A. α(ω) = 20 log10|T(jω)| dB, so a larger α means more output B. α(ω) = -10 log10|T(jω)| dB C. α(ω) = -20 log10|T(jω)| dB, so a larger α means a smaller output D. α(ω) = |T(jω)| in volts Answer: C. Attenuation is minus the gain in dB. (Key point: Attenuation) 2. How are wp and ws defined for a low-pass filter? A. wp is the last frequency with α ≤ αmax; ws is the frequency from which α ≥ αmin B. wp is the frequency where α = αmin; ws where α = αmax C. wp is the 3 dB frequency and ws is twice wp D. wp is the stopband edge and ws is the passband edge Answer: A. wp is the passband edge and ws the stopband edge; between them lies the transition band. (Key point: wp and ws) 3. Which change requires a higher filter order? A. Larger αp and smaller αs B. Smaller αp, larger αs, and a narrower transition ws/wp C. A wider transition band D. Lower cut-off frequency only Answer: B. Tighter specifications need a steeper, higher-order response. (Key point: Order dependence) 4. For frequency scaling by kf = w_new/w_old, how do R, L and C change? A. R' = kf R, L' = L/kf, C' = C/kf B. R' = R, L' = kf L, C' = kf C C. R' = R, L' = L/kf, C' = C kf D. R' = R, L' = L/kf, C' = C/kf Answer: D. Equal impedance at kf*w requires kf*L' = L and kf*C' = C; R does not depend on frequency. 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