--- title: "How transformation reduces design steps; band stop design" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:bandstop-design-steps-reduction" source: https://www.bctengineer.com/q/semester_5/filter_design/bandstop-design-steps-reduction site: BCT Engineering --- # How transformation reduces design steps; band stop design Filter Design · Chapter 3: Frequency Transformation - Asked in: 2080 Bhadra, 1+4 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/bandstop-design-steps-reduction - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question How frequency transformation reduces the design steps required to design a filter? Design a band stop filter having center frequency 2000 rad/s and bandwidth 400 rad/s from a 3rd order Butterworth low pass filter. (Refer Table) ## Short answer Frequency transformation eliminates separate approximation for each filter type; band-stop specs map to a low-pass equivalent, then standard tables give elements, which are transformed by fixed formulas. For ω₀=2000 rad/s, B=400 rad/s, prototype L1=1 H, C2=2 F, L3=1 H becomes series arms 0.1 mH∥2.5 mF, shunt 1.25 mH in series with 0.2 mF. ## Answer #### How frequency transformation reduces design steps Without frequency transformation, a band-stop filter would need its own approximation (finding a suitable $T(s)$ with the required notch band) and its own synthesis, which is long and error-prone. With frequency transformation: 1. The band-stop specification is converted to an equivalent **normalized low-pass** specification. 2. The order and the ladder element values are read directly from the **standard low-pass table** (no new approximation). 3. Each element of the low-pass ladder is replaced by a simple LC pair using fixed formulas (no new synthesis). So one set of low-pass tables, plus a few substitution rules, covers HP, BP and BS designs. #### Design of the band-stop filter The LP → BS transformation is $s \to \dfrac{Bs}{s^2+\omega_0^2}$. Element rules: - series $L$ → parallel tank in the series arm: $L' = \dfrac{LB}{\omega_0^2}$, $C' = \dfrac{1}{LB}$; - shunt $C$ → series LC in the shunt arm: $L' = \dfrac{1}{CB}$, $C' = \dfrac{CB}{\omega_0^2}$. Prototype from the table ($n = 3$, $R_s = R_L = 1\ \Omega$): series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H. $$ \begin{aligned} L_1 = 1\ \text{H}:&\quad L_1' = \frac{L_1B}{\omega_0^2} = \frac{400}{4\times10^6} = 0.1\ \text{mH},\quad C_1' = \frac{1}{L_1B} = \frac{1}{400} = 2.5\ \text{mF} \\ C_2 = 2\ \text{F}:&\quad L_2' = \frac{1}{C_2B} = \frac{1}{800} = 1.25\ \text{mH},\quad C_2' = \frac{C_2B}{\omega_0^2} = \frac{800}{4\times10^6} = 0.2\ \text{mF} \\ L_3 = 1\ \text{H}:&\quad L_3' = 0.1\ \text{mH},\quad C_3' = 2.5\ \text{mF} \end{aligned} $$ ```text Rs=1 [ Z1 ] [ Z3 ] o-/\/\--[ ]---+------[ ]---+-----o | | | + Vs [ Y2 ] RL=1 Vo | | | - o----------------+---------------+-----o ``` | Arm | Prototype | Band-stop element | |---|---|---| | Z1 (series) | $L_1 = 1$ H | 0.1 mH ∥ 2.5 mF (tank) | | Y2 (shunt) | $C_2 = 2$ F | 1.25 mH in series with 0.2 mF | | Z3 (series) | $L_3 = 1$ H | 0.1 mH ∥ 2.5 mF (tank) | Check: $1/\sqrt{0.1\text{ mH}\times2.5\text{ mF}} = 1/\sqrt{1.25\text{ mH}\times0.2\text{ mF}} = 2000$ rad/s. Band edges: $\omega_{1,2} = \sqrt{\omega_0^2 + (B/2)^2} \mp B/2 = 1810.0$ and $2210.0$ rad/s (3 dB points). These values are for $1\ \Omega$ terminations. For practical values, magnitude-scale by $k_m$ (e.g. $k_m = 1000$ for 1 kΩ terminations): multiply every L by $k_m$ and divide every C by $k_m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does frequency transformation reduce the design steps for a band-stop filter? A. A new BS approximation is derived, then tabulated B. The BS spec is converted to a normalized LP spec, values come from LP tables, and each element is replaced by a fixed LC pair C. It avoids using any ladder circuit D. It only needs magnitude scaling of a BS table Answer: B. No new approximation or synthesis is needed. (Key point: How steps are reduced) 2. Which substitution gives a band-stop filter? A. s -> (s^2 + w0^2)/(B s) B. s -> w0/s C. s -> s/w0 D. s -> B s/(s^2 + w0^2) Answer: D. The BS substitution is the reciprocal of the BP one. (Key point: LP to BS substitution) 3. In the LP to BS transformation, what does a series inductor L become? A. A series LC with L' = L/B and C' = B/(L w0^2) B. A parallel LC tank in the series arm, L' = LB/w0^2 and C' = 1/(LB) C. A series LC in the shunt arm D. A single capacitor 1/L Answer: B. Z = sL -> L B s/(s^2+w0^2), which is a parallel tank in the series arm. (Key point: BS element rule) 4. For the 3rd-order Butterworth prototype (1 H, 2 F, 1 H) with w0 = 2000 rad/s and B = 400 rad/s, what is the shunt arm? A. 5 mF in parallel with 50 uH B. 1.25 mH in parallel with 0.2 mF C. 0.1 mH in series with 2.5 mF D. 1.25 mH in series with 0.2 mF Answer: D. L2' = 1/(C B) = 1/800 = 1.25 mH and C2' = C B/w0^2 = 800/(4x10^6) = 0.2 mF in a series branch. (Key point: BS numerical result) ## Related in Filter Design - [Band stop filter (2000 rad/s, 400 rad/s) from 3rd order](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandstop-2000-400-third-order) — Filter Design, chapter 3, asked 4× - [Band pass (2000 rad/s, 400 rad/s) from 3rd order](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-2000-400-third-order) — Filter Design, chapter 3, asked 2× - [Band pass (2000 rad/s, 400 rad/s) from 4th order](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-2000-400-fourth-order) — Filter Design, chapter 3, asked 2× - [Low pass to band stop transformation](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Alowpass-to-bandstop-transformation) — Filter Design, chapter 3, asked 10× - [Band pass (1 krad/s, 100 rad/s) from 4th order ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-1k-100-fourth-order) — Filter Design, chapter 3 - [Band pass (1500 rad/s, 300 rad/s) from 4th order ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-1500-300-fourth-order) — Filter Design, chapter 3, asked 4× ## Related in other subjects - [Noise equivalent bandwidth of RC low pass filter](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aneb-rc-lowpass) — Communication Systems, chapter 9