--- title: "Low pass to band stop transformation" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:lowpass-to-bandstop-transformation" source: https://www.bctengineer.com/q/semester_5/filter_design/lowpass-to-bandstop-transformation site: BCT Engineering --- # Low pass to band stop transformation Filter Design · Chapter 3: Frequency Transformation - Asked in: 2074 Chaitra, 2+4 marks; 2072 Kartik, 4 marks; 2071 Shrawan, 5 marks; 2069 Chaitra, 4 marks; 2081 Bhadra, 4 marks (asked 5 times) - Video: [Filter Design #31. Lowpass to Highpass/ Bandpass/ Bandstop: The Complete Frequency Transform Guide.](https://www.youtube.com/watch?v=Trot13fNFGc) — Technologies Discussion - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/lowpass-to-bandstop-transformation - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question What is frequency transformation? Describe (with necessary derivations) the frequency transformation from low pass to band stop filter with a suitable example. ## Short answer Frequency transformation converts normalized LP to BS via s → Bs/(s²+ω₀²) with ω₀=√(ω₁ω₂), B=ω₂−ω₁. Series L becomes parallel LC tank; shunt C becomes series LC pair, both resonate at ω₀. Example: 3rd-order Butterworth (1 H, 2 F, 1 H), ω₀=2000 rad/s, B=400 rad/s yields Z₁: 0.1 mH‖2.5 mF; Y₂: 1.25 mH + 0.2 mF (series); Z₃: 0.1 mH‖2.5 mF. ## Answer #### Frequency transformation **Frequency transformation** is a change of the complex frequency variable that converts a normalized low-pass prototype filter (cutoff 1 rad/s) into a high-pass, band-pass or band-stop filter (or a low-pass with a different cutoff). The magnitude response of the prototype is mapped onto the new frequency axis, so the same attenuation values appear at the corresponding new frequencies. #### Low-pass to band-stop transformation **Transformation:** replace $s$ in the low-pass prototype by $$ s \;\rightarrow\; \frac{B\,s}{s^2 + \omega_0^2}, \qquad \omega_0 = \sqrt{\omega_1\omega_2},\quad B = \omega_2 - \omega_1 $$ It is the band-pass transformation applied to the high-pass ($s \to 1/s$) version. On the $j\omega$ axis, $\Omega_{LP} = \dfrac{B\omega}{\omega_0^2 - \omega^2}$, so: - $\omega = 0$ and $\omega = \infty \Rightarrow \Omega_{LP} = 0$ (passbands at low and high frequencies); - $\omega = \omega_0 \Rightarrow \Omega_{LP} = \infty$ (maximum attenuation, the notch); - $\omega = \omega_1, \omega_2 \Rightarrow |\Omega_{LP}| = 1$ (band edges). **Element transformation:** Series inductor $L$: $$ Z = \frac{LBs}{s^2+\omega_0^2} \;\Rightarrow\; Y = \frac{s}{LB} + \frac{\omega_0^2}{LBs} $$ becomes a **parallel LC tank in the series arm**: $C' = \dfrac{1}{LB}$, $L' = \dfrac{LB}{\omega_0^2}$. Shunt capacitor $C$: $$ Y = \frac{CBs}{s^2+\omega_0^2} \;\Rightarrow\; Z = \frac{s}{CB} + \frac{\omega_0^2}{CBs} $$ becomes a **series LC in the shunt arm**: $L' = \dfrac{1}{CB}$, $C' = \dfrac{CB}{\omega_0^2}$. Resistors are unchanged; every LC pair resonates at $\omega_0$. At $\omega_0$ the series-arm tanks are open and the shunt-arm series LCs are shorts, so no signal reaches the load. ```text LP element BS element series L --> series arm: L' || C' shunt C --> shunt arm: L'--C' in series R --> R ``` **Example:** third-order Butterworth prototype ($R_s = R_L = 1\ \Omega$, $L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H) converted to a band-stop filter with $\omega_0 = 2000$ rad/s and $B = 400$ rad/s ($\omega_0^2 = 4\times10^6$). $$ \begin{aligned} L_1 = 1\ \text{H}:&\quad L_1' = \frac{L_1B}{\omega_0^2} = \frac{400}{4\times10^6} = 0.1\ \text{mH},\quad C_1' = \frac{1}{L_1B} = \frac{1}{400} = 2.5\ \text{mF} \\ C_2 = 2\ \text{F}:&\quad L_2' = \frac{1}{C_2B} = \frac{1}{800} = 1.25\ \text{mH},\quad C_2' = \frac{C_2B}{\omega_0^2} = \frac{800}{4\times10^6} = 0.2\ \text{mF} \\ L_3 = 1\ \text{H}:&\quad L_3' = 0.1\ \text{mH},\quad C_3' = 2.5\ \text{mF} \end{aligned} $$ ```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. What does a frequency transformation do? A. Converts a normalized LP prototype into HP, BP or BS (or another cutoff) by changing s B. Changes the filter order C. Converts digital to analog D. Adds gain Answer: A. The prototype magnitude is mapped onto the new frequency axis. (Key point: Frequency transformation) 2. Which substitution gives low-pass to band-stop? A. s -> (s^2 + w0^2)/(B s) B. s -> B s/(s^2 + w0^2) C. s -> 1/s D. s -> s/w0 Answer: B. Bs/(s^2+w0^2) maps w0 to infinite Omega (the notch). (Key point: BS substitution) 3. A series inductor L in the prototype becomes... A. A series LC B. A parallel LC tank in the series arm, L' = LB/w0^2, C' = 1/(LB) C. A shunt LC D. A single capacitor Answer: B. Z = LBs/(s^2+w0^2) is a parallel resonant tank. (Key point: Series L element) 4. For L1 = 1 H, B = 400, w0 = 2000 rad/s, C1' is... A. 0.1 mF B. 1.25 mH C. 2.5 mF D. 0.2 mF Answer: C. C' = 1/(LB) = 1/400 = 2.5 mF and L' = 0.1 mH. (Key point: Example value) ## Close variants set in other exams - [Low pass to band pass transformation](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Alowpass-to-bandpass-transformation) — Filter Design, chapter 3, asked 10× ## Related in Filter Design - [Importance of frequency transformation; obtaining band pass](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afrequency-transformation-importance-bandpass) — Filter Design, chapter 3, asked 3× - [How transformation reduces design steps; band stop design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandstop-design-steps-reduction) — Filter Design, chapter 3 - [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× - [Low pass to high pass transformation](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Alowpass-to-highpass-transformation) — Filter Design, chapter 3, asked 2× - [Band pass (10000 rad/s, 1000 rad/s) from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-10000-1000-from-ladder) — Filter Design, chapter 3 - [Band pass (2000 rad/s, 400 rad/s) from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-2000-400-from-ladder) — Filter Design, chapter 3 ## Related in other subjects - [RC low pass filter operation](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Arc-low-pass-filter) — Basic Electronics Engineering, chapter 1, asked 5× - [Impulse response of ideal low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-lpf-impulse-response) — Signals and Systems, chapter 5, asked 5× - [Ideal band pass and low pass filters (DT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aideal-bandpass-lowpass-dt) — Signals and Systems, chapter 6 - [Low-pass FIR window design, αs = 40 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afir-lowpass-window-0-3-0-5-40) — Digital Signal Analysis and Processing, chapter 5, asked 2×