--- title: "Low pass to band pass transformation" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:lowpass-to-bandpass-transformation" source: https://www.bctengineer.com/q/semester_5/filter_design/lowpass-to-bandpass-transformation site: BCT Engineering --- # Low pass to band pass transformation Filter Design · Chapter 3: Frequency Transformation - Asked in: 2082 Bhadra, 4 marks; 2082 Baisakh, 4 marks; 2081 Baisakh, 5 marks; 2078 Bhadra, 1+4 marks; 2079 Bhadra, 5 marks (asked 5 times) - Video: [Analog Filter Design: Lowpass Prototype Transformation Explained (LP to HP, BP, BS)](https://www.youtube.com/watch?v=l16VmQhWSrA) — EC Academy - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/lowpass-to-bandpass-transformation - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question What is frequency transformation? Describe the frequency transformation from (prototype) low pass to band pass filter with necessary derivations (expressions for R, L and C) and a suitable example. ## Short answer Frequency transformation maps a low-pass prototype to another filter by replacing s. For band-pass, s→(s²+ω₀²)/(B s), ω₀=√(ω₁ω₂), B=ω₂−ω₁. Series L becomes series L'=L/B, C'=B/(L ω₀²); shunt C becomes parallel C'=C/B, L'=B/(C ω₀²). Example: third-order Butterworth with ω₀=2000 rad/s, B=400 rad/s yields series 2.5 mH/100 μF and shunt 50 μH/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-pass transformation **Transformation:** replace $s$ in the low-pass prototype by $$ s \;\rightarrow\; \frac{s^2 + \omega_0^2}{B\,s}, \qquad \omega_0 = \sqrt{\omega_1\omega_2},\quad B = \omega_2 - \omega_1 $$ On the $j\omega$ axis this gives $\Omega_{LP} = \dfrac{\omega^2 - \omega_0^2}{B\omega}$. So: - $\omega = \omega_0 \Rightarrow \Omega_{LP} = 0$ (centre of passband maps to DC of the LP); - $\omega = \omega_1, \omega_2 \Rightarrow \Omega_{LP} = \mp1$, since $\omega_1\omega_2 = \omega_0^2$ and $\omega_2 - \omega_1 = B$; - $\omega = 0$ and $\omega = \infty \Rightarrow \Omega_{LP} = \mp\infty$ (stopbands). The band-pass response is geometrically symmetric about $\omega_0$, and its order is twice that of the prototype. **Element transformation:** Series inductor $L$: $$ Z = L\,\frac{s^2+\omega_0^2}{Bs} = \frac{L}{B}s + \frac{1}{\left(\dfrac{B}{L\omega_0^2}\right)s} $$ becomes a **series LC**: $L' = \dfrac{L}{B}$, $C' = \dfrac{B}{L\omega_0^2}$. Shunt capacitor $C$: $$ Y = C\,\frac{s^2+\omega_0^2}{Bs} = \frac{C}{B}s + \frac{1}{\left(\dfrac{B}{C\omega_0^2}\right)s} $$ becomes a **parallel LC** to ground: $C' = \dfrac{C}{B}$, $L' = \dfrac{B}{C\omega_0^2}$. Resistor $R$: unchanged (frequency independent). Every new LC pair resonates at $\omega_0$ ($L'C' = 1/\omega_0^2$). ```text LP element BP element series L --> series L'--C' shunt C --> shunt L' || C' R --> R ``` **Example:** third-order Butterworth prototype ($R_s = R_L = 1\ \Omega$, series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H) converted to a band-pass filter with $\omega_0 = 2000$ rad/s and $B = 400$ rad/s ($\omega_0^2 = 4\times10^6$). $$ \begin{aligned} L_1 = 1:&\quad L_1' = \frac{1}{400} = 2.5\ \text{mH},\quad C_1' = \frac{400}{1\times4\times10^6} = 100\ \mu\text{F} \\ C_2 = 2:&\quad C_2' = \frac{2}{400} = 5\ \text{mF},\quad L_2' = \frac{400}{2\times4\times10^6} = 50\ \mu\text{H} \\ L_3 = 1:&\quad L_3' = 2.5\ \text{mH},\quad C_3' = 100\ \mu\text{F} \end{aligned} $$ ```text Rs=1 [ Z1 ] [ Z3 ] o-/\/\--[ ]---+------[ ]---+-----o | | | + Vs [ Y2 ] RL=1 Vo | | | - o----------------+---------------+-----o ``` | Arm | Prototype | Band-pass element | |---|---|---| | Z1 (series) | $L_1 = 1$ H | 2.5 mH in series with 100 μF | | Y2 (shunt) | $C_2 = 2$ F | 5 mF ∥ 50 μH | | Z3 (series) | $L_3 = 1$ H | 2.5 mH in series with 100 μF | Check: $1/\sqrt{2.5\text{ mH}\times100\ \mu\text{F}} = 2000$ rad/s. The 3 dB band edges are $\omega_{1,2} = \sqrt{\omega_0^2+(B/2)^2} \mp B/2 = 1810.0$ and $2210.0$ rad/s, so $\omega_2 - \omega_1 = 400$ rad/s. The result is a 6th-order band-pass filter. 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-pass? A. s -> (s^2 + w0^2)/(B s) B. s -> B s/(s^2 + w0^2) C. s -> w0/s D. s -> s + w0 Answer: A. Omega = (w^2 - w0^2)/(B w) maps w0 to 0 and w1,w2 to -1,+1. (Key point: BP substitution) 3. A series inductor L becomes... A. A parallel LC in series arm B. A series LC with L' = L/B, C' = B/(L w0^2) C. A shunt capacitor D. A resistor Answer: B. Z = (L/B)s + 1/((B/(L w0^2))s). (Key point: Series L element) 4. For C2 = 2 F, B = 400, w0 = 2000 rad/s, the shunt parallel LC has... A. C' = 5 mF, L' = 50 uH B. C' = 2.5 mH, L' = 100 uF C. C' = 0.2 mF, L' = 1.25 mH D. C' = 50 uH, L' = 5 mF Answer: A. C' = C/B = 5 mF; L' = B/(C w0^2) = 50 uH. Order doubles to 6. (Key point: Example) ## Close variants set in other exams - [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× ## 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× - [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× - [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 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 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 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 - [Matched filter approximation using RC low pass filter](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amatched-filter-rc-approximation) — Communication Systems, chapter 9, asked 2×