--- title: "Low pass to high pass transformation" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:lowpass-to-highpass-transformation" source: https://www.bctengineer.com/q/semester_5/filter_design/lowpass-to-highpass-transformation site: BCT Engineering --- # Low pass to high pass transformation Filter Design · Chapter 3: Frequency Transformation - Asked in: 2080 Baisakh, 1+4 marks; 2072 Chaitra, 6 marks; 2080 Baisakh, 5 marks (asked 2 times) - Video: [Lowpass Prototype Transformation Solved Problem | Analog Filter Design Example](https://www.youtube.com/watch?v=eMueNgEzJFQ) — 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-highpass-transformation - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question What is frequency transformation and what are its applications in filter design? How can you obtain a high pass filter from a given (normalized) low pass filter? Explain with a suitable example. ## Short answer Frequency transformation maps a normalized low-pass prototype to other types by replacing s, reducing design effort. For high-pass, s→ω_c/s. Series inductor L becomes capacitor C'=1/(L ω_c); shunt capacitor C becomes inductor L'=1/(C ω_c). Example: third-order Butterworth (1 H, 2 F) with ω_c=1000 rad/s, 1 kΩ yields series 1 μF, shunt 0.5 H. ## Answer #### Frequency transformation and its applications **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. Applications in filter design: - Only **one set of approximation tables** (low-pass poles and ladder element values) is needed; HP, BP and BS filters are derived from it. - The difficult approximation step (finding $T(s)$ for given $\alpha_{max}$, $\alpha_{min}$) is done once, for the simpler low-pass case; specifications of the other types are converted to an equivalent LP specification. - It can be applied either to the **transfer function** (substitute for $s$) or directly to **each element** of an LC ladder, so the circuit is obtained without re-deriving it. - It **reduces design time and errors**, and keeps the properties of the prototype (equiripple, maximally flat, etc.). #### High-pass filter from a normalized low-pass filter **Transformation:** replace $s$ in the normalized low-pass prototype by $$ s \;\rightarrow\; \frac{\omega_c}{s} \quad (\text{normalized: } s \to 1/s) $$ On the $j\omega$ axis $\Omega_{LP} = -\omega_c/\omega$: low frequencies map to the LP stopband and high frequencies to the LP passband, and $\omega = \omega_c$ maps to the LP band edge $\Omega = 1$. So the attenuation the LP has at $\Omega$ appears in the HP at $\omega_c/\Omega$. **Element transformation** (normalized, $\omega_c = 1$): - inductor $L$: $Z = sL \to L/s = \dfrac{1}{s(1/L)}$, i.e. a **capacitor** $C' = 1/L$; - capacitor $C$: $Y = sC \to C/s = \dfrac{1}{s(1/C)}$, i.e. an **inductor** $L' = 1/C$; - resistors unchanged. Then frequency-scale to $\omega_c$ and magnitude-scale to the required impedance. **Example:** third-order Butterworth low-pass prototype ($R_s = R_L = 1\ \Omega$; series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H). Required: high-pass, $\omega_c = 1000$ rad/s, terminations 1 kΩ. 1. LP → HP (normalized): series $C_1 = 1/1 = 1$ F, shunt $L_2 = 1/2 = 0.5$ H, series $C_3 = 1$ F. 2. Scale with $k_f = 1000$, $k_m = 1000$ ($C' = C/(k_mk_f)$, $L' = Lk_m/k_f$): $$ C_1' = C_3' = \frac{1}{10^3\times10^3} = 1\ \mu\text{F}, \qquad L_2' = \frac{0.5\times10^3}{10^3} = 0.5\ \text{H} $$ ```text Rs=1k 1uF 1uF o-/\/\---||-----+--------||-----+-----o | | | + Vs 0.5 H RL=1k Vo | | | - o---------------+---------------+-----o ``` Transfer function check: $T_{LP}(s) = \dfrac{1}{s^3+2s^2+2s+1}$ (for equal terminations the passband gain is 1/2, omitted here); $s \to 1000/s$ gives $T_{HP}(s) = \dfrac{s^3}{s^3 + 2000s^2 + 2\times10^6 s + 10^9}$, a third-order Butterworth high-pass with 3 dB frequency 1000 rad/s. ## 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. Changes the filter order to meet attenuation B. Changes the frequency variable so a normalized LP prototype becomes HP, BP, BS or another LP C. Converts analog filters to digital filters D. Scales only the impedance level of the circuit Answer: B. It is a change of variable that maps the prototype response onto a new frequency axis. (Key point: Meaning of frequency transformation) 2. Which substitution gives a high-pass filter from a normalized low-pass filter? A. s -> s/wc B. s -> (s^2+wc^2)/(wc s) C. s -> wc s D. s -> wc/s (normalized s -> 1/s) Answer: D. Replacing s by wc/s swaps low and high frequencies. (Key point: LP to HP substitution) 3. In LP to HP transformation, what happen to a series inductor L and a shunt capacitor C (normalized)? A. L becomes a series capacitor L and C becomes a shunt inductor C B. L becomes a series capacitor 1/L and C becomes a shunt inductor 1/C C. L becomes a shunt capacitor and C a series inductor D. Both stay unchanged and only R changes Answer: B. Z = sL -> L/s = 1/(s(1/L)), a capacitor of 1/L; Y = sC -> C/s, an inductor of 1/C. (Key point: LP to HP element changes) 4. For the 3rd-order Butterworth LP (1 H, 2 F, 1 H) made HP with wc = 1000 rad/s and 1 kOhm terminations, what are C1 and L2? A. C1 = 1 mF and L2 = 0.5 mH B. C1 = 1 uF and L2 = 2 H C. C1 = 1 F and L2 = 0.5 H D. C1 = 1 uF and L2 = 0.5 H Answer: D. Normalized C1 = 1 F, L2 = 0.5 H; C' = 1/(10^3*10^3) = 1 uF and L' = 0.5*10^3/10^3 = 0.5 H. 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