--- title: "Importance of frequency transformation; obtaining band pass" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:frequency-transformation-importance-bandpass" source: https://www.bctengineer.com/q/semester_5/filter_design/frequency-transformation-importance-bandpass site: BCT Engineering --- # Importance of frequency transformation; obtaining band pass Filter Design · Chapter 3: Frequency Transformation - Asked in: 2070 Chaitra, 1+3 marks; 2070 Asar, 5 marks; 2081 Baisakh, 1+1+5 marks (asked 3 times) - Video: [Frequency transformation in analog domain # DSP# KTU ECT 303](https://www.youtube.com/watch?v=K_K1x0eOsxU) — Anu's Classroom ANUPAMA - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/frequency-transformation-importance-bandpass - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question What is the significance (importance) of frequency transformation in filter design? How can a band pass filter be obtained from a (normalized) prototype low pass filter? Explain with example. ## Short answer Frequency transformation allows using a single low-pass prototype table for HP, BP, BS designs, reducing complexity. For band-pass, s→(s²+ω₀²)/(B s); series L becomes series L'=L/B, C'=B/(L ω₀²); shunt C becomes parallel C'=C/B, L'=B/(C ω₀²). Example: first-order LP function 1/(s+1) yields BP B s/(s²+B s+ω₀²). ## Answer #### Significance of 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. - 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.). #### Band-pass filter from the normalized prototype low-pass filter **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 (transfer function method):** first-order prototype $T_{LP}(s) = \dfrac{1}{s+1}$. Substituting $s \to \dfrac{s^2+\omega_0^2}{Bs}$: $$ T_{BP}(s) = \frac{1}{\dfrac{s^2+\omega_0^2}{Bs} + 1} = \frac{Bs}{s^2 + Bs + \omega_0^2} $$ This is the standard second-order band-pass function with centre frequency $\omega_0$, bandwidth $B$ and $Q = \omega_0/B$. For $\omega_0 = 1000$ rad/s and $B = 100$ rad/s: $T_{BP}(s) = \dfrac{100s}{s^2 + 100s + 10^6}$, $Q = 10$. **Example (element method):** the third-order Butterworth ladder (1 H, 2 F, 1 H) with $\omega_0 = 2000$ rad/s, $B = 400$ rad/s gives series arms of 2.5 mH in series with 100 μF and a shunt arm of 5 mF in parallel with 50 μH. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is frequency transformation important in filter design? A. Each filter type needs its own separate approximation and table B. Only the low-pass approximation is solved once; tables of LP poles and ladder values serve HP, BP and BS filters C. It removes the need for any element scaling D. It converts a passive ladder into an active filter Answer: B. Specifications of other filter types are converted to an LP one, so only the LP approximation and tables are needed. (Key point: Why frequency transformation is used) 2. Which substitution converts a normalized low-pass prototype into a band-pass filter? A. s -> B s/(s^2 + w0^2) B. s -> w0/s C. s -> s/B D. s -> (s^2 + w0^2)/(B s) Answer: D. The BP substitution is (s^2 + w0^2)/(B s); B s/(s^2+w0^2) is the band-stop one and w0/s is high-pass. (Key point: LP to BP substitution) 3. Under s -> (s^2+w0^2)/(B s), where does the LP point Omega = 0 appear in the band-pass filter? A. At w = 0 B. At the centre frequency w = w0 C. At the band edge w1 D. At w = infinity Answer: B. Omega_LP = (w^2 - w0^2)/(B w) is zero at w = w0; w1 and w2 map to -1 and +1. (Key point: Mapping of frequencies) 4. How does the order of the band-pass filter compare with that of the LP prototype? A. It is the same B. It is half the prototype order C. It is one more than the prototype order D. It is twice the prototype order Answer: D. Each s becomes a ratio of quadratic to linear, so every pole gives a pair and the order doubles. (Key point: Order of BP filter) ## Related in Filter Design - [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× - [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 (10 kHz, 4 kHz) from normalized LPF TF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-10khz-from-transfer-function) — Filter Design, chapter 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× - [Band pass (ω1 = 100, ω2 = 10000) from normalized ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-100-10000-normalized-ladder) — Filter Design, chapter 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 ## Related in other subjects - [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× - [Hanning window low-pass FIR design](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afir-hanning-window-design) — Digital Signal Analysis and Processing, chapter 5 - [Low-pass FIR window design, αs = 41 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Afir-lowpass-window-0-2-0-5-41) — Digital Signal Analysis and Processing, chapter 5 - [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×