--- title: "Band pass (10⁴ rad/s, 9.9×10⁴ rad/s) from given ladder" subject: "Filter Design" chapter: 3 key: "semester_5/filter_design:bandpass-from-figure1-ladder" source: https://www.bctengineer.com/q/semester_5/filter_design/bandpass-from-figure1-ladder site: BCT Engineering --- # Band pass (10⁴ rad/s, 9.9×10⁴ rad/s) from given ladder Filter Design · Chapter 3: Frequency Transformation - Asked in: 2073 Chaitra, 2+4 marks - Video: [LC Ladder Bandpass Filter Design 🌟 Chebyshev Response - 3rd Order 💡 Example 3](https://www.youtube.com/watch?v=D0jadjCNHvI) — CAN Education - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/bandpass-from-figure1-ladder - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3 ## Question What is frequency transformation? Obtain the bandpass filter from lowpass filter given in figure 1 having center frequency 10⁴ rad/s and bandwidth of 9.9 × 10⁴ rad/s. [Figure 1: source V1, R1 = 1 Ω, series L1 = 2.024 H, shunt C1 = 0.994 F, series L2 = 2.024 H, load R2 = 1 Ω] ## Short answer Frequency transformation maps a normalized LP prototype to other filter types by substituting s. For BP, s→(s²+ω₀²)/(Bs). With ω₀=10⁴ rad/s, B=9.9×10⁴ rad/s: series arms become 20.44 µH in series with 489.13 µF; shunt arm becomes 10.04 µF ∥ 995.98 µH; terminations 1 Ω. ## 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. For band-pass: $s \to \dfrac{s^2+\omega_0^2}{Bs}$. #### Band-pass filter from the given low-pass filter Given: $\omega_0 = 10^4$ rad/s ($\omega_0^2 = 10^8$), $B = 9.9\times10^4$ rad/s (values used as given; they correspond to band edges of about $\omega_1 \approx 1000$ and $\omega_2 \approx 10^5$ rad/s, since $\omega_1\omega_2 = 10^8$ and $\omega_2 - \omega_1 = 9.9\times10^4$). Prototype (Figure 1): $R_1 = 1\ \Omega$, series $L_1 = 2.024$ H, shunt $C_1 = 0.994$ F, series $L_2 = 2.024$ H, $R_2 = 1\ \Omega$. Element rules: series $L$ → series $L/B$ and $B/(L\omega_0^2)$; shunt $C$ → parallel $C/B$ and $B/(C\omega_0^2)$. $$ \begin{aligned} L_1 = 2.024:&\quad L_1' = \frac{2.024}{9.9\times10^4} = 20.444\ \mu\text{H},\quad C_1' = \frac{9.9\times10^4}{2.024\times10^8} = 489.13\ \mu\text{F} \\ C_1 = 0.994:&\quad C' = \frac{0.994}{9.9\times10^4} = 10.040\ \mu\text{F},\quad L' = \frac{9.9\times10^4}{0.994\times10^8} = 995.98\ \mu\text{H} \\ L_2 = 2.024:&\quad L_2' = 20.444\ \mu\text{H},\quad C_2' = 489.13\ \mu\text{F} \end{aligned} $$ ```text Rs=1 [ Z1 ] [ Z3 ] o-/\/\--[ ]---+------[ ]---+-----o | | | + Vs [ Y2 ] RL=1 Vo | | | - o----------------+---------------+-----o ``` | Arm | Band-pass element | |---|---| | Z1 (series) | 20.444 μH in series with 489.13 μF | | Y2 (shunt) | 10.040 μF ∥ 995.98 μH | | Z3 (series) | 20.444 μH in series with 489.13 μF | Check: $1/\sqrt{20.444\ \mu\text{H}\times489.13\ \mu\text{F}} = 10^4$ rad/s. 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. The ladder is series L, shunt C, series L. What are the BP rules for the shunt C? A. Series C' = C/B and L' = B/(C w0^2) B. Parallel C' = C/B and L' = B/(C w0^2) C. Parallel C' = C B and L' = 1/(C B) D. Parallel C' = C/w0 and L' = 1/B Answer: B. A shunt capacitor becomes a shunt parallel LC. (Key point: Rules for this ladder) 2. With w0 = 10^4 and B = 9.9x10^4 rad/s, what does L = 2.024 H become? A. 20.444 uH in parallel with 489.13 uF B. 200 kH in series with 489.13 uF C. 20.444 mH in series with 489.13 mF D. 20.444 uH in series with 489.13 uF Answer: D. L' = 2.024/9.9x10^4 = 20.444 uH and C' = 9.9x10^4/(2.024x10^8) = 489.13 uF. (Key point: Series arm value) 3. What does C = 0.994 F become? A. 10.040 uF in series with 995.98 uH B. 10.040 uF in parallel with 995.98 uH C. 995.98 uF in parallel with 10.040 uH D. 98.4 kF in parallel with 1 H Answer: B. C' = 0.994/9.9x10^4 = 10.04 uF and L' = 9.9x10^4/(0.994x10^8) = 995.98 uH. 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