---
title: "Transform given low pass ladder to 2 kHz high pass"
subject: "Filter Design"
chapter: 3
key: "semester_5/filter_design:highpass-2khz-from-ladder"
source: https://www.bctengineer.com/q/semester_5/filter_design/highpass-2khz-from-ladder
site: BCT Engineering
---

# Transform given low pass ladder to 2 kHz high pass

Filter Design · Chapter 3: Frequency Transformation

- Asked in: 2073 Shrawan, 4 marks
- 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/highpass-2khz-from-ladder
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-3

## Question

The following low pass filter has passband frequency ωp of 1 rad/s. Transform it into a highpass filter having passband frequency of 2KHz. [Figure: source Vs, series 3 H inductor, shunt 2/3 F capacitor, series 1 H inductor, 2 Ω load resistor]

## Short answer

LP to HP transformation substitutes s→ωp/s with ωp=2π×2000≈12566 rad/s. Series inductor L becomes series capacitor C=1/(Lωp): 3 H → 26.53 µF, 1 H → 79.58 µF. Shunt capacitor becomes shunt inductor L=1/(Cωp): 2/3 F → 119.4 µH. Load 2 Ω unchanged.

## Answer

For LP → HP we substitute $s \to \omega_p/s$. Then a series inductor $L$ (impedance $sL$) becomes $\omega_p L/s$, i.e. a series capacitor, and a shunt capacitor becomes a shunt inductor. Resistors are unchanged.

$$
C_{new} = \frac{1}{L\,\omega_p}, \qquad L_{new} = \frac{1}{C\,\omega_p}
$$

**Given:** prototype $\omega_p = 1$ rad/s with series $L_1 = 3$ H, shunt $C_2 = 2/3$ F, series $L_3 = 1$ H, $R_L = 2\ \Omega$. New passband edge $f_p = 2$ kHz:

$$
\omega_p = 2\pi\times2000 = 12566.37\ \text{rad/s}
$$

**Element values**

$$
\begin{aligned}
C_1 &= \frac{1}{3\times12566.37} = 26.53\ \mu\text{F} \\
L_2 &= \frac{1}{(2/3)\times12566.37} = 119.37\ \mu\text{H} \\
C_3 &= \frac{1}{1\times12566.37} = 79.58\ \mu\text{F} \\
R_L &= 2\ \Omega\ \text{(unchanged)}
\end{aligned}
$$

| LP element | HP element | Value |
|---|---|---|
| series $L_1 = 3$ H | series $C_1$ | 26.53 µF |
| shunt $C_2 = 2/3$ F | shunt $L_2$ | 119.37 µH |
| series $L_3 = 1$ H | series $C_3$ | 79.58 µF |
| $R_L = 2\ \Omega$ | $R_L$ | 2 Ω |

```text
o-(Vs)-[C1]-+-[C3]-+
            |      |
          [L2]   [RL]
            |      |
o-----------+------+
```

**Answer:** $C_1 = 26.53\ \mu$F, $L_2 = 119.37\ \mu$H, $C_3 = 79.58\ \mu$F, $R_L = 2\ \Omega$; the circuit passes frequencies above 2 kHz. (These values can be impedance scaled later if a larger load resistance is needed.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In LP to HP with s -> wp/s, what happens to a series L and a shunt C?
   A. Series L becomes a shunt inductor; shunt C a series capacitor
   B. Series L becomes a series capacitor; shunt C becomes a shunt inductor
   C. Series L and shunt C swap values only
   D. Both become resistors
   Answer: B. sL -> wp L/s is a capacitor impedance, and sC -> wp C/s is an inductor admittance. (Key point: HP substitution and elements)

2. What wp is used for 2 kHz and what is the formula for the new capacitor from series L?
   A. wp = 2000 rad/s; C = 1/(L wp)
   B. wp = 12566.37 rad/s; C = L wp
   C. wp = 2 pi x 2000 rad/s; C = L/wp
   D. wp = 2 pi x 2000 = 12566.37 rad/s; C = 1/(L wp)
   Answer: D. Hz must be converted to rad/s and C = 1/(L wp). (Key point: Frequency in rad/s)

3. What are C1 (from 3 H) and L2 (from 2/3 F)?
   A. C1 = 79.58 uF and L2 = 119.37 uH
   B. C1 = 26.53 uF and L2 = 119.37 uH
   C. C1 = 26.53 uF and L2 = 79.58 uH
   D. C1 = 3 F and L2 = 2/3 H
   Answer: B. C1 = 1/(3x12566.37) = 26.53 uF; L2 = 1/((2/3)x12566.37) = 119.37 uH. (Key point: Element values)

## Related in Filter Design

- [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
- [Band pass (400 rad/s, 150 rad/s) from 1 dB ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-400-150-from-ladder) — Filter Design, chapter 3
- [Scaling a ladder to 1000 rad/s and 75 Ω load](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ascaling-ladder-75-ohm-numerical) — Filter Design, chapter 1
- [Leapfrog simulation of given ladder at 6 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-ladder-6khz) — Filter Design, chapter 8
- [Band pass (10⁴ rad/s, 9.9×10⁴ rad/s) from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-from-figure1-ladder) — Filter Design, chapter 3

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