---
title: "RC-CR transformation with examples"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:rc-cr-transformation-examples"
source: https://www.bctengineer.com/q/semester_5/filter_design/rc-cr-transformation-examples
site: BCT Engineering
---

# RC-CR transformation with examples

Filter Design · Chapter 6: Active Filters

- Asked in: 2073 Chaitra, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/rc-cr-transformation-examples
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Explain RC-CR transformation with suitable examples.

## Short answer

RC-CR transformation converts a low-pass to high-pass filter by replacing each resistor Ri with a capacitor Ci'=1/Ri, and each capacitor Cj with a resistor Rj'=1/Cj, leaving gain-setting resistors unchanged. The transfer function becomes THP(s)=TLP(1/s). Examples: first-order LP becomes HP; Sallen-Key LP becomes HP with same Q and reciprocal normalized ωo.

## Answer

The **RC-CR transformation** converts an active RC low-pass filter into a high-pass filter (and vice versa) by interchanging resistors and capacitors.

**Rule (normalized values, $\omega_o = 1$):**
- Each resistor $R_i$ becomes a capacitor $C_i' = 1/R_i$.
- Each capacitor $C_j$ becomes a resistor $R_j' = 1/C_j$.
- Resistors that only set a dimensionless gain ratio (op-amp gain resistors $R_A, R_B$) are **not** changed.

**Why it works:** a resistor's admittance $1/R_i$ becomes $sC_i' = s/R_i$, and a capacitor's admittance $sC_j$ becomes $1/R_j' = C_j$. In both cases the new admittance is $s\,Y(1/s)$, where $Y(s)$ is the old one. The transfer function depends only on ratios of admittances, so the extra factor $s$ cancels and only the replacement $s \to 1/s$ remains:
$$
T_{HP}(s) = T_{LP}\left(\frac{1}{s}\right)
$$
Every pole $p$ moves to $1/p$. So $Q$ is unchanged, and $\omega_o$ becomes $1/\omega_o$ (for a normalized filter with $\omega_o = 1$ it stays 1). After the transformation, frequency scaling moves the response to the wanted frequency.

**Example 1: first-order RC low-pass.** Series $R = 2\ \Omega$, shunt $C = 1$ F:
$$
T_{LP} = \frac{1}{1 + sRC} = \frac{0.5}{s + 0.5}
$$
Transformation: the series resistor becomes a series capacitor $C' = 1/R = 0.5$ F, and the shunt capacitor becomes a shunt resistor $R' = 1/C = 1\ \Omega$:
$$
T_{HP} = \frac{sR'C'}{1 + sR'C'} = \frac{0.5s}{1 + 0.5s} = \frac{s}{s + 2} = T_{LP}\left(\frac{1}{s}\right)
$$
The pole moves from $-0.5$ to $-2$, the reciprocal of its magnitude. For the normalized case $R = C = 1$ the result is $1/(s+1) \to s/(s+1)$.

**Example 2: Sallen-Key low-pass to high-pass**
```text
LP:  Vin-R1-+-R2-+-->(+)       C1 from node to Vo
            |    C2 to ground

HP:  Vin-C1'-+-C2'-+-->(+)     R1' from node to Vo
             |     R2' to ground
```
For the normalized circuit:
$$
\frac{K}{s^2 + \frac{1}{Q}s + 1} \;\Rightarrow\; \frac{K s^2}{s^2 + \frac{1}{Q}s + 1}
$$
With equal-element values ($R = C = 1$, $K = 3 - 1/Q$), the HP circuit also has $C' = R' = 1$ and the same $K$. Frequency scaling then gives $\dfrac{Ks^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$.

**Example 3: band-pass.** For $T_{BP}(s) = \dfrac{(\omega_o/Q)s}{s^2 + (\omega_o/Q)s + \omega_o^2}$, replacing $s$ by $1/s$ gives
$$
\frac{\frac{1}{Q\omega_o}\,s}{s^2 + \frac{1}{Q\omega_o}s + \frac{1}{\omega_o^2}}
$$
This is another band-pass response with the same $Q$ and the same peak gain, but its centre frequency is $1/\omega_o$. For a normalized circuit ($\omega_o = 1$) the response is unchanged, so a band-pass circuit maps into another band-pass circuit.

## Check yourself

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

1. What does the RC-CR transformation do to each element (normalized)?
   A. R and C are both doubled
   B. R becomes inductor 1/R
   C. R becomes capacitor 1/R; C becomes resistor 1/C
   D. R becomes C = R; C becomes R = C
   Answer: C. Elements are swapped with inverted values. (Key point: Rule)

2. What does the transformation do to s and the filter type?
   A. Replaces s by s^2: low-pass becomes band-pass
   B. Replaces s by -s: no change
   C. Changes wo and Q
   D. Replaces s by 1/s: low-pass becomes high-pass
   Answer: D. T_HP(s) = T_LP(1/s) with the same wo and Q. (Key point: Effect)

3. Which elements are NOT transformed?
   A. Gain-ratio resistors RA, RB
   B. Capacitors
   C. Input resistors
   D. The op-amp supply
   Answer: A. They set only a dimensionless ratio. (Key point: Unchanged elements)

4. What happens to a band-pass circuit?
   A. Becomes high-pass
   B. Maps to a band-pass with same centre frequency and Q
   C. Becomes low-pass
   D. Becomes all-pass
   Answer: B. BP is unchanged under s -> 1/s. (Key point: Bandpass)

## Related in Filter Design

- [RC-CR transformation of Sallen-Key LPF to HPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Arc-cr-transformation-sallen-key) — Filter Design, chapter 6, asked 2×