RC-CR transformation with examples

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

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.