3rd order Butterworth HPF from LPF transfer function

Given: $T_{LP}(s) = \dfrac{1}{(s+1)(s^2+s+1)} = \dfrac{1}{s^3+2s^2+2s+1}$ (3rd order Butterworth, $\omega_c = 1$ rad/s), $R_1 = R_2 = 1\ \Omega$.

Step 1: High-pass transmission coefficient

Apply the LP to HP transformation $s \to 1/s$:

$$t_{HP}(s) = \frac{1}{(1/s)^3 + 2(1/s)^2 + 2(1/s) + 1} = \frac{s^3}{s^3+2s^2+2s+1}, \quad |t_{HP}|^2 = \frac{\omega^6}{1+\omega^6}$$

Step 2: Reflection coefficient

$$|\rho|^2 = 1 - |t_{HP}|^2 = \frac{1}{1+\omega^6} \;\Rightarrow\; \rho(s) = \frac{1}{s^3+2s^2+2s+1}$$

Step 3: Input impedance

$$Z_1 = R_1\frac{1+\rho}{1-\rho} = \frac{s^3+2s^2+2s+2}{s^3+2s^2+2s} = \frac{s^3+2s^2+2s+2}{s(s^2+2s+2)}$$

Step 4: Continued fraction about $s = 0$ (removing poles at $s = 0$)

$$\begin{aligned} Z_1 &= \frac{1}{s} + \frac{s(s+1)}{s^2+2s+2} && \Rightarrow \text{series } C_1 = 1\ \text{F} \\ Y' &= \frac{s^2+2s+2}{s(s+1)} = \frac{2}{s} + \frac{s}{s+1} && \Rightarrow \text{shunt } L_2 = \tfrac12 = 0.5\ \text{H} \\ Z'' &= \frac{s+1}{s} = \frac{1}{s} + 1 && \Rightarrow \text{series } C_3 = 1\ \text{F},\ R_2 = 1\ \Omega \end{aligned}$$

So

$$Z_1 = \frac{1}{s} + \cfrac{1}{\dfrac{2}{s} + \cfrac{1}{\dfrac{1}{s} + 1}}$$

The same result follows by designing the LP ladder ($L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H) and replacing each $L$ by a capacitor $1/L$ and each $C$ by an inductor $1/C$.

Step 5: Circuit

  R1=1    C1=1 F        C3=1 F
 o-/\/\----||----+----||----+
 |               |          |
(~) V1        L2=0.5 H    R2=1
 |               |          |
 o---------------+----------+

Answer: $R_1 = 1\ \Omega$, series $C_1 = 1$ F, shunt $L_2 = 0.5$ H, series $C_3 = 1$ F, $R_2 = 1\ \Omega$; $V_2/V_1 = \dfrac{0.5\,s^3}{s^3+2s^2+2s+1}$.