Transform given low pass ladder to 2 kHz high pass

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

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

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.)