GIC HPF design at 5 kHz with 10 nF

A generalized impedance converter (GIC) is an active two-port, usually Antoniou's two-op-amp, five-impedance circuit. Its input impedance is the load impedance $Z_5$ multiplied by a conversion function $k(s) = Z_1Z_3/(Z_2Z_4)$. By choosing which positions hold capacitors, a GIC can make an inductor or an FDNR from only R, C and op-amps.

Simulating a grounded inductor with a GIC

   Iin  node 1
 o-->-----+--------------(+) A1
 Vin      |
         [Z1]
          +--- node 2 ---- output of A1
         [Z2]
          +--- node 3 ---- (-) A1 and (+) A2
         [Z3]
          +--- node 4 ---- output of A2
         [Z4]
          +--- node 5 ---- (-) A2
         [Z5]
          |
         GND

Each op-amp has negative feedback (A1 through $Z_2$, A2 through $Z_4$). With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$.

$$\begin{aligned} \text{Node 5: } & \frac{V_4 - V_{in}}{Z_4} = \frac{V_{in}}{Z_5} \;\Rightarrow\; V_4 = V_{in}\left(1 + \frac{Z_4}{Z_5}\right) \\ \text{Node 3: } & \frac{V_2 - V_{in}}{Z_2} = \frac{V_{in} - V_4}{Z_3} = -\frac{Z_4}{Z_3 Z_5}V_{in} \;\Rightarrow\; V_2 = V_{in}\left(1 - \frac{Z_2 Z_4}{Z_3 Z_5}\right) \\ \text{Node 1: } & I_{in} = \frac{V_{in} - V_2}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3 Z_5}V_{in} \end{aligned}$$ $$Z_{in} = \frac{V_{in}}{I_{in}} = \frac{Z_1 Z_3 Z_5}{Z_2 Z_4}$$

Put a capacitor in position 4 ($Z_4 = 1/sC$) and resistors elsewhere:

$$Z_{in} = \frac{R_1 R_3 R_5}{R_2}\, sC = sL, \qquad L = \frac{C R_1 R_3 R_5}{R_2}$$

With all resistors equal to $R$: $L = CR^2$. One end of this inductor is ground, so it replaces a grounded (shunt) inductor.

Design

The ladder is the 5th-order Butterworth highpass prototype (1 rad/s, 1 Ω): series 1.618 F, shunt 0.618 H, series 0.5 F, shunt 0.618 H, series 1.618 F. The shunt inductors are grounded and are simulated by GICs.

Frequency scale to $\omega_0 = 2\pi(5000) = 31{,}416\ \text{rad/s}$ and impedance scale by $k_z$ (chosen so that the largest capacitor, 1.618 F, becomes 10 nF: $k_z = 1.618/(\omega_0 \times 10^{-8})$):

$$R_{new} = k_z R, \qquad L_{new} = \frac{k_z L}{\omega_0}, \qquad C_{new} = \frac{C}{k_z\,\omega_0}$$

$k_z = 5150.3\ \Omega$, so the source and load resistors become $R_s = R_L$ = 5.15 kΩ.

Each grounded inductor is replaced by a GIC with $Z_4 = C$ and $R_1 = R_2 = R_3 = R_5 = R$, so $L = CR^2$ and $R = \sqrt{L/C}$.

Final circuit (each GIC realises the grounded inductor in that position):

Vi o--[Rs]--[C1]-+----[C3]-+----[C5]-+----o Vo
                 |         |         |
              [GIC-L2]  [GIC-L4]    [RL]
GND ------------------------------------------

Answer: $R_s = R_L$ = 5.15 kΩ; capacitors 10 nF, 3.09 nF, 10 nF; each 101.3 mH inductor = GIC with C = 10 nF and R = 3.183 kΩ; half-power frequency 5 kHz.