GIC simulation of HPF with Kf = 2000

Generalized impedance converter (GIC)

The GIC (Antoniou's generalized impedance converter) is a two-op-amp circuit with five impedances $Z_1 \dots Z_5$ in a chain. Its input impedance is

$$Z_{in} = \frac{Z_1Z_3Z_5}{Z_2Z_4}$$

By choosing which $Z$'s are resistors and which are capacitors, it can simulate a grounded inductor or an FDNR.

 Iin -->
 V1 o-----+---------- A1 input
          |
          Z1
          |
          +---------- A1 output (Va)
          Z2
          |
     (3)  +---------- A1 and A2 inputs
          Z3
          |
          +---------- A2 output (Vb)
          Z4
          |
     (5)  +---------- A2 input
          Z5
          |
         GND

Op-amp A1 has its inputs at nodes 1 and 3; A2 has its inputs at nodes 3 and 5.

Derivation (ideal op-amps: virtual short gives $V_1 = V_3 = V_5 = V$; no input current):

$$\begin{aligned} I_{Z5} &= \frac{V}{Z_5} \quad\text{(flows through } Z_4 \text{ too)} \\ V_b &= V + Z_4\frac{V}{Z_5} \\ I_{Z3} &= \frac{V - V_b}{Z_3} = -\frac{Z_4}{Z_3Z_5}V \quad\text{(flows through } Z_2 \text{ too)} \\ V_a &= V + Z_2 I_{Z3} = V - \frac{Z_2Z_4}{Z_3Z_5}V \\ I_{in} &= \frac{V - V_a}{Z_1} = \frac{Z_2Z_4}{Z_1Z_3Z_5}V \\ Z_{in} &= \frac{V}{I_{in}} = \frac{Z_1Z_3Z_5}{Z_2Z_4} \end{aligned}$$

Simulating a grounded inductor with a GIC

Grounded inductor simulation: make $Z_4 = \dfrac{1}{sC_4}$ and $Z_1, Z_2, Z_3, Z_5$ resistors:

$$Z_{in} = \frac{R_1R_3R_5}{R_2}\,sC_4 = sL, \qquad L = \frac{C_4R_1R_3R_5}{R_2}$$

(Equally, $Z_2$ can be the capacitor, giving $L = C_2R_1R_3R_5/R_4$.) With all resistors equal to $R$, $L = R^2C$. One end of the simulated inductor is ground, so the GIC replaces the shunt (grounded) inductors of an LC ladder, e.g. in a high-pass ladder.

Example: $L = 0.5$ H with $C_4 = 0.1\ \mu$F and $R_1 = R_2 = R_3 = R_5 = R$: $R = \sqrt{L/C} = \sqrt{0.5/10^{-7}} = 2.236\ \text{k}\Omega$.

Active simulation of the given high-pass ladder

The values (C 1.618, L 0.618, C 0.5, L 0.618, C 1.618) are the 5th-order Butterworth high-pass ladder obtained from the low-pass prototype 0.618, 1.618, 2, 1.618, 0.618. Assumption: the prototype is normalized with $R_s = R_L = 1\ \Omega$ (the source resistor is not legible); the 1 kΩ in the figure matches the impedance scaling chosen below.

Scaling: $k_f = 2000$ (rad/s), and choose $k_m = 1000$:

$$R' = k_mR,\qquad L' = \frac{k_m}{k_f}L,\qquad C' = \frac{C}{k_fk_m}$$

Replace each grounded 0.309 H inductor by a GIC with $Z_4 = C_4$: $L = C_4R_1R_3R_5/R_2$. Choose $C_4 = 0.1\ \mu$F and $R_2 = R_3 = R_5 = 1\ \text{k}\Omega$:

$$R_1 = \frac{L R_2}{C_4R_3R_5} = \frac{0.309 \times 1000}{10^{-7}\times10^{6}} = 3.09\ \text{k}\Omega$$

Vs --[1k]--||-----+-----||-----+-----||-----+---- Vout
         0.809uF      0.25uF       0.809uF
                  |            |            |
            [GIC 0.309H] [GIC 0.309H]     [1k]
                  |            |            |
GND --------------+------------+------------+

Final circuit: $R_s = R_L = 1\ \text{k}\Omega$; series capacitors 0.809 μF, 0.25 μF, 0.809 μF; two GICs, each with $R_1 = 3.09\ \text{k}\Omega$, $R_2 = R_3 = R_5 = 1\ \text{k}\Omega$, $C_4 = 0.1\ \mu$F. The filter is a 5th-order Butterworth high-pass with cutoff $\omega_c = 2000$ rad/s (318.3 Hz) and no inductors.