Design of High-Order Active Filters

GIC (Antoniou's 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

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 $R_1 = R_2 = R_3 = R_5 = R$: $R = \sqrt{L/C} = \sqrt{0.5/10^{-7}} = 2.236\ \text{k}\Omega$.

Advantages: the realized inductor has high Q and low sensitivity to op-amp gain; it uses only R, C and op-amps, so the low-sensitivity property of the passive LC ladder is kept.

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-+-0.809uF-+-0.25uF-+-0.809uF-+--+
      |         |        |         |  |
              [GIC]            [GIC] 1k
              0.309H           0.309H |
                |                |    |
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.

Ideal gyrator

An ideal gyrator is a lossless two-port that converts a voltage at one port into a proportional current at the other. With gyration conductance $g$ (gyration resistance $r = 1/g$):

$$I_1 = g V_2, \qquad I_2 = -g V_1$$

 I1 ->                    <- I2
 o------+  +---------+  +------o
  V1    |--| gyrator |--|    V2
 o------+  |   g     |  +------o
           +---------+

It is non-reciprocal and inverts impedance: a load $Z_L$ at port 2 appears at port 1 as $Z_{in} = \dfrac{1}{g^2Z_L} = \dfrac{r^2}{Z_L}$.

Simulating an inductor

Terminate port 2 with a capacitor $C$. The load current is $sCV_2$, and it leaves port 2, so $I_2 = -sCV_2$.

$$\begin{aligned} -sCV_2 &= -gV_1 \;\Rightarrow\; V_2 = \frac{g}{sC}V_1 \\ I_1 &= gV_2 = \frac{g^2}{sC}V_1 \\ Z_{in} &= \frac{V_1}{I_1} = s\frac{C}{g^2} = s\,r^2C \end{aligned}$$

So the input looks like an inductor

$$L_{eq} = \frac{C}{g^2} = r^2C$$

 o----+  +---------+  +----+
      |--| gyrator |--|   === C   => L = r^2 C
 o----+  +---------+  +----+

Example: $r = 10\ \text{k}\Omega$, $C = 0.1\ \mu$F gives $L = 10^8 \times 10^{-7} = 10$ H.

Practical realization: a gyrator is built with two voltage-controlled current sources of opposite sign (transconductance $\pm g$), or with op-amps (e.g. the Riordan circuit, a form of GIC).

Grounded vs floating: one gyrator with a grounded C gives a grounded inductor. A floating inductor needs two gyrators connected back to back with a grounded capacitor between them, giving $L = r^2C$ between the two outer terminals.

Simulated inductors are small, have high Q and can be integrated, so a passive LC ladder can be converted to an active RC filter while keeping its low sensitivity.

Bruton transformation

The Bruton transformation scales every impedance of an LC ladder by $1/s$ (i.e. by $k/s$). The transfer function (a ratio of impedances) does not change, but the element types change:

The FDNR (frequency-dependent negative resistance) has impedance $Z = \dfrac{1}{s^2D}$, which is $-\dfrac{1}{\omega^2D}$ on the $j\omega$ axis. All inductors become resistors and the shunt capacitors become grounded FDNRs, which are easy to build with a GIC.

Design: 4th-order Butterworth LPF, $\omega_c = 2000$ rad/s

Prototype (table, $R_s = R_L = 1\ \Omega$): series $L_1 = 0.7654$, shunt $C_2 = 1.848$, series $L_3 = 1.848$, shunt $C_4 = 0.7654$.

After the Bruton transformation (normalized):

• $R_s = 1\ \Omega \to C_s = 1$ F; $R_L = 1\ \Omega \to C_L = 1$ F
• $L_1 \to R_1 = 0.7654\ \Omega$; $L_3 \to R_3 = 1.848\ \Omega$
• $C_2 \to D_2 = 1.848$; $C_4 \to D_4 = 0.7654$

Scaling: $k_f = 2000$, choose $k_m = 5000$ (so $C_s$ becomes 0.1 μF):

$$\begin{aligned} C' &= \frac{C}{k_fk_m} = \frac{1}{2000 \times 5000} = 0.1\ \mu\text{F} \\ R_1' &= 0.7654 \times 5000 = 3.827\ \text{k}\Omega, \quad R_3' = 1.848 \times 5000 = 9.24\ \text{k}\Omega \\ D' &= \frac{D}{k_f^2k_m}: \quad D_2' = \frac{1.848}{2\times10^{10}} = 9.24\times10^{-11},\quad D_4' = 3.827\times10^{-11}\ \text{F·s} \end{aligned}$$

FDNR by GIC: in $Z_{in} = \dfrac{Z_1Z_3Z_5}{Z_2Z_4}$, make $Z_1 = \dfrac{1}{sC_1}$, $Z_3 = \dfrac{1}{sC_3}$ and the rest resistors:

$$Z_{in} = \frac{R_5}{s^2C_1C_3R_2R_4} \;\Rightarrow\; D = \frac{C_1C_3R_2R_4}{R_5}$$

Take $C_1 = C_3 = 0.1\ \mu$F and $R_2 = R_5 = 10\ \text{k}\Omega$, so $D = C^2R_4$:

$$R_4 = \frac{D}{C^2}: \quad D_2' \Rightarrow R_4 = \frac{9.24\times10^{-11}}{10^{-14}} = 9.24\ \text{k}\Omega, \quad D_4' \Rightarrow R_4 = 3.827\ \text{k}\Omega$$

Vin-+-||-+-3.83k-+--9.24k-+--+-- Vout
    |0.1uF       |        |  |
             [FDNR D2] [FDNR D4] === 0.1uF
                 |        |  |
GND -------------+--------+--+

DC path: the series 0.1 μF source capacitor blocks DC bias for the op-amps, so add large bleeding resistors (e.g. 1 MΩ) across the source capacitor and from output to ground. With equal values they keep the DC gain at about 0.5, as in the original ladder.

Answer: the inductorless filter has half-power frequency 2000 rad/s, a 4th-order Butterworth response and only practical R, C and op-amp elements.

Leapfrog (operational) simulation realises the doubly terminated LC ladder by simulating its equations rather than its elements: each reactive element is replaced by an op-amp integrator, and the integrators are coupled exactly as the ladder voltages and currents are coupled. The result keeps the low sensitivity of the passive ladder.

From the table (n = 4), the doubly terminated Butterworth prototype (1 Ω terminations, $\omega_0 = 1$ rad/s) is $R_s = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_L = 1\ \Omega$ (T-ladder: series L first, shunt C to ground).

Write the state equations of the doubly terminated ladder (normalised, $R_s = R_L = 1\ \Omega$). Every current is turned into a voltage by multiplying by a 1 Ω resistance, so each equation becomes an integrator:

$$\begin{aligned} I_1 &= \frac{V_{in} - V_2}{sL_1 + 1} \\ V_2 &= \frac{I_1 - I_3}{sC_2} \\ I_3 &= \frac{V_2 - V_4}{sL_3} \\ V_4 &= \frac{I_3}{sC_4 + 1} \\ V_o &= V_4\cdot R_L \end{aligned}$$

The first and last equations are lossy integrators (the terminations give the "+1"); the middle ones are lossless integrators. Neighbouring integrators feed each other in alternate directions, which gives the "leapfrog" structure.

Using inverting op-amp integrators, $V_o = -\frac{1}{sRC}\sum V_i$ (lossless) or $V_o = -\frac{\sum V_i/R}{sC + 1/R}$ (lossy, resistor $R$ across $C$), the signs are arranged as follows. Inverters supply the signals of opposite sign:

Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(1.848)}$, which is the ladder equation for $V_2$.

Denormalisation. Frequency scale by $\omega_0 = 1000\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF} \\ C_{A2} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\ C_{A4} &= \frac{1.848}{1000 \times 10^4} = \text{184.8 nF} \\ C_{A5} &= \frac{0.7654}{1000 \times 10^4} = \text{76.54 nF} \end{aligned}$$

Final element values:

 Lossy integrator (A1, last stage)
           +----[ R ]----+
           +----| C |----+
 Va--[R]--+|             |
 Vb--[R]--++----(-)      |
                (+) A ---+---- Vo
                 |
                GND
 Lossless integrator: same circuit without the
 feedback resistor (only C from output to (-)).

The passband gain is 1/2 (−6 dB), the same as the doubly terminated prototype; the response is that of the passive ladder but with no inductors, and the low sensitivity of the ladder is kept.

Answer: six op-amps (four integrators, two inverters), all resistors 10 kΩ, integrator capacitors 76.54 nF, 184.8 nF, 184.8 nF and 76.54 nF; half-power frequency 1000 rad/s.

Leapfrog (operational) simulation realises the doubly terminated LC ladder by simulating its equations rather than its elements: each reactive element is replaced by an op-amp integrator, and the integrators are coupled exactly as the ladder voltages and currents are coupled. The result keeps the low sensitivity of the passive ladder.

Given prototype (normalised to 1 rad/s, 1 Ω terminations): $R_1 = 1$, $L_1 = 0.7654$ H, $C_2 = 1.848$ F, $L_3 = 1.848$ H, $C_4 = 0.7654$ F, $R_2 = 1\ \Omega$.

 Vin  1 Ohm  L1=0.7654   L3=1.848
 o---[R1]---(LLL)--+---(LLL)--+------o Vo
                   |          |    |
               C2=1.848  C4=0.7654 [R2=1]
                   |          |    |
 GND --------------+----------+----+

Write the state equations of the doubly terminated ladder (normalised, $R_s = R_L = 1\ \Omega$). Every current is turned into a voltage by multiplying by a 1 Ω resistance, so each equation becomes an integrator:

$$\begin{aligned} I_1 &= \frac{V_{in} - V_2}{sL_1 + 1} \\ V_2 &= \frac{I_1 - I_3}{sC_2} \\ I_3 &= \frac{V_2 - V_4}{sL_3} \\ V_4 &= \frac{I_3}{sC_4 + 1} \\ V_o &= V_4\cdot R_L \end{aligned}$$

The first and last equations are lossy integrators (the terminations give the "+1"); the middle ones are lossless integrators. Neighbouring integrators feed each other in alternate directions, which gives the "leapfrog" structure.

Using inverting op-amp integrators, $V_o = -\frac{1}{sRC}\sum V_i$ (lossless) or $V_o = -\frac{\sum V_i/R}{sC + 1/R}$ (lossy, resistor $R$ across $C$), the signs are arranged as follows. Inverters supply the signals of opposite sign:

Check of one stage: A2 gives $-\frac{(-I_1) + I_3}{sC_{A2}} = \frac{I_1 - I_3}{s(1.848)}$, which is the ladder equation for $V_2$.

Denormalisation. Frequency scale by $\omega_0 = 2\pi(1000) = 6283.2\ \text{rad/s}$ and choose all resistors $R$ = 10 kΩ (impedance scale $k_z = 10^4$). Each capacitor becomes $C = g/(\omega_0 R)$:

$$\begin{aligned} C_{A1} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \\ C_{A2} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A4} &= \frac{1.848}{6283.19 \times 10^4} = \text{29.41 nF} \\ C_{A5} &= \frac{0.7654}{6283.19 \times 10^4} = \text{12.18 nF} \end{aligned}$$

Final element values:

 Lossy integrator (A1, last stage)
           +----[ R ]----+
           +----| C |----+
 Va--[R]--+|             |
 Vb--[R]--++----(-)      |
                (+) A ---+---- Vo
                 |
                GND
 Lossless integrator: same circuit without the
 feedback resistor (only C from output to (-)).

The passband gain is 1/2 (−6 dB), the same as the doubly terminated prototype; the response is that of the passive ladder but with no inductors, and the low sensitivity of the ladder is kept.

Since no frequency is given, the normalised design (all R = 1 Ω, C = 0.7654, 1.848, 1.848, 0.7654 F) is the answer; the 1 kHz values above show how it is made practical.

Answer (1 kHz example): R = 10 kΩ everywhere, capacitors 12.18 nF, 29.41 nF, 29.41 nF, 12.18 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.

GIC circuit and input impedance

   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

With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$.

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

Avoiding shunt inductors

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.

In an LC ladder the shunt inductors (for example in a highpass ladder, where every inductor goes from a node to ground) are exactly such grounded inductors. The procedure is:

1. Obtain the passive ladder (prototype values, then frequency and impedance scaling).
2. Keep the resistors and capacitors as they are.
3. Replace each shunt inductor $L$ by a GIC: choose $C$ (e.g. 10 nF) and $R = \sqrt{L/C}$.

 o--[Rs]--||--+--||--+--||--o Vo
              |      |      |
            [GIC]  [GIC]   [RL]
              |      |      |
 GND ---------+------+------+

Example: a shunt inductor of 0.1 H with $C$ = 10 nF needs $R = \sqrt{0.1/10^{-8}}$ = 3.162 kΩ in positions 1, 2, 3 and 5.

Advantages: no bulky, lossy inductors; the ladder's low sensitivity is kept; elements are easy to integrate. Floating (series) inductors cannot be replaced by one grounded GIC; for those, use two GICs or the Bruton/FDNR method.

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.

Input impedance of the 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

With ideal op-amps the two input terminals of each op-amp are at the same voltage and draw no current, so $V_1 = V_3 = V_5 = V_{in}$.

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

Grounded inductor

With $Z_4 = 1/sC$ and all other positions resistors $R$: $Z_{in} = sCR^2$, i.e. $L = CR^2$. This works only when one end of the inductor is grounded.

Floating inductor using two GICs

A series (floating) inductor, as in a lowpass ladder, has neither end grounded. It is simulated by two identical GICs connected back to back with a resistor $R_x$ between their port-2 terminals:

  a o--[GIC 1]--o x --[ Rx ]-- y o--[GIC 2]--o b
      port1  port2             port2  port1
      (GIC 2 is the same circuit, reversed)

Treat the GIC with $Z_5$ removed as a two-port (port 2 at node 5). From the analysis above, $V_1 = V_2$ and $I_1 = I_2/k(s)$, with $k(s) = Z_1Z_3/(Z_2Z_4)$. With $Z_4 = 1/sC$ and equal resistors, $k(s) = sCR$. In transmission (ABCD) form:

$$\text{GIC 1: } \begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}, \quad \text{series } R_x: \begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}, \quad \text{GIC 2 reversed: } \begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 1/k \end{bmatrix}\begin{bmatrix} 1 & R_x \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} = \begin{bmatrix} 1 & kR_x \\ 0 & 1 \end{bmatrix}$$

This is the matrix of a single series (floating) impedance $Z = k(s)R_x = s\,(CRR_x)$, so between a and b we get a floating inductor

$$L = C R R_x$$

Example: $C$ = 10 nF, $R = R_x$ = 10 kΩ gives $L = 10^{-8} \times 10^8$ = 1 H. The two GICs must be well matched; a mismatch leaves a small unwanted grounded element. (The Bruton/FDNR method is the usual alternative for lowpass ladders.)

GIC circuit

The generalized impedance converter (GIC) (Antoniou circuit) uses two op-amps and five impedances in a chain from the input to ground:

   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

Relation between input and output current

Treat node 1 as port 1 ($V_1$, $I_1$) and node 5 as port 2, where the load $Z_5$ draws current $I_2$ ($V_2 = I_2 Z_5$). Ideal op-amps force node 1, node 3 and node 5 to the same voltage and draw no input current, so $V_1 = V_2$. Let $V_{N2}$ and $V_{N4}$ be the voltages of nodes 2 and 4 (op-amp outputs).

$$\begin{aligned} \text{Node 5: } & V_{N4} = V_2 + Z_4 I_2 \\ \text{Node 3: } & \frac{V_{N2} - V_1}{Z_2} = \frac{V_1 - V_{N4}}{Z_3} = -\frac{Z_4}{Z_3}I_2 \\ \text{Node 1: } & I_1 = \frac{V_1 - V_{N2}}{Z_1} = \frac{Z_2 Z_4}{Z_1 Z_3}\, I_2 \end{aligned}$$

So the GIC passes voltage unchanged and scales current:

$$V_1 = V_2, \qquad I_1 = \frac{Z_2Z_4}{Z_1Z_3}\, I_2, \qquad Z_{in} = \frac{V_1}{I_1} = \frac{Z_1Z_3}{Z_2Z_4}\,Z_5 = k(s)\,Z_5$$

Grounded FDNR

Put capacitors in positions 1 and 5 ($Z_1 = 1/sC_1$, $Z_5 = 1/sC_5$) and resistors in 2, 3, 4:

$$Z_{in} = \frac{R_3}{s^2 C_1 C_5 R_2 R_4} = \frac{1}{s^2 D}, \qquad D = \frac{C_1 C_5 R_2 R_4}{R_3}$$

With $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R$: $D = C^2 R$. At $s = j\omega$, $Z_{in} = -1/(\omega^2 D)$, a negative resistance whose value depends on frequency: an FDNR (also called a D-element or supercapacitor).

Practical design: choose $C$ (e.g. 10 nF), then $R = D/C^2$. Example: $D = 10^{-12}\ \text{F}^2\Omega$ with $C$ = 10 nF gives $R = 10^{-12}/10^{-16}$ = 10 kΩ. Because port 1 is referred to ground, this FDNR is grounded, which is exactly what the Bruton-transformed lowpass ladder needs (shunt capacitors become grounded FDNRs).

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

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 given ladder is a 5th-order Butterworth highpass prototype ($\omega_0 = 1$ rad/s, 1 Ω terminations): series 1.618 F, shunt 0.618 H, series 0.5 F, shunt 0.618 H, series 1.618 F. Both inductors are grounded, so each is replaced by a GIC.

Frequency scale to $\omega_0 = 4000\ \text{rad/s}$ and impedance scale by $k_z$ (chosen as $10^4$ so that capacitors are in the nF range):

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

$k_z = 10000\ \Omega$, so the source and load resistors become $R_s = R_L$ = 10 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: $R_s = R_L$ = 10 kΩ; series capacitors 40.45 nF, 12.5 nF, 40.45 nF; each simulated inductor 1.545 H = GIC with C = 10 nF and four resistors of 12.43 kΩ; $\omega_0$ = 4000 rad/s.

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

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.

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

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 given ladder is a 5th-order Butterworth highpass (1 rad/s, 1 Ω): shunt 1.618 H, series 0.618 F, shunt 0.5 H, series 0.618 F, shunt 1.618 H. All three inductors are grounded, so three GICs are needed.

Frequency scale to $\omega_0 = 2\pi(4500) = 28{,}274\ \text{rad/s}$ and impedance scale by $k_z$ (chosen so that the largest capacitor, 0.618 F, becomes 0.1 μF: $k_z = 0.618/(\omega_0 \times 10^{-7})$):

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

$k_z = 218.57\ \Omega$, so the source and load resistors become $R_s = R_L$ = 218.6 Ω.

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]-+----[C2]-+----[C4]-+-------+----o Vo
           |         |         |       |
        [GIC-L1]  [GIC-L3]  [GIC-L5]  [RL]
GND --------------------------------------------

Answer: $R_s = R_L$ = 218.6 Ω; series capacitors 0.1 μF each; simulated inductors 12.51 mH, 3.865 mH, 12.51 mH, realised by GICs with C = 0.1 μF and R = 353.7 Ω, 196.6 Ω, 353.7 Ω; half-power frequency 4.5 kHz.