FDNR 4th order Butterworth LPF at 4000 rad/s

The design uses the Bruton transformation to remove the inductors of the doubly terminated Butterworth ladder; the shunt capacitors become grounded FDNRs ($Z = 1/s^2D$), each realised with a GIC.

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, $R_L = 1\ \Omega$ (series L first, shunt C to ground).

FDNR from a GIC. Each FDNR is made with an Antoniou GIC (two op-amps, five impedances $Z_1$ to $Z_5$ in a chain with $Z_5$ grounded, so that $Z_{in} = Z_1Z_3Z_5/(Z_2Z_4)$). 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).

Step 1: Bruton transformation (divide all impedances by $s$): source and load resistors (1 Ω) become 1 F capacitors, each inductor $L$ becomes a resistor of $L$ Ω, each shunt capacitor $C$ becomes a grounded FDNR with $D = C$.

Step 2: Scaling. Frequency scaling factor $\omega_0 = 4000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 100 nF, so

$$k_z = \frac{1}{\omega_0 C_t} = \frac{1}{4000 \times 1 \times 10^{-7}} = 2500\ \Omega$$ $$\begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned}$$

Step 3: Element values. Each FDNR is a GIC with $C_1 = C_5 = C$ and $R_2 = R_3 = R_4 = R_D$, so $D = C^2R_D$. Taking $C = C_t$ gives $R_D = D_{new}/C_t^2 = k_z D$.

Step 4: DC path. The source and load capacitors block DC, and the op-amps in the GICs need a DC bias path. Put $R_a$ across the source capacitor and $R_b$ across the load capacitor. They must be much larger than $k_z$ (the capacitor impedance at $\omega_0$). At DC the FDNRs are open, so the DC gain is $R_b/(R_a + \Sigma R + R_b)$. To keep the prototype DC gain of 1/2, set $R_b = R_a + \Sigma R$:

$R_a$ = 100 kΩ, $\Sigma R$ = 6.534 kΩ, so $R_b$ = 106.5 kΩ.

Final circuit ($C_s \| R_a$ in series from the source, $C_L \| R_b$ as the load, each $D_k$ a GIC FDNR to ground):

Vi o--[Cs||Ra]--[R1]-+----[R3]-+-------+----o Vo
                     |         |       |
                    [D2]      [D4]  [CL||Rb]
GND --------------------------------------------

Answer: terminating capacitors 0.1 μF; series resistors 1.914 kΩ and 4.62 kΩ; FDNRs: $D_2$ = GIC (C = 0.1 μF, $R_D$ = 4.62 kΩ), $D_4$ = GIC (C = 0.1 μF, $R_D$ = 1.914 kΩ); $R_a$ = 100 kΩ, $R_b$ = 106.5 kΩ.