FDNR RLC equivalent; 4th order LPF at 1 kHz

RLC elements using FDNR (Bruton transformation)

Dividing every impedance by $s$ keeps the voltage transfer function the same and changes each element as follows:

 Element     Impedance    /s          New element
 --[ R ]--     R          R/s         --| C=1/R |--
 --(LLL)--     sL         L           --[ R=L ]--
 --| C |--     1/sC       1/(s^2 C)   --[FDNR D=C]--

So a series R–L–C branch becomes a series C–R–FDNR branch:

 Before:  o--[ R ]--(LLL)--| C |--o
 After:   o--| 1/R |--[ L ]--[D = C]--o

In a lowpass ladder the capacitors are grounded, so the FDNRs are grounded and each is one Antoniou GIC (two op-amps, five impedances $Z_1$ to $Z_5$ in a chain with $Z_5$ grounded, so that $Z_{in} = Z_1Z_3Z_5/(Z_2Z_4)$) with capacitors in positions 1 and 5 and resistors in positions 2, 3, 4, so $D = C^2R_D$.

Design: 4th-order Butterworth LPF, 1 kHz, capacitors 0.01 μF

Given ladder (1 rad/s, 1 Ω): shunt $C_1 = 0.7654$ F, series $L_2 = 1.848$ H, shunt $C_3 = 1.848$ F, series $L_4 = 0.7654$ H, load 1 Ω.

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 $\omega_0 = 2\pi(1000) = 6283.2\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF (all capacitors in the final design), so

$$k_z = \frac{1}{\omega_0 C_t} = \frac{1}{6283.19 \times 1 \times 10^{-8}} = 15915\ \Omega$$ $$\begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned}$$

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 $R_b = R_a + \Sigma R$:

$R_a$ = 1 MΩ, $\Sigma R$ = 41.59 kΩ, so $R_b$ = 1.042 MΩ.

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]-+----[R2]-+----[R4]-+----o Vo
               |         |         |
              [D1]      [D3]    [CL||Rb]
GND ----------------------------------------

Answer: all capacitors 0.01 μF; series resistors 29.41 kΩ and 12.18 kΩ; FDNR GICs with C = 0.01 μF and $R_D$ = 12.18 kΩ ($D_1$) and 29.41 kΩ ($D_3$); $R_a$ = 1 MΩ, $R_b$ = 1.042 MΩ.