Bruton transformation; FDNR 4th order LPF at 20,000 rad/s

Bruton transformation

Bruton transformation: divide every impedance of the RLC ladder by $s$ (more generally by $ks$). A voltage transfer function is a ratio of impedances, so it does not change:

In a lowpass ladder the inductors are in the series arms (often floating) and the capacitors are shunt (grounded). After the transformation the inductors become plain resistors and the grounded capacitors become grounded FDNRs, which one GIC can realise. So no inductor is left.

Design: 4th-order Butterworth LPF, $\omega_0$ = 20,000 rad/s

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).

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 $\omega_0 = 20{,}000\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so

$$k_z = \frac{1}{\omega_0 C_t} = \frac{1}{20000 \times 1 \times 10^{-8}} = 5000\ \Omega$$ $$\begin{aligned} R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2} \end{aligned}$$

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, $\Sigma R$ = 13.07 kΩ, so $R_b$ = 113.1 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 --------------------------------------------

Building each FDNR with a GIC

In the Antoniou GIC, $Z_{in} = \dfrac{Z_1Z_3Z_5}{Z_2Z_4}$. 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).

Answer: source and load capacitors 10 nF; series resistors 3.827 kΩ and 9.24 kΩ; FDNRs $D_2$ (GIC with C = 10 nF, $R_D$ = 9.24 kΩ) and $D_4$ (C = 10 nF, $R_D$ = 3.827 kΩ); $R_a$ = 100 kΩ, $R_b$ = 113.1 kΩ.