---
title: "FDNR to avoid inductors; design from Figure 1 at 10⁴ rad/s"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:fdnr-figure1-ladder-1e4"
source: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-figure1-ladder-1e4
site: BCT Engineering
---

# FDNR to avoid inductors; design from Figure 1 at 10⁴ rad/s

Filter Design · Chapter 8: Design of High-Order Active Filters

- Asked in: 2073 Chaitra, 5+5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/fdnr-figure1-ladder-1e4
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

What is frequency dependent negative resistor (FDNR)? How can it be used to avoid inductors in Lowpass LC ladder circuit? Explain. From the circuit given in figure 1 design the lowpass filter having ωo = 10⁴ rad/s and practical element values using FDNR. [Figure 1: source V1, R1 = 1 Ω, series L1 = 2.024 H, shunt C1 = 0.994 F, series L2 = 2.024 H, load R2 = 1 Ω]

## Short answer

Bruton transform the Chebyshev ladder (Rs=1 Ω, L1=2.024 H, C1=0.994 F, L2=2.024 H, RL=1 Ω) and scale to10⁴ rad/s with Ct=10 nF (kz=10 kΩ). Series resistors20.24 kΩ each. Shunt FDNR D=0.994 via GIC with C=10 nF, RD=9.94 kΩ. DC paths: Ra=1 MΩ, Rb≈1.026 MΩ.

## Answer

#### FDNR

A **frequency dependent negative resistor (FDNR)** is a one-port with impedance $Z(s) = \dfrac{1}{s^2D}$. At $s = j\omega$, $Z = -\dfrac{1}{\omega^2 D}$: a real, negative resistance whose size falls as $1/\omega^2$. $D$ (unit F²Ω) is its constant. It cannot be made from passive elements; a GIC with two capacitors realises it.

#### Avoiding inductors in a lowpass LC ladder

**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:

| Element | Impedance | After dividing by $s$ | New element |
|---|---|---|---|
| Resistor $R$ | $R$ | $R/s$ | capacitor $C = 1/R$ |
| Inductor $L$ | $sL$ | $L$ | resistor $R = L$ |
| Capacitor $C$ | $1/sC$ | $1/(s^2C)$ | FDNR $D = C$ |

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.

In the 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).

#### Design from Figure 1, $\omega_0$ = 10⁴ rad/s

Prototype (Figure 1): $R_1 = 1\ \Omega$, series $L_1 = 2.024$ H, shunt $C_1 = 0.994$ F, series $L_2 = 2.024$ H, $R_2 = 1\ \Omega$ (a 3rd-order Chebyshev ladder). Numbering the elements 1, 2, 3 from the source:

**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 $\omega_0 = 10^4\ \text{rad/s}$. Choose the terminating capacitors as $C_t$ = 10 nF, so

$$
k_z = \frac{1}{\omega_0 C_t} = \frac{1}{10000 \times 1 \times 10^{-8}} = 10000\ \Omega
$$

$$
\begin{aligned}
R_{new} &= k_z R, \qquad C_{new} = \frac{C}{k_z\,\omega_0}, \qquad D_{new} = \frac{D}{k_z\,\omega_0^2}
\end{aligned}
$$

**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 $R_D = D_{new}/C_t^2 = k_z D$.

| Prototype | After Bruton | Scaled | Final element |
|---|---|---|---|
| $R_s = 1\ \Omega$ | $C_s = 1$ F | $1/(k_z\omega_0)$ | **10 nF** |
| $L_1 = 2.024$ H (series) | $R_1 = 2.024\ \Omega$ | $R_1 = 2.024 \times 10000$ | **20.24 kΩ** |
| $C_2 = 0.994$ F (shunt) | FDNR $D_2 = 0.994$ | $D_2 = 0.994/(k_z\omega_0^2) = 9.94 \times 10^{-13}$ | GIC: C = 10 nF, $R_D$ = **9.94 kΩ** |
| $L_3 = 2.024$ H (series) | $R_3 = 2.024\ \Omega$ | $R_3 = 2.024 \times 10000$ | **20.24 kΩ** |
| $R_L = 1\ \Omega$ | $C_L = 1$ F | $1/(k_z\omega_0)$ | **10 nF** |

**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, $\Sigma R$ = 40.48 kΩ, so $R_b$ = **1.04 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):

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

**Answer:** terminating capacitors 10 nF; series resistors 20.24 kΩ each; FDNR = GIC with C = 10 nF and $R_D$ = 9.94 kΩ; $R_a$ = 1 MΩ, $R_b$ = 1.04 MΩ.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The Figure 1 ladder has two series inductors and one shunt capacitor. After the Bruton transformation what is it made of?
   A. Two resistors and one grounded FDNR, with capacitors at the ends
   B. Two FDNRs and one resistor
   C. Two capacitors and one inductor
   D. Three resistors
   Answer: A. Series L becomes resistor, shunt C becomes grounded FDNR, and the 1 Ω terminations become capacitors. (Key point: Elements after Bruton)

2. With kz = 10 kΩ, the series L = 2.024 H becomes which resistor?
   A. 2.024 kΩ
   B. 202.4 kΩ
   C. 20.24 kΩ
   D. 10 kΩ
   Answer: C. R = L x kz = 2.024 x 10000 = 20.24 kΩ. (Key point: Scaled resistor value)

3. For shunt C = 0.994 F, what GIC resistor RD (C = 10 nF) is needed?
   A. 9.94 kΩ
   B. 99.4 kΩ
   C. 20.24 kΩ
   D. 0.994 kΩ
   Answer: A. RD = kz x D = 10000 x 0.994 = 9.94 kΩ. (Key point: FDNR GIC resistor)

4. With Ra = 1 MΩ and ΣR = 40.48 kΩ, Rb is about:
   A. 1 MΩ
   B. 40.48 kΩ
   C. 2.04 MΩ
   D. 1.04 MΩ
   Answer: D. Rb = Ra + ΣR = 1 MΩ + 40.48 kΩ, about 1.04 MΩ, keeping DC gain at 1/2. (Key point: Bias resistor Rb)

## Related in Filter Design

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