Chapter 9.6, Problem 6E

Interpretation Introduction

Interpretation:

To derive the circuit equation for the transmitter circuit shown below

Concept Introduction:

• ➢ The Lorenz equations are

  $\begin{array}{l}\dot{\text{x}}=\sigma (\text{y}-\text{x})\\ \dot{\text{y}}=\text{rx}-\text{y}-\text{xz}\\ \dot{\text{z}}=\text{xy}-\text{bz}\\ \text{Here σ, r, b > 0}\end{array}$

• ➢ The governing equation for the above system is written by using Kirchhoff’s laws and proper non-dimensionalization as follows

  $\begin{array}{l}\dot{\text{u}}=\sigma \left(\text{v}-\text{u}\right)\\ \dot{\text{v}}=\text{ru}-\text{v}-\text{20uw}\\ \dot{\text{w}}\text{ = 5uv}-\text{bw}\end{array}$

  The relation between u, v, w and Lorenz variable is

  $\text{u}=\frac{1}{10}\text{x},\text{v}=\frac{1}{10}\text{y},\text{w}=\frac{1}{20}\text{z}$

  The receiver variables are

  $\begin{array}{l}{\dot{\text{u}}}_{\text{r}}=\sigma \left({\text{v}}_{\text{r}}-{\text{u}}_{\text{r}}\right)\\ {\dot{\text{v}}}_{\text{r}}=\text{ru(t)}-{\text{v}}_{\text{r}}-{\text{20u(t)w}}_{\text{r}}\\ {\dot{\text{w}}}_{\text{r}}{\text{ = 5u(t)v}}_{\text{r}}-{\text{bw}}_{\text{r}}\end{array}$

• ➢ The output of the voltage subtractor using Op-Amp is given as

  ${\text{V}}_{\text{out}}\text{=}\frac{{\text{R}}_{\text{f}}}{{\text{R}}_{\text{1}}}\left[{\text{V}}_{\text{2}}-{\text{V}}_{\text{1}}\right]$

• ➢ The output of integrator circuit using Op-Amp is given as

  ${\text{V}}_{\text{out}}\text{= -}\frac{\text{1}}{\text{RC}}{\displaystyle \int {\text{V}}_{\text{in}}}\text{dt}$