Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 7.EX.3

Bridged Tee Circuit in State-Variable Form
Determine the state-space equations for the circuit shown in Fig. 2.25.

## Verified Solution

In order to write the equations in the state-variable form (i.e., a set of simultaneous first-order differential equations), we select the capacitor voltages $v_1 \text{ and } v_2$ as the state elements (i.e., $x = [v_1 v_2 ]^T$ ) and $v_i$ as the input (i.e., $u = v_i$ ). Here $v_1 = v_2, v_2 = v_1 − v_3,$ and still $v_1 = v_i.$ Thus $v_1 = v_i, v_2 = v_1 ,$ and $v_3 = v_i − v_2 .$ In terms of $v_1$ and $v_2$ , Eq. (2.34) is

$\frac{v_1 − v_i}{R_1} + \frac{v_1 − (v_i − v_2 )}{R_2} + C_1 \frac{dv_1}{dt} = 0.$

$-\frac{v_{1}-v_{2}}{R_{1}}+\frac{v_{2}-v_{3}}{R_{2}}+C_{1} \frac{d v_{2}}{d t}=0 ,$           (2.34)

Rearranging this equation into standard form, we get

$\frac{dv_1}{dt} = − \frac{1}{C_1} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) v_1 − \frac{1}{C_1} \left( \frac{1}{R_2} \right) v_2 + \frac{1}{C_1} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) v_i.$                  (7.6)

In terms of $v_1$ and $v_2$, Eq. (2.35) is

$\frac{v_i − v_2 − v_1}{R_2} + C_2 \frac{d}{dt}(v_i − v_2 − v_i) = 0.$

$\frac{v_{3}-v_{2}}{R_{2}}+C_{2} \frac{d\left(v_{3}-v_{1}\right)}{d t}=0$             (2.35)

In standard form, the equation is

$\frac{dv_2}{dt} = − \frac{v_1}{C_2R_2} − \frac{v_2}{C_2R_2} + \frac{v_i}{C_2R_2}.$            (7.7)

Equations (2.34)–(2.35) are entirely equivalent to the state-variable form, Eqs. (7.6) and (7.7), in describing the circuit. The standard matrix definitions are

$\pmb F = \left[\begin{matrix} -\frac{1}{C_1}\left(\frac{1}{R_1}+ \frac{1}{R_2} \right) & -\frac{1}{C_1} \left( \frac{1}{R_2}\right)\\ -\frac{1}{C_2R_2} & -\frac{1}{C_2R_2} \end{matrix} \right] , \\ \pmb G = \left[\begin{matrix} \frac{1}{C_1}\left( \frac{1}{R_1} + \frac{1}{R_2}\right) \\ \frac{1}{C_2R_2} \end{matrix} \right] , \\ \pmb H = \left[\begin{matrix} 0 & -1 \end{matrix} \right] , \ \ \ J =1 .$