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## Homework Statement:

- Derive the ordinary differential equation (ODE) for the network with the capacitor voltage vC4(t) as the dependent variable, time t as the independent variable, and Eg(t) as the network excitation described by a 10V step function. Find the resistor values that will give you a damping ratio of .1.

## Relevant Equations:

- The damping ratio is require to be .1

The circuit to be analyzed is shown below:

Since initial conditions are zero (from the instructions) I will use laplace transforms for the cirucit and I will use the MAME method to solve this circuit. The laplace transforms that are required will give me:

$$E_g(s) = \frac{10}{s}$$

$$ L_3 = sL_3$$

$$ C_4 = \frac{1}{sC_4}$$

again, the initial conditions are zero so i did not include them in the transforms.

Now solving the MAME (mesh analysis matrix equation), by setting the first loop as Im1 and the second loop as Im2, I get:

$$

\begin{bmatrix}

R_1+ sL_3 + \frac{1}{sC4} & -(\frac{1}{sC4} + sL_3) \\

-(\frac{1}{sC4} + sL_3) & R_2+ sL_3 + \frac{1}{sC4}

\end{bmatrix}

*

\begin{bmatrix}

I_{m1}\\

I_{m2}

\end{bmatrix}

=

\begin{bmatrix}

\frac{10}{s}\\

0

\end{bmatrix}

$$

Which is of the form:

$$Z*I = V$$

I can use Cramer's rule to solve for I_m1 and I_m2. Doing so gives me the following two formulas:

$$I_{m1} = \frac{10(R_2+ sL_3 + \frac{1}{sC4})}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})}$$

and

$$I_{m2} = \frac{10(sL_3 + \frac{1}{sC4})}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})}$$

from the circuit I can see that

$$V_{C4}(t) = \frac{I_{m1}-I_{m2}}{sC_4}$$

thus i can use my equations to solve for Vc4 as:

$$\frac{10(R_2)}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})} * \frac{1}{sC_4}$$

am i doing this right? Even if i turn this into the time domain I would have a third order differential equation in the denominator. What am I doing wrong?

Since initial conditions are zero (from the instructions) I will use laplace transforms for the cirucit and I will use the MAME method to solve this circuit. The laplace transforms that are required will give me:

$$E_g(s) = \frac{10}{s}$$

$$ L_3 = sL_3$$

$$ C_4 = \frac{1}{sC_4}$$

again, the initial conditions are zero so i did not include them in the transforms.

Now solving the MAME (mesh analysis matrix equation), by setting the first loop as Im1 and the second loop as Im2, I get:

$$

\begin{bmatrix}

R_1+ sL_3 + \frac{1}{sC4} & -(\frac{1}{sC4} + sL_3) \\

-(\frac{1}{sC4} + sL_3) & R_2+ sL_3 + \frac{1}{sC4}

\end{bmatrix}

*

\begin{bmatrix}

I_{m1}\\

I_{m2}

\end{bmatrix}

=

\begin{bmatrix}

\frac{10}{s}\\

0

\end{bmatrix}

$$

Which is of the form:

$$Z*I = V$$

I can use Cramer's rule to solve for I_m1 and I_m2. Doing so gives me the following two formulas:

$$I_{m1} = \frac{10(R_2+ sL_3 + \frac{1}{sC4})}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})}$$

and

$$I_{m2} = \frac{10(sL_3 + \frac{1}{sC4})}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})}$$

from the circuit I can see that

$$V_{C4}(t) = \frac{I_{m1}-I_{m2}}{sC_4}$$

thus i can use my equations to solve for Vc4 as:

$$\frac{10(R_2)}{s^2(L_3R_2+R_1L_3)+sR_1R_2+(\frac{R_1+R_2}{C_4})} * \frac{1}{sC_4}$$

am i doing this right? Even if i turn this into the time domain I would have a third order differential equation in the denominator. What am I doing wrong?