# Help finding the damping ratio formula for this circuit

• Engineering
• Boltzman Oscillation
In summary: The circuit to be analyzed is shown below:Since initial conditions are zero (from the instructions), I will use laplace transforms for the circuit 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
Boltzman Oscillation
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?

Hi BO.
the standard form of second order systems is like this:
$$x(t) = y''(t) + 2 \delta y'(t) + \omega_{r}^{2} y(t)$$
I would do a thevenin transform from the point of view of the branch of the inductor and capacitor in series, you get the following thevenin parameters:
$$v_{th}(t) = E_{g}(t) \cdot \dfrac{R_{2}}{R_{1} + R_{2} }$$

The thevenin equivalent impedence in this case:
$$Z_{th} = \dfrac{R1 \cdot R_{2} }{R_{1} + R_{2} }$$
Redraw the circuit, with Vth and Rth and the L and C in series, apply the standard KVL in differential equation form:
$$V_{thevenin}(t) = V_{R thevenin}(t) + V_{L}(t) +V_{c}(t)$$
Which becomes:
$$V_{Th}(t) = i(t)\cdot R_{Th} + L \cdot i'(t) + \dfrac{1}{C} \displaystyle \int i(t)$$
Remove the integral by taking the first derivative:
$$V_{Th}'(t) = i''(t) \cdot L + i'(t) R_{Th} + \dfrac{i(t)}{C}$$
Multiply throughout with
$$\dfrac{1}{L}$$
THen
$$\dfrac{V_{Th}'(t)}{L} = i''(t) + i'(t) \dfrac{R_{Th}}{L} + \dfrac{i(t)}{LC}$$
Then the damping factor is defined as:
$$\delta = \dfrac{R_{Th}}{2 L}$$
And the square of the resonant angular velocity:
$$\omega_{r}^{2} = \dfrac{1}{LC}$$
Then the damping factor is simply:
$$\delta = \bigg( \dfrac{R1 \cdot R2 }{R1 + R2} \bigg) \cdot \dfrac{1}{2L}$$
The reason the damping factor is multiplied by 2 in the above equation is due to the quadratic formula:
$$\sqrt{ 4 \cdot( \delta^{2} - \omega_{r}^{2} ) }$$
And now you can find out the value of the thevenin resistance:
$$R_{th} = 7.4 \,\, \text{m} \Omega$$
What values of $$R_{1} ,R_{2}$$ give this? My answer is to use $$14.8 \,\, \text{m} \Omega$$ for both. Or to decide on one value for R1 and then substitute to find the value for R2.
Boltzman Oscillation said:
Original post above.

Last edited:

## 1. What is the damping ratio formula for this circuit?

The damping ratio formula for a circuit is given by the equation: ζ = R/2√(L/C), where ζ is the damping ratio, R is the resistance, L is the inductance, and C is the capacitance.

## 2. How do I calculate the damping ratio for a circuit?

To calculate the damping ratio for a circuit, you will need to know the values of the resistance, inductance, and capacitance. Then, you can plug these values into the equation ζ = R/2√(L/C) to determine the damping ratio.

## 3. What is the significance of the damping ratio in a circuit?

The damping ratio is a measure of how much energy is dissipated in a circuit. A higher damping ratio indicates that more energy is being dissipated, resulting in a slower response and less oscillation in the circuit. A lower damping ratio means that less energy is being dissipated, resulting in a faster response and more oscillation in the circuit.

## 4. How does the damping ratio affect the stability of a circuit?

The damping ratio affects the stability of a circuit by determining whether the circuit will oscillate or not. A damping ratio of less than 1 indicates an underdamped circuit, which will oscillate before settling to a steady state. A damping ratio of greater than 1 indicates an overdamped circuit, which will not oscillate and will take longer to reach a steady state. A damping ratio of exactly 1 indicates a critically damped circuit, which will reach a steady state without oscillation in the shortest amount of time.

## 5. Can the damping ratio be adjusted in a circuit?

Yes, the damping ratio can be adjusted in a circuit by changing the values of the resistance, inductance, and capacitance. Increasing the resistance or decreasing the capacitance will result in a higher damping ratio, while decreasing the resistance or increasing the capacitance will result in a lower damping ratio. The inductance has an indirect effect on the damping ratio, as it is used in the calculation of the damping ratio formula.

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