# Deriving Second Order Circuit Equation Using Natural Response

• Engineering
• p75213
In summary, the natural response of the circuit can be found by using nodal analysis and mesh analysis to derive a second order differential equation. Solving this equation and using the initial conditions, the voltage at v2 can be determined. Then, using v2 and the step input, the voltage at v1 can be found by subtracting v1 from v2. The final result is in the form of a decaying exponential function.
p75213

## Homework Statement

Refer Attachment.
I am trying to derive the second order equation using the natural response of the circuit

Refer Attachment

## The Attempt at a Solution

Nodal Analysis:
v1+$\frac{1}{2}$$\frac{dv1}{dt}$+$\frac{1}{3}$$\frac{dv2}{dt}$=0
Mesh Analysis LHS:
i+v1=0

I can't see how to get everything in terms of v1 or v2

#### Attachments

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First I would label the components from left to right: R1, R2, C1, C2.

Then I would write equations summing currents at v1 and v2.

Then solve the diff. eq's (zero initial conditions), which gets you v1 and v2. Then your answer is v1 - v2.

The reason you should label the components is otherwise you get all mixed up as to which resistor is which, for example. And you wind up with equations like your
i + v1 = 0 which seemingly mixes volts and amps in terms within the same equation, making dimensional checking impossible. The greatest tool for error checking of equations is dimensional checking!

Thanks for getting the ball rolling rude man. The most challenging part is coming up with the differential equation.

$$\begin{array}{l} {\rm{Find natural response:}} \\ \frac{{{v_1}}}{1} + \frac{1}{2}\frac{{d{v_1}}}{{dt}} + \frac{{{v_1} - {v_2}}}{1} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\frac{1}{2}\frac{{d{v_1}}}{{dt}} + 2{v_1} - {v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\ \frac{{{v_2} - {v_1}}}{1} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {v_1} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\ {\rm{Substitute }}{v_1}{\rm{ into equation 1:}} \\ \frac{1}{2}\left[ {\frac{{d{v_2}}}{{dt}} + \frac{1}{3}\frac{{{d^2}{v_2}}}{{d{t^2}}}} \right] + 2{v_2} + \frac{2}{3}\frac{{d{v_2}}}{{dt}} - {v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \frac{1}{6}\frac{{{d^2}{v_2}}}{{d{t^2}}} + \frac{1}{2}\frac{{d{v_2}}}{{dt}} + \frac{2}{3}\frac{{d{v_2}}}{{dt}} + {v_2} = 0 \\ \frac{{{d^2}{v_2}}}{{d{t^2}}} + 7\frac{{d{v_2}}}{{dt}} + 6{v_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {\rm{let }}{v_2} = {e^{st}}\,\,\,\,\,\,\,\,\,\,\,\,\, \to {s^2}{e^{st}} + 7s{e^{st}} + 6{e^{st}} = 0 \\ {\rm{characteristic equation = }}{s^2} + 7s + 6 \to \,{\rm{ roots }} - 6{\rm{ and }} - 1 \\ {v_{2{\rm{natural}}}}(t) = A{e^{ - 6t}} + B{e^{ - t}} \\ {v_2}(t) = 5 + A{e^{ - 6t}} + B{e^{ - t}} \\ @t = 0\,\,{v_2}(0) = 0 = 5 + A + B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to A + B = - 5 \\ \frac{{d{v_2}}}{{dt}} = - 6A{e^{ - 6t}} - B{e^{ - t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to @t = 0\,\,\frac{{d{v_2}(0)}}{{dt}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0 = - 6A - B \\ - 6A - B = 0 \\ A + B = - 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to A = 1,\,\,\,B = - 6 \\ {v_2}(t) = 5 + {e^{ - 6t}} - 6{e^{ - t}} \\ {\rm{ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - }} \\ {\rm{Find }}{v_0}{\rm{:}} \\ {{\rm{v}}_0} = {v_1} - {v_2} \\ {\rm{From equation 2: }}{v_1} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} \\ {v_0} = {v_2} + \frac{1}{3}\frac{{d{v_2}}}{{dt}} - {v_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to {v_0} = \frac{1}{3}\frac{{d{v_2}}}{{dt}} \\ {v_0}(t) = \frac{1}{3}\left( { - 6{e^{ - 6t}} + 6{e^{ - t}}} \right)\,\,\,\,\,\, \to {v_0}(t) = 2\left( {{e^{ - t}} - {e^{ - 6t}}} \right)V \\ \end{array}$$

You certainly seem to know what you're doing. I don't intend to check your math fully. What I did check was correct. I can help you verify your answer though since I have a 'cheat sheet' which lists the transfer function of your network, and with apologies I must use the Laplace transform method, so if you don't understand much of the following, don't be concerned. I am just trying to give you the answer I got & maybe something for you to look at later when you do hit transform methods.

V2/Vin = 1/[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2]

so your characteristic equation should be the denominator above set to zero, and it looks like it might just be!

Then a step input transforms to 5/s so
V2 = 5/s[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2].

However, you're asked to find V1-V2 which conveniently happens to be the current flowing into C2 multiplied by R2, so i = sC2V2 and

V1-V2 = iR2 = sC2R2V2 = 5R2C2/[1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s^2].

This inverse-transforms to
V1-V2 = 5R2C2{1/(T1-T2)}[exp(-t/T1) - exp(-t/T2)] volts.

where T1 and T2 are the coefficients in the chas. equation:
(T1+s)(T2+s) = 1 + (R1C1 + R2C2 + R1C2)s + R1C1R2C2s2), T1, T2 > 0 and real. I'm leaving it up to you to solve the appropriate quadratic if you want.

Your answer is in the right form so other than a math mistake you're looking very good.

No worries. I typed out the maths for my own benefit more than anything else. Just as a learning tool.

## 1. What is a general second order circuit?

A general second order circuit is an electrical circuit that contains two energy storage elements, such as capacitors or inductors, and two independent sources, such as voltage or current sources. It can be represented by a second order differential equation and exhibits both resistive and reactive elements.

## 2. What are the key components of a general second order circuit?

The key components of a general second order circuit are two energy storage elements, two independent sources, and resistive and reactive elements. These elements work together to create a complex circuit that can exhibit various behaviors, such as overdamped, underdamped, or critically damped responses.

## 3. How is a general second order circuit analyzed?

A general second order circuit is analyzed using techniques such as Kirchhoff's laws, Ohm's law, and nodal or mesh analysis. The circuit's differential equation can also be solved to determine the circuit's behavior over time.

## 4. What are the applications of general second order circuits?

General second order circuits have various applications in electronic devices, such as filters, oscillators, amplifiers, and power supplies. They are also commonly used in communication systems, control systems, and signal processing.

## 5. How do the parameters of a general second order circuit affect its behavior?

The parameters of a general second order circuit, such as the values of the energy storage elements and the sources, can significantly impact its behavior. For example, changing the values of the energy storage elements can alter the circuit's natural frequency and damping factor, while changing the source values can affect the amplitude and phase of the circuit's response.

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