# General Second Order Circuit

#### p75213

1. The problem statement, all variables and given/known data
Refer Attachment.
I am trying to derive the second order equation using the natural response of the circuit

2. Relevant equations
Refer Attachment

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attachments

• 10.1 KB Views: 336
Related Engineering and Comp. Sci. Homework News on Phys.org

#### rude man

Homework Helper
Gold Member
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!

#### p75213

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}$$

#### rude man

Homework Helper
Gold Member
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.

#### p75213

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

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving