Second-order circuit derivations

[DigiTan]

Are there any good sites or web articles that do step-by-step differential equations for second-order circuits. I'm trying to brush up on this stuff for a class and all the textbooks are wild about skipping steps. Mainly, I'm trying to remember what to do with all the differential terms left after applying the Kirchoff Laws. I know I'm supposed to wind up with a linear differential equation roughly of the form:

A (d²v/dt²) + B (dv/dt) + C v(t) = D

but I don't know the steps to get there. I've googled but this mostly turns up EE course syllabi.

If it helps, here's an attachment of the circuit I'm stuck on. Just so everyone's clear, you aren't doing my homework here, I got it out of Schaum's Outline "Basic Electrical Engineering" but they skipped a lot of steps in the example. :shock:

 

[_3iMaJ]

I find it easiest to use Laplace and just take the inverse transform. Rather than writing the DE directly.

 

[Styx]

I find it easiest to use Laplace and just take the inverse transform. Rather than writing the DE directly.

Second that, with Laplace the result just drops out

Or you could download MuPAD and use the diff() function

 

[DigiTan]

Okay, I guess that means I have to use the property where L{y''(t)} = s²Y(s) - s*y(0) + y'(0) for the second-order terms, and L{y'(t)} = sY(s) - y'(0) for first orders. Since I need to get my equation in the form of:

A y''(t) + B y'(t) + C y(t) = D
or essentially...
A (d²v/dt) + B (dv/dt) + C v(t) = D

...how to I combine the differential equations I got from the voltage and current laws? So far, from the Kirchoff Laws, I get:

Voltage at point "v"

1. (1) v = R1*i2 + L*(di2/dt)
   (2) v = R2*i3 + (1/C) * ∫ i3 dt

Currents:

1. (3) i1 = i2 + i3 (also, i1 is a constant)

Then the book goes on to remove the integral from equation 2 by differentiating it...

1. (4) dv/dt = R2*(di3/dt) + (1/C)*i3
   (5) d²v/dt² = R2*(d²i3/dt²) + (1/C)*(di3/dt)

Then it differentiates equation 1...

1. (6) dv/dt = R1*(di2/dt) + L*(d²i2/dt²)

Then it differentiates the current equation twice...

1. (7) (di1/dt) = (di2/dt) + (di3/dt) = 0 (since i1 is constant)
   (8) (d²i1/dt²) = (d²i2/dt²) + (d²i3/dt²) = 0 (since i1 is constant)

-------
...From all this, they somehow get:

L*(d²v/dt²) + (R1+R2)*(dv/dt) + (1/C)*v = (R1*I)/C

How did they use those 8 equations, to get that final form? The book didn't explain squat about that part.

 

[DigiTan]

*Bump*

Okay, I'm still stuck on this one. Can anyone at least skip through the Laplace version of this?

 

[i_build_stuff]

Just from skimming over it, they probably back-substituted the equations into each other.

Here's how you'd do it in the s-domain:

First, get the loop equations:
V(s) - R1 * I2(s) - L1 * s * I2(s) = 0
V(s) - R2 (I3)s - (1 / (c1 * s)) * I3(s) = 0

And the relation:
I1(s) = I2(s) + I3(s) = 1/s (transform of 1A * u(t))

--note the u(t); this isn't *technically* true, but it works because you're only modeling it from outside the current source - switch "black box"

--also note that we now have three linearly independant equations in three variables (I2, I3, and V). I1 was eliminated as it is already known.

Get the two loop equations equal to each other:


V(s) = R1 * I2(s) + L1 * s * I2(s) = R2 * I3(s) + (1 / (c1 * s)) * I3(s)

--all I did was pull V(s) to one side in both equations, and set them equal to each other.

--now, use the known value of I1 to solve for I2 in terms of I3:

I1(s) = I2(s) + I3(s)

=> I2(s) = 1 / s - I3(s)

--substitute that into the I2 side:

V(s) = R1 * (1 / s - I3(s)) + L1 * s * (1 / s - I3(s)) = R2 * I3(s) + (1 / (c1 * s)) * I3(s)

--and collect terms to get one side with only things multiplied by I3(s):

R1 / s + L1 = R2 * I3(s) + I3(s) / (C1 * s) + R1 * I3(s) + L1 * s * I3(s)

--just pull I3 out as a factor, and divide the other stuff out of the other side to solve for it explicitly:

(R1 / s + L1) / ((R2 + 1 / (C1 * s) + R1 + L1 * s)) = I3(s)

--now that you have I3(s), you repeat the operation from where you solved for I3 in terms of I2 (this step: I2(s) = 1 / s - I3(s)), but solve for I2(s) instead. At this point, you can also solve either equation for V(s), and use that with either eq. to get I2(s).

Also, it might be worthwhile to point out that doing this in the s-domain takes you through exactly the same operations you'd use to solve the linear diff. eq's by hand--the function of s is producing an "auxiliary equation" that you solve algebraically...

*disclaimer: I did the math really fast, and didn't check it afterward...

 

[DigiTan]

Whew! Thanks for giving me the low-down on this one, i_build_stuff.