# Second order RLC circuit

1. ### sandy.bridge

793
1. The problem statement, all variables and given/known data
Hello all,
Representing RLC circuits via differential equations is rather new to me, in fact differentials as an entirety are new for me. Essentially I am looking for clarification regarding this type of question and whether or not I am doing it right. Thanks in advance.

$$I=i_C+i_L+i_R\rightarrow{I}=C\frac{dv}{dt}+\frac{1}{L}\int_0^tvdt+\frac{v}{R}$$
Differentiating both sides, we have
$$I=C\frac{d^2v}{dt^2}+\frac{v}{L}+R\frac{dv}{dt} \rightarrow{0}=v''+\frac{v'}{CR}+\frac{v}{LC}$$
let
$$v=e^{st}, v'=se^{st}, v''=s^2e^{st}$$
$${0}=s^2e^{st}+\frac{se^{st}}{CR}+\frac{e^{st}}{LC}$$
dividing out e^{st} and solving for the roots we have:
$$0=s^2+s/(CR)+1/(LC)\rightarrow{s=(-1/(CR)+/-\sqrt{(1/(CR))^2-4(1/LC)})/2}$$
Let C=1, R=1, L=0.5, also note that for t<0 the current source is grounded and there is absolutely no current or voltage in the network.
Thus,
$$(s_1, s_2)=((-1+i\sqrt{7})/2, (-1-i\sqrt{7})/2)$$
The general solution for second order differential can be expressed as
$$v(t)=C_1e^{s_1t}+C_2e^{s_2t}=C_1e^{-0.5t}e^{i\sqrt{7}/2t}+C_2e^{-0.5t}e^{-i\sqrt{7}/2t}$$
$$v(t)=C_1e^{-0.5t}(cos(\sqrt{7}/2t)+sin(\sqrt{7}/2t)i)+C_2e^{-0.5t}(cos(-\sqrt{7}/2t)-sin(-\sqrt{7}/2t)i)$$
at t=0-, v(0)=0
thus,
$$0=C_1(cos(0)+sin(0)i)+C_2(cos(0)-sin(0)i)\rightarrow{C_1+C_2=0}$$
however, I am a little unsure of how to determine the other equation for determining the constants..

Last edited: Jan 18, 2012
2. ### sandy.bridge

793
I just had a brain-wave for determining the constants of integration:
$$i_C+i_R+i_L=I$$
$$C\frac{dv(t)}{dt}+v(t)/R+i_L(t)=I$$
and at t=0, we have:
$$dv(0)/dt=I/C$$
and hence I can therefore differentiate my expression
$$v(t)=C_1e^{-0.5t}(cos(\sqrt{7}/2t)+sin(\sqrt{7}/2t)i)+C_2e^{-0.5t}(cos(-\sqrt{7}/2t)-sin(-\sqrt{7}/2t)i)$$
to acquire another expression for C1 and C2 which would be solvable due to having two equations. ?
or could I merely differentiate this expression, which appears to be less ugly:
$$v(t)=C_1e^{s_1t}+C_2e^{s_2t}$$

3. ### rude man

5,501
What happened to iL(t)? How did you get from your 2nd eq. to the final one? And, your time solution v(t) can't contain imaginary numbers!

I would add that if you're going to try to understand this you need a stronger background in elementary calculus and differential equations.

4. ### sandy.bridge

793
Well, it's a little late for that. At my school students do not take differential equations for an electrical engineering degree.. We're supposed to "pick it up" as we go.

iL disappeared because at t=0 there is no current through the inductor. We were told for t<0, the current source is grounded

5. ### sandy.bridge

793
Here is my physical work. at t<0 the switch is closed. It opens at t>0. The final steps end up canceling out, and i cannot figure out where I went wrong.

File size:
17.5 KB
Views:
16
6. ### rude man

5,501
What did the instructor say about I? Is it zero for t < 0 and = I for t > 0? Is I constant for t > 0?

I will assume I(0-)= 0, I(0+) = I constant

It often helps to invoke one of the basic circuit theorems, in this case Thevenin. So redraw with a voltage source V = IR in series with R, then C and L dropping from the r/h/s of R to ground.

Now you can perhaps see what the initial conditions are:

1: vC(0+) = 0 (initial charge on C = 0 so vC can't change instantaneously)

2: L looks like an infinite impedance initially, so all current goes into C, which is I/R. But i = Cdv/dt so dv/dt(0+) = I/C.

Now your equation is the one you came up with, almost:

Cv'' + v'/R + v/L = 0.

Note the rhs is 0, not I. That's because when you differentiated your original equation you didn't differentiate both sides.

The rest is routine sol'n of a 2nd order diff eq. with constant coefficients and given initial conditions.

NOTE: A 2nd order diff eq has to have 2 initial conditions (not more, not less), a 3rd three, etc.

Last edited: Jan 18, 2012
7. ### sandy.bridge

793
I=1A. When t<0, the switch is closed and 1 A of current is grounded since the wire containing the switch ideally has zero impedance. Therefore, no voltage across the capacitor, inductor and resistor exists until the switch opens and the source current is no longer ground. At that point, I still equals 1A.

Somehow when I am simplifying my final answer it is canceling out, however, wolfram shows me that there is indeed a waveform and the equation is not equal to zero.

http://www.wolframalpha.com/input/?i=%28-i%2Fsqrt%287%29%29*e^%28-0.5*t%2B1.3*t*i%29%2B%28i%2Fsqrt%287%29%29e^%28-0.5*t-1.3*t*i%29

8. ### rude man

5,501
Well, trust Wolfram! But does Wolfram give you a closed-form solution?

9. ### sandy.bridge

793
It gives me a solution with imaginary numbers, so my solution (attached in post 5) must have an error in it somewhere. I am applying the same logic that gets me the right answer at other times when dealing with parallel RLC circuits.

10. ### rude man

5,501
It's hard to read, but my biggest problem is you substituted actual values for L, R and C. This should never be done until the end. Why? Because doing otherwise deprives you of the ability to track dimensions in each term of an equation. I find lots of math mistakes that way.

11. ### rude man

5,501
If you got an imaginary sol'n from Wolfram you can always combine imaginary terms to get a real solution.

In other words, your imag. sol'n is mathematically correct but makes no physical sense.

12. ### sandy.bridge

793
This class is seriously not well thought out. Almost the entire class was unable to get the final solution, less the imaginary numbers. In fact, anyone I talked to had complex numbers in the time domain. Furthermore, nowhere in the textbook (which assumes one is savvy in respect to DE's) does it give the general equations for second order DE's.

I was able to get a final solution without complex numbers after searching to find the general solution for a second order DE with complex numbers.

13. ### rude man

5,501
Looks like you did the best you could. It reflects poorly on the state of our educational system and discourages prospective engineers & scientists from entering the professions. I would consider a "class-action" against the instructor. This is nothing new and usually brings results, eventuaslly if not immediately. Meantime you'll have to work on your own to make up the gaps as best as you can.

Remember that imaginary terms in a time expression always appear in "complex-conjugate" form. That means the "j"s can always be eliminated by the Euler relation or simple addition.

14. ### sandy.bridge

793
I've never learnt The Laplace Transform and we're applying it to circuits; although doing it blindly. Any suggestions or study recommendations? I feel as though this semester is going to be a rollercoaster.

5,501