Mathematics 304 – Ordinary Differential Equations Lab 3 – RLC Circuits and 2nd Order Equations October 26, 2000

Background and Goals

Today we will apply our knowledge of 2nd order linear ODE to some questions about electrical circuits. An “RLC circuit” is one that contains a voltage source v_{T} (t) (time dependent), a resistor (i.e. something that dissipates electrical energy) with resistance R, an inductor (a device or component that stores magnetic energy) with inductance L, and a capacitor (a device or component that stores electrical energy) with capacitance C. Many basic electrical circuits have this form.

An RLC circuit

From physics, it can be derived that the voltage change across the capacitor, written v_{C}, satisfies the 2nd order ODE

(1)  LC \frac{d^{2}v_{C}}{dt^{2}} + RC \frac{dv_{C}}{dt} + v_{C} = v_{T} (t)

We will always consider a periodic, sinusoidal voltage source (“forcing”) term

v_{T} (t) = a sin(ωt).

Lab Questions

A) (A worked example – you supply the details, graph, etc.) For instance, suppose we want to understand the voltage in an RLC circuit with R = 1000, L = 2, and C = 10^{−6}, and a voltage source v_{T} = 20 sin(500t). From class on Tuesday, we know that the general solution of (1) in this case:

(2)  (2 × 10^{−6})\frac{d^{2}v_{C}}{dt^{2}} + (10^{−3})\frac{dv_{C}}{dt} + v_{C} = 20 sin(50t)

looks like v_{C} = c_{1}v_{1} + c_{2}v_{2} + v_{p}, where {v_{1}, v_{2}} is a fundamental set of solutions for the homogeneous equation

(2 × 10^{−6})\frac{d^{2}v_{C}}{dt^{2}} + (10^{−3})\frac{dv_{C}}{dt} + v_{C} = 0

and v_{p} is some particular solution.

1) To find v_{1}, v_{2}, we find the roots of the auxiliary equation of the ODE. What are they here? What does the corresponding general solution of the homogeneous equation look like?

2) Now, we need to find a particular solution v_{p}. This is where the method of undetermined coefficients that we discussed in class on Tuesday comes in. The 20 sin(50t) term on the right is a very simple special case of the third general case we discussed in class. (It’s also a special case of class V in the table on page 208 of the text.) We guess that a particular solution will have the form v_{p} = A cos(50t)+B sin(50t). (Note, no term here is a solution of the homogeneous equation, so we do not need the t^{s} factor. Take s = 0.) Plug v_{p} into (2) and determine the values of A, B to yield a solution. You should get:

\begin{matrix}.995A + .05B = 0 \\ −.05A + .995B = 20 \end{matrix}

So A ≐ −1.007 and B ≐ 20.05.

3) Plot the solution of (2) with v_{C}(0) = 0, and v′_{C}(0) = 1. (Note: You’ll need to think about an appropriate interval of t values for your plot. A term like cos(661.437t) is oscillating really fast!)

B) Suppose L, C > 0, but the resistance R = 0. (This is a very, very idealized situation, of course, even better than “superconductivity.”) What do the solutions of (1) look like in this case? Derive analytic formulas by our undetermined coefficients technique for inhomogeneous 2nd order equations. Be sure to distinguish between the two different cases:

1) iω is not a root of the auxiliary equation, and

2) iω is a root of the auxiliary equation. How is the analytic form of the solution different in these two cases?

C) Using Maple, plot the analytic solutions of (1) with R = 0, C = 10^{−6}, L = 1, a = 1, v_{C}(0) = 1, v′_{C}(0) = 0, and ω = 900, 950, 1000, 1050, 1100 (five graphs in all). (You’ll need to chose your t interval carefully to get informative graphs!) The name for the ω = 1000 case is resonance. Describe what a resonant solution does as t → ∞.

D) Next, we will study the more realistic case R > 0. By considering the form of the roots of the auxiliary equation, show that if R > 0, every solution of the corresponding homogeneous equation

LC \frac{d^{2}v_{C}}{dt^{2}} + RC \frac{dv_{C}}{dt} + v_{C} = 0

tends to zero as t → ∞. These terms in the general solution of (1) are called “transients” for this reason.

E) Plot and describe the solutions for various values of a and ω when R = 2000, C = 2 × 10^{−7}, and L = 1.5.

F) What types of long term behavior are possible for solutions of (1) when R > 0? Does resonance occur in this situation?

Maple Notes

All of the plots necessary for this lab will be generated using the basic Maple plot command. However, if you want to check your results, you can also use the DEplot command from Lab 1 to plot approximate numerical solutions to 2nd order ODE as follows. For example, say we want to see the solution of the initial value problem

\{ \begin{matrix}4\frac{d^{2}y}{dt^{2}} + 3\frac{dy}{dt} + 10y = 3 cos(2t) \\ y(0) = 5 \\ y′(0) = −3 \end{matrix}

Then the following commands could be used. First load the DEtools package with:

                              with(DEtools):

Then define the ODE with:

        eqn:=4*diff(y(t),t$2)+3*diff(y(t),t)+10*y(t) = 3*cos(2*t);

and plot the solution for t ∈ [0, 10] with

              DEplot(eqn,y(t),t=0..10,[[y(0)=5,D(y)(0)=-3]]);

As always, adjusting the linecolor and stepsize options may be necessary to get reasonable, printable graphs.

Assignment

Individual write-ups giving answers to the above questions. Due: In class, November 2.