\magnification=\magstep1
\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Lab 3 -- RLC Circuits and 2nd Order Equations}
\centerline{October 26, 2000}
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\noindent
{\it Background and Goals}
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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 {\it resistor} (i.e. something that {\it dissipates electrical 
energy}) with resistance $R$, an {\it inductor} (a device or component 
that {\it stores magnetic energy}) with inductance $L$, and a {\it capacitor} 
(a device or component that {\it stores electrical energy}) with capacitance 
$C$.  Many basic electrical circuits have this form.
\vglue 2.0truein
\centerline{An RLC circuit}
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From physics, it can be derived that the voltage change across the capacitor,
written $v_C$, satisfies the 2nd order ODE
$$LC{d^2 v_C\over dt^2} + RC {d v_C\over dt} + v_C = v_T(t)\leqno(1)$$
  We will always consider
a {\it periodic, sinusoidal} voltage source (``forcing'') term
$$v_T(t) = a\sin(\omega t).$$
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\noindent
{\it Lab Questions}
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\item{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 $v_T = 20\sin(500 t)$.  From class on Tuesday, we know that
the general solution of (1) in this case:
$$(2\times 10^{-6}){d^2 v_C\over dt^2} + (10^{-3}) {d v_C\over dt} + v_C = 20\sin(50 t)
\leqno(2)$$
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
$$(2\times 10^{-6}){d^2 v_C\over dt^2} + (10^{-3}) {d v_C\over dt} + v_C = 0$$
and $v_p$ is some particular solution.
\itemitem{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?  
\itemitem{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 $20\sin(50 t)$ 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 $v_p = A\cos(50 t) + B\sin(50 t)$.  (Note, no term here is a
solution of the homogeneous equation, so we {\it 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:
$$\eqalign{.995A + .05 B = 0\cr
           -.05A + .995 B = 20\cr}$$
So $A \doteq -1.007$ and $B \doteq 20.05$.
\itemitem{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 $\cos(661.437 t)$ is oscillating really fast!)
\bigskip
\item{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 {\it undetermined coefficients} technique 
for inhomogeneous 2nd order
equations.   Be sure to distinguish between the two different cases:
\itemitem{1)}  $i \omega$ is not a root of the auxiliary equation, and 
\itemitem{2)}  $i \omega$ is a root of the auxiliary equation.
\item{} How is the analytic form of the solution different in these two 
cases?
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\item{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 $\omega = 900, 950, 1000, 1050, 1100$  (five graphs in all).
(You'll need to chose your $t$ interval carefully to get informative graphs!)
The name for the $\omega = 1000$ case is {\it resonance}.  Describe
what a resonant solution does as $t\to\infty$.
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\item{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
$$LC{d^2 v_C\over dt^2} + RC {d v_C\over dt} + v_C = 0$$
tends to {\it zero} as $t\to \infty$.  These terms in the general solution 
of (1) are called ``transients'' for this reason.
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\item{E)} Plot and describe the solutions for various values of $a$ and 
$\omega$ when $R = 2000$, $C = 2\times 10^{-7}$, and $L = 1.5$.
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\item{F)}  What types of long term behavior are possible for solutions
of (1) when $R > 0$?  Does resonance occur in this situation?
\bigskip
\noindent
{\it Maple Notes}
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All of the plots necessary for this lab will be generated using the
basic Maple {\tt plot} command.  However, if you want to check your
results, you can also use the {\tt 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
$$\cases{ 4{d^2 y\over dt^2} + 3 {dy\over dt} + 10y = 3\cos(2t)\cr
           y(0) = 5\cr
           y'(0) = -3\cr}$$
Then the following commands could be used.  First load the {\tt DEtools}
package with:
\medskip
\centerline{\tt with(DEtools):}
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\noindent
Then define the ODE with:
\medskip
\centerline{\tt eqn:=4*diff(y(t),t\$2)+3*diff(y(t),t)+10*y(t) = 3*cos(2*t);}
\medskip
\noindent
and plot the solution for $t\in [0,10]$ with
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\centerline{\tt DEplot(eqn,y(t),t=0..10,[[y(0)=5,D(y)(0)=-3]]);}
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\noindent
As always, adjusting the {\tt linecolor} and {\tt stepsize} options
may be necessary to get reasonable, printable graphs.
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\noindent
{\it Assignment}
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Individual write-ups giving answers to the above questions.  {\bf Due:}
In class, November 2. 
\bye
