1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
|
\def\rload{R_{\rm load}}
\input ../format
\titlesub{Part 3: Executive Summary}{Mystery Circuit}
\input ../com
\section{Matrix Representation and Homogeneous Solution}
To determine the relevant properties of the linear system, matrix form
is useful (this form was chosen to reduce fractions' usage):
\def\x{{\bf x}}
$$\x' =
{1\over R_1C_1C_2\rload}
\pmatrix{0&-C_2\rload &0 \cr
0&-C_2(R_1+\rload)&C_1R_1\cr
0&C_2R_1 &-C_1R_1} \x +
{1\over R_1}
\pmatrix{\omega\cos(\omega t)\cr
\omega\cos(\omega t)\cr
0}
.$$
The characteristic polynomial is
$$-\lambda( (-\lambda-C_2(R_1+\rload))(-\lambda-C_1R_1)
- C_2C_1R_1^2).$$
Expanded,
$$-\lambda( \lambda^2 + \lambda(C_2(R_1+\rload)+C_1R_1)
+ C_1C_2R_1\rload).$$
In terms of its roots (with $b=C_2(R_1+\rload)+C_1R_1$ and
$c = C_1C_2R_1\rload,$
$$-\lambda{(\lambda-{-b+\sqrt{b^2-4c}\over2})
(\lambda-{-b-\sqrt{b^2-4c}\over 2})}$$
%For reference,
%$$b^2-4c = C_2^2(R_1+\rload)^2 + C_1^2\rload^2
% - 2C_1C_2\rload(3R_1+\rload).$$
Let $r_1$ and $r_2$ designate these two non-zero roots.
In the original matrix $A$, this being the transformed matrix $A/c$,
$Av = cr_1v$ because $A/c * v = r_1.$
The trivial zero eigenvalue corresponds to a unit x-direction vector
by inspection. The two remaining roots, in the general case of a non-%
degenerate system, which hasn't been explicitly ruled out, the middle
row can be ``ignored'' because it is linearly dependent in the
following system:
$$
\pmatrix{-r&-C_2\rload &0\cr
* &* &*\cr
0 &C_2R_1 &-r-C_1R_1}
$$
With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and
$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$
\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
\noindent The eigenvalues and respective eigenvectors are:
\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$
\def\num#1{\pmatrix{1\cr
-{r_#1\over C_2\rload}\cr
-{R_1r_#1\over \rload(r_#1+C_1R_1)}}}
\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$
\bu $\displaystyle\lambda_3 = {r_2C_1C_2R_1\rload}, v_3 = \num2.$
This corresponds to a solution of the form $f = Ce^{\lambda t}v,\cdots$
where $C\in {\bf C},$ $f\in {\bf R}\to{\bf R}$. If the non-trivial
eigenvectors are complex, %% TRY TO PROVE THIS!!
\def\re{{\rm Re}}\def\im{{\rm Im}}
their exponential solutions form, in the reals,
$g = C_1\cos(\re(\lambda)t)\re(v) + C_2\sin(\re(\lambda)t)\im(v).$ %% DOUBLE CHECK.
\section{Nonhomogeneous System}
Extending to the nonhomogeneous system will take slightly different
paths depending on if the system has complex roots or has real roots.
But in either case, $\cos x*{\rm polynomial}+\sin x*{\rm polynomial}$
should be a particular solution.
\bye
|
