Lecture S9
Muddiest Points
General
Comments
In this lecture,
we began
learning
how
to
find
the time response of RC circuits to
initial
conditions.
To
do
this,
we used
the node method, and
used
the constitutive relations for
capacitors and
resistors.
This
led
to
coupled, first
order differential equations, which
we can
solve
by
guessing
that the solution
is
exponential, proportional to
e
st
. We then
find
that
solutions are
possible only
for
some exponentials.
This gives the characteristic values,
s
, of the system.
Responses
to MuddiestPartoftheLecture
Cards
(57
cards)
1.
Why do we assume that the guess
we made (
e
st
) is
right
and
go
ahead
as if it
could not be wrong? (1)
We assume the exponential as a
trial answer, and
show that
it
indeed
works.
That is,
the trial
solution
does indeed
solve the differential equations, and
you can
choose constants
(the characteristic vector)
to
solve the initial conditions. It
could
have been
wrong,
and
we would
have found
out
if it
were.
2.
When
doing row
reduction, can
you
alter
columns
at
all?
(1)
The row
reduction
approach seems
to have no rules, i.e., you
can
do
anything to
the rows.
The
legal
row
operations
are (1) scaling
of a
row; (2)
exchanging
rows; and
(3)
adding
the
multiple of
one row
to
another
row. You
can’t
do
column
operations.
3.
Why is
it necessary to do row
reduction
when
solving for
the [characteristic
vectors]? (1)
Row
reduction
is
the most
direct
way
to
find
solutions to
the equation
M
(
s
)
E
= 0
(8)
4.
When
solving for
characteristic vectors, why can
we choose an
arbitrary value
for
E
3
? (2)
If
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 Fall '05
 MarkDrela
 Differential Equations, Linear Algebra, Laplace, Laplacians

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