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18.03
Muddy
Card
responses,
April
14,
2006
1.
A
number
of
people
brought
up
the
point
made
at
the
end
of
Lecture
25,
on
April
w
12:
how
do
we
know
what
initial
conditions
yield
the
unit
step
or
impulse
responses?
This
is
a
tricky
point
and
I
did
not
explain
it
completely.
One
point
to
be
made
is
this:
in
the
case
of
the
unit
impulse
response,
when
we
have
a
n
x
(
n
)
+
· · ·
+
a
0
x
=
δ
(
t
),
the
solution
x
should
not
be
too
wild
at
t
=
0
since
I
have
to
be
able
to
diﬀerentiate
it
n
times
in
order
to
apply
the
diﬀerential
operator
to
it.
If
I
wind
up
with
a
discontinuity
in
the
(
n
−
2)nd
derivative,
for
example,
then
when
I
diﬀerentiate
once
more
I
get
a
delta
function,
and
we
have
not
tried
to
understand
what
happens
when
you
diﬀerentiate
a
delta
function.
So
the
derivatives
up
to
the
(
n
−
2)nd
should
be
zero
at
t
=
0
(since
they
are
zero
for
t
<
0).
The
(
n
−
1)st
should
be
such
that
when
I
diﬀerentiate
once
more
and
multiply
by
a
n
I
get
the
delta
function:
so
it
should
increase
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.
 Winter '08
 Staff

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