, then both of the above problems
can be solved in one single command:
>>
x
=(0:
s i z e
(
y
)

1)/
S
;
2
Underdamped
The underdamped plot is a bit more complex than previous weeks. We’re trying to model the
data with the following equation:
V
(
t
) =
V
0
e

t
/
2
τ
sin
ωt
τ
=
L
R
ω
=
r
1
LC

R
2
4
L
2
>>
modelUnder
=
@
(
p
,
x
)
p
(1) .
exp
(

x
./(2
p
(2) ) ) .
s i n
(
p
(3) .
x
+
p
(4) )
+
p
(5) ;
1
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p(1):
V
0
; notice this is not simply
V
(
t
= 0) = 0. You may guess the value of your first
peak, but this isn’t quite right either (although it should be close enough).
p(2):
τ
, it can be estimated as in labs 2 and 4 by looking at the peaks in your data as
an exponential decay (note that the actual decay time is 2
τ
, but this factor should be
irrelevant for your guessing).
p(3):
ω
(=
2
π
T
) where
T
is the period. The period is easily estimated as the time difference
between consecutive peaks in your data.
p(4): This is the phase
φ
, it is approximately zero if you selected your data to have (0,0)
as the first data point.
p(5): A vertical shift in the data that is approximately zero.
3
Overdamped
This is almost exactly the same as the previous model, except the hyperbolic sine is used. There
is not simply one set of criteria to use for guessing the following, it depends on where you ran
the selection over the data. Let’s narrow it to two cases:
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 Spring '11
 shpyrko
 Statistics, Frequency, Time series analysis, Exponential decay

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