prof. david r. jackson ece dept. spring 2014 notes 32 ece 6341 1
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Prof. David R. JacksonECE Dept.
Spring 2014
Notes 32
ECE 6341
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Laplace’s MethodConsider
b g x
aI f x e dx
0 0
0
0, ,
, ,
g x x a b
g x g x x a b
Assumptions:
Note: If there is no point where the derivative of the g function vanishes, then integration by parts may be used, in exactly the same manner as was done for the case when there was a j in the exponent (Notes 28).
(“saddle point”)
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Laplace’s Method (cont.)
b g x
aI f x e dx
0x
g x
x
0g x
a b
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Laplace’s Method (cont.)
00b g x g xg x
aI e f x e dx
1 23
0x
1 0 g x g xe
x
The exponential function behaves similar to a function as .
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Laplace’s Method (cont.)
00
0~
b g x g xg x
aI f x e e dx
2
0 0 0 0 0
1
2g x g x g x x x g x x x
2
0 0 0
2
0 0
1
21
2
g x g x g x x x
g x x x
Hence
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Laplace’s Method (cont.)
Hence
20 0
0
1
20~
b g x x xg x
aI f x e e dx
20 0
0
1
20~
g x x xg xI f x e e dx
Since only the local neighborhood of x0 is important, we can write
Next, let
00 2
g xs x x
0
2
g xds dx
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Laplace’s Method (cont.)
Next use
2
0
00
2~ g x sI f x e e ds
g x
2
se ds
Then
Hence
0
00
2~ g xI f x e
g x
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Laplace’s Method (cont.)
Summary
0
00
2~ g xI f x e
g x
b g x
aI f x e dx
0 0
0
0, ,
, ,
g x x a b
g x g x x a b
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Complete Asymptotic Expansion (Using Watson’s Lemma)
00b g x g xg x
aI e f x e dx
Let 2
0 s g x g x
0x
0s
x
0s
a b
2s2s g x
We adopt the convention of positive and negative s as shown in the figure, in order to make the mapping x(s) unique.
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Complete Asymptotic Expansion (cont.)
Let
20
b
a
Sg x s
S
dxI e f x s e ds
ds
dx sh s f x s
ds
20
b
a
Sg x s
SI e h s e ds
aS bS
0s s
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Assume 0,1,2
~ 0nn
n
h s a s s
as
(This is Watson’s Lemma.)
20
0,1,2
~b
a
Sg x n sn S
n
I e a s e ds
Then
20
b
a
Sg x s
SI e h s e ds
Complete Asymptotic Expansion (cont.)
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Since only the local neighborhood of s = 0 is important,
Because of symmetry,
20
0,1,2
~ g x n sn
n
I e a s e ds
20
00,2,4
~ 2 g x n sn
n
I e a s e ds
n even
Complete Asymptotic Expansion (cont.)
(The error made in extending the limits is exponentially small.)
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Denote
Use
2
0
n snI s e ds
2
2
t s
dt s ds
12
210 0
2
1
2 2
nn
t tn n
t dtI e t e dt
t
Complete Asymptotic Expansion (cont.)
Then we have
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Now use 1
01 ! x tx x t e dt
1
2
1 1
22
n n
nI
1
21 0
2
1
2
nt
n nI t e dt
Hence
Complete Asymptotic Expansion (cont.)
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20
00,2,4
~ 2 g x n sn
n
I e a s e ds
2
102
1 1
22
n sn n
nI s e ds
0
10,2 2
1~
2g x n
nn
a nI e
Hence
Recall that
Complete Asymptotic Expansion (cont.)
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Summary
0
10,2,4 2
1~
2g x n
nn
a nI e
b g x
aI f x e dx
0,2,4...
~ 0nn
n
h s a s s
as
20
dx ss g x g x h s f x s
ds
Complete Asymptotic Expansion (cont.)
Note: The hard part is determining the an coefficients!
0 0
0
0, ,
, ,
g x x a b
g x g x x a b
Assume
Then
Note: Integer powers are
assumed, since h(s) is assumed to
be analytic
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Leading term:
so that
0~ 0h s a h
0 01
2
1~
2
g x aI e
0
0~
g xI a e
1
2
Complete Asymptotic Expansion (cont.)
Note
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0 0a h
dx
h s f x sds
20
2
s g x g x
dss g x
dx
Recall that
and that
To find h(0), we must evaluate the derivative term. To do this, we
take the derivative with respect to x:
00
0s
dxh f x
ds
Note: At s = 0 (x = x0), this yields 0 = 0 (not useful).
Complete Asymptotic Expansion (cont.)
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2 2
22 2
ds d ss g x
dx dx
0 , ( 0)x x s At
2
02ds
g xdx
Take one more derivative:
2ds
s g xdx
Complete Asymptotic Expansion (cont.)
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Hence
We then have
0
2dx
ds g x
00
20h f x
g x
Therefore 0
00
2~ g xI f x e
g x
Complete Asymptotic Expansion (cont.)
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0
00
2~ g xI f x e
g x
or
Complete Asymptotic Expansion (cont.)
This agrees with the result from Laplace’s method.
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Watson’s Lemma (Alternative Form)
Here we do not necessarily assume integer powers in the
expansion of the function, and we also start the integral at s = 0.
0
sI h s e ds
~ 0nn
n
h s a s s asAssume
This one-sided form occurs when integrating along branch cuts in the complex plane (discussed later).
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Watson’s Lemma (Alternative Form) (cont.)
Then
Let
0
sI h s e ds
~ 0nn
n
h s a s s as
0
~ n sn
n
I a s e ds
t s
dt ds
(This is Watson’s Lemma.)
Assume
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Hence
0
0
1 0
1
1
11
n
n
n
n
n
sn
t
t
n
I s e ds
t dte
t e dt
1
1~ 1
nn n
n
I a
t s
dt ds
Watson’s Lemma (Alternative Form) (cont.)
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Summary
1
1~ 1
nn n
n
I a
0
sI h s e ds
~ 0nn
n
h s a s s as
Watson’s Lemma (Alternative Form) (cont.)
Assume
Then