Post on 05-Jan-2016
1
Prof. David R. JacksonECE Dept.
Spring 2014
Notes 32
ECE 6341
2
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”)
3
Laplace’s Method (cont.)
b g x
aI f x e dx
0x
g x
x
0g x
a b
4
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 .
5
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
6
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
7
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
8
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
9
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.
10
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
11
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.)
12
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.)
13
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
14
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.)
15
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.)
16
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
17
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
18
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.)
19
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.)
20
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.)
21
0
00
2~ g xI f x e
g x
or
Complete Asymptotic Expansion (cont.)
This agrees with the result from Laplace’s method.
22
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).
23
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
24
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.)
25
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