1

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

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

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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.)

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.)

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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.)

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.

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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