Fall 2019 David R. Jackson - University of Houstoncourses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes...

1

ECE 6382

Green’s Functions

Notes are from D. R. Wilton, Dept. of ECE

David R. Jackson

Fall 2019

Notes 19

Green’s Functions

2

Green's Mill in Sneinton (Nottingham), England, the mill owned by Green's father. The mill was renovated in 1986 and is now a science centre.

George Green (1793-1841)

The Green's function method is a powerful and systematic method for determining a solution to a problem with a known forcing function on the RHS.

The Green’s function is the solution to a “point” or “impulse” forcing function.

It is similar to the idea of an “impulse response” in circuit theory.

Green’s Functions

3

Consider the following second-order linear differential equation:

( ) ( ) ( ) ( ) ( )1 ( )d dP x u x Q x u x f xw x dx dx

− + =

u f=or

( ) ( ) ( )1 d dP x Q xw x dx dx

≡− +

where

( f is a “forcing” function.)

Green’s Functions

4

Problem to be solved:

u f=

( ) ( )0, 0 ( )u a u b= = Dirichlet BCs

( ) ( )0, 0 ( )u a u b′ ′= = Neuman BCs

or

Green’s Functions (cont.)

5

We can think of the forcing function f(x) as being broken into many small rectangular pieces.

Using superposition, we add up the solution from each small piece.

Each small piece can be represented as a delta function in the limit as the width approaches zero.

Note:The ∆(x) function becomes a δ (x) function in

the limit as ∆x → 0.

( ) ( )1 , / 2, / 2

0,

x x xx x

∈ ∆ ∆∆ ≡ ∆ otherwise

( ) ( ) ( )P x x x= ∆ ∆1.0 x∆

ix

( )( ) ( )

( )

( )i i

i i

f x P x x

f x x x x

−

= ∆ ∆ −

x

( )f x

a

x∆ b

( )f x


6

( ) ( ) ( ) ( ) ( )1 1

, ,N N

i i i ii i

u x f x P x x f x x G x x∆ ∆= =

≈ = ∆∑ ∑

( ) ( ), i i iG x x x x x∆ ∆≡ −solution from single pulse centered at

From superposition:

( ) ( ) ( ) ( )1

, ,bN

i ii a

f x x G x x f x G x x dx∆=

′ ′ ′∆ →∑ ∫

( ) ( ), i i iPP x x x x x∆ ≡ −solution from single pulse centered at

0x∆ →Let

( ) ( ), G x x x x xδ′ ′ ′≡ −solution from centered at


7

The solution to the original differential equation (from superposition) is then

( ) ( ) ( ),b

a

u x f x G x x dx′ ′ ′= ∫

( ) ( ),G x x x xδ′ ′= −

The Green’s function G(x, x′) is defined as the solution with adelta-function at x = x′ for the RHS.

xx′

( )G x,x′

a a


8

There are two general methods for constructing Green’s functions.

Method 1:Find the solution to the homogenous equation to the left and right of the delta function, and then enforce boundary conditions at the location of the delta function.

( ) ( )( )

1

2

,,

,Au x x x

G x xBu x x x

′ ≤′ = ′≥

The Green’s function is assumed to be continuous. The derivative of the Green’s function is allowed to be discontinuous.

The functions u1 and u2 are solutions of the homogenous equation.

xx′

( )2Bu x( )1Au x

a b


9

Method 2:Use the method of eigenfunction expansion.

n n nψ λψ=

( ) ( ), n nn

G x x a xψ′ =∑

Eigenvalue problem:

We then have:

( ) ( )0, 0 ( )a bψ ψ= = Dirichlet BCs

( ) ( )0, 0 ( )a bψ ψ′ ′= = Neuman BCsor

Note:The eigenfunctions

are orthogonal.


10

Method 1

( ) ( ) ( )

( )( ) ( )

0 0

0

1lim lim

( , ) ( , )1 lim ( ) 1

x x

x x

x

x

d dGG dx P x Q x G dxw x dx dx

P x dG x x P x dG x xx x dx

w x dx dx

ε ε

ε εε ε

ε

εε

δ

′ ′+ +

→ →′ ′− −

+ + − − ′+

→′−

= − + ′ ′ ′ ′ ′ ′

′ = − − = − = ′

∫ ∫

∫

( ) ( ),G x x x xδ′ ′= −

( ) ( ) ( )( )2 1

w xBu x Au x

P x′

′ ′ ′ ′− = −′

( ) ( )( )

1

2

,,

,Au x x x

G x xBu x x x

′ ≤′ = ′≥

Integrate both sides over the delta function:

( )( )

( , ) ( , ) w xdG x x dG x xdx dx P x

+ − ′′ ′ ′ ′ − = − ′

xx′

( )2Bu x( )1Au x

a b


11

Method 1 (cont.)

Also, we have (from continuity of the Green’s function):

( ) ( )1 2Au x Bu x′ ′=

( ) ( )( )

1

2

,,

,Au x x x

G x xBu x x x

′ ≤′ = ′≥

xx′

( )2Bu x( )1Au x

a b


12

Method 1 (cont.)

We then have:

( ) ( )( ) ( )

( )( )1 2

1 2 0

w xu x u x A

P xu x u x B

′ ′ ′ ′ ′ −− ′= ′ ′−

( )( )

( ) ( )( )

( )2 1

1 2 1 2

,[ , ] [ , ]

w x u x w x u xA B

P x W u u P x W u u ′ ′ ′ ′

= − = − ′ ′

( )1 2 1 2 1 2[ , ] ( ) ( ) ( ) ( )W u u W x u x u x u x u x′ ′ ′ ′ ′ ′ ′= = − (Wronskian)

where

xx′

( )2Bu x( )1Au x

a b

( )W x′∆ = = −determinantNote :


13

Method 1 (cont.)

We then have:

( )

( )( )

( ) ( )( )

( )( )

( ) ( )( )

2 1

1 2

,

,

,

w x u x u xx x

P x W xG x x

w x u x u xx x

P x W x

′ ′′− < ′ ′ ′ =

′ ′ ′− > ′ ′

xx′

( )2Bu x( )1Au x

a b


14

Method 2

xa bx′

n=1n=2

n=3

n=4

The Green’s function is expanded as a series of eigenfunctions.

n n nψ λψ=

( ) ( ), n nn

G x x a xψ′ = ∑

where

The eigenfunctions corresponding to distinct eigenvalues are orthogonal (from Sturm-Liouville theory).


15

Method 2 (cont.)

xa bx′

n=1n=2

n=3

n=4

( ) ( ),G x x x xδ′ ′= −

( ) ( )n nn

a x x xψ δ ′= −∑

( ) ( )n nn

a x x xψ δ ′= −∑

( ) ( )n n nn

a x x xλ ψ δ ′= −∑

( ) ( ) ( ) ( ), ,n n n m mn

a x x x x xλ ψ ψ δ ψ′= −∑

( ) ( ) ( ) ( )*,n n n m mn

a x x w x xλ ψ ψ ψ′ ′=∑

( ) ( ) ( ) ( )*,m m m m ma x x w x xλ ψ ψ ψ′ ′=

Orthogonality

Delta-function property

Multiply both sides by ψm*(x) w(x) and then integrate from a to b.


16

Method 2 (cont.)

xa bx′

n=1n=2

n=3

n=4

( ) ( )( ) ( )

*m

mm m m

w x xa

x xψ

λ ψ ψ′ ′

=

Therefore, we have

( ) ( ) ( )( ) ( )

( )*

, nn

n n n n

w x xG x x x

x xψ

ψλ ψ ψ

′ ′′ =

∑

Hence

( ) ( ) ( ) ( )2b

n n na

x x x w x dxψ ψ ψ≡ ∫Note :

Application: Transmission Line

17

A short-circuited transmission line with a distributed current source:

The distributed current source is a surface current.

( )I z

+- ( )V z

( )dsI z0z = z h=

( ) ( ) [ ]A/mds sxI z J z=

x

z

( ) ( ) 1Ads sI z z z Iδ ′= − ⇒ =

Green’s function:

18

An illustration of the Green’s function:

Application: Transmission Line (cont.)

( ) ( ) ( )0

,h

dsV z I z G z z dz′ ′ ′= ∫

The total voltage due to the distributed current source is then

+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

z z′=1AsI =

19

Take the derivative of the first and substitute from the second:

ds

dV j LIdzdI I j CVdz

ω

ω

= −

= −

Telegrapher’s equations for a distributed current source:

(Please see the Appendix.)


( )2

2ds

d V j L I j CVdz

ω ω= − −

Note: j is used instead of i here.

L = inductance/meterC = capacitance/meter

20

Hence

zk LCω=where


( )2

22

dz s

d V k V j L Idz

ω+ = −

( ) ( ) ( )2

22

1 dz s

d V zk V z I z

j L dzω

+ = −

or

( ) ( ) ( )2

22

,1 ,zd G z z

k G z z z zj L dz

δω

′ ′ ′+ = − −

Therefore:

21


Compare with:

( ) ( ) ( ) ( ) ( )1 ( )d dP z u z Q z u z f zw z dz dz

− + =

( ) ( ) ( )2

22

1 dz s

d V zk V z I z

j L dzω

+ = −

Therefore:

( ) ( ) ( ) 2, 1, zw z j L P z Q z kω= = =

Note: The self-adjoint property required the operator to be real, which is not the case here since w(z) is not real. However, we can multiply both sides of the equation by j to make the operator real, and then divide the final answer by jat the end. This process does not affect the final result, so it is not done here.

22


( ) ( ) ( )1 1 sin zV z Au z A k z= =

( ) ( ) ( )( )2 2 sin zV z Bu z B k z h= = −

The general solution of the homogeneous equation is:

Method 1

+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

1A

22

2 0zd V k Vdz

+ =

Homogeneous equation:

23


( )

( )( )

( ) ( )( )

( )( )

( ) ( )( )

2 1

1 2

,

,

,

w z u z u zz z

P z W zG z z

w z u z u zz z

P z W z

′ ′′− < ′ ′ ′ =

′ ′ ′− > ′ ′

The Green’s function is:

( ) ( )1 sin zu z k z=

( ) ( )( )2 sin zu z k z h= −

+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

1A

24


The final form of the Green’s function is:

( )( ) ( )( ) ( )

( )

( ) ( ) ( )( )( )

sin sin,

,sin sin

,

z z

z z

k z h k zj L z z

W zG z z

k z k z hj L z z

W z

ω

ω

′−′− <

′′ = ′ − ′− > ′

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )1 2[ , ] sin cos cos sin sinz z z z z z zW z W u z u z k k z k z h k z k z h k k h ′ ′ ′ ′ ′ ′ ′= = − − − =

where

+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

1A

25


The eigenvalue problem is

Method 2

+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

1A

22

21

zd k

j L dzψ ψ λψ

ω

− + =

22

2ddzψ λ ψ′= − ( )2 2

zj L kλ ω λ′ ≡ +

This may be written as

where

Note: A minus sign in introduced in the

eigenvalue problem for convenience.

26

Application: Transmission Line (cont.)We then have:

( ) ( )2

22 , 0 0d h

dzψ λ ψ ψ ψ′= − = =

( ) ( )sinn

n

z znh

ψ λπλ

′=

′ =

The solution is:

( ) ( ) ( ) ( ) ( ) ( ) ( )* 2

0

sin2

b h

n n n na

n z hz z z z w z dz j L dz j Lhπψ ψ ψ ψ ω ω = = =

∫ ∫Note :

( )2

2

sin

1

n

n z

n zzh

n kj L h

πψ

πλω

=

= −

( )2 21n n zk

j Lλ λ

ω′= −Recall :

27


( ) ( ) ( )( ) ( )

( )*

, nn

n n n n

w z zG z z z

z zψ

ψλ ψ ψ

′ ′′ =

∑

We then have:

where

( )2

2 2 21 1n n z z

nk kj L j L h

πλ λω ω

′= − = −

( ) sinnn zz

hπψ =

( ) ( ) ( )2n nhz z j Lψ ψ ω=

( )w z j Lω′ =

28


The final solution is then:

( ) ( ) 21 2

sin2, sin

nz

n zn zhG z z j L

h hnkh

ππω

π

∞

=

′ ′ = − −

∑

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Summary


( ) ( ) 21 2

sin2, sin

nz

n zn zhG z z j L

h hnkh

ππω

π

∞

=

′ ′ = − −

∑

( )( ) ( )( ) ( )

( )

( ) ( ) ( )( )( )

sin sin,

sin,

sin sin,

sin

z z

z z

z z

z z

k z h k zj L z z

k k hG z z

k z k z hj L z z

k k h

ω

ω

′−′− <

′ = ′ − ′− >

zk LCω=+- ( ) ( ),V z G z z′=

z z′=

( )I z0z = z h=

1A

30


Other possible Green’s functions for the transmission line:

We solve for the Green’s function that gives us the current I(z)due to the 1A parallel current source.

We solve for the Green’s function giving the voltage due a 1V series voltage source instead of a 1A parallel current source.

We solve for the Green’s function giving the current to due a 1V series voltage source instead of a 1A parallel current source.

Cavity Model for Patch Antenna

Next we use the cavity model and the method of eigenfunction expansion to solve for the input impedance of the rectangular

microstrip patch antenna.

31

xL

z

h rε

Cavity Model

0 0

0 0

e

e

x x Ly y W= + ∆

= + ∆

Note: ∆L is often chosen from Hammerstad’s formula∆W is often chosen from Wheeler’s formula

32

The coordinates (x0, y0) are measured from the corner of the physical patch.

Accounting for fringing:

0 0( , )e ex yPMC

eL

eW

x

y

PEC

22

e

e

L L LW W W

= + ∆= + ∆

( )

( )

0.3 0.264/ 0.412

0.258 0.8

effr

effr

WhL h

Wh

ε

ε

+ + ∆ = − +

ln 4/W hπ

∆ =

Cavity Model (cont.)

eff0e rck k ε=Let

( )effeff1rc r jlε ε= −

eff eff1 1 1 1 1tan

d c sp sw

lQ Q Q Q Q

δ= = = + + +

Assume no z variation (the probe current is constant in the z direction.)

33

0 0( , )e ex yPMC

eL

eW

x

y

PEC

1tan ddQ

δ =Note :

Cavity Model (cont.)We first derive the Helmholtz equation for Ez.

Substituting Faraday’s law into Ampere’s law, we have

efficH J j E

E j Hωε

ωµ∇× = +∇× = −

( )

( )( )

eff

2

2 2

2 2

1 ic

ie

ie

ie

E J j Ej

E j J k E

E E j J k E

E k E j J

ωεωµ

ωµ

ωµ

ωµ

− ∇× ∇× = +

⇒∇× ∇× = − +

⇒∇ ∇⋅ −∇ = − +

⇒∇ + =34

Cavity Model (cont.)Hence

Denote

2 2 iz e z zE k E j Jωµ∇ + =

( , ) ( , )zx y E x yψ =

( , ) ( , )izf x y j J x yωµ=

Then

35

2 2 ( , )ek f x yψ ψ∇ + =

Introduce eigenfunctions of the 2-D Laplace operator:

For rectangular patch we have, from separation of variables,

( , )mn x yψ2 2( , ) ( , )mn mn mnx y x yψ λ ψ∇ = −

0mnCn

ψ∂=

∂

2 22

( , ) cos cosmne e

mne e

m x n yx yL W

m nL W

π πψ

π πλ

=

= +

36

2mnλ− = eigenvalue


Note:The eigenvalues are

real and the eigenvalues are

orthogonal.

Cavity Model (cont.)Assume an “eigenfunction expansion”:

Hence

,( , ) ( , )mn mn

m nx y A x yψ ψ=∑

2 2 ( , )ek f x yψ ψ∇ + =

2 2

, ,( , )mn mn e mn mn

m n m nA k A f x yψ ψ∇ + =∑ ∑

Using the properties of the eigenfunctions, we have

( )2 2

,( , ) ( , )mn e mn mn

m nA k x y f x yλ ψ− =∑

This must satisfy

37

( ), 0,1, 2m n =

Cavity Model (cont.)Multiply the previous equation by and integrate.

Note that the eigenfunctions are orthogonal, so that

Denote

( , )m n x yψ ′ ′

*( , ) ( , ) 0 ( , ) ( , )mn m nS

x y x y dS m n m nψ ψ ′ ′ ′ ′= ≠∫

2, ( , )mn mn mnS

x y dSψ ψ ψ< > = ∫

( )2 2 , ,mn e mn mn mn mnA k fλ ψ ψ ψ− < > = < >

We then have

38

Note: The eigenfunctions are real, so we can drop the conjugate here.

Cavity Model (cont.)Hence, we have

Using

2 2

, 1,

mnmn

mn mn e mn

fAk

ψψ ψ λ

< >= < > −

2 2

, 1,

iz mn

mnmn mn e mn

JA jk

ψωµψ ψ λ

< >= < > −

The field inside the patch cavity is then given by

,( , ) ( , )z mn mn

m nE x y A x yψ=∑

39

( , ) ( , )izf x y j J x yωµ=


2

2

2

2

*

*

*

,

*

,

2

,

1 ( , )21 ( , )212

1 ,2

,1

1

12

,

,

,

iin z z

V

iz z

S

imn z

m nS

imn z

m n

imn z

m n mn mn

mn

imn z

mn

e

mn e

m

mn

n

A

P E x y J dV

h E x y J dS

h J dS

h J

Jh j

k

Jjk

λ

ψω

ψ

ψ

ψωµ

ψ

µψ ψ λ

ψ

< >

= −

= −

= −

= − < >

< > = −

−

< >

< >

−

∫

∫

∑∫

∑

∑

To calculate the input impedance, we first calculate the complex power going into the patch as

40

The integral is over the surface of the probe current.


Also, from circuit theory,

212in in inP Z I=

so

22 in

inin

PZI

=

41

2

2 2 2,

,1 1,

imn z

inm n mn mn e mnin

JZ j h

kI

ψωµ

ψ ψ λ

< > = − < > − ∑

Hence we have

, 0 0m n m n

∞ ∞

= =

=∑ ∑ ∑

Cavity Model (cont.)For the rectangular patch:

where

2 22

eff0

cos cosmne e

mne e

e rc

m x n yL W

m nL W

k k

π πψ

π πλ

ε

=

= +

=

( )effeff1rc r jlε ε= −

2 2

0 0

, cos cose eL W

mn mne e

m x n ydx dyL Wπ πψ ψ

=

∫ ∫

42

We need:


so

( )( )0 0, 1 12 2

e emn mn m n

W Lψ ψ δ δ = + +

0

1, 00, 0m

mm

δ=

= ≠

43


We also need . , imn zJψ

44

To calculate this, assume a strip model for the probe feed:

0 0( , )e ex y

eL

eW

x

y

0 0( , )e ex ypWpa

0 0( , )e ex y

( )( )

0 022

0

, ,2 2

2

4

p pe einsz

p e

p p

W WIJ y y y yW

y y

W a

π

= ∈ − +

− −

=

For a “Maxwell” strip current assumption, we have:

Note: The total probe current is Iin.

45

0 0( , )e ex ypW


( ) 0 0

32

, ,2 2

4.482

p pe einsz

p

p p p

W WIJ y y y yW

W a e a

= ∈ − +

=

For a uniform strip current assumption, we have:

Note: The total probe current is Iin.

46

0 0( , )e ex ypW


0

0

20

2

20

0

2

0 0 0

, cos cos

cos cos '

cos cos cos sin

pe

pe

p

p

Wy

ei in

mn zW e e p

y

We

ein

Wp e e

e e ein

p e e e e

m x In yJ dyL W W

I m x n y y dyW L W

I m x n y n yn yW L W W W

π πψ

π π

π π ππ

+

−

+

−

=

′ = +

′= −

∫

∫

2

2

0 0

sin

cos cos sinc2

p

p

W

W e

e epin

pp e e e

n y dyW

n WI m x n y WW L W W

π

ππ π

+

−

′′

=

∫

Assume a uniform strip current model:

0ey y y′= +

Use

Integrates to zero

47


Uniform Model (cont.)

Hence

0 0, cos cos sinc2

e epi

mn z ine e e

n Wm x n yJ IL W W

ππ πψ

=

sinc2

p

e

n WWπ

Note: It is the term that causes the series for Zin to converge.

Note:We cannot assume a probe of zero radius,

or else the series will not converge – the input reactance will be infinite.

48

Cavity Model (cont.)Summary

2

2 2 20 0

,1 1,

imn z

inm n mn mn e mnin

JZ j h

kI

ψωµ

ψ ψ λ

∞ ∞

= =

< > = − < > − ∑∑

49

( )( )0 0, 1 12 2

e emn mn m n

W Lψ ψ δ δ = + +

0 0, cos cos sinc2

e epi

mn z ine e e

n Wm x n yJ IL W W

ππ πψ

=

eff0e rck k ε= ( )eff

eff1rc r jlε ε= −

where

2 2

mne e

m nL Wπ πλ

= +

eff 1/l Q=

32 4.482p p pW a e a=

Cavity Model (cont.)CAD Formulas for Q Factors*

50

( )effeff1rc r jlε ε= − eff eff

1 1 1 1 1tand c sp sw

lQ Q Q Q Q

δ= = = + + +

1tand

d

Qδ

=

where

0 0ave

( )2c

s

k hQR

η =

( )ave patch ground / 2s s sR R R= +

1 0

3 116 /

ersp

e

LQpc W hε

λ

≈

hed

hed1r

sw spr

eQ Qe

= − *Derived in ECE 6345.

1sR

σδ=

2δωµσ

=

Cavity Model (cont.)CAD Formulas for Q Factors (cont.)

51

( )

hed3

01

1

3 1 11 14

r

r

e

k hc

πε

=

+ −

1 2

1 2 / 51r r

cε ε

= − +

( ) ( ) ( ) ( )

( ) ( )

2 4 2220 2 4 0 2 0

2 22 2 0 0

3 11 210 560 5

170

ap k W a a k W c k L

a c k W k L

= + + + +

+

2 0.16605a = −

4 0.00761a =

2 0.0914153c =−

Appendix

52

Allow for distributed sources

+−L z∆

C z∆dsi z∆

dsv z∆

v+v -

+

-

+

-

∆z

i+i-

This appendix presents a derivation of the telegrapher’s equations with distributed sources.

53

( ) ( )d ds sv v v z L z i i z

t+ − −∂− = ∆ − ∆ + ∆

∂

ds

v iv Lz t∂ ∂

= −∂ ∂

0 :z∆ →

+−L z∆

C z∆dsi z∆

dsv z∆

v+v -

+

-

+

-

∆z

i+i-

Appendix (cont.)

54

0 :z∆ →

+−L z∆

C z∆dsi z∆

dsv z∆

v+v -

+

-

+

-

∆z

i+i-

( )ds

vi i i z C zt

++ − ∂− = ∆ − ∆

∂

ds

i vi Cz t∂ ∂

= −∂ ∂

Appendix (cont.)

55

In the phasor domain:

ds

ds

v iv Lz ti vi Cz t

∂ ∂= −

∂ ∂∂ ∂

= −∂ ∂

ds

ds

dV V j LIdzdI I j CVdz

ω

ω

= −

= −

jt

ω∂→

∂

Appendix (cont.)

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Fall 2019 David R. Jackson - University of Houstoncourses.egr.uh.edu/ECE/ECE6382/Class Notes/Notes...

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