Finite Element Method

1. Finite Element Method

2. FEM Approaches
There are two main methods that are applied to solving
problems using the FEM method.
Variational methods that use the classical Rayleigh-Ritz
technique
Galerkin’s methods which is one technique in a family of
methods referred to as method of weighted residuals.

3. Variational Approach
In solving problems arising in physics and engineering it is
often possible to replace the problem of integrating a
differential equation by the equivalent problem of seeking
a function that gives a minimum value of some integral.
Problems of this type are called variational problems.
The methods that allow us to reduce the problem of
integrating a differential equation to the equivalent
variational problem are usually called variational
methods.

4. Variational Approach
What is a functional?
dx
y
y
x
F
y
I
b
a
)
,
,
(
)
(  

subjected to the boundary conditions
B
b
y
A
a
y 
 )
(
,
)
(
functional
Goal: Find a function F(x,y,y’) for which the functional
I(y) has an extremum (usually a minimum)

5. Variational Approach
Some problems:
dx
y
y
x
F
y
I
b
a
)
,
,
(
)
(  

NOT:
Question: Are there situations in which the function
F(x,y,y’) for which the functional I(y) in minimized is
ALSO a solution to the PDE and BCs??
.....
0
)
,
( 2
2
or
t
U
U
or
y
x
f
U 





 

6. Variational Approach
Question: Are there situations in which the function
F(x,y,y’) for which the functional I(y) in minimized is
ALSO a solution to the PDE and BCs??
Answer: Yes, all PDEs typically found in physics and
engineering have functionals or variational equations
whose solution is equivalent to solving the PDE
directly.

7. Variational Approach
Deriving those equations can be done using the calculus
of variations (beyond the scope of this class).
Answer: Yes, all PDEs typically found in physics and
engineering have functionals or variational equations
whose solution is equivalent to solving the PDE
directly.

8. Variational Approach
Name of equations PDE Variational principle
Homogeneous
wave equation with
sources
Homogeneous
wave equation
without sources
Diffusion equation
Homogenous
Poisson’s equation
Homogenous
Laplace’s equation
g
k 



 2
2
0
2
2




 k
0
2







t
k
g


2
0
2



 dv
g
k
I
v
 






 2
2
1
)
( 2
2
2
 dv
k
I
v
 




 2
2
2
2
1
)
(
dvdt
t
k
I
t v
 













 2
2
2
1
)
(
 dv
g
I
v
 




 2
2
1
)
(
2
 dv
I
v
 



2
2
1
)
(

9. Finite Element Method
The finite element method (FEM) has its origin in the field of
structural analysis. However, since then the method has
been employed in nearly all areas of computational
physics and engineering.
The FEM method, while more difficult to program than
either the finite difference (FD) or method of moments
(MOM), is a more powerful and versatile numerical
technique for handling problems involving complex
geometries and inhomogeneous media.

10. Basic concept
Although the behaviour may be complex when viewed over a large
region, a simple approximation may suffice over a small subregion.
The region is divided up into finite elements.
(usually, triangles or squares,
but can be more complicated)
Regardless of the shape the field is
approximated by a different
expression over each element,
maintaining continuity at adjoining
elements.

11. Solution Strategy: Variational Approach
The equations to be solved are usually stated not in terms of
field the variables but in terms of an integral-type functional
such as energy.
The functional is chosen such that the field solution makes
the functional stationary
The total functional is the sum of the integral over each element

12. Finite Element Method
The finite element method (FEM) involves basically four
steps:
(1) Discretize the solution region into a finite number of
subregions or elements
(2) Derive the governing equations for each element based
on either a variational approach or Galerkin’s method
(3) Assemble all the elements together in the solution
space.
(4) Solve the resulting system of equations

13. Finite Element Method
CREATING THE MESH
(1) Discretize the solution region into a finite number of
subregions or elements

14. Finite Element Method in 2D
CREATING THE MESH
(1) Discretize the solution region into a finite number of
subregions or elements
• divide the geometry into elements
(in 2D triangular elements are common)
• each element has a number of nodes
attached.
In this figure there are 8 elements (E1-E8)
and 9 nodes (N1-N9)

15. Finite Element Method
The mesh is often described using two tables
(element table and a node table)
Element Table
Element# Node1 Node2 Node3
1 4 7 8
2 4 8 5
3 8 9 5
......
8 2 6 3
Node Table
Node# x y
1 0 1
2 0.5 1
3 1 1
......
9 1 0

16. 2D FEM: Right Triangular Single Element
The simplest element is the right triangle (0,1,2)
0
The potential can be
calculated as
 
,
x y a bx cy
   
a,b,c unknown
! Smart to specify the
interpolation at the vertices!
Then:   1 0 2 0
0
,
x y x y
h h
   
 
 
  

17. Other elements
Square element: consider as two triangles. This is ok, but does
not lead to a smooth interpolation over the square. Better to use
a higher order interpolation that uses four nodal values
 
,
x y a bx cy dxy
    
This scheme leads to:
0 1
2
8
4 5
3
6 7 8
0
1
1
8
i
i
 

 

18. Element expansion
In general we wish to approximate the unknown function, (x,y),
Inside an element by an interpolation of its values at the vertices.
element shape
functions or basis functions. In FEM these
are usually interpolation functions
unknown anywhere
within the element
Provided the values
at the vertices are known



N
i
i
i y
x
y
x
1
)
,
(
)
,
( 


Number of vertices

19. Generalized Triangular
Develop the Governing Equations for a Single Element
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
cy
bx
a
y
x
Ve 


)
,
(
Assume the unknown variable (V), in each
element can be found as a linear interpolation
of its value at the three nodes.
FIND a, b and c































c
b
a
y
x
y
x
y
x
V
V
V
e
e
e
3
3
2
2
1
1
3
2
1
1
1
1

20. Finite Element Method
Develop the Governing Equations for a Single Triangular Element
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)































c
b
a
y
x
y
x
y
x
V
V
V
e
e
e
3
3
2
2
1
1
3
2
1
1
1
1
Solve for the unknowns (a, b and c) and then
substituting that result into
cy
bx
a
y
x
Ve 


)
,
(

21. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)



3
1
)
,
(
)
,
(
i
ei
i
e V
y
x
y
x
V 
where
     
 
     
 
     
 
     
 
1
2
1
3
1
3
1
2
1
2
2
1
1
2
2
1
3
3
1
1
3
3
1
1
3
2
2
3
3
2
2
3
3
2
1
2
1
2
1
)
,
(
2
1
)
,
(
2
1
)
,
(
y
y
x
x
y
y
x
x
A
y
x
x
x
y
y
y
x
y
x
A
y
x
y
x
x
x
y
y
y
x
y
x
A
y
x
y
x
x
x
y
y
y
x
y
x
A
y
x



























Shape or basis functions
(only function of geometry)

22. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
     
 
     
 
     
 
     
 
1
2
1
3
1
3
1
2
1
2
2
1
1
2
2
1
3
3
1
1
3
3
1
1
3
2
2
3
3
2
2
3
3
2
1
2
1
2
1
)
,
(
2
1
)
,
(
2
1
)
,
(
y
y
x
x
y
y
x
x
A
y
x
x
x
y
y
y
x
y
x
A
y
x
y
x
x
x
y
y
y
x
y
x
A
y
x
y
x
x
x
y
y
y
x
y
x
A
y
x



























Properties of the shape functions
1
)
,
(
,
0
,
1
)
,
(
3
1









i
i
i
y
x
j
i
j
i
y
x



23. 2D Example:
Laplace’s Equation
When the quantity being sought (here the electrostatic potential V)
Recall:
The true potential is known to minimize the electrostatic
field energy.
 dv
V
V
I
v
 

2
2
1
)
(
0
)
,
(
2

 y
x
V

24. Create the Mesh
Step #1: Discretize the surface/volume to be solved into small
finite elements. Label each element and its associated nodes
1
2
3
4
1 2 3

25. Basic concept: Review
Step #2: Approximate the complex solution within the whole region
in terms of a sum of solutions found within each element
1
2
3
4
1 2 3



e
N
e
e y
x
V
y
x
V
1
)
,
(
)
,
(

26. FEM Laplace’s Equation
Approximate the solution within each element in terms of
each value at the corresponding nodes. (i.e. interpolation)
(x1,y1)
(x2,y2)
(x3,y3) Element        
       
       
1 2 3 3 2 2 3 3 2
2 3 1 1 3 3 1 1 3
3 1 2 2 3 1 2 2 1
1
,
2
1
,
2
1
,
2
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A



 
     
 
 
     
 
 
     
 



3
1
)
,
(
)
,
(
i
ei
i
e V
y
x
y
x
V 

27. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
To derive an equation for each element we plug in our
linear approximation into an functional expression.
dv
y
x
V
dv
V
y
x
V
I
v i
i
ei
v i
ei
i
e  
  

































2
3
1
2
3
1
)
,
(
2
1
)
,
(
2
1
)
( 




3
1
)
,
(
)
,
(
i
ei
i
e V
y
x
y
x
V 
 dv
V
V
I
v
e
e  

2
2
1
)
(
ej
i j
j
v
i
ei
e V
ds
y
x
y
x
V
V
I  
 









3
1
3
1
)
,
(
)
,
(
2
1
)
( 


28. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
To derive an equation for each element we plug in our
linear approximation into an functional expression.
dv
y
x
V
dv
V
y
x
V
I
v i
i
ei
v i
ei
i
e  
  

































2
3
1
2
3
1
)
,
(
2
1
)
,
(
2
1
)
( 




3
1
)
,
(
)
,
(
i
ei
i
e V
y
x
y
x
V 
 dv
V
V
I
v
e
e  

2
2
1
)
(
ej
i j
j
v
i
ei
e V
ds
y
x
y
x
V
V
I  
 









3
1
3
1
)
,
(
)
,
(
2
1
)
( 

       
       
       
1 2 3 3 2 2 3 3 2
2 3 1 1 3 3 1 1 3
3 1 2 2 3 1 2 2 1
1
,
2
1
,
2
1
,
2
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A



 
     
 
 
     
 
 
     
 

29. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
Example: Laplace’s Equation
ds
y
x
y
x
V
V
I j
v
i
i j
ei
e )
,
(
)
,
(
2
1
)
(
3
1
3
1

 

 

 
dxdy
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 
 

 
Let
    
e
e
t
e
e V
C
V
V
I
2
1
)
( 
where
 











3
2
1
e
e
e
e
V
V
V
V and  











e
e
e
e
e
e
e
e
e
e
C
C
C
C
C
C
C
C
C
C
33
32
31
23
22
21
13
12
11
element coefficient matrix

30. Finite Element Method
Example: Laplace’s Equation
dxdy
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 
 

 
    
e
e
t
e
e V
C
V
V
I
2
1
)
( 
 
2
3
3
2
2
1
Q
P
Q
P
A 

       
       
       
1 2 3 3 2 2 3 3 2
2 3 1 1 3 3 1 1 3
3 1 2 2 3 1 2 2 1
1
,
2
1
,
2
1
,
2
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A
x y x y x y y y x x x y
A



 
     
 
 
     
 
 
     
 
 
j
i
j
i
e
ij Q
Q
P
P
A
C 

4
1
x
y
Ve(
x,y
)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
 
 
 
2
1
3
1
3
2
3
2
1
y
y
P
y
y
P
y
y
P





  
 
 
1
2
3
2
1
2
2
3
1
x
x
Q
x
x
Q
x
x
Q






Solution for triangular elements

31. Finite Element Method
x
y
Ve(x,y)
1
2
3
Ve1
(x1, y1)
Ve2
(x2, y2)
Ve3
(x3, y3)
dxdy
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 
 

 
Main property of the coefficient matrix:
e
ji
e
ij C
C 

32. Finite Element Method
Assembling of All Elements Together
Solve for the unknowns (a, b and c) and then
substituting that result into



N
e
e y
x
I
y
x
I
1
)
,
(
)
,
(
    
V
C
V
y
x
I
y
x
I
t
N
e
e
2
1
)
,
(
)
,
(
1

 

where
 

















N
V
V
V
V
V
....
3
2
1
and  

















NN
N
N
e
C
C
C
C
C
C
C
C
C
C
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1
31
22
21
1
13
12
11
Global coefficient matrix

33. Finite Element Method
Assembling of All Elements Together
 

















NN
N
N
e
C
C
C
C
C
C
C
C
C
C
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1
31
22
21
1
13
12
11
Coefficients are only non-zero for
elements that touch each other
1
2 4
3
5
E1
1-4-2
(1-2-3) E2
1-3-4
(1-2-3)
E3
3-5-4
(1-2-3)
 

























3
22
3
23
3
21
3
32
3
33
2
33
1
22
3
31
2
32
1
23
2
31
1
21
3
12
3
13
2
23
3
11
1
22
2
21
1
32
1
33
1
31
2
13
1
12
2
12
1
13
2
11
1
11
0
0
0
0
0
0
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C

34. Finite Element Method
Setting up and solving the resulting equations
The solution for the node values is the ones that result in the functional
obtaining its minimum value!
    
V
C
V
y
x
I
y
x
I
t
N
e
e
2
1
)
,
(
)
,
(
1

 

N
k
V
y
x
I
V
y
x
I
V
y
x
I
V
y
x
I
k
N
,....,
2
,
1
0
)
,
(
0
)
,
(
......
)
,
(
)
,
(
2
1















35. Finite Element Method
Setting up and solving the resulting equations
N
k
V
y
x
I
k
,....,
2
,
1
0
)
,
(




Example for five nodes:
 

































5
4
3
2
1
55
54
53
52
51
45
44
43
42
41
35
34
33
32
31
25
24
23
22
21
15
14
13
12
11
5
4
3
2
1
V
V
V
V
V
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
V
V
V
V
V
I
0
2
51
5
41
4
31
3
21
2
15
5
14
4
13
3
12
2
11
1
1












C
V
C
V
C
V
C
V
C
V
C
V
C
V
C
V
C
V
V
I

36. Finite Element Method
Setting up and solving the resulting equations
N
k
V
y
x
I
k
,....,
2
,
1
0
)
,
(




Example for five nodes:
0
2
51
5
41
4
31
3
21
2
15
5
14
4
13
3
12
2
11
1
1












C
V
C
V
C
V
C
V
C
V
C
V
C
V
C
V
C
V
V
I
In general:
N
k
C
V
N
i
ik
i ,....
3
,
2
,
1
0
1





37. Iterative solution method
At node k in a mesh with n nodes, we have the solution
e.g. see previous
slide
We note that Cki = 0 if node k is not directly connected to node i,
only nodes that are directly connected to node k contribute to Vk.
We apply this iteratively to all free nodes (not at fixed boundary
value) where the potential is unknown. Initially we assign either
0, random or some average value to each free node and then
iterate on the above equation until convergence is met.





n
k
i
i
ki
i
kk
k C
V
C
V
,
1
1

38. Finite Element Method
Example: Poisson’s Equation
dv
V
y
x
g
y
x
V
y
x
V
I
v i
ei
i
i
ei
i
i
ei
i
e  

 



















3
1
3
1
2
3
1
)
,
(
)
,
(
2
)
,
(
2
1
)
( 





3
1
)
,
(
)
,
(
i
ei
i
e V
y
x
y
x
V 
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
e g
ds
y
x
y
x
V
V
ds
y
x
y
x
V
V
I 














 


  
 
)
,
(
)
,
(
)
,
(
)
,
(
2
1
)
(
3
1
3
1
3
1
3
1




)
,
(
2
y
x
g
V 

 dv
gV
V
V
I
v
 

 2
2
1
)
(
2



3
1
)
,
(
)
,
(
i
ei
i g
y
x
y
x
g 
Equations for single element:

39. Finite Element Method
Example: Poisson’s Equation
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
e g
ds
y
x
y
x
V
V
ds
y
x
y
x
V
V
I 














 


  
 
)
,
(
)
,
(
)
,
(
)
,
(
2
1
)
(
3
1
3
1
3
1
3
1




)
,
(
2
y
x
g
V 

 dv
gV
V
V
I
v
 

 2
2
1
)
(
2
Equations for single element:
         
e
e
t
e
e
e
t
e
e g
T
V
V
C
V
V
I 

2
1
)
(
where
 











3
2
1
e
e
e
e
V
V
V
V and








  ds
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 

ds
y
x
y
x
T j
v
i
e
ij )
,
(
)
,
( 




40. Finite Element Method
Example: Poisson’s Equation )
,
(
2
y
x
g
V 

Equations for single element:          
e
e
t
e
e
e
t
e
e g
T
V
V
C
V
V
I 

2
1
)
(








  ds
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 
 ds
y
x
y
x
T j
v
i
e
ij )
,
(
)
,
( 



 
2
3
3
2
2
1
Q
P
Q
P
A 

 
j
i
j
i
e
ij Q
Q
P
P
A
C 

4
1
 
 
 
2
1
3
1
3
2
3
2
1
y
y
P
y
y
P
y
y
P





  
 
 
1
2
3
2
1
2
2
3
1
x
x
Q
x
x
Q
x
x
Q






Solution for triangular elements
 
2
3
3
2
2
1
Q
P
Q
P
A 







j
i
A
j
i
A
Te
ij
6
/
12
/
 
 
 
2
1
3
1
3
2
3
2
1
y
y
P
y
y
P
y
y
P





  
 
 
1
2
3
2
1
2
2
3
1
x
x
Q
x
x
Q
x
x
Q






Solution for triangular elements

41. Finite Element Method
Example: Poisson’s Equation )
,
(
2
y
x
g
V 

 dv
gV
V
V
I
v
 

 2
2
1
)
(
2
Put all the elements together:
         
g
T
V
V
C
V
V
I
V
I
t
t
N
e
e 

 
 2
1
)
(
)
(
1
Construct matrix and then solve:
       
N
k
V
g
T
V
V
C
V
V
V
I
k
t
t
k
,.....
3
,
2
,
1
0
2
1
)
(















42. Finite Element Method
Example: Poisson’s Equation )
,
(
2
y
x
g
V 

Example:
       
N
k
V
g
T
V
V
C
V
V
V
I
k
t
t
k
,.....
3
,
2
,
1
0
2
1
)
(














    0
2
1
5
4
3
2
1
55
54
53
52
51
45
44
43
42
41
35
34
33
32
31
25
24
23
22
21
15
14
13
12
11
5
4
3
2
1
5
4
3
2
1
55
54
53
52
51
45
44
43
42
41
35
34
33
32
31
25
24
23
22
21
15
14
13
12
11
5
4
3
2
1
1






















































































g
g
g
g
g
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
T
V
V
V
V
V
V
V
V
V
V
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
V
V
V
V
V
V

 




5
1
1
11
5
2
1
11
1
1
1
i
i
i
i
i
i g
T
C
C
V
C
V
Result in:
In General: 
 





N
i
i
ki
kk
N
k
i
i
ki
i
kk
k g
T
C
C
V
C
V
1
)
(
1
1
1

43. Finite Element Method
Example: Homogenous
Wave Equation
dv
y
x
g
y
x
y
x
k
y
x
I
v i
ei
i
i
ei
i
i
ei
i
i
ei
i
e  


 































3
1
3
1
2
3
1
2
2
3
1
)
,
(
)
,
(
2
)
,
(
)
,
(
2
1
)
( 








3
1
)
,
(
)
,
(
i
ei
i
e y
x
y
x 
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
e
g
ds
y
x
y
x
ds
y
x
y
x
k
ds
y
x
y
x
I



































 
 
 
)
,
(
)
,
(
)
,
(
)
,
(
2
)
,
(
)
,
(
2
1
)
(
3
1
3
1
3
1
3
1
2
3
1
3
1






)
,
(
)
,
(
)
,
( 2
2
y
x
g
y
x
k
y
x 







3
1
)
,
(
)
,
(
i
ei
i g
y
x
y
x
g 
Equations for single element:
 dv
g
k
I
v
 






 2
2
1
)
( 2
2
2

44. Finite Element Method
Example: Wave Equation
Equations for single element:
              
e
e
t
e
e
e
t
e
e
e
t
e
e g
T
T
k
C
I 








2
2
1
)
(
2
where
 















3
2
1
e
e
e
e and








  ds
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 

ds
y
x
y
x
T j
v
i
e
ij )
,
(
)
,
( 



)
,
(
)
,
(
)
,
( 2
2
y
x
g
y
x
k
y
x 




ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
e
g
ds
y
x
y
x
ds
y
x
y
x
k
ds
y
x
y
x
I



































 
 
 
)
,
(
)
,
(
)
,
(
)
,
(
2
)
,
(
)
,
(
2
1
)
(
3
1
3
1
3
1
3
1
2
3
1
3
1







45. Finite Element Method
Example: Wave Equation
with no sources
Equations for single element:
         
e
e
t
e
e
e
t
e
e T
k
C
I 






2
2
1
)
(
2
where
 















3
2
1
e
e
e
e and








  ds
y
x
y
x
C j
v
i
e
ij )
,
(
)
,
( 

ds
y
x
y
x
T j
v
i
e
ij )
,
(
)
,
( 



0
)
,
(
)
,
( 2
2




 y
x
k
y
x
ej
j
v
i
i j
ei
ej
j
v
i
i j
ei
e ds
y
x
y
x
k
ds
y
x
y
x
I 



















 


  
 
)
,
(
)
,
(
2
)
,
(
)
,
(
2
1
)
(
3
1
3
1
2
3
1
3
1





46. Finite Element Method
Example: Wave Equation
Put all the elements together:
       








 

T
k
C
I
I
t
t
N
e
e
2
2
1
)
(
)
(
2
1
0
)
,
(
)
,
( 2
2




 y
x
k
y
x

47. x
y
z
Finite Element Method: Waveguide
TM Modes
2
2
t
z k
k 
 

z
jk
z
z
e
y
x
z
y
x
E 

 )
,
(
)
,
,
(
TE Modes
z
jk
z
z
e
y
x
z
y
x
H 

 )
,
(
)
,
,
(
where 0
)
,
(
)
,
( 2
2




 y
x
k
y
x t
and

48. Finite Element Method: Waveguide
TM Modes
Put all the elements together:
       








 

T
k
C
I
I
t
t
t
N
e
e
2
2
1
)
(
)
(
2
1
0
)
,
(
)
,
( 2
2




 y
x
k
y
x t
Band Matrix Solution: Break the matrices up into sub-matrices that distinguishes
free-nodes from those that are known or prescribed nodes (i.e. boundary values):
    

































p
f
pp
pf
fp
ff
p
f
t
p
f
pp
pf
fp
ff
p
f
T
T
T
T
k
C
C
C
C
I
2
2
1 2
z
jk
z
z
e
y
x
z
y
x
E 

 )
,
(
)
,
,
(

49. Finite Element Method: Waveguide
TM Modes
0
)
,
(
)
,
( 2
2




 y
x
k
y
x t
Minimize the functional:
    

































p
f
pp
pf
fp
ff
p
f
t
p
f
pp
pf
fp
ff
p
f
T
T
T
T
k
C
C
C
C
I
2
2
1 2
z
jk
z
z
e
y
x
z
y
x
E 

 )
,
(
)
,
,
(
    0
0
2






















p
f
fp
ff
t
p
f
fp
ff
f
T
T
k
C
C
I

50. Finite Element Method: Waveguide
TM Modes
0
)
,
(
)
,
( 2
2




 y
x
k
y
x t
z
jk
z
z
e
y
x
z
y
x
E 

 )
,
(
)
,
,
(
For TM modes: 0

p
since tangential E field must vanish on PEC boundary
    0
2


















p
f
fp
ff
t
p
f
fp
ff T
T
k
C
C
  0
2


 f
ff
t
ff T
k
C

51. Finite Element Method: Waveguide
TM Modes 0
)
,
(
)
,
( 2
2




 y
x
k
y
x t
z
jk
z
z
e
y
x
z
y
x
E 

 )
,
(
)
,
,
(
  0
2


 f
ff
t
ff T
k
C
Pre-multiply both sides by
1

ff
T
 
 
  0
0
0
2
1
1
2
1












f
f
t
ff
ff
f
ff
ff
t
ff
ff
I
A
I
k
C
T
T
T
k
C
T

Where:
2
1
t
ff
ff
k
C
T
A





52. Finite Element Method
Project #2: Write a FEM program that calculates the mode shapes and
cutoff frequencies of the TM modes for the air filled ridge waveguide
shown below.
h
w w
d
t
1
1


r
r


PEC Walls