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Finite Element Method in 1-D*ECE 222c, UCSD

Instructor: Vitaliy Lomakin

*Used material: Notes by J. Jin, ECE, UIUCand J. Jin, The Finite Element Method in EM, Wiley 2002

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Finite Element Method in 2-D*ECE 222c, UCSD

Instructor: Vitaliy Lomakin

*Based on the notes by J. Jin, ECE, UIUCand J. Jin, The Finite Element Method in EM, Wiley 2002

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Boundary-Value Problem

ΩΓ

n̂

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Equivalent Variational Problem

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Equivalent Variational Problem

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Equivalent Variational Problem

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Equivalent Variational Problem

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Basic FEM Steps

1. Discretization/subdivision of solution domain

2. Selection of interpolation schemes

3. Formulation of the system of equations

4. Solution of the system of equations

1-D: 2-D: 3-D:

Linear or higher-order polynomials

Using either the Ritz or Galerkin method:Formulate elemental equations and assemble

Using either a direct or iterative method

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FEM Analysis – Domain subdivision

Step 1: Domain Discretization

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FEM Analysis – Domain subdivision

Step 1: Domain Discretization

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FEM Analysis – Domain subdivision

Step 1: Domain Discretization

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FEM Analysis – Domain subdivision

Step 1: Domain Discretization

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FEM Analysis – Element interpolation

Step 2: Element Interpolation

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3

e

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FEM Analysis – Element interpolation

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FEM Analysis – Element formulation

Step 3: Formulation of the System of EquationsA. Elemental equations

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FEM Analysis – Element formulation

Elemental functional:

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FEM Analysis – Element formulation

Integration formula:

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FEM Analysis – Element formulation

Use matrix notation:

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FEM Analysis – Assembly

B. Assembly

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FEM Analysis – Assembly

Apply the stationarity condition:

Carry out the summation:

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FEM Analysis – Assembly

How to carry out the summation?

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FEM Analysis – Assembly

Example: 1

2

3

4

5

6

1

2

3

4

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FEM Analysis

1. Start from a null matrix and add in the first element:

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FEM Analysis

2. Add in the second element:

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FEM Analysis

3. Add in the third element:

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FEM Analysis

4. Add in the fourth element:

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FEM Analysis

5. Follow a similar procedure:

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FEM Analysis – Apply BC

C. Impose the Dirichlet Boundary Condition:

To impose , simply set:

To maintain symmetry, set:

Approach #1:

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FEM Analysis – Apply BC

After imposing , , :

RemainsSymmetric!

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FEM Analysis – Apply BC

Can be made smaller:

Worthwhile when there are many prescribed nodes.

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FEM Analysis – Apply BC

Approach #2 (Simple one):

To impose , simply set:

After imposing , , :

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FEM Analysis – Sample Program

c.....Input data descriptionc c nn total number of nodes c For i = 1 to nn, input:c x(i) x-coordinate c y(i) y-coordinate c end forc c ne total number of elements c For e = 1 to ne, input:c alpha(e) value of alphac beta(e) value of beta c f(e) value of f c For i = 1 to 3, input:c n(i,e) global node number c end forc end for

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FEM Analysis – Sample Program

c c n1 number of nodes with prescribed valuesc For i = 1 to n1, input:c p(i) prescribed value of phi c nd(i) global node number c end forc c.....Initialize the matrix [K]

do 1 i = 1, nndo 1 j = 1, nn

1 k(i,j) = 0.Cc.....Start to assemble all area elements in Omega

do 4 e = 1, ne

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FEM Analysis – Sample Program

c..... Calculate b^e_i and c^e_i (i=1,2,3) i = n(1,e)j = n(2,e)m = n(3,e)be(1) = y(j) - y(m) be(2) = y(m) - y(i) be(3) = y(i) - y(j) ce(1) = x(m) - x(j) ce(2) = x(i) - x(m) ce(3) = x(j) - x(i)

Cc..... Calculate Delta^e

deltae = 0.5*(be(1)*ce(2)-be(2)*ce(1))

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FEM Analysis – Sample Program

c..... Generate the elemental matrix [K^e]do 2 i = 1, 3 do 2 j = 1, 3

if (i.eq.j) then del_ij = 1.0

else del_ij = 0.0

endif2 ke(i,j) = alphax(e)*(be(i)*be(j)

& + ce(i)*ce(j))/(4.0*deltae) & + beta(e)*(1.+del_ij)*deltae/12.

cc..... Add [K^e] to [K]

do 3 i = 1, 3 do 3 j = 1, 3

3 k(n(i,e),n(j,e)) = k(n(i,e),n(j,e))+ke(i,j) c

4 continue

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FEM Analysis – Sample Program

c.....Impose the Dirichlet boundary condition do 8 i = 1, n1

b(nd(i)) = p(i) k(nd(i),nd(i)) = 1. do 7 j = 1, nn

if(j.eq.nd(i)) go to 7 b(j) = b(j) - k(j,nd(i))*p(i) k(nd(i),j) = 0.k(j,nd(i)) = 0.

7 continue 8 continue

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Electrostatic Problems

Partial differential equation:

Boundary conditions:

Continuity conditions:

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Electrostatic Problems

Example:

Problem: To compute the characteristic impedance

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Electrostatic Problems

Mesh: Equi-potential:

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Electrostatic Problems

Axisymmetric (body of revolution):

Along the z-axis:

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Electrostatic Problems

Example:

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Electrostatic Problems

Mesh:

Equi-potential:

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Magnetostatic Problems

2-D:

Continuity conditions:

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Magnetostatic Problems

Axisymmetric (body of revolution):

Continuity conditions:

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Electrodynamic Problems

2 mm resolution14 kinds of tissue

Currents:RF shield

Head:

Currentelements

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64 MHz 128 MHz

171 MHz 256 MHz

Example – Electric field (no load)

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64 MHz 128 MHz

171 MHz 256 MHz

Example – Magnetic field (no load)

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64 MHz 128 MHz

171 MHz 256 MHz

Example – Magnetic field (with load)

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Electrodynamic ProblemsWaveguide device analysis:

Question: How to find the boundary condition at AB and CD?

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Electrodynamic Problems

At AB:

At CD:

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Electrodynamic ProblemsExample:

Equi-potential:

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Electrodynamic Problems

Scattering analysis:

Question: How to find the boundary condition on theartificial boundary?

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Electrodynamic Problems

Asymptotic expansion:

First-order ABC:

Take the derivative:

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Electrodynamic Problems

From a circular to an non-circular boundary:

First-order ABC:

For total field:

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Electrodynamic Problems

Example:

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Electrodynamic Problems

Example:

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Electrodynamic Problems

Example:

Amplitude of the scattered field