lecture 4 -- fd analysis of anisotropic transmission lines

8/13/2019 Lecture 4 -- FD Analysis of Anisotropic Transmission Lines

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ECE 5390 Special Topics:21 st Century Electromagnetics

Instructor:

Office:Phone:EMail:

Dr. Raymond C. Rumpf

A337(915) 747 [email protected] Spring 2012

Finite-Difference Analysis

of AnisotropicTransmission Lines

Lecture #4

Lecture 4 1

Lecture Outline RF transmission lines Finite-difference analysis and transmission lines Generalization to transmission lines embedded in

anisotropic materials Formulation using yeeder()

Lecture 4 Slide 2


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RF TransmissionLines

What are Transmission Lines?

Lecture 4 Slide 4

Transmission lines are metallic structures that guide electromagnetic waves from DC to very high frequencies.

Microstrip Coaxial Cable

Stripline Slotline


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Characteristic Impedance

Lecture 4 Slide 5

The ratio of the voltage to current as well as the electric to magnetic field is a constant along the length of the transmission line. This ratio is the characteristic impedance .

The value of the characteristic impedance alone has little meaning. Reflections are caused whenever the impedance changes.

c L

Z C

1

377 21 1 ln 2.230 4.454 4.464 3.89c r r r

r r

H H H H Z

W W W W

S. Y. Poh, W. C. Chew, J. A. Kong, Approximate Formulas for Line Capacitance and Characteristic Impedance of Microstrip Line, IEEE Trans. Microwave Theory Tech. , vol. MTT29, no. 2, pp. 135 142, May 1981.

For small H/W,

2

2 2

1ln 8 32377

2 1 11ln 8 0.041 0.454

16 1

c

r r

r r

H W W H H

Z H W W W H H

For large H/W,

Analysis: Electrostatic Approximation

Lecture 4 Slide 6

Transmission lines are waveguides. To be rigorous, they should be modeled as such. This can be rather computationally intensive. An alternative is to analyze transmission lines using Laplaces equation.

The dimensions of a transmission are typically much smaller than the operating wavelength so the wave nature of electromagnetics is less important to consider. Therefore, we are essentially solving Maxwells equations as 0. Setting =0 is called the electrostatic approximation .

0

0

B

D

H J D t

E B t

0

0

0

B

D

H J

E

All of

the

field

components

are

separable. The transmission line is TEM.

Rigorously, the transmission line is quasi TEM.


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Analysis: Calculating DistributedInductance

Lecture 4 Slide 9

The voltage along the transmission line travels at the same velocity as the electric field so we can write

Assuming there are no magnetic materials, r =1 and this equation can be written as

20

1 1 r r V E v v LC c LC

20

r Lc C

This means that for the dielectric only case, we can calculate the distributed inductance directly from the distributed capacitance.

Dielectric materials should not alter the inductance. However if we use the value of C calculated previously, it will. This is incorrect. The solution is to

calculate capacitance with a homogeneous dielectric and then calculate inductance from this.

20

r

H

Lc C

Analysis: Calculating TLParameters

Lecture 4 Slide 10

The characteristic impedance is calculated from the distributed inductance and capacitance through

The velocity of a signal travelling along the transmission line is

c

L Z

C

1v

LC The effective refractive index is therefore

00

cn c LC

v Recall, both Z c and n are needed to analyze transmission line circuits.


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Two Step Model

Lecture 4 Slide 11

outer conductor

inner conductor

outer conductor

inner conductor

1

air

air

air

2

Homogeneous Case

Inhomogeneous Case

distributed inductance

distributed capacitance

L

C

How We Will AnalyzeTransmission Lines

Lecture 4 Slide 12

Step 1Construct homogeneous TL

2

src

1src

0h

h h

h h

V

L v v

V L v

202

0

20

1

h h A

h

C V dAV

Lc C

c

L Z

C 0n c LC

Step 4Construct inhomogeneous TL

Step 2Construct and solve matrix equation

Step 3Calculate distributed inductance

Step 5Construct and solve matrix equation

2

src

1src

0r r

V V

Lv v

V L v

Step 6Calculate distributed capacitance

202

0r

A

C V dAV

Step 7Calculate TL parameters

outer conductor

inner conductor

outer conductor

inner conductor

1

air

air

air

2


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Finite-DifferenceAnalysis of

Transmission

Lines

Reduce 3D Problem to 2D

Slide 14Lecture 4

Real devices are inherently three dimensional. Ignoring loss, the mode(s) in a transmission line are uniform in the direction of propagation except for phase accumulated.

, , , z E x y z A x y e

, A x y

x

y

z

progation constant j

complex amplitude,mode shape

accumulation of phase in z direction

z e

This means we can solve the problem by just analyzing the cross section. This is a two dimensional problem.


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Step 1: Choose a CoaxialTransmission Line

Slide 15Lecture 4

Choose a transmission line. Look at only the cross section.

1r 2r

3r

1 2 % TRANSMISSION LINE PARAMETERS

r1 = 0.1; %radius of inner conductor (0.1 mm)r2 = 1.5; %inner radius of outer conductor (1.5 mm)r3 = 1.6; %outer radius of outer conductor (1.6 mm)

er1 = 3.0; %dielectric constant of material 1er2 = 9.0; %dielectric constant of material 2

Step 2: Build arrays for the

conductors and dielectrics

Slide 16Lecture 4

Build device on a grid by creating separate arrays for the conductors and the dielectrics.

% GRID PARAMETERSNx = 64; %number of points along xNy = 64; %number of points along ySx = 3.2; %physical size of grid along y (3.2 mm)Sy = 3.2; %physical size of grid along y (3.2 mm)

% GRIDxa = Sx*linspace(-0.5,+0.5,Nx); %x axisya = Sy*linspace(-0.5,+0.5,Ny); %y axis[Y,X] = meshgrid(ya,xa); %gridRSQ = X.^2 + Y.^2; %radius squared

% BUILD INNER CONDUCTORCIN = (RSQ


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Step 3: Calculate the ( r / r ) Term

Slide 17Lecture 4

1 1 , r r r r r r

G x y x y x y

1 r r x

1 r r y

r

% COMPUTE GRADIENT

dx = xa(2) - xa(1) ;dy = ya(2) - ya(1) ;[Gy,Gx] = gradient(ER,dy,dx);

% DIVIDE BY ERGx = Gx ./ ER;Gy = Gy ./ ER;

Step 4: Identify the Component

Matrices

Slide 18Lecture 4

2

2 2

2 2 1

2 2

1

0

0

0 x y x x y y

r

r

x y

V V

V V G V G V x y x y

D D G D G Dv v v v

2

2 2

2 2

2 2

0

0

0

h

h h

h x h y

V

V V x y

D vDv

Homogeneous Case Inhomogeneous Case

1 1 2 2We need , , , , , and x y x y x yG G D D D D


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Step 5: Diagonalize the MaterialsMatrices

Slide 19Lecture 4

,1 ,1

,2 ,2

, ,

0 0

0 0 x y x y

x y

x y x y

x N N y N N

g g

g g

g g

G G

% DIAGONALIZE GGx = diag(sparse(Gx(:)));Gy = diag(sparse(Gy(:)));

Step 6: Construct the Derivative

Matrices

Slide 20Lecture 4

% COMPUTE DERIVATIVE OPERATORS[DX,D2X,DY,D2Y] = fdder(NS,RES,BC);


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Step 7: Compute the LaplaceEquation Operators

Slide 21Lecture 4

% COMPUTE LAPLACE EQUATIONSLh = D2X + D2Y; %homogeneous caseL = Lh + Gx*DX + Gy*DY; %inhomogeneous case

2 2 1 1

1 1

0 x y x x y y

h x x y y

D D G D G D v

L L G D G D

2 2

2 2

0 x y h

h x y

D D v

L D D

Homogeneous Case

Inhomogeneous Case

L

We Need a Source Term

Slide 22Lecture 4

1

Lv 0

v L 0 01

h h

h h

L v 0

v L 0 0


These are trivial solutions. We need a source term so our equations look like

src

rc1

s

L v

v vL

vsrc

sr 1

c

h h

h h

L v

v vL

v



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Step 8: Define the known potentials

Slide 23Lecture 4

On the conductors, the potential function is known because we are viewing this as a DC circuit.

V =0 everywhere in the outer conductor.

V =1 everywhere in the inner conductor.0V

1V

Step 9: Compute the Source Term

Slide 24Lecture 4

We must modify the matrix equations to enforce the known potentials.

% FORCE POTENTIALSVsrc = zeros(Nx*Ny,1);for ny = 1 : Ny

for nx = 1 : Nx% Force Inner Conductorif CIN(nx,ny)

m = (ny-1)*Nx + nx;Lh(m,:) = 0;Lh(m,m) = 1;L(m,:) = 0;L(m,m) = 1;Vsrc(m) = 1;

end% Force Outer Conductorif COUT(nx,ny)

m = (ny-1)*Nx + nx;Lh(m,:) = 0;Lh(m,m) = 1;L(m,:) = 0;L(m,m) = 1;

endend

end

1

2

metal

1

src

# # # # # # 0

# # # # # # 0

0 0 1 0 0 0 or 1

# # # # # # 0

# # # # # # 0 x y

x y

m

N N

N N

V

V

V

V

V

vvL


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Step 10: Solve the Matrix Equation toCompute V and V h

Slide 25Lecture 4

We now have two complete matrix equations ready to be solved.

src

1src

L v v

v L v

src

1src

h h

h h

L v v

v L v

% COMPUTE POTENTIAL FUNCTIONSVh = Lh\Vsrc;V = L\Vsrc;

!!! Note !!!Do NOT compute the inverse of L!!!!!!!!!!!!Use backward division!!!!!!!!

V = inv(L)*Vsrc; BAD BAD BAD

Step 11: Reshape to the 2D grid

Slide 26Lecture 4

We need to reshape the column vectors back to a 2D grid.

% RESHAPE THE POTENTIAL FUNCTIONS TO A 2D GRIDVh = reshape(Vh,Nx,Ny);V = reshape(V,Nx,Ny);

#

#

#

#

v

,V x y


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Step 12: Compute the ElectricField

Lecture 4 Slide 27

Next we calculate the electric field from the scalar potential. We can use this to visualize the fields as well as well as computing the transmission line parameters.

% COMPUTE THE ELECTRIC FIELD[Ey,Ex] = gradient(V,dx,dy);Ex = - Ex ;Ey = - Ey;

E V x E y E E

Step 13: Calculate thedistributed capacitance

Lecture 4 Slide 28

First we calculate the distributed capacitance for both the homogeneous case and the inhomogeneous case.

Note that,

Numerically, the integrations are approximated by

We set the potential of the inner conductor to 1 so V 0=1.

202

0r

A

C V dAV

202

0h

A

C V dAV

22 2 2

x yV E E E

2

0 , ,

y x

x y x y

y x

N N

x y n n n nn n

C E 20 , y x

x y

y x

N N

h x y n nn n

C E

% CALCULATE THE DISTRIBUTED CAPACITANCEEhsq = abs(Ehx). 2 + abs(Ehy). 2;Esq = abs(Ex).^2 + abs(Ey).^2;Ch = e0*dx*dy*sum(Ehsq(:));C = e0*dx*dy*sum(ER(:).*Esq(:));


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Step 14: Calculate theTransmission Line Parameters

Lecture 4 Slide 29

Second, we calculate the distributed inductance using the following equation.

Third, we calculate the characteristic impedance and effective refractive index of the transmission line.

% CALCUALTE THE DISTRIBUTED INDUCTANCEL = 1/c0^2/Ch;

20

1

h

Lc C

0 c L

Z n c LC C

% CALCUALTE Zc AND nZc = sqrt(L/C);nc = c0*sqrt(L*C);

Step 15: Show the results!

Lecture 4 Slide 30

Problem definition:

Problem results:

1

2

3

1

2

0.1 mm

1.5 mm

1.6 mm

3.0

9.0

r

r

r

pFnmeff mm mm554 , 121 , 6.0, 67.78c L C Z


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Generalization toAnisotropic Dielectrics

New Governing Equation

Lecture 4 Slide 32

The divergence condition for the electric field is

0 D But, we know that D=[ E , so this equation becomes

0r E The electric field E is related to the scalar potential V as follows.

E V

The inhomogeneous Laplaces equation is derived by substituting Eq. (3) into Eq. (2).

Eq. (1)

Eq. (2)

Eq. (3)

0r V


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Some Algebra (1 of 3)

Lecture 4 Slide 33

Since this is a scalar fields we cannot multiply common derivatives. We must expand the governing equation.

0

0

r

xx xy xz

yx yy yz

zx zy zz

V

x

y V x y z

z

0 z

The device is uniform in the zdirection.

There is no phase accumulation in an electrostatic model.


Lecture 4 Slide 34

Over governing equation is then

0

0

xx xy xz

yx yy yz

zx zy zz

x

y V x y z

This reduces to

2 2

2 2

0

0

xx xy

yx yy

xx yy

xy yx yy xx

xy yx

xV

y x y

x y

V x x x y y x y y

x y y x


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Lecture 4 Slide 35

The final form of the governing equation is then2 2

2 2

0

xx yy

yx xy yy xx

xy yx

x y

V x y x x y y

x y y x

Matrix Form on Scalar Grid

Lecture 4 Slide 36

The governing equation in matrix form is

Lv 0

2 2 xx x yy y xx yx x xy yy y xy x y yx y x L D D G G D G G D D D D D

diag diag

diag diag

xy xx xx xy

yx yy yx yy

x x

y y

G G

G G

The spatial derivatives here should be calculated explicitly to construct the G matrices.


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Formulation Usingyeeder()

The Grid Scheme

Lecture 4 Slide 38

V E x

E y

, 0 xx xyT r T T yx yy

xV V

y x y

Governing equation

2D Grid (Think of H mode)


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Matrix Form Using yeeder()

Lecture 4 Slide 39

0 0e er x y E D D e

Combining these matrix equations yields

h xh y

E V

De v

D

0 0h

xx xy xe er x y h

yx yy yV

DD D v

D

ER = [ERxx ERxy ; ERyx ERyy];L = [DEX,DEY]*ER*[DHX;DHY];

Think of v as being the magnetic field in Hmode analysis.

lecture 4 -- fd analysis of anisotropic transmission lines

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