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ECE 5322 21st Century Electromagnetics
Instructor:Office:Phone:E‐Mail:
Dr. Raymond C. RumpfA‐337(915) 747‐[email protected]
Transmission Lines Embedded in Arbitrary
Anisotropic Media
Lecture #4
Lecture 4 1
Lecture Outline
• RF transmission lines
• Transmission lines embedded in anisotropic media
• Finite-difference analysis of transmission lines embedded in arbitrary anisotropic media– Formulation
– Implementation
– Example
• Generalization of the method
Lecture 4 Slide 2
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RF Transmission Lines
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
LZ
C
1 2
377 21 1 ln 2.230 4.454 4.464 3.89c r r r
rr
H H H HZ
W W WW
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. MTT‐29, 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 HHZ
H W W
W H H
For large H/W,
21st Century Electromagnetics 6
Approximate Equations
0
60ln
r
bZ
a
Coaxial
10
120cosh
2r
dZ
ab
Twin‐Lead
0
120
r
dZ
w
Parallel Plate
0
60 2 8 ln 1
4
120 1
2 3 ln 1.444 1.393
1 1
2 2 1 12 2r r
d w
D d w dZ
w
D w d w d d
Dd
Microstrip
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Analysis: Electrostatic Approximation
Lecture 4 Slide 7
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 Laplace’s equation.
The dimensions of transmission lines are typically much smaller than the operating wavelength so the wave nature of electromagnetics is less important to consider. Therefore, we are essentially solving Maxwell’s equations as 0. Setting =0 is called the electrostatic approximationand the solution approximates the TEM solution of a waveguide.
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.
Analysis: Inhomogeneous Isotropic Laplace’s Equation
Lecture 4 Slide 8
The divergence condition for the electric field is
0D
But, we know that D=rE, so this equation becomes
0r E
The electric field E is related to the scalar potential V as follows.
E V
The inhomogeneous Laplace’s equation is derived by substituting Eq. (3) into Eq. (2).
Eq. (1)
Eq. (2)
Eq. (3)
2 0r
r
V V
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Analysis: Calculating Distributed Capacitance
Lecture 4 Slide 9
In the electrostatic approximation, the transmission line is a capacitor. The total energy stored in a capacitor is
12
A
U D E dA
The integral is performed over the entire cross section of device. For open devices like microstrips, this is infinite area. In practice, we integrate over a large enough area to incorporate as much of of the electric field as possible.
This equation is valid in anisotropic media.
The capacitance is related to the total stored energy through
20
2
CVU V0 is the voltage across the capacitor.
If we set the above equations equal and substitute into the expression, we derive the equation for the distributed capacitance.
20
1
A
C D E dAV
E V
Analysis: Calculating Distributed Inductance
Lecture 4 Slide 10
The voltage along the transmission line travels at the same velocity as the electric field so we can write
Assuming the surrounding medium is homogeneous, we can calculate the distributed inductance from the above equation as
20
1 1 r r
V Ev v LCcLC
20
r rLc 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 Ch and then calculate inductance from this. It is simplest to use just air.
20
1
h
Lc C
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Analysis: Calculating TL Parameters
Lecture 4 Slide 11
The characteristic impedance is calculated from the distributed inductance and capacitance through
The velocity of a signal travelling along the transmission line is
c
LZ
C
1v
LC
The effective refractive index neff is therefore
0eff 0
cn c LC
v Both Zc and neff are needed to analyze
transmission line circuits using network theory.
Two Step Modeling Approach
Lecture 4 Slide 12
outer conductor
inner conductor
outer conductor
inner conductor
1
air
air
air
2
Homogeneous Case
Inhomogeneous Case
distributed inductance
distributed capacitance
L
C
Step 1
Step 2
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How We Will Analyze Transmission Lines
Lecture 4 Slide 13
Step 1Construct homogeneous TL
2
src
1src
0h
h h
h h
V
L v v
v L v
02
0
20
1
h h h
A
h
C D E dAV
Lc C
c
LZ
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
02
0 A
C D E dAV
Step 7Calculate TL parameters
outer conductor
inner conductor
outer conductor
inner conductor
1
air
air
air
2
Transmission Lines Embedded
in Anisotropic Media
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Effect of Strength of Anisotropy
Lecture 4 15
• Dielectric anisotropy has no effect on the distributed inductance.
• Capacitance increases and impedance lowers with increasing yy.
• Field develops most strongly in the direction with highest permittivity.
• The degree to which field follows highest is proportional to the strength of the anisotropy.
Effect of Spatially Varying the Tensor Orientation
Lecture 4 16
• Impedance changes in the presences of the anisotropic medium.
• Field actually “follows” the anisotropy.
• Impedance changes slightly with tilt.
• Impedance relatively constant regardless of spatial variance.
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Isolation of a Microstrip
Lecture 4 17
Ordinary microstrip with metal ball in proximity. Near‐field is perturbed affecting the signal in the line.
Same microstrip embedded in an anisotropic material with dielectric tensor tilted away from ball. Near‐field is less perturbed, affecting the
line much less.
Experimental Results
Lecture 4 18
Ordinary Microstrip
Embedded Microstrip
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Finite-Difference Analysis of Transmission Lines Embedded in Arbitrary
Anisotropic Media
Formulation
The Grid
Lecture 4 20
We have two unknowns, V and E, so we adopt a staggered grid.
We can use our yeeder() function to construct the derivative operators.
V Ex
Ey
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Governing Equations
Lecture 4 21
0 0
x
y
z
x xx xy xz x
y yx yy yz y
z zx zy zz z
x
y
z
D
D Dx y z
D
D E
D E D E
D E
E x
E V E y V
E z
Reduction to Two Dimensions
Lecture 4 22
x y z
0
00
zx
xy
yz
x
y
z
DD
DD
DD x y
D
D
D
0zxx xy xz x
x xx xy xyx yy yz y
y yx yy yzx zy zz z
x
y
z
EE
D EE
D EE
E
E
E
x
y
z
x
y
E xV V
E y
Transmission line properties are independent of xz, yz, zx, zy, and zz.
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Finite-Difference Approximation
Lecture 4 23
? ?
0
?
x xe ex
xx xy
y
yy y
x xx xy x x x
y yx y x yyy y y y
vx x x
vy y y
D
Dx y
D E
D E
E xV
E y
dD D 0
d
d e
d e
e Dv
e D
ε ε
ε ε
The Problem with the Tensor
Lecture 4 24
In the first finite‐difference equation, Ey is in a physically different position than Ex and Dx.
Similarly, in the second finite‐difference equation, Ex is in a physically different position than Ey and Dy.
The finite‐difference equations above are incorrect because not all of the terms exist at the same position in space.
,???
, , , ,
???, , ,, ,
x xx x xy yx xx xy x
y yx yy y y yx x yy y
i j i j i j i jx xx x xy
i j i j i j i jy
i jy
i jxyx yy y
D E ED E
D E D E E
D E
D E
E
E
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The Correction
Lecture 4 25
, , 1, 1, , 1 , 1 1, 1 1, 1, , ,
, , 1, 1, , 1 , 1 1, 1 1, 1, , ,
4
4
i j i j i j i j i j i j i j i jxy y xy y xy y xy yi j i j i j
x xx x
i j i j i j i j i j i j i j i jyx x yx x yx x yx xi j i j i j
y yy y
x xx x x y xy y
y x y yx x
E E E ED E
E E E ED E
d ε e R R ε e
d R R ε e xy yε e
Recall the interpolation matrices from Lecture 10 in “Computational Electromagnetics.”
, , , , , and x y z x y z R R R R R R
interpolation matrix
axis along which average is taken
direction of adjacent cell
a
a
R
Revised Finite-Difference Approximation
Lecture 4 26
0x xe e
x yy y
x xx xy x x xxx x y xy
y yx yy y y yx y yx yy
vx x x
vy y y
D
Dx y
D E
D E
E xV
E y
dD D 0
d
d eε R R ε
d eR R ε ε
e Dv
e D
We modify our tensor by incorporating the interpolation matrices.
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Matrix Form of the Inhomogeneous Laplace’s Equation
Lecture 4 27
Eq. 1
Eq. 2
Eq. 3
xe ex y
y
x xxx x y xy
y yx y yx yy
vx x
vy y
dD D 0
d
d eε R R ε
d eR R ε ε
e Dv
e D
We now derive the matrix form of the inhomogeneous Laplaces’ equation.
Step 1: Substitute Eq. (2) into Eq. (1)
xxx x y xye ex y
yx y yx yy
eε R R εD D 0
eR R ε ε
Step 2: Substitute Eq. (3) into the above
vxx x y xy xe e
x y vx y yx yy y
ε R R ε DD D v 0
R R ε ε D
Step 3: Drop the negative sign
vxx x y xy xe e
x y vx y yx yy y
ε R R ε DD D v 0
R R ε ε D
Matrix Form of the Homogeneous Laplace’s Equation
Lecture 4 28
For the homogeneous case, our tensor reduces to
xx x y xy
x y yx yy
ε R R ε I 0
R R ε ε 0 I
Substituting this into the inhomogeneous Laplace’s equation gives us
vxe e
x y vy
vxe e
x y vy
DI 0D D v 0
D0 I
DD D v 0
D
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The Excitation Vector b
Lecture 4 29
We have Lv = 0. This is not solvable because we have not stated the applied potentials. When we do this, our equation becomes Lv = b.
? +‐
V0 V z
Lv 0 Lv b
1
2
metalmetal applied applied
1
'
# # # # # # 0
# # # # # # 0
0 0 1 0 0
# # # # # # 0
# # # # # # 0x y
x y
mm m m
N N
N N
V
V
VV V V
V
V
bvL
Enforcing the Known Potentials
Lecture 4 30
It turns out we must also modify L in addition to constructing b.
Each point on the grid containing a metal must be forced to some known potential. We do this by modifying the corresponding row in the matrix as follows:
1. Replace entire row in L with all zeroes.
2. Place a 1 in the positionalong the center diagonal.
3. Place the value of the applied potential inthe same row in b.
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Fancy Way of Forcing Potentials
Lecture 4 31
We construct two terms:
Force matrixF Diagonal matrix containing 1’s in the positions corresponding to values we wish to force.
forced potentialsf v Column vector containing the forced potentials.Numbers in positions not being forced are ignored.
1. Replace forced rows in L with all zeros.
L I F L
2. Place a 1 in the diagonal element.
L F I F L
2. Place the forced potentials in b.
fb Fv Use same b for both homogeneous and inhomogeneous cases.
1
2
metalmetal applied applied
1
'
# # # # # # 0
# # # # # # 0
0 0 1 0 0
# # # # # # 0
# # # # # # 0x y
x y
mm m m
N N
N N
V
V
VV V V
V
V
bvL
Calculating Transmission Line Parameters
Lecture 4 32
Calculate the field quantities
,
,
, ,
, ,
v vx x x h x
v hvy y y h y
x x x h x hxx x y xy
y y y h y hx y yx yy
e D e Dv v
e D e D
d e d eε R R ε
d e d eR R ε ε
0
T
x x
y y
C x y
d e
d e, ,
2, 0
0,
1
T
x h x hh
y h y h h
C x y Lc C
d e
d e
Calculate the transmission line parameters
c
LZ
C eff 0n c LC 2
,eff effr n
CAUTION:Be careful that you are consistent with your units. For example, it is incorrect to set centimeters=1 and then set e0=8.854e-12 because 0 has units of F/m.
Note: we have moved the constant 0 to the capacitance equation for convenience.
Calculate the inductance and capacitance using numerical integration
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Finite-Difference Analysis of Transmission Lines Embedded in Arbitrary
Anisotropic Media
Implementation
Step 1: Choose a Transmission Line
Lecture 4 34
We will choose to model a microstrip transmission line with the following parameters:
wh
r,sub
w = 4.0 mmh = 3.0 mm
r,sup
,sup
1.0 0 0
0 1.0 0 air superstrate
0 0 1.0r
,sub
9.0 0 0
0 9.0 0 dielectric substrate
0 0 9.0r
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Step 2:Build Device on Grid (1 of 2)
Lecture 4 35
Nx unit cells
Nyunit cells
We will use Dirichletboundary conditions so there must be sufficient space between the microstrip and boundaries.
Rule‐of‐Thumb:Spacer regions should be around three times the width of the line.
Step 2:Build Device on Grid (2 of 2)
Lecture 4 36
We use six arrays to describe the transmission line.
1× grid
ER2xx ER2xy ER2yx ER2yy
9’s 0’s 0’s 9’s
0’s 0’s
2× grid
We do not need to build ER2xz, ER2yz, ER2zx, ER2zy, or ER2zz.
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Step 3:Construct Matrix Operators
Lecture 4 37
We use yeeder() to calculate the derivative operators.
% CALL FUNCTION TO CONSTRUCT DERIVATIVE OPERATORSNS = [Nx Ny]; %grid sizeRES = [dx dy]; %grid resolutionBC = [0 0]; %Dirichlet BCs[DVX,DVY,DEX,DEY] = yeeder(NS,RES,BC); %Build matrices
Since we are using Dirichlet boundary conditions, we can construct our interpolation matrices directly from the derivative operators.
2 2
v ex x y y x y
x y R D R D R R R
% FORM INTERPOLATION MATRIX FROM DERIVATIVE OPERATORSRXP = (dx/2)*abs(DVX);RYM = (dy/2)*abs(DEY);R = RXP*RYM;
11
2 2i i i if f f f
1i if f
Step 4:Construct Diagonalized Tensor
Lecture 4 38
Step 1: Parse materials from 2 grid to 1 grid.
ERxx = ER2xx(2:2:Nx2,1:2:Ny2);ERxy = ER2xy(1:2:Nx2,2:2:Ny2);ERyx = ER2yx(2:2:Nx2,1:2:Ny2);ERyy = ER2yy(1:2:Nx2,2:2:Ny2);
Step 2: Diagonalize all four tensor elements.
ERxx = diag(sparse(ERxx(:)));ERxy = diag(sparse(ERxy(:)));ERyx = diag(sparse(ERyx(:)));ERyy = diag(sparse(ERyy(:)));
Step 3: Construct composite tensor
ER = [ ERxx , R*ERxy ; R’*ERyx , ERyy ];
Note the complex transpose here.
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Step 5:Build L and Lh
Lecture 4 39
Step 1. Build the inhomogeneous matrix L
L = [ DEX DEY ] * ER * [ DVX ; DVY ];
Step 2. Build the homogeneous matrix Lh
Lh = [ DEX DEY ] * [ DVX ; DVY ];
L
Step 6:Force Known Potentials
Lecture 4 40
Step 1. Build the force matrix F that identifies the locations of the forced potentials.
F = SIG | GND;F = diag(sparse(F(:)));
Step 2: Build the column vector containing the values of the forced potentials.
vf = 1*SIG + 0*GND;
Step 3: Force the known potentials.
L = (I - F)*L + F;Lh = (I - F)*Lh + F;b = F*vf(:);
For multiple conductors…
F = SIG1 | SIG2 | … | GND;
1 volt applied to SIG
For two conductors…
vf = 0.5*SIG1 – 0.5*SIG2 + 0*GND;
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Step 7:Compute the Fields
Lecture 4 41
Step 1. Compute the potentials. BIG COMPUTATIONAL STEP!!!!
Step 2: Calculate the E fields.
Step 3: Compute the D fields.
1v L b
,
,
vx h x
h hvy h y
e De v
e D
x xx xy xH
y yx yy y
d ε Rε ed
d R ε ε e
1h h
v L b
vx x
vy y
e De v
e D
, ,
, ,
x h x h
hy h y h
d ed
d e
Step 8:Compute TL Parameters
Lecture 4 42
Step 1. Compute the distributed capacitance C.
Step 2: Calculate the distributed inductance L.
Step 3: Calculate characteristic impedance impedance
20
1
A
C D E dAV
0
LZ
C
C = (e0*dx*dy)*dxy'*exy;
2
0
20
1
1
h h h
A
h
C D E dAV
L c C
Ch = (e0*dx*dy)*dxyh'*exyh;L = 1/(c0^2*Ch);
Recall that we moved the 0
term to this equation for convenience.
Be sure you are consistent with your units!
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Step 9:Visualize the Fields
Lecture 4 43
Step 1. Extract the x and y components of E.
Step 2: Reshape from column vectors to 2D arrays.
v = reshape(v,Nx,Ny);Ex = reshape(Ex,Nx,Ny);Ey = reshape(Ey,Nx,Ny);
Step 3: Visualize in MATLAB using imagesc(), pcolor(), and/or quiver().
Ex = exy(1:M);Ey = exy(M+1:2*M);
x
y
ee
e
For day‐to‐day data analysis, I use imagesc(). For generating pictures for publications and presentations, I use pcolor().
Finite-Difference Analysis of Transmission Lines Embedded in Arbitrary
Anisotropic Media
Examples
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Example #1:Ordinary Microstrip (1 of 2)
Lecture 4 45
wh
r,sub
w = 4.0 mmh = 3.0 mm
r,sup
,sup
1.0 0 0
0 1.0 0 air
0 0 1.0r
,sub
9.0 0 0
0 9.0 0 dielectric
0 0 9.0r
Example #1:Ordinary Microstrip (2 of 2)
Lecture 4 46
Nx = 200Ny = 150dx = 0.13793dy = 0.13636Z0 = 44.3436 ohms
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Example #2:Anisotropic Microstrip (1 of 3)
Lecture 4 47
w = 4.0 mmh = 3.0 mm
wh
r,sub
r,sup
,sup
3.0000 3.4641
3.4641 7.0000r
,sub
3.0000 3.4641
3.4641 7.0000r
x
y
z
Example #2:Anisotropic Microstrip (2 of 3)
Lecture 4 48
1.0 0 0
0 9.0 0 30
0 0 1.0r z
To generate correct tensors, we start with the diagonalized tensor and the angle that it should be rotated around the z‐axis.
The rotation matrix is
cos 30 sin 30 0 0.8660 0.5000 0
30 sin 30 cos 30 0 0.5000 0.8660 0
0 0 1 0 0 1zR R
The rotated tensor is then
1
1
0.8660 0.5000 0 1.0 0 0 0.8660 0.5000 0 3.0000 3.4641 0
0.5000 0.8660 0 0 9.0 0 0.5000 0.8660 0 3.4641 7.0000 0
0 0 1 0 0 1.0 0 0 1 0 0 1rR R
x
y
z
z
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Example #2:Anisotropic Microstrip (3 of 3)
Lecture 4 49
Nx = 200Ny = 150dx = 0.13793dy = 0.13636Z0 = 49.9482 ohms
Generalizationof the Method
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Method Can Do More than Transmission Lines!
Lecture 4 51
1 5 VV
2 7 VV
2 0 VV
5 V
7 V
0 V
The model we just developed simply calculates the potential in the vicinity of multiple forced potentials. We used it to model a transmission line, but the method has other applications.
Potential
Think of Laplace’s Equation as a Linear “Number Filler Inner”
Lecture 4 52
Given known values at certain points (boundary conditions), Laplace’s equation calculates the numbers everywhere else so they vary linearly.
Map of known values (boundary values) Laplace’s equation is solved to fill in the values away from the boundaries.
