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Electromagnetics:
Microwave Engineering
Numerical Analysis of Transmission Lines
Outline
•Governing equations•Numerical representation
• Finite‐difference approximations
•Matrix solution of electric potential 𝑉 𝑥, 𝑦•Calculating the transmission line parameters
2
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Governing Equations
Slide 3
Maxwell’s Equations
4
0
0
B
D
H J D t
E B t
Start with Maxwell’s equations.
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Electrostatic Approximation
5
The dimensions of a transmission line are typically much smaller than the operating wavelength so the wave nature of electromagnetics is less important to consider. Therefore, Maxwell’s equations are essential be solved in the limit as 𝑑 𝑑𝑡⁄ → 0.
0
0
B
D
H J D t
E B t
Electrostatics & Magnetostatics
6
0
0
0
D
B
H J
E
Maxwell’s equations have decoupled into two sets of equations. One describes electrostatics while the other describes magnetostatics.
Electrostatics
0
0
D
E
Magnetostatics
0B
H J
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Governing Equations
7
Maxwell’s equations:
0 Eq. (1)
0 Eq. (2)
D
E
In addition, there is the constitutive relation
r Eq. (3)D E
It is not preferred to solve vector equations if there is a way to avoid it. Electrostatic fields are completely characterized by the electric potential 𝑉 𝑥,𝑦 .
Eq. (4)E V
Differential Equation to Solve
8
2. Substitute Eq. (4) into Eq. (5) to eliminate 𝐸.
r
0 Eq. (1)
0 Eq. (2)
Eq. (3)
Eq. (4)
D
E
D E
E V
r 0 Eq. (5)E
r 0 Eq. (6)V
1. Substitute Eq. (3) into Eq. (1) to eliminate 𝐷.
Inhomogeneous Laplace’s Equation
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0V r 0V
Differential Equation in a Homogeneous Medium
9
When the dielectric is homogeneous, our differential equation simplifies to
In Cartesian coordinates, this expands to
2 2 2
2 2 20
V V V
x y z
2 0V
Transmission lines are uniform in the z direction so and Laplace’s equation reduces to
2 2 0z
2 2
2 20
V V
x y
Numerical Representations
Slide 10
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Discrete Function for Electric Potential 𝑉 𝑥,𝑦
11
Analytical functions contain an infinite amount of information.
In order to store 𝑉 𝑥,𝑦 on a computer, it is stored in an array where function values are only known at discrete points.
x
y
Start with the analytical function 𝑉 𝑥, 𝑦 .
,V x y
Discrete Function for Electric Potential 𝑉 𝑥,𝑦
12x
y
Divide space into a grid.
,V x y
x
y
Analytical functions contain an infinite amount of information.
In order to store 𝑉 𝑥,𝑦 on a computer, it is stored in an array where function values are only known at discrete points.
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Discrete Function for Electric Potential 𝑉 𝑥,𝑦
13i
j
Store the analytical function only at a discrete points inside of each cell on the grid.
,V i j
Analytical functions contain an infinite amount of information.
In order to store 𝑉 𝑥,𝑦 on a computer, it is stored in an array where function values are only known at discrete points.
Discrete Function for Electric Potential 𝑉 𝑥,𝑦
14i
j
The discrete function tends to be thought of as being uniform throughout the cell, but this is technically incorrect.
,V i j
Analytical functions contain an infinite amount of information.
In order to store 𝑉 𝑥,𝑦 on a computer, it is stored in an array where function values are only known at discrete points.
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Discrete Function for Electric Potential 𝑉 𝑥,𝑦
15x
y
The discrete function 𝑉 𝑖, 𝑗 is usually visualized this way.
, ,V i j V x y
Analytical functions contain an infinite amount of information.
In order to store 𝑉 𝑥,𝑦 on a computer, it is stored in an array where function values are only known at discrete points.
A Fundamental Tradeoff for Discrete Functions
16
There is always a fundamental trade‐off between accuracy and speed of simulation. The purpose of almost all work in computational methods is to get improved accuracy with fewer points.
Fewer pointsLess memory
Faster simulationsPoorer accuracy
More pointsMore memorySlower simulationsBetter accuracy“Sweet spot”
Best compromise between accuracy, memory, and speed of simulation.
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Finite‐Difference Approximations
Slide 17
Finite‐Difference Approximations (1 of 2)
Slide 18
Suppose the first‐order derivative of 𝑓 𝑥 is to be numerically calculated at 𝑥 𝑥 .
The first‐order derivative is slope. The slope can be estimated as riserun using information from surrounding points.
3 12
rise
run 2
f ff x
x
1x 2x 3x
1f
2f3f
x
run = 2x
rise
= f 3
–f 1
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Finite‐Difference Approximations (2 of 2)
Slide 19
The derivative at the midpoints between data points can be estimated.
1x 2x 3x
1f
2f3f
2 11.5
f ff x
x
3 22.5
f ff x
x
1.5f
2.5f
The second‐order derivative is the slope of the slope.
2.5 1.52
3 2 2 1
3 2 12
2
f x f xf x
xf f f f
x xx
f f f
x
Finite‐Difference Approximation of Laplace’s Equation
20
2 2
2 20
V V
x y
2 2
1, 2 , 1, , 1 2 , , 10
V i j V i j V i j V i j V i j V i j
x y
2
22
1, 2 , 1,V i j V i j V i jV
x x
2
22
, 1 2 , , 1V i j V i j V i jV
y y
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Rearrange Terms in the Finite‐Difference Equation
21
2 2
1, 2 , 1, , 1 2 , , 10
V i j V i j V i j V i j V i j V i j
x y
2 2 2 2 2 2
1, 2 , 1, , 1 2 , , 10
V i j V i j V i j V i j V i j V i j
x x x y y y
2 2 2 2 2 2
1 1 2 2 1 11, 1, , , 1 , 1 0V i j V i j V i j V i j V i j
x x x y y y
Matrix Solution of Electric Potential
Slide 22
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Write Large Set of Equations
23
This equation is written once for every point on the grid.
2 2 2 2 2 2
1 1 2 2 1 11, 1, , , 1 , 1 0V i j V i j V i j V i j V i j
x x x y y y
The final form of the finite‐difference equation is
4
4x
y
N
N
0.5
0.5
x
y
Consider Boundary Conditions
24
4
4x
y
N
N
0.5
0.5
x
y
All of the highlighted terms will be set to zero.
Dirichlet Boundary Conditions
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Build Matrix Equation 𝐿 𝑣 0
25
16 4 0 0 4 0 0 0 0 0 0 0 0 0 0 0
4 16 4 0 0 4 0 0 0 0 0 0 0 0 0 0
0 4 16 4 0 0 4 0 0 0 0 0 0 0 0 0
0 0 4 16 0 0 0 4 0 0 0 0 0 0 0 0
4 0 0 0 16 4 0 0 4 0 0 0 0 0 0 0
0 4 0 0 4 16 4 0 0 4 0 0 0 0 0 0
0 0 4 0 0 4 16 4 0 0 4 0 0 0 0 0
0 0 0 4 0 0 4 16 0 0 0 4 0 0 0 0
0 0 0 0 4 0 0 0 16 4 0 0 4 0 0 0
0 0 0 0 0 4 0 0 4 16 4 0 0 4 0 0
0 0 0 0 0 0 4 0 0 4 16 4 0 0 4 0
0 0
1,1
2,1
3,1
4,1
1,2
2, 2
3, 2
4, 2
1,3
2,3
3,3
4,30 0 0 0 0 4 0 0 4 16 0 0 0 4
1,40 0 0 0 0 0 0 0 4 0 0 0 16 4 0 0
2, 40 0 0 0 0 0 0 0 0 4 0 0 4 16 4 0
3, 40 0 0 0 0 0 0 0 0 0 4 0 0 4 16 4
40 0 0 0 0 0 0 0 0 0 0 4 0 0 4 16
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
, 4 0
Is the Matrix Equation Solvable?
26
16 4 0 0 4 0 0 0 0 0 0 0 0 0 0 0
4 16 4 0 0 4 0 0 0 0 0 0 0 0 0 0
0 4 16 4 0 0 4 0 0 0 0 0 0 0 0 0
0 0 4 16 0 0 0 4 0 0 0 0 0 0 0 0
4 0 0 0 16 4 0 0 4 0 0 0 0 0 0 0
0 4 0 0 4 16 4 0 0 4 0 0 0 0 0 0
0 0 4 0 0 4 16 4 0 0 4 0 0 0 0 0
0 0 0 4 0 0 4 16 0 0 0 4 0 0 0 0
0 0 0 0 4 0 0 0 16 4 0 0 4 0 0 0
0 0 0 0 0 4 0 0 4 16 4 0 0 4 0 0
0 0 0 0 0 0 4 0 0 4 16 4 0 0 4 0
0 0
1,1
2,1
3,1
4
0 0 0 0 0 4 0 0 4 16 0 0 0 4
0 0 0 0 0 0 0 0 4 0 0 0 16 4 0 0
0 0 0 0 0 0 0 0 0 4 0 0 4 16 4 0
0 0 0 0 0 0 0 0 0 0 4 0 0 4 16 4
0 0 0 0 0 0 0 0 0 0 0 4 0 0 4 16
L
V
V
V
V
0
0
0
0
,1 0
1, 2 0
2, 2 0
3, 2 0
4, 2 0
1,3 0
2,3 0
3,3 0
4,3 0
1, 4 0
2, 4 0
3, 4 0
4, 4 0
v
V
V
V
V
V
V
V
V
V
V
V
V
2 0 0V L v 1 0 0v L
Trivial Solution
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Force Known Potentials
27
16 4 0 0 4 0 0 0 0 0 0 0 0 0 0 0
4 16 4 0 0 4 0 0 0 0 0 0 0 0 0 0
0 4 16 4 0 0 4 0 0 0 0 0 0 0 0 0
0 0 4 16 0 0 0 4 0 0 0 0 0 0 0 0
4 0 0 0 16 4 0 0 4 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 4 0 0 4 16 0 0 0 4 0 0 0 0
0 0 0 0 4 0 0 0 16 4 0 0 4 0 0 0
0 0 0 0 0 4 0 0 4 16 4 0 0 4 0 0
0 0 0 0 0 0 4 0 0 4 16 4 0 0 4 0
0 0
1,1
2,1
3,1
4,1
1,2
2, 2
3, 2
4, 2
1,3
2,3
3,3
4,30 0 0 0 0 4 0 0 4 16 0 0 0 4
1,40 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
2, 40 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
3, 40 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
40 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
0
0
0
0
0
5
5
0
0
0
0
0
2
2
2
, 4 2
Force values here to 2.0
Force values here to 5.0
1,4 2
2,4 2
3,4 2
4,4 2
V
V
V
V
2,2 5
3,2 5
V
V
Solve for Electric Potential 𝑣
28
16 4 0 0 4 0 0 0 0 0 0 0 0 0 0 0
4 16 4 0 0 4 0 0 0 0 0 0 0 0 0 0
0 4 16 4 0 0 4 0 0 0 0 0 0 0 0 0
0 0 4 16 0 0 0 4 0 0 0 0 0 0 0 0
4 0 0 0 16 4 0 0 4 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 4 0 0 4 16 0 0 0 4 0 0 0 0
0 0 0 0 4 0 0 0 16 4 0 0 4 0 0 0
0 0 0 0 0 4 0 0 4 16 4 0 0 4 0 0
0 0 0 0 0 0 4 0 0 4 16 4 0 0 4 0
0 0
1,1
2,1
3,1
4,1
0 0 0 0 0 4 0 0 4 16 0 0 0 4
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
L
V
V
V
V
0
0
0
0
1,2 0
2, 2 5
3,2 5
4, 2 0
1,3 0
2,3 0
3,3 0
4,3 0
1,4 2
2, 4 2
3,4 2
4, 4 2
v b
V
V
V
V
V
V
V
V
V
V
V
V
1 L v b v L b
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Calculating the TransmissionLine Parameters
Slide 29
Calculating the Electric Fields
30
, ,E x y V x y
Once the scalar potential 𝑉 𝑥,𝑦 is found, the electric field intensity 𝐸 𝑥,𝑦 is
It follows that the electric flux density 𝐷 𝑥,𝑦 is
r, , ,D x y x y E x y
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Distributed Capacitance 𝐶
31
In the electrostatic approximation, the transmission line is a capacitor. The total energy U stored in a capacitor is
12
A
U D E dA
The capacitance C is related to the total stored energy U through
20
2
CVU V0 is the voltage across the capacitor.
If the above equations are set equal and solved for C, the answer is
20
1
A
C D E dAV
Distributed Inductance 𝐿
32
The voltage signal 𝑣 along the transmission line travels at the same velocity as the electromagnetic field 𝑣 . This means
Solving this for L gives
r r20
1 1 V Ev v LC
cLC
r r20
Lc C
This means the distributed inductance 𝐿 can be calculated
directly from the distributed capacitance 𝐶.
Dielectric materials should not alter the inductance. However if the value of C calculated on the previous slide is used, it will. This is incorrect. The solution is to calculate the distributed capacitance Ch with a homogeneous dielectric and then calculate the distributed inductance 𝐿 from this.
20 h
1L
c C
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Calculating the Transmission Line Parameters
Slide 33
The characteristic impedance Zc is calculated from the distributed inductance L and distributed capacitance C through
c
LZ
C
It follows that the phase constant is
0 effLC k n Recall, both Zc and are needed to analyze transmission line circuits.
Note: This simple model did not consider loss, so 𝑅 𝐺 0 (lossless transmission line)
00
2k
