Topic 4 -- Numerical Analysis of Transmission Linesemlab.utep.edu/ee4380microwave/Topic 4 --...

12/5/2017

1

ECE 4380/5390Instructor:

Office:E-Mail:

Dr. Raymond [email protected]

Topic 4

Numerical Analysis of Transmission Lines

1Numerical Analysis of Transmission Lines

Outline

• Solution Approach

– Do not need to understand completely

• Using tlcalc.p to Analyze Transmission Lines

–Must understand completely

• Examples

2Numerical Analysis of Transmission Lines

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Solution Approach

Maxwell’s Equations

Numerical Analysis of Transmission Lines 4

0

0

B

D

H J D t

E B t

We start with Maxwell’s equations.

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Small Size Approximation


0

0

B

D

H J D t

E B t

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 Maxwell’s equations as d/dt0.

Electrostatics & Magnetostatics


0

0

0

D

E

B

H J

Maxwell’s equations have decoupled into two sets of equations. Once describes electrostatics while the other describes magnetostatics.

Electrostatics

0

0

D

E

Magnetostatics

0B

H J

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Governing Equations


Maxwell’s equations:

0 Eq. (1)

0 Eq. (2)

D

E

In addition, we have the constitutive relation

Eq. (3)rD E

We do not like to solve vector equations if we do not have to. Electrostatic fields are completely characterized by the scalar potential V.

Eq. (4)E V

Differential Equation to Solve


2. Substitute Eq. (4) into Eq. (5) toeliminated E.

0 Eq. (1)

0 Eq. (2)

Eq. (3)

Eq. (4)

r

D

E

D E

E V

0 Eq. (5)r E

0 Eq. (6)r V

1. Substitute Eq. (3) into Eq. (1) to eliminate D.

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Derive Finite‐Difference Equation


20 0r V V

When the dielectric is homogeneous, our differential equation simplifies to

In Cartesian coordinates, this expands to

2 2

2 20

V V

x y

Now, the function V(x,y) is made discrete and the derivatives are approximated using finite‐differences.

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

Write Large Set of Equations


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

We rearrange our finite‐difference equation to the following form

4

4x

y

N

N

0.5

0.5

x

y

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Consider Boundary Conditions


4

4x

y

N

N

0.5

0.5

x

y

We will set all of the highlighted terms to zero

Dirichlet Boundary Conditions

Build Matrix


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

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Is Our Matrix Equation Solvable?


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

Force Known Potentials


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

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Solve for Scalar Potential


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

Calculating the Fields


, ,E x y V x y

Once the scalar potential is solved, the electric field intensity is

The electric flux density is then

, , ,rD x y x y E x y

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Distributed Capacitance


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 we set the above equations equal and solve for C, we get

20

1

A

C D E dAV

Distributed Inductance


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

Solving this for L we get

20

1 1 r r

V Ev v LCcLC

20

r rLc C

This means we can calculate the distributed inductance L

directly from the distributed capacitance.

Dielectric materials should not alter the inductance. However if we use the value of C calculated on the previous slide, it will. This is incorrect. The solution is to calculate the distributed capacitance CH with air dielectric and then calculate the distributed inductance from this.

20

r

H

Lc C

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Calculating the Transmission Line Parameters

Numerical Analysis of Transmission Lines Slide 19

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

c

LZ

C

The phase constant is

0 effLC k n Recall, both Zc and are needed to analyze transmission line circuits.

Note: It was assumed R = G = 0 (lossless transmission line)

00

2k

Using tlcalc.p to Analyze Transmission Lines

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The Short Story


[Z0,n,L,C,V,Ex,Ey] = tlcalc(ER,C1,C2,dx,dy);

tlcalc() (1 of 2)


tlcalc Transmission Line Calculator

[Z0,n,L,C,V,Ex,Ey] = tlcalc(ER,C1,C2,dx,dy);

This MATLAB function calculates the properties of anarbitrary transmission.

ER is a 2D array describing the relative permittivity.C1 is a 2D array identifying the points of conductor 1.C2 is a 2D array identifying the points of conductor 2.dx and dy are the grid resolution parameters.

To simulate the line C1 is set to 1 volt and C2 is set to 0 volts.

EE 4380/5390 Microwave EngineeringDr. Raymond C. RumpfUniversity of Texas at El Paso

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Number Values for Inputs


Permittivity Function

1 ,r x y

r should be purely real.

Loss would be accounted for differently.

Conductor #1

0 dielectric

1 metal

Array should be all 0’s except 1’s in the positions that describe the first conductor.

Conductor #2

0 dielectric

1 metal

Array should be all 0’s except 1’s in the positions that describe the second conductor.

Grid Resolution


Pick your smallest dimension and resolve this with at least 5 to 10 points.

dx = w/10;dy = dx;

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Space Around Transmission Line


Add enough space between transmission line and the boundary to ensure the electric field decays sufficiently. Remember, we have used Dirichlet boundary conditions.

Thickness of Conductors


Conductors are usually very thin (20‐50 micrometers).

This is not usually feasible to resolve in this simple analysis tool.

Just make the conductors one cell thick, unless you have very thick conductors.

One cell thick

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Scalability


These TLs will have the exact same properties.

Electrostatics has no fundamental size scale.

Examples

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Coplanar Transmission Line


1’s

2.5’s

1 mm 1.38 mm

0

eff

317.3 nH m

59.9 pF m

72.8

1.3066

L

C

Z

n

Microstrip Transmission Line


1’s

4.4’s

1.93 mm

1.38 mm0

eff

304.7 nH m

116.8 pF m

51.1

1.7881

L

C

Z

n

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Parallel Plate Transmission Line


2.5’s

2.85 mm

1.0 mm

0

eff

263.3 nH m

105.7 pF m

49.9

1.5811

L

C

Z

n

Coaxial Transmission Line


2.2’s

0.5 mm

3.2 mm

0

eff

379.5 nH m

64.5 pF m

76.7

1.4832

L

C

Z

n

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Symmetric Stripline


4.0’s0.36 mm

0.9 mm

0

eff

381.1 nH m

116.8 pF m

57.1

2.00

L

C

Z

n

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