lecture 4a -- transmission lines · all two‐conductor transmission lines either support a tem...

9/27/2017

1

Transmission Lines Slide 1

EE 4347

Applied Electromagnetics

Topic 4a

Transmission Lines

These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited

Course InstructorDr. Raymond C. RumpfOffice: A‐337Phone: (915) 747‐6958E‐Mail: [email protected]

Lecture Outline


• Introduction• Transmission Line Equations• Transmission Line Wave Equations• Transmission Line Parameters

– and – Characteristic Impedance, Z0

• Special Cases of Transmission Lines– General transmission lines– Lossless lines– Weakly absorbing lines– Distortionless lines

• Examples– RG‐59 coaxial cable– Microstrip design

9/27/2017

2


Introduction

Slide 4

Map of Waveguides (LI Media)

Transmission Lines

• Contains two or more conductors.• No low‐frequency cutoff.• Thought of more as a circuit clement

• Confines and transports waves.• Supports higher‐order modes.

• Has TEM mode.• Has TE and TM modes.

stripline

coaxial microstrip

slotline

coplanar

Transmission Lines

“Pipes”

• Has one or less conductors.• Usually what is implied by the label “waveguide.”

Metal Shell Pipes Dielectric Pipes

Inhomogeneous

Homogeneous

• Enclosed by metal.• Does not support TEM mode.• Has a low frequency cutoff.

• Supports TE and TM modes

• Supports TE and TM modes only if one axis is uniform.

• Otherwise supports quasi‐TM and quasi‐TE modes.

rectangular circular

Channel Waveguides

Slab Waveguides

• Composed of a core and a cladding.• Symmetric waveguides have no low‐frequency cutoff.

• Confinement only along one axis.• Supports TE and TM modes.• Interfaces can support surface waves.

• Confinement along two axes.• TE & TM modes only supported in circularly symmetric guides.

dielectric Slab interface

optical Fiber rib

dual‐ridge

no uniform axis(no TE or TM)

Waveguides

Homogeneous Inhomogeneous• Supports only quasi‐(TEM, TE, & TM) modes.

Single‐Ended

Differential

buried parallel plate

coplanar strips

photonic crystal

shielded pairlarge‐area

parallel plate

uniform axis(has TE and TM)

9/27/2017

3


Transmission Line Parameters RLGC

We can think transmission lines as being composed of millions of tiny little circuit elements that are distributed along the length of the line.

In fact, these circuit element are not discrete, but continuous along the length of the transmission line.


RLGC Circuit Model

It is not technically correct to represent a transmission line with discrete circuit elements like this.

However, if the size of the circuit zis very small compared to the wavelength of the signal on the transmission line, it becomes an accurate and effective way to model the transmission line.

z

9/27/2017

4


L‐Type Equivalent Circuit Model

Distributed Circuit Parameters

R (/m)Resistance per unit length. Arises due to resistivity in the conductors.

L (H/m)Inductance per unit length. Arises due to stored magnetic energy around the line.

G (1/m)Conductance per unit length. Arises due to conductivity in the dielectric separating the conductors.

C (F/m)Capacitance per unit length. Arises due to stored electric energy between the conductors.

z z z

R z L z

G z C z

There are many possible circuit models for transmission lines, but most produce the same equations after analysis.

1G

R


Relation to Electromagnetic Parameters

LC , ,

G

C

Every transmission line with a homogeneous fill has:

9/27/2017

5


Fundamental Vs. Intuitive Parameters

Fundamental Parameters Intuitive Parameters

Electromagnetics Electromagnetics

Transmission Lines Transmission Lines

, , , , , , tann

, , , R L G C 0 , , , VSWRZ

The fundamental parameters are the most basic parameters needed to solve a transmission line problem.

However, it is difficult to be intuitive about how they affect signals on the line.

An electromagnetic analysis is needed to determine R, L, G, and C from the geometry of the transmission line.

The intuitive parameters provide intuitive insight about how signals behave on a transmission line.

They isolate specific information to a single parameter.

The intuitive parameters are calculated from R, L, G, and C .


Example RLGC Parameters

0

36 mΩ m

430 nH m

10 m

69 pF m

75

R

L

G

C

Z

0

176 mΩ m

490 nH m

2 m

49 pF m

100

R

L

G

C

Z

Surprisingly, almost all transmission lines have parameters very close to these same values.

0

150 mΩ m

364 nH m

3 m

107 pF m

50

R

L

G

C

Z

RG‐59 Coax CAT5 Twisted Pair Microstrip

9/27/2017

6


Transmission Line Equations


E & H V and I

Fundamentally, all circuit problems are electromagnetic problems and can be solved as such.

All two‐conductor transmission lines either support a TEM wave or a wave very closely approximated as TEM.

An important property of TEM waves is that E is uniquely related to Vand H and uniquely related to E.

L

V E d

L

I H d

This let’s us analyze transmission lines in terms of just V and I. This makes analysis much simpler because these are scalar quantities!

9/27/2017

7


Transmission Line Equations

The transmission line equations do for transmission lines the same thing as Maxwell’s curl equations do for unguided waves.

Maxwell’s Equations Transmission Line Equations

HE

t

EH

t

V IRI L

z t

I VGV C

z t

Like Maxwell’s equations, the transmission line equations are rarely directly useful. Instead, we will derive all of the useful equations from them.

,V z t ,I z t

L zt

,I z t R z ,V z z t


Derivation of First TL Equation (1 of 2)

z z z

R z L z

G z C z

+

‐

,V z t

+

‐

,V z z t

Apply Kirchoff’s voltage law (KVL) to the outer loop of the equivalent circuit:

1

2 3

4

,I z t

1 23

4

0

9/27/2017

8


Derivation of First TL Equation (2 of 2)

We rearrange the equation by bringing all of the voltage terms to the left‐hand side of the equation, bringing all of the current terms to the right‐hand side of the equation, and then dividing both sides by z.

,, , , 0

, , ,,

I z tV z t I z t R z L z V z z t

t

V z z t V z t I z tRI z t L

z t

In the limit as z 0, the expression on the left‐hand side becomes a derivative with respect to z.

, ,,

V z t I z tRI z t L

z t

,0

V z z tC z

t

,G zV z z t ,I z z t


Derivation of Second TL Equation (1 of 2)

z z z

R z L z

G z C z

+

‐

,V z t

+

‐

,V z z t

1 2

3 4

Apply Kirchoff’s current law (KCL) to the main node the equivalent circuit:

,I z t

,I z t

1 2 34

,I z z t

9/27/2017

9


Derivation of Second TL Equation (2 of 2)

We rearrange the equation by bringing all of the current terms to the left‐hand side of the equation, bringing all of the voltage terms to the right‐hand side of the equation, and then dividing both sides by z.

,, , , 0

, , ,,

V z z tI z t I z z t G zV z z t C z

t

I z z t I z t V z z tGV z z t C

z t

In the limit as z 0, the expression on the left‐hand side becomes a derivative with respect to z.

, ,,

I z t V z tGV z t C

z t


Transmission Line Wave Equations

9/27/2017

10


Starting Point – Telegrapher Equations

We start with the transmission line equations derived in the previous section.

, ,,

V z t I z tRI z t L

z t

, ,

,I z t V z t

GV z t Cz t

time‐domain

For time‐harmonic (i.e. frequency‐domain) analysis, we Fourier transform the equations above.

dV zR j L I z

dz

dI zG j C V z

dz frequency‐domain

Note: Our derivative d/dz became an ordinary derivative because z is the only independent variable left.

These last equations are commonly referred to as the telegrapher equations.


Wave Equation in Terms of V(z)

To derive a wave equation in terms of V(z), we first differentiate Eq. (1) with respect to z.

dV zR j L I z

dz

dI zG j C V z

dz Eq. (1) Eq. (2)

2

2

d V z dI zR j L

dz dz Eq. (3)

Second, we substitute Eq. (2) into the right‐hand side of Eq. (3) to eliminate I(z) from the equation.

2

2

d V zR j L G j C V z

dz

Last, we rearrange the terms to arrive at the final form of the wave equation.

2

20

d V zR j L G j C V z

dz

9/27/2017

11


Wave Equation in Terms of I(z)

To derive a wave equation in terms of just I(z), we first differentiate Eq. (2) with respect to z.

dV zR j L I z

dz

dI zG j C V z

dz Eq. (1) Eq. (2)

2

2

d I z dV zG j C

dz dz Eq. (3)

Second, we substitute Eq. (1) into the right‐hand side of Eq. (3) to eliminate V(z) from the equation.

2

2

d I zG j C R j L I z

dz

Last, we rearrange the terms to arrive at the final form of the wave equation.

2

20

d I zG j C R j L I z

dz


Propagation Constant,

Define the propagation constant to be

G j C R j L

j G j C R j L

Given this definition, the transmission line equations are written as

2

22

0d V z

V zdz

2

22

0d I z

I zdz

In our wave equations, we have a common term .

9/27/2017

12


Solution to the Wave Equations

If we hand the wave equations off to a mathematician, they will return with the following solutions.

2

22

0d V z

V zdz

2

22

0d I z

I zdz

0 0 z zV z V e V e

0 0 z zI z I e I e

Both V(z) and I(z) have the same differential equation so it makes sense they have the same solution.

Forward wave Backward wave


Transmission Line Parameters:

Attenuation Coefficient, Phase Constant,

9/27/2017

13


Derivation and (1 of 7)

Step 1 – Start with our expression for .

j G j C R j L

2j G j C R j L

2 2 22j RG j RC j LG LC

2 2 22j RG LC j RC LG

Square this expression to get rid of square‐root on right‐hand side.

Expand this expression.

Collect real and imaginary parts on the left‐hand and right‐hand sides.



Step 2 – Generate two equations by equating real and imaginary parts.

2 2 22j RG LC j RC LG

2 2 2RG LC

2 RC LG

We now have two equations and two unknowns.

9/27/2017

14



Step 3 – Derive a quadratic equation for 2.

2 2 2

2 Eq. (1a)

Eq. (1b)

RC LG

RG LC

Solve Eq. (1a) for .

Eq. (2)2

RC LG

Substitute Eq. (2) into Eq. (1b) and simplify.

22 2

222 2

2

24 2 2 2 2

24 2 2

2

4

4 4 4

02

RC LG RG LC

RC LGRG LC

RC LG RG LC

LC RG RC LG



Step 4 – Solve for 2 using the quadratic formula.

Recall the quadratic formula:

2

4 2 2 0 2

LC RG RC LG

22 4

0 2

b b acax bx c x

a

Our equation for is in the form of a quadratic equation where

2

2

2

1

2

a

b LC RG

c RC LG

x

The solution is

22 2

2

2 2 2 2 2 2 2

42

2

2

LC RG LC RG RC LG

RG LC R L G C

9/27/2017

15



Step 5 – Resolve the sign of the square‐root.

In order for this expression to always give a real value for , the sign of the square‐root must be positive.

The final expression is

2 2 2 2 2 2 2

2

2

RG LC R L G C

2 2 2 2 2 2 2

2

2

RG LC R L G C



Step 6 – Solve for 2 using our expression for 2.

Recall Eq. (1b):

We obtain an equation for 2 by substituting our expression for 2 into Eq. (1b).

2 2 2RG LC

2 2 2 2 2 2 2

2 2

2 2 2 2 2 2 2

2

2

2

RG LC R L G CRG LC

RG LC R L G C

2 2 2 2 2 2 2

2

2

RG LC R L G C

9/27/2017

16



Step 7 – We arrive at our final expressions for and in terms of the fundamental parameters R, L, G, and C by taking the square‐root of our latest expressions for 2 and 2.

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2

2

RG LC R L G C

RG LC R L G C

Both and must be positive quantities for passive materials. This means we take the positive sign for the square‐root.

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2

2

RG LC R L G C

RG LC R L G C


Transmission Line Parameters:

Characteristic Impedance, Z0

9/27/2017

17


Characteristic Impedance, Z0 ()

The characteristic impedance Z0 of a transmission line is defined as the ratio of the voltage to the current at any point of a forward travelling wave.

0 00

0 0

V VZ

I I

Definition for a forward travelling wave.

Definition for a backward travelling wave. Notice the negative sign!

Most characteristic impedance values fall in the 50 to 100 range. The specific value of impedance is not usually of importance. What is important is when the impedance changes because this causes reflections, standing waves, and more.


Derivation of Z0 (1 of 5)

Step 1 – Substitute our solution into the transmission line equations.

dV zR j L I z

dz

dI zG j C V z

dz

0 0

0 0

z z

z z

V z V e V e

I z I e I e

0 0

0 0

z z

z z

d

dz

R j L

V e V e

I e I e

0 0

0 0

z z

z z

d

dz

G j C

I e I e

V e V e

9/27/2017

18



Step 2 – Expand the equations and calculate the derivatives.

0 0

0 0

z z

z z

dV e V e

dz

R j L I e I e

0 0

0 0

z z

z z

dI e I e

dz

G j C V e V e

0 0

0 0

z z

z z

V e V e

R j L I e R j L I e

0 0

0 0

z z

z z

I e I e

G j C V e G j C V e



Step 3 – Equate the expressions multiplying the common exponential terms.

0 0 0 0z z z zV e V e R j L I e R j L I e

0 0 0 0z z z zI e I e G j C V e G j C V e

0 0V R j L I

0 0V R j L I

0 0I G j C V

0 0I G j C V

9/27/2017

19



Step 4 – Solve each of our four equations for V0/I0 to derive expressions for Z0.

0 0

0 0

0 0

0 0

V R j L I

V R j L I

I G j C V

I G j C V

00

0

00

0

00

0

00

0

V R j LZ

I

V R j LZ

I

VZ

I G j C

VZ

I G j C



Step 5 – Put Z0 in terms of just R, L, G, and C.

0

R j LZ

G j C

Recall our expression for : j G j C R j L

We can substitute this into either of our expressions for Z0.

Proceed with the first expression.

2

0

R j LR j L R j LZ

G j C R j L G j CG j C R j L

9/27/2017

20


Final Expression for Z0 ()

We have derived a general expression for the characteristic impedance Z0 of a transmission line in terms of the fundamental parameters R, L, G, and C.

0 00

0 0

V VZ

I I

Definition:

Expression: 0

R j L R j LZ

G j C G j C


Dissecting the Characteristic Impedance, Z0

The characteristic impedance describes the amplitude and phase relation between voltage and current along a transmission line. With this picture in mind, the characteristic impedance can be written as

00 0 ZZ Z

The characteristic impedance can also be written in terms of its real and imaginary parts.

0

0

00 0 0

0

Z

z

jz z z

V z V e

VI z I e e Z V e e

Z

0 0 0Z R jX

Reactive part of Z0. This is not equal to jL or 1/jC.

Resistive part of Z0. This is not equal to R or G.

9/27/2017

21


Special Cases of Transmission Lines:

General Transmission Line


Parameters for General TLs


j G j C R j L

Attenuation Coefficient,

2 2 2 2 2 2 2

2

RG LC R L G C

Phase Constant,

2 2 2 2 2 2 2

2

RG LC R L G C


0 0 0

R j LZ R jX

G j C

9/27/2017

22



Lossless Lines


Definition of Lossless TL

For a transmission line to be lossless, it must have

When we think about transmission lines, we tend to think of the special case of the lossless line because the equations simplify considerably.

0R G

9/27/2017

23


Parameters for Lossless TLs


j j LC


0

Phase Constant,

LC


0 0 0

LZ R jX

C

0 0 0L

R XC



Weakly Absorbing Line

9/27/2017

24


Definition of Weakly Absorbing TL

Most practical transmission lines have loss, but very low loss making them weakly absorbing.

We will define a weakly absorbing line as

and R L G C

Ensures very little conduction between the lines through the dielectric.

Ensures low ohmic loss for signals propagating through the line.


Parameters for Weakly Absorbing TLs


00

1

2

RGZ

Z

Conductance through the dielectric dominates attenuation in high‐impedance transmission lines.

Resistivity in the conductors dominates attenuation in low‐impedance transmission lines.

In weakly absorbing transmission lines, there usually exists a “sweep spot” for the impedance where attenuation is minimized.

9/27/2017

25



Distortionless Lines


Definition of Distortionless TL

In a real transmission line, different frequencies will be attenuated differently because is a function of . This causes distortion in the signals carried by the line.

2 2 2 2 2 2 2

2

RG LC R L G C

To be distortionless, there must be a choice of R, L, G, and C that eliminates from the expression of , effectively making independent of frequency .

The necessary condition to be distortionless is

R G

L C

9/27/2017

26


Parameters for Distortionless TLs


j RG j LC


RG

Phase Constant,

LC


0 0 0

R LZ R jX

G C

0 0 0R L

R XG C

To be distortionless, we must have . is a measure of how quickly a signal accumulates phase. Different frequencies have different wavelengths and therefore must accumulate different phase through the same length of line.


Example:

Properties of RG-59 Coax

9/27/2017

27


The Lossless Circular Coax

Attenuation Coefficient, 0

Phase Constant,


0 0 0 ln 2

bZ R jX a b

a

0 0ln 02

bR X

a

Fundamental Parameters (derived in EE 3321)

2 F m

ln

1ln H m

2 4

Cb a

bL

a

a

b

and


Typical RLGC for RG‐59 Coax at 2 GHz

The typical RG‐59 coaxial cable operating at 2.0 GHz has the following RLGC parameters:

36 mΩ m

430 nH m

10 m

69 pF m

R

L

G

C

Calculate the transmission line parameters , , , and Z0.

Classify the line as lossless, weakly absorbing, distortionless, etc.

9/27/2017

28


Solution (1 of 3)

Our equations mostly utilize the angular frequency instead of the ordinary frequency f.

9 1 92 2 2.0 10 s 12.5664 10 rad sf

The characteristic impedance Z0 is

0

9

9

4

36 mΩ m 12.5664 10 rad s 430 nH m

10 m 12.5664 10 rad s 69 pF m

78.94 1.92 10

R j LZ

G j C

j

j

j

Note the imaginary part of Z0 is very small indicating that our line is very low loss.


Solution (2 of 3)

The complex propagation constant is

9

9

4 1

36 mΩ m 12.5664 10 rad s 430 nH m

10 m 12.5664 10 rad s 69 pF m

6.23 10 68.45 m

R j L G j C

j

j

j

From this result, we read off and .4 16.23 10 68.45 mj j

46.23 10 Np m

68.45 rad m

Np is Nepers

rad is radians

9/27/2017

29


Solution (3 of 3)

Is the line lossless? NO

No because R ≠ 0 and G ≠ 0.Also, we can determine this because ≠ 0 .

Is the line weakly absorbing? YES

?

?9

?

36 mΩ m 12.5664 10 rad s 430 nH m

0.036 5403.5

Yes

R L

?

?9

?6

10 m 12.5664 10 rad s 69 pF m

10 10 0.8671

Yes

G C

Is the line distortionless? NO, but close

?

?

?12 12

36 mΩ m 69 pF m 430 nH m 10 m

2.48 10 4.30 10

No, but close

RC LG

Cable Loss Vs. Characteristic Impedance

As we adjust the cable dimensions (i.e. b/a), we change both its impedance and its loss characteristics. This let’s us plot the cable loss vs. characteristic impedance for a coax with different dielectric fills.

For the air‐filled coax, we observe minimum loss at around 77 , where b/a 3.5.

A coaxial cable filled with polyethelene (r = 2.2), the minimum loss occurs at 51.2 (b/a = 3.6).


https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms

9/27/2017

30

Power Handling Vs. Characteristic Impedance

As we adjust the cable dimensions (i.e. b/a), we affect the peak voltage handling capability (breakdown) and its power handling capability (heat).

We observe the lowest peak voltage at just over 50 which we interpret as the point of best voltage handling capability.

Transmission Lines Slide 59https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms

We observe the lowest peak current at around 30 which we interpret as the point of best power handling capability.


Why 50 Impedance is Best?

Two researchers, Lloyd Espenscheid and Herman Affel, working at Bell Labs produced this graph in 1929. They needed to send 4 MHz signals hundreds. Transmission lines capable of handling high voltage and high power were needed in order to accomplish this.

Best for High Voltage: Z0 = 60 Best for High Power: Z0 = 30 Best for Attenuation: Z0 = 75

50 seems like the best compromise.

Data to the right was generated for an air‐filled coaxial cable.

9/27/2017

31


Why 75 Impedance Standard for Coax?

Nobody really knows!!

The ideal impedance is closer to 50 , however this requires a thicker center conductor. Maybe 75 is a compromise between low loss and mechanical flexibility?


Example:

Microstrip Design

9/27/2017

32


The Lossless Microstrip

Attenuation Coefficient, 0

Phase Constant,

,eff

0 ,eff

1 1

2 2 1 12r r

r

r

h w

k


eff

0 0 0

eff

60 8 ln 1 thin lines

4

1 120 1 wide lines

1.393 0.667 ln 1.444

h ww h

w hZ R jX

w hw h w h

rh

w


Problem Description

Typically, the manufacturing process fixes the value of dielectric constant r. This means the impedance of microstrips is controlled solely through the ratio w/h.

For this example, design a 50 microstrip transmission line in FR‐4, which as a dielectric constant of 4.5, to operate at 2.4 GHz.

?w

h

9/27/2017

33


Design Equations

To solve this problem, we must first derive some design equations. To do this, we solve our microstrip equations for w/h. This gives

0

2

0

2

1 1 0.110.23

60 2 1

60

8 2 thin lines

2

12 0.611 ln 2 1 ln 1 0.39

2

r r

r r

r

A

A

r

r r

ZA

BZ

ew h

ew

hB B B

2 wide linesw h


Design Solution (1 of 2)

Applying our design equations, we get

1.5438

5.5831

1.8799 2 thin lines

1.8812 2 wide lines

A

B

w hw

w hh

Since the above numbers for w/h are essentially the same, we conclude that

1.88w

h

9/27/2017

34


Design Solution (2 of 2)

We learn from our manufacturing engineer that a convenient choice for substrate thickness h is 0.5 mm. From this, to get 50 the width w of the microstrip should be

The phase constant for this line will be

1.88 1.88 0.5 mm 0.94 mmw h

eff

9 1

10

0 0

1 1

3.3941

2 2.4 10 s250.3 m

299792458 m s

50.3 m 3.3941 92.67 m

fk

c c

lecture 4a -- transmission lines · all two‐conductor transmission lines either support a tem...

Documents