Prof. Ji Chen

Notes 6 Transmission Lines

(Time Domain)

ECE 3317

1

Spring 2014

Note:

Transmission lines is the subject of Chapter 6 in the Shen & Kong book. However, the subject of wave propagation in the time domain is not treated very thoroughly there. Chapter 10 of the Hayt book has a more thorough discussion.

Note about Notes 6

2

A transmission line is a two-conductor system that is used to transmit a signal from one point to another point.

Transmission Lines

Two common examples:

Coaxial cable Twin lead

r a

b

z

A transmission line is normally used in the balanced mode, meaning equal and opposite currents (and charges) on the two conductors.

3

Transmission Lines (cont.)

Coaxial cable

Here’s what they look like in real-life.

Twin lead

4

Transmission Lines (cont.)

5

CAT 5 cable(twisted pair)

Transmission Lines (cont.)

Some practical notes:

Coaxial cable is a perfectly shielded system (no interference).

Twin line is not a shielded system – more susceptible to noise and interference.

Twin line may be improved by using a form known as “twisted pair” (e.g., CAT 5 cable). This results in less interference.

l

-l

+

++

+

++

-

---

- -

E

Coax Twin lead

E

6

Another common example (for printed circuit boards):

Microstrip line

w

hr

Transmission Lines (cont.)

7

Transmission lines are commonly met on printed-circuit boards.

A microwave integrated circuit (MIC)

Microstrip line

8

Transmission Lines (cont.)

Microstrip line

Transmission Lines (cont.)

9

Microstrip

h

w

er

er

w

Stripline

h

Transmission lines commonly met on printed-circuit boards

Coplanar strips

her

w w

Coplanar waveguide (CPW)

her

w

10

Transmission Lines (cont.)

4 parameters

Symbol for transmission line:

z

Note: We use this schematic to represent a general transmission line, no matter what the actual shape of the conductors.

Transmission Lines (cont.)

11

+-

,V z t

,I z t

4 parameters

C = capacitance/length [F/m]

L = inductance/length [H/m]

R = resistance/length [/m]

G = conductance/length [S/m]

Four fundamental parameters that characterize any transmission line:

z

These are “per unit length” parameters.

Capacitance between the two wires

Inductance due to stored magnetic energy

Resistance due to the conductors

Conductance due to the filling material between the wires

Transmission Lines (cont.)

12

Circuit Model

Circuit Model:

Dz13

z

Dz

+-

,V z t

,I z t

RDz LDz

GDz CDz

z

Coaxial Cable

Example (coaxial cable)

r a

bz

0

0

2F/m

ln

ln H/m2

rCba

bL

a

2S/m

ln

1 1/m

2 2

d

a ma a mb

Gba

Ra b

2

m

(skin depth of metal)

d = conductivity of dielectric [S/m]

m = conductivity of metal [S/m]

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,,

2a b

ma mb

ma

mb

Coaxial Cable (cont.)

Overview of derivation: capacitance per unit length

1QC

V z V

0 0

ln2 2

b b

r ra a

bV E d d

a

02

ln

rCba

15

l

-l

+

++

+

++

-

---

- -

E

a

b

Coaxial Cable (cont.)

Overview of derivation: inductance per unit length

1L

I z

0 0

0

2

ln2

b b

a a

Iz H d z d

I bz

a

0 ln2

bL

a

16

y

E

x

Js

b

a

Coaxial Cable (cont.)

Overview of derivation: conductance per unit length

2

ln

dGba

d

C G

RC Analogy:

2

lnC

ba

17

da

b

Coaxial Cable (cont.)

Relation Between L and C:

0 02F/m ln H/m

2ln

r bC L

b aa

0 0r rLC

Speed of light in dielectric medium: 1

dc

2

1

d

LCc

Hence:This is true for ALL transmission lines.

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( A proof will be seen later.)

Telegrapher’s Equations

Apply KVL and KCL laws to a small slice of line:

+ V (z,t)-

RDz LDz

GDz CDz

I (z,t)

+ V (z+Dz,t)-

I (z+Dz,t)

z

z z+Dz

( , )KVL : ( , ) ( , ) ( , )

( , )KCL : ( , ) ( , ) ( , )

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

tV z z t

I z t I z z t V z z t G z C zt

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Hence

( , ) ( , ) ( , )( , )

( , ) ( , ) ( , )( , )

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

z tI z z t I z t V z z t

GV z z t Cz t

Now let Dz 0:

V IRI L

z tI V

GV Cz t

“Telegrapher’s Equations (TEs)”

Telegrapher’s Equations (cont.)

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To combine these, take the derivative of the first one with

respect to z:

2

2

2

2

V I IR L

z z z t

I IR L

z t z

V V VR GV C L G C

t t t

Take the derivative of the first TE with respect to z.

Substitute in from the second TE.

Telegrapher’s Equations (cont.)

21

I VGV C

z t

2 2

2 2( ) 0

V V VRG V RC LG LC

z t t

The same equation also holds for i.

Hence, we have:

2 2

2 2

V V V VR GV C L G C

z t t t

There is no exact solution to this differential equation, except for the lossless case.

Telegrapher’s Equations (cont.)

22

2 2

2 2( ) 0

V V VRG V RC LG LC

z t t

The same equation also holds for i.

Lossless case:

2 2

2 20

V VLC

z t

0R G

Note: The current satisfies the same differential equation.

Telegrapher’s Equations (cont.)

23

“wave equation”

The same equation also holds for i.

2 2

2 2 2

10

d

V V

z c t

Hence we have

Solution:

,d d

V z t f z c t g z c t

where f and g are arbitrary functions.

Solution to Telegrapher's Equations

This is called the D’Alembert solution to the wave equation (the solution is in the form of traveling waves).

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Traveling Waves

2 2

2 2 2

1

d

V V

z c t

,d d

V z t f z c t g z c t

Proof of solution

2

2

22 2

2

,

,

d d

d d d d

V z tf z c t g z c t

zV z t

c f z c t c g z c tt

It is seen that the differential equation is satisfied by the general solution.

General solution:

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Example(square pulse):

z

t = 0 t = t1 > 0 t = t2 > t1

z0 + cd t1z0 z0 + cd t2

V(z,t)

,d

V z t f z c t

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z

f (z)

z0

t = 0

Traveling Waves (cont.)

“snapshots of the wave”

z

t = 0t = t1 > 0t = t2 > t1

z0 - cd t1 z0z0 - cd t2

V(z,t)

,d

V z t g z c t

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z

g (z)

z0

t = 0

Traveling Waves (cont.)

Example(square pulse):

“snapshots of the wave”

Loss causes an attenuation in the signal level, and it also causes distortion (the pulse changes shape and usually gets broader).

z

t = 0 t = t1 > 0t = t2 > t1

z0 + cd t1z0 z0 + cd t2

V(z,t)

Traveling Waves (cont.)

(These effects can be studied numerically.)

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Current

V IRI L

z t

V I

Lz t

Lossless

,d d

V z t f z c t g z c t

,d d

V z tf z c t g z c t

z

,d d

I z t u z c t v z c t

,d d d d

I z tc u z c t c v z c t

t

(first Telegrapher’s equation)

Our goal is to now solve for the current on the line.

Assume the following forms:

The derivatives are:

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Current (cont.)

d d d d d df z c t g z c t L c u z c t c v z c t

V IL

z t

d d df z c t L c u z c t

d d dg z c t L c v z c t

1

1

d d

d

d d

d

u z c t f z c tLc

v z c t g z c tLc

This becomes

Equating like terms, we have

Hence we have

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Note: There may be a constant of integration, but this would correspond to a DC current, which is ignored here.

1 1d

LLc L L

CLC

Then

0

0

1

1

d d

d d

u z c t f z c tZ

v z c t g z c tZ

Define the characteristic impedance Z0 of the line:

0

LZ

C The units of Z0 are Ohms.

Current (cont.)

Observation about term:

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General solution:

,d d

V z t f z c t g z c t

0

1, d dI z t f z c t g z c t

Z

For a forward wave, the current waveform is the same as the voltage, but reduced in amplitude by a factor of Z0.

For a backward traveling wave, there is a minus sign as well.

Current (cont.)

32

Current (cont.)

z

Picture for a forward-traveling wave:

0

,

1,

d

d

V z t f z c t

I z t f z c tZ

+

- ,V z t

,I z t

forward-traveling wave

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0

,

,

V z tZ

I z t

z

Physical interpretation of minus sign for the backward-traveling wave:

0

,

1,

d

d

V z t g z c t

I z t g z c tZ

+

- ,V z t

,I z t

backward-traveling wave

,I z t

The minus sign arises from the reference direction for the current.

Current (cont.)

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0

,

,

V z tZ

I z t

0

,

,

V z tZ

I z t

Coaxial Cable

Example: Find the characteristic impedance of a coax.

0

0

2F/m

ln

ln H/m2

rCba

bL

a

0

00

ln2

2

ln

r

bL a

ZC

ba

00

0

1 1ln

2r

bZ

a

r a

bz

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Coaxial Cable (cont.)

0 0

1 1ln

2r

bZ

a

00

0

0 376.7303

-120

-70

8.8541878 10 [F/m]

= 4 10 [H/m] ( )µ

exact

(intrinsic impedance of free space)

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r a

bz

Twin Lead

10 0

1 1cosh

2r

dZ

a

d

a = radius of wires

r

10 0

1

F/m cosh H/m2

cosh2

r dC L

d aa

0 0

1 1ln

r

dZ

a

a d

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Twin Line (cont.)

Coaxial cable

0 300Z

Twin line

0 75Z

These are the common values used for TV.

75-300 [] transformer

300 [] twin line 75 [] coax

38

w

hr

Microstrip Line

0

0

,

,

r

wC w h

hh

L w hw

0 0

1,

r

hZ

w

w h

Parallel-plate formulas:

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Microstrip Line (cont.)

More accurate CAD formulas:

w

hr

w

hr

0

120

/ 1.393 0.667 ln / 1.444effr

Zw h w h

1 1 11 /

2 2 4.6 /1 12 /

eff r r rr

t h

w hh w

( / 1)w h

( / 1)w h

Note: The effective relative permittivity accounts for the fact that some of the fields are outside of the substrate, in the air region. The effective width w' accounts for the strip thickness.

21 ln

t hw w

t

t = strip thickness

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Some Comments

Transmission-line theory is valid at any frequency, and for any type of waveform (assuming an ideal straight length of transmission line).

Transmission-line theory is perfectly consistent with Maxwell's equations (although we work with voltage and current, rather than electric and magnetic fields).

Circuit theory does not view two wires as a "transmission line": it cannot predict effects such as single propagation, reflection, distortion, etc.

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One thing that transmission-line theory ignores is the effects of discontinuities (e.g. bends). These may cause reflections and possibly also radiation at high frequencies, depending on the type of line.