I. Introduction to Wave Propagation - NYU Tandon...

July, 2003 © 2003 by H.L. Bertoni 1

I. Introduction to WavePropagation

• Waves on transmission lines• Plane waves in one dimension• Reflection and transmission at junctions• Spatial variations for harmonic time

dependence• Impedance transformations in space• Effect of material conductivity


Waves on Transmission Lines

• Equivalent circuits using distributed C and L

• Characteristic wave solutions

• Power flow


Examples of Transmission Lines

I(z,t) +

V(z,t) - z

I(z,t) +

V(z,t) -

Dielectric

Conductors

Strip Line

Coaxial Line

Two-Wire Line(Twisted Pair)


Properties of Transmission Lines (TL’s)

• Two wires having a uniform cross-section in one(z) dimension

• Electrical quantities consist of voltage V(z,t) andcurrent I(z,t) that are functions of distance z alongthe line and time t

• Lines are characterized by distributed capacitanceC and inductance L between the wires– C and L depend on the shape and size of the conductors

and the material between them


Capacitance of a Small Length of Line

The two wires act as a capacitor. Voltage applied to the wires

induces a charge on the wires, whose time derivative is the current.

Since the total charge, and hence the current, is proportional to

the length of the wires. Let the constant of proportionality be

Farads/meter. Then

l

C

I t CldV t

dt( )

( )=

I(t) +

V(t) -

l

Open circuitE


Inductance of a Small Length of Line

The wire acts as a one - turn coil. Current applied to the wires induces

a magnetic field throught the loop, whose time derivative generates the

voltage. The amount of magnetic flux (magnetic field area), and hence

the voltage, is proportional to the length of the wires. Let the constant

of proportality be Henrys/meter. Then

×

=

l

L

V t LldI t

dt( )

( )

I(t) +

V(t) -

l

Short circuitB


C and L for an Air Filled Coaxial Line

a

b

Permittivity of vacuum: Farads/m

Permeability of vacuum: Henrys/m

Cb a

L b ao o

o

o

=( )

= ( )

≈

≡ ×

−

−

22

10364 10

9

7

πε µπ

επ

µ π

lnln

Suppose that mm and mm. Then

pF/m and H/m

Note that

m/s and

a b

C L

LC

L

C

b a

o o

o o

o

o

= =

= = = =

= = × =( )

= =

0 5 2

24

40 14

20 277

1 13 10

24

2377 83 28

.

ln.

ln.

ln ln.

πε µπ

µ

µ ε πµε π

Ω


C and L for Parallel Plate Line

w

h

z

Note that for air between the plates and so that

m/s

Cw

hL

h

w

LC

L

C

h

w

h

w

o o

o o

o

o

= =

= =

= = × = =

ε µ

ε ε µ µ

µ εµε

1 13 10 3778 Ω


Two-Port Equivalent Circuit of Length ∆z

I(z,t) +

V(z,t)

-z z+∆z z

L∆z C ∆z

I(z,t) +

V(z,t)

-

+ I(z +∆z,t)

V(z+∆z,t)

-Kirchhoff circuit equations

or

V z t L zI z t

tV z z t I z t C z

V z z t

tI z z t

V z z t V z t

zL

I z t

t

I z z t I z t

z

( , )( , )

( , ) ( , )( , )

( , )

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

= + + =+

+ +

+ −= −

+ −= −

∆ ∆ ∆∆

∆

∆∆

∆∆

∂∂

∂∂

∂∂

CCV z z t

t

∂∂

( , )+ ∆


Transmission Line Equations

Taking the limit as gives the Transmission Line Equations

These are coupled, first order, partial differential equations whose solutions

are in terms of functions ( - / ) and that are determined by

the sources. The solutions for voltage and current are of the form

( - / ) +

∆z

V z t

zL

I z t

t

I z t

zC

V z t

t

F t z v G t z v

V z t F t z v G t z v I

→

= − = −

+

= +

0

∂∂

∂∂

∂∂

∂∂

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

( / )

( , ) ( / ) (zz tZ

F t z v G t z v

vF t z v G t z v L

ZF t z v G t z v

vZF t z v G t z v C F

, ) ( / )

' '( / ) ' '( / )

' '( / ) '

= +[ ]

− +[ ] = − +[ ]

− +[ ] = −

1

1 1

1

( - / ) -

Direct substitution into the TL Equations, and using the chain rule gives

( - / ) - ( - / ) -

( - / ) + (tt z v G t z v

F G

- / ) +

where the prime (' ) indicates differentiation with respect to the total variable

inside the parentheses of or .

'( / )+[ ]


Conditions for Existence of TL Solution

For the two equations to be satisfied

and

Multiplying both sides of the two equations gives or

m/s

Dividing both sides of the two equations gives or

and are interpreted as the wave velocity and wave impedance.

1 1

1

1

2

v

L

Z vZC

v Z

LC

Z

vLC

vZ

v

L

ZC

ZL

Cv Z

= =

=

=

=

= Ω


F(t-z/v) Is a Wave Traveling in +z Direction

Assume that

Then the voltage and current are

represents a wave disturbance

traveling in the positive direction with

velocity .

Note that the current in the conductor at

positive potential flows in the direction of

wave propagation.

G t z v

V z t F t z v F v z vt

I z tZ

F t z vZ

F v z vt

F t z v

z

v

( / )

( , ) ( / ) ( )( )

( , ) ( / ) ( )( )

( / )

+ =

= − = − −[ ]

= − = − −[ ]

−

0

1

1 11

V(z,0)=F[(-1/v)(z)]

V(z,t)=F[(-1/v)(z-vt)]

a z

-a

a+vt z

-a+vt

vt

t = 0

t > 0


G(t+z/v) Is a Wave Traveling in -z Direction

Assume that

Then the voltage and current are

represents a wave disturbance

traveling in the negative direction with

velocity .

Because of the minus sign in the

physical current in the conductor at positive

potential flows in the direction of wave propagation.

F t z v

V z t G t z v G v z vt

I z tZ

G t z vZ

G v z vt

G t z v

z

v

I z t

( / )

( , ) ( / ) ( )( )

( , ) ( / ) ( )( )

( / )

( , ),

− =

= + = +[ ]

=−

+ =−

+[ ]

+

0

1

1 11

V(z,0)=G[(1/v)(z)]

a 2a z

t = 0

V(z,t)=G[(1/v)(z+vt)]

2a-vt z

-vt

a-vt

t > 0


Example of Source ExcitationExcitation at one end of a semi - infinite length of transmission line.

Source has open circuit voltage and internal resistance

Radiation condition requires that excited waves travel away from source.

Terminal conditions at

or

V t R

z

V t R I t V t

RZ

F t F t

F tZ

Z RV t

V t R I t V t

RZ

G

S S

S S

S

SS

S S

S

( ) .

:

( ) ( , ) ( , )

( ) ( )

( ) ( )

( ) ( , ) ( , )

(

=

= +

= +

=+

= − +

= −−

0

0 0

1

0 0

1tt G t

G tZ

Z RV t

SS

) ( )

( ) ( )

+

=+

or

∞

z

VS(t) +

0

RS I(0,t)

V(0,t)

+ VS(t)

I(0,t) RS

V(0,t)

0 z

∞


Receive Voltage Further Along Line

+ VS(t)

∞

z

VS(t) +

0 l

RS

V(l,t)

Scope

RS

V(-l,t)

-l 0 z

∞

Scope

Voltage observed on a high impedance scope at a distance from source.

Delayed version of the source voltage

with the semi - infinite line acting as a

load resisor for the source.

Delayed version of the source voltage

with the semi - infinite line acting as a

load resisor for the source.

l

V l t F t l vZ

Z RV t l v

V l t G t l vZ

Z RV t l v

SS

SS

( , ) ( ) ( )

( , ) ( ) ( )

= − =+

−

− = + −[ ] =+

−


Power Carried by Waves

P(z,t)

I(z,t)

V(z,t)

z

Instantaneous power ( , ) carried past plane

perpendicular to .

The two waves carry power independently in the direction of wave

propagation

For each wave, a transmission line extending to acts as a resistor

of value Z, even though the wires were assumed to have no resistance.

P z t

z

P z t V z t I z t

F t z v G t z vZ

F t z v G t z v

ZF t z v G t z v

z

( , ) ( , ) ( , )

( ) ( ) ( ) ( )

( ) ( )

=

= − + +[ ] − − +[ ]

= − − +[ ]

→∞

1

1 2 2


Summary of Solutions for TL’s

• Solutions for V and I consists of the sum of thevoltages and current of two waves propagating in±z directions

• For either wave, the physical current flows in thedirection of propagation in the positive wire

• Semi-infinite segment of TL appears at itsterminals as a resistance of value Z (even thoughthe wires are assumed to have no resistance)

• The waves carry power independently in thedirection of wave propagation


Plane Waves in One Dimension

• Electric and magnetic fields in terms of voltageand current

• Maxwell’s equations for 1-D propagation

• Plane wave solutions

• Power and polarization


Electric Field and Voltage for Parallel Plates

The electric field goes from the positive plate to the negative plate. If

>> , the electric field outside of the plates is very small. Between

the plates it is nearly constant over the cross - section with value

Volts/m or

Recall that .

w h

E z th

V z t V z t hE z t

Cw

h

x x( , ) ( , ) ( , ) ( , ).= − = −

=

1

ε

w

h

z

y

Ex(z,t) + V(z,t)-

x


Magnetic Field and Current for Parallel Plates

w

h

z

y Hy(z,t) or By(z,t)

I(z,t)x

The magnetic field links the currents in the plates. If >> , the magnetic

field outside of the plates is very small. Between the plates it is nearly

constant over the cross - section, as if in a solenoid, with value

Amps/m or

Recall that .

w h

H z t B z tw

I z tw

I z t I z t wH z t

Lh

w

y y y( , ) ( , ) ( , ) ( , ) ( , ) ( , ).≡ = −

= − = −

=

1 1 1µ µ

µ

µ


Maxwell’s Equations in 1-D

Inserting the foregoing expressions for ( , ), , ( , ) and into the

Transmission Line equations

or

V z t C I z t L

zhE z t

h

w twH z t

zwH z t

w

h thE z t

zE z t

x y y x

x

∂∂

µ∂∂

∂∂

ε∂∂

∂∂

−[ ] = −

−[ ] −[ ] = −

−[ ]

= −

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

( , ) µµ∂∂

∂∂

ε∂∂t

H z tz

H z tt

E z t

h w

x y

y y x( , ) ( , ) ( , )

These are the two Maxwell equations for linearly polarized wave propagating in

1- D. They are independent of ( , ) and refer to the fields.

We may think of the plates as being taken to ( , ) so they need not be

considered.

The field are in the form of a plane wave, which covers all space and is a simple

approximation for fields in a limited region of space, such as a laser beam.

= −

→∞


Plane Waves: Solutions to Maxwell Equations

Maxwell's equations are formally equivalent to the Transmission Line Equations

The solution is therefore in terms of two wave traveling in opposite directions

along .

In air 3 10 m/s is the speed of light and

is the wave impedance.

For waves in simple dielectric medium, is multiplied by the relative dielectric

constant

8

z

E z t F t z v G t z v H z t F t z v G t z v

v c

x y

o o

o

o

o

r

( , ) ( / ) ( / ) ( , ) ( / ) ( / )= − + + = − − +[ ]

= ≡ = × = =

1

1377

η

µ εη

µε

ε

ε

Ω

..

,

,

. .

For normal media but it can be a function of frequency. As and example,

in water at radio frequencies (below 20 GHz) but at optical

frequencies

ε

ε

ε

r

r

r

>

=

=

1

81

1 78


Power Density Carried by Plane Waves

Total instantaneous power carried in parallel plate line

watts

Power density crossing any plane perpendicular

to is

watt/m

2

P z t V z t I z t hE z t wH z t

hwE z t H z t

z

p z t P z t hw E z t H z t

F t z v

x y

x y

x y

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

( , ) ( , )

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

( / )

= = −[ ] −[ ]=

= =

= −1 2

η−− +[ ]G t z v

H

E H

y

x y

2( / )

Direction of is such that turning a right hand screw in the

direction from to advances the screw in the direction of

propagation

E

Direction of

propagation

H


Polarization

The physical properties of a plane wave are independent of the

coordinate system.

For a plane wave traveing in one direction :

Electric field vector must be perpedicular to the direction of

propagation.

Magnetic field vector must be perpedicular to and to the

direction of propagation.

The vector cross product watt/m is in the direction

of propagation.

The ratio is the wave impedance

2

E

H E

p E H

E H

= ×

η.


Examples of Polarization

Linear polaization of along Linear polaization of along

E x E y

E a t z v E a t z v

H a t z v H a t z v

x y

y x

= −[ ] = −[ ]

= −[ ] = − −[ ]

cos ( / ) sin ( / )

cos ( / ) sin ( / )

ω ω

ηω

ηω

1 1

Circular polarization

E a t z v a t z v

H a t z v a t z v

x y

y x

= −[ ] + −[ ]

= −[ ] − −[ ]

cos ( / ) sin ( / )

cos ( / ) sin ( / )

ω ω

ηω ω

1

E

x z

H

y

x z

E

y H

a x

a yx

y

= unit vector along

= unit vector along


Summary of Plane Waves

• Plane waves are polarized with fields E and Hperpendicular to each other and to the direction ofpropagation

• Wave velocity is the speed of light in the medium

• ExH watts/m2 is the power density carried by aplane wave


Reflection and Transmission atJunctions

• Junctions between different propagation media• Reflection and transmission coefficients for 1-D

propagation• Conservation of power, reciprocity• Multiple reflection/transmission


Junctions Between Two Regions

0 z

I(0-,t) I(0+,t)

TL 1 V(0-,t) + V(0+,t) TL 2

Ex(0-,t) Ex(0+,t)

Hy(0-,t) Hy(0+,t)

Medium 1 Medium 2

x

z

Terminal condtions for the

Junction of two TL's

Boundary conditions at the

interface of two media

Plane wave propagation and

boundary conditions are analogus

to junctioning of two TL's

y y

V t V t

I t I t

E t E t

H t H t

x x

( , ) ( , )

( , ) ( , )

( , ) ( , )

( , ) ( , )

0 0

0 0

0 0

0 0

− +

− +

− +

− +

=

=

=

=


Reflection and Transmission

Incident wave

ExIn(z,t)=F1(t-z/v1)

HyIn(z,t) Transmitted wave

Reflected wave

v1 and η1 v2 and η2

x

z

A source creates an incident wave whose electric field is given by the known

function (t - z/v ). Using the boundary conditions we solve for the unknown

functions (t +z/v ) and (t - z/v ) for the electric fields of the reflected

and transmitted waves : (t) +G (t) (t)

(t) - G (t) (t)

1

1 2

1

1

F

G F

E t F F E t

H t F F H t

x x

y y

1

1 2

1 2

11

22

0 0

01 1

0

( , ) ( , )

( , ) ( , )

− +

− +

= = =

= [ ] = =η η


Reflection and Transmission Coefficients

Solution of the boundary condition equations for and in terms of

The reflection coefficient and transmission coefficient are given by :

Examples :

I. Suppose medium 1 is air so that 377 and medium 2 has

relative dielectric constant

G t F t F t

G t F t F t F t

o o

r

1 2 1

1 1 2 1

2 1

2 1

2

2 1

1

12

( ) ( ) ( )

( ) ( ) ( ) ( )= =

=−+

= + =+

= ≡ =

=

Γ Τ

Γ Τ

Γ Τ Γη ηη η

ηη η

η η µ ε

ε 44 0 5

0 50 5

13

113

23

2 so that Then going

from air - to - dielectric and

η µ ε ε η

η ηη η

= =

=−+

= − = − =

o r o

ad ad

. .

.

.Γ Τ


Reflection and Transmission, cont.

II. Now suppose the wave is incident from the dielectric onto air so that medium 1

is the dielectric and medium 2 is air Then going from

dielectic - to - air, and

Note that :

1.

2. Since T is the ratio of fields, not power, it can be greater than 1.

2η η η η

η ηη η

1 0 5

0 50 5

13

113

43

=( ) =( )

=−+

= + = + =

= −

. .

.

.Γ Τ

Γ Γ

da ad

da ad


Reflected and Transmitted Power

Instantaneous power carried by the incident wave the reflected wave

, and the transmitted wave

p z t

p z t p z t

p z t E z t H z t F t z v

p z t E z t H z t G t z v

p

In

Tr

InxIn

yIn

x y

Tr

( , ),

( , ) ( , )

( , ) ( , ) ( , ) ( )

( , ) ( , ) ( , ) ( )

(

Re

Re Re Re

= = −

= =−

+

1

11

12

1

112

1

η

η

zz t E z t H z t F t z v

p tZ

F t

p t G t F t p t F t F

xTr

yTr

In

Tr

, ) ( , ) ( , ) ( )

( , ) ( )

( , ) ( ) ( ) ( , ) ( )Re

= = −

=

= = = =

−

− +

1

01

01 1

01 1

222

1

112

112 2

112

222 2

2

η

η η η η

Just on either side of the interface as well as

and Γ Τ 112( )t


Conservation of Power and Reciprocity

Conservation of power requires that so that

or 1

This relation is easily shown to be satisfied from the expressions for , .

For waves going from medium 2 to medium 1, the reflection coefficient is

the negative of going from medium 1 to medium 2. Thus for either

case the ratios

21

p t p t p t

F t F t F t

p t

p

In Tr( , ) ( , ) ( , )

( ) ( ) ( )

( , )

Re

Re

0 0 0

1 1 1

0

11

2 2

11

2 2

21

2 2 2 1

2

12

− − +

−

− =

− = − =η η η

ηη

Γ Τ Γ Τ

Γ Τ

Γ

Γ

InIn

Tr

Int

p t

p t( , )( , )( , )000

2 2−

+

−= = −Γ Γ and 1 are the same.

Therefore the same fraction of the incident power is reflected from and

transmitted through the interface for waves incident from either medium.

This result is an example of a very general wave property called reciprocity.


Termination of a Transmission Line

I(0-,t)

TL V(0-,t) + RL

0 z

Terminal condtions

Solving for ( ) in terms of ( ),

( ) where the reflection

coefficient is

Special cases :

1. Matched termination, and Simulates a semi - infinite TL

2. Open circuit, and Total reflection with

V t R I t

F t G tR

ZF t G t

G t F t

G t F t

R Z

R Z

R Z

R V

L

L

L

L

L

L

( , ) ( , )

( ) ( ) ( ) ( )

( )

.

. ( ,

0 0

0

1 0

=

+ = −

=

=−+

= =

→∞ =

Γ

Γ

Γ

Γ tt F t

R V tL

) ( ).

. ( , ) .

=

= = − =

2

0 1 0 0 3. Short circuit, and Total reflection with Γ


Reflections at Multiple Interfaces

Incident wave

ExIn(z,t)=F1(t-z/v1)

TransmittedHy

In (z,t) waves

Reflected waves Multiple

internal reflections

v1 and η1 v2 and η2 v3 and η3

x

0 l z

Multiple internal reflections occur within the finite thickness layer. These

internal waves generate multiple reflected waves in medium 1 and multiple

transmitted waves in medium 3.


Scattering Diagram for a Layer

1

Γ12

Τ12 Γ23 Τ21

Τ12 (Γ23)2 Γ21 Τ21

Τ12Τ23

Τ12 Γ23 Γ21 Τ23

Τ12 (Γ23)2 (Γ21)2 Τ23

l z

2l/v2

4l/v2

t

Τ12

Τ12 Γ23

Τ12 Γ23 Γ21

Τ12 (Γ23)2 Γ21

Τ12 (Γ23)2 (Γ21)2

Τ12 (Γ23)3 (Γ21)2

Space - time diagram indicates the relative amplitudes of the electric field of

the individual components of the multiply reflected waves. In adding fields,

account must be taken of the relative delay between the different components.


Summary of Reflection and Transmission

• The planar interface between two media is analogous to thejunction of two transmission lines

• At a single interface (junction) the equation T = 1 + Γ is astatement of the continuity of electric field (voltage)

• The ratio of reflected to incident power = Γ2

• Power is conserved so that the ratio of transmitted to incidentpower = 1 - Γ2

• The reciprocity condition implies that reflected and transmittedpower are the same for incidence from either medium

• At multiple interfaces, delayed multiple interactions complicatethe description of the reflected and transmitted fields forarbitrary time dependence


Spatial Variations for HarmonicTime Dependence

• Traveling and standing wave representations ofthe z dependence

• Period average power

• Impedance transformations to account for layeredmaterials

• Frequency dependence of reflection from a layer


Harmonic Time Dependence at z = 0

Suppose that the voltage and current (or and fields) have harmonic time

dependence ( ) at Then

where and are the complex voltage and current at .

The functions and can satisfy these equations only if they too have

E H

j t z

V t V e F t G t

I t I eZ

F t G t

V I z

F t G t

x y

j t

j t

exp .

( , ) ( ) ( ) ( )

( , ) ( ) ( ) ( )

( ) ( )

( ) ( )

ωω

ω

=

= = +

= = −[ ]

=

0

0 0

0 01

0 0 0

harmonic time dependence. Hence harmonic time dependence. Hence

( ) and G(t) =

where and are the complex

voltage amplitudes of the waves traveling in the directions.

F t V e V e

V V ZI V V ZI

z

j t j t=

= +[ ] = −[ ]±

+ −

+ −

ω ω

12

120 0 0 0( ) ( ) ( ) ( )


Traveling Wave Representation

At other locations

z

V z t F t z v G t z v V j t z v V j t z v

V e V e e V z e

I z tZ

F t z v G t z v

j z v j z v j t j t

≠

= − + + = −[ ] + +[ ]= + =

= − − + =

+ −

+ − − +

0

1

( , ) ( ) ( ) exp ( ) exp ( )

( )

( , ) ( ) ( )

ω ωω ω ω ω

11

1

1

ZV j t z v V j t z v

ZV e V e e I z e

V z I z

j t

k v

V z V

j z v j z v j t j t

+ −

+ − − +

−

+

−[ ] − +[ ]

= − =

≡

=

exp ( ) exp ( )

( )

( ) ( )

exp

.

( )

ω ω

ω

ω

ω ω ω ω

Here is the phasor voltage and is the phasor current, which give the

spatial variation for the implied time dependence ( ).

Define the wave number (propagation constant) m Then

ee V e I zZ

V e V ejkz jkz jkz jkz− − + + − − ++ = − and

is the traveling wave representation of phasor voltage and current.

( )1


Standing Wave Representation

Substituting the expressions for and in terms of and ( ),

and rearranging terms gives the standing wave representation of the phasor

voltage and current :

V V V I

V z V e e ZI e e V kz jZI kz

I zZ

V e e I e e

jkz jkz jkz jkz

jkz jkz jkz

+ −

− + − +

− + − +

= +[ ] + −[ ] = −

= −[ ] + +

( )

( ) ( ) ( ) ( )cos ( )sin

( ) ( ) ( )

0 0

0 0 0 0

0 0

12

12

12 1

2jkzjkz I kz j

ZV kz[ ] = −( )cos ( )sin0

10

The wavenumber is where is the

wavelength 2 k

For plane waves in a dielectric medium

k v f v

v f

k

= = =

= =

=

ω π π λ λ

λ π

ω µε

2 2


Variation of the Voltage Magnitude

For we have a pure traveling

wave . The magnitude

= is independent

of

V

V z V e

V z V e V

z

jkz

jkz

−

+ −

+ − +

=

=

=

0

( )

( )

.

|V+|

z

V z( )

For (0) we have a pure standing

wave . Its magnitude

cos is periodic with

period 2.

I

V z V kz

V z V kz

k

=

=

=

=

0

0

0

( ) ( )cos

( ) ( )

π λ

0 λ/2 z

V z( )

V ( )0


Standing Wave Before a Conductor

ISC

η, v short

0 z

Incident wave

ExIn(z)

HyIn(z)

ExRe(z)

Reflected wave

x

Perfect

conductor

0 z

Plane wave incident on a perfectly

conduticng plate and the equivalent

circuit of a shorted TL

and

The standing wave field is

Two waves of equal amplitude and

traveling in opposite directions create

a standing wave.

E H I

E z I e e

j I kz

x y SC

x SCjkz jkz

SC

( ) ( )

( )

sin

0 0 0

12

= =

= −[ ]= −

− +η

η


Standing Wave Before a Conductor, cont.

Plot of the magnitude of the standing wave field

E z I kzx SC( ) sin=η

Since the nodes (zeros) of the field are

separated by a distance or

k v f v

k z z k

= = =

= = =

ω π π λ

π π λ

2 2

2∆ ∆

ηISC

-λ −3λ/4 −λ/2 −λ/4 0 z


Period Averaged Power

For harmonic time dependence on a TL, the time average over one period

of the instantaneous power is watts

Using the traveling wave representation

Note that the average power is the algebraic sum of the power carried by

the incident and reflected waves, and it is independent of

For harmonic plane waves

P z V z I z

P z V e V eZ

V e V eZ

V V

z

p z E

jkz jkz jkz jkz

x

( ) Re ( ) ( )

( ) Re

.

( ) Re (

=

= +[ ] −[ ]

= −

=

∗

+ − − + + − − + ∗ + −

12

12

2 2

12

1 12

zz H z

p z E E

y

xIn

x

) ( )

( ) Re

∗ = −

watts/m

In terms of traveling waves

2

12

2 2

η


Reflection From a Load Impedance

V+

V- ZL

0 z

I(0)

V(0) + ZL

0 z

For a complex load impedance

Solving for in terms of gives

where the complex

reflection coefficient is

Reflected power

Z

V V V Z IZ

ZV V

V V

V V

Z Z

Z Z

PZ

VZ

V P

L

LL

L

L

In

( ) ( )

Re

0 0

12

12

2 2 2

= + = = −( )

=

=−+

= = =

+ − + −

− +

− +

− +

Γ

Γ

Γ

Γ Γ


Summary of Spatial Variation for HarmonicTime Dependence

• Field variation can be represented by two traveling wavesor two standing waves

• The magnitude of the field for a pure traveling wave isindependent of z

• The magnitude of the field for a pure standing wave isperiodic in z with period λ/2

• The period average power is the algebraic sum of thepowers carried by the traveling waves

• The period average power is independent of z no matter ifthe wave is standing or traveling

• The fraction of the incident power carried by a reflectedwave is |Γ|2


Impedance Transformationsin Space

• Impedance variation in space

• Using impedance for material layers

• Frequency dependence of reflection from a brickwall

• Quarter wave matching layer


Defining Impedance Along a TL

I(0)

ZIN V(0) + ZL

-l 0 z

At the ratio of voltage to current

can have some value ( )

Using the formulas for and

we can compute their ratio at

Defining this ratio as we have

( ) (-

(- ( ) (- )

Dividing numerator and denominator by and rearranging gives

z

V I Z

V z I z

z l

Z l

Z lV l

I l

V kl jZI kl

I kl jZ

V kl

I

L

IN

IN

=

≡

= −

≡−−

=− −

−

0

0 0

0 0

01

0

0

( )

( ) ( )

.

( )

( )( )( )

cos ) ( )sin( )

( )cos ) sin

( )

(

( ( ) ( )Z l Z

Z kl jZ kl

Z kl jZ klZ

Z jZ kl

Z jZ klINL

L

L

L

( )cos ) sin( )

cos ) sintan( )tan

=+

+=

++


Properties of the Impedance Transform

The impedance formula

(

( ( ) ( )

shows that a length TL (or region of space) transforms an impedance

to a different value.

Some properties of the transformation :

1. For a matched load the imput impedace is matched

2. The impedance repeats for or

Z l ZZ kl jZ kl

Z kl jZ klZ

Z jZ kl

Z jZ kl

Z Z Z Z

Z l Z l l k l

l k

INL

L

L

L

L IN

IN IN

( )cos ) sin( )

cos ) sintan( )tan

,

( ) ( )

=+

+=

++

= =

= + =

= =

∆ ∆

∆

π

π λ 2

3. For quarter wave displacement 3. For quarter wave displacement and impedance

inverts

4. If then

l kl

Z Z Z

Z Z l jZ klIN L

L IN

= =

=

= =

λ π

λ

4 2

4

0

2

,

( )

, ( ) tan( )


Using Transform for Layered Media

Incident wave

ExIn(z) Ex

TR(z) Transmitted

HyIn (z) wave

ExRe(z)

Reflected wave

v1 , η1 v2 , η2 v3 , η3

x

0 l z

ZIN(l) ZL = η3

Z= η2


Circuit Solution for Reflection Coefficient

Medium 3 acts as a load on the layer to the left. A semi- infinite TL (medium)

at its terminals (accessible surface) acts as a resistor so that

Impedance of the finite segment of TL is . Wavenumber of this

segment is

where is the wavenumber of free space.

Input impedance at left surface of the layer is then

(( ( )

Z

Z

k v k

k

Z lk l j k l

k l j k l

L

r o o o r

o o o

IN

=

=

= = =

=

=++

η

η

ω ω ε ε µ ε

ω ε µ

ηη ηη η

3

2

2 2 2 2

23 2 2 2

2 2 3 2

.

( )cos ) sin( )cos ) sin

Reflection coefficient for the wave incident from medium 1 is Reflection coefficient for the wave incident from medium 1 is

(

(Γ =

−+

=−( ) + −

+( ) + +Z l

Z l

k l j k l

k l j k lIN

IN

( )( )

cos ) ( )sin( )

cos ) ( )sin( )ηη

η η η η ηηη η η η ηη

1

1

2 3 1 2 22

1 3 2

2 3 1 2 22

1 3 2


Example 1: Reflection at a Brick Wall

H yIN

E xIN

w

Medium 1 and medium 3 are air

Medium 2 is brick with

and

η η ηµε

ε

ηµε ε

η

1 3

2

2 22

12

4

2

= = ≡

≈

= = =

oo

o

r

oo

r ook k

Reflection coefficient for the wave incident from air is

(

(

Γ =−( ) + −

+( ) + +

=−( )+ +( )

η η η η ηηη η η η ηη

η η

η η η

2 3 1 2 22

1 3 2

2 3 1 2 22

1 3 2

14

2 2

2 14

2 2

2

2 2

cos ) ( )sin( )

cos ) ( )sin( )

sin( )

cos( ) sin(

k w j k w

k w j k w

j k w

k w jo o o

o o o o 22

2

2 2 2

34

54k w

j k w

k w j k wo

o

o o)

sin( )

cos( ) sin( )=

+


Example 1: Reflection at a Brick Wall, cont.

Let the wall thickness be cm so that

Then

w k wf

f

p pf

f f

o GHz

in GHz

GHz GHz

= =×

=

= =+

30 24

3 100 3 4

9 4

64 4 25 4

8

22

2 2

ππ

ππ π

.

sin ( )

cos ( ) sin ( )Re Γ

Since there is no conductivity in the brick wall, the fraction of the incident

power transmitted through the wall is 2

1− Γ

0 0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0 fGHz

|Γ|2

9/25


Example 2: Quarter Wave Layers

Incident wave

ExIn(z) Ex

TR(z) Transmitted

HyIn (z) wave

ExRe(z)

Reflected wave

v1 , η1 v2 , η2 v3 , η3

l=π/(2k2)=λ2/4

x

0 z

cos( ) cos( ) cos( / ) ( ) sin( / )

( / ) /

k l k k l

ZIN

2 2 2 2

2 22

3

4 2 0 2 1

4

= = = = =

=

λ π π

λ η η

and sin

so that


Example 2: Quarter Wave Layers, cont.

For this value of we have

If we choose the layer material such that then and no

reflection takes place.

Suppose that medium 1 is air and medium 3 is glass with relative

dielectric constant

For no reflection : or

Note that the layer thickness is

g

Z

lv

f f

IN

o

r o

o

o

o

g or g

r o o

Γ

Γ

=−+

= =

= = = =

= = =

η ηηη ηη

η ηη

ε

ηµε ε

ηηµε

µε ε

ε ε

λε ε µ

22

1 3

22

1 3

22

1 3

22

21 3 2

22

2

0

44

1

4

,

/ ==

=

v

f

l

o

r

o

g

o

4

4

2

4

ε

λε

λ or where is the wavelength in air.


Summary of Impedance Transformation

• The impedance repeats every half wavelength inspace, and is inverted every quarter wavelength

• Impedances can be cascaded to find the impedanceseen by an incident wave

• Reflection from a layer has periodic frequencydependence with minima (or maxima) separatedby ∆f = v2/(2w)

• Quarter wave layers can be used impedancematching to eliminate reflections


Effect of Material Conductivity

• Equivalent circuit for accounting for conductivity

• Conductivity of some common dielectrics

• Effect of conductivity on wave propagation


G, C, L for Parallel Plate Line

w

h

z

If the material between the plate conducts electricity, there will be a

conductance mho/m in addition to the capacitance farads/m

and inductance henry/m.

The conductivity of a material is give by the parameter mho/m

Expressions for the circuit quantities are :

G C

L

Gw

hC

w

hL

h

w

σ

σ ε µ= = =


Equivalent Circuit for Harmonic Waves

In the limit as the Kirchhoff circuit equations for the phasor

voltage and current give the TL equations for harmonic time dependence

∆z

dV z

dzj LI z

dI z

dzG j C V z

→

= − = − +( )

0

( )( )

( )( )ω ω

+

I(z) V(z)

-z z+∆z z

I(z) +

V(z)

+ I(z +∆z)

V(z+∆z)jωL∆z jω C ∆z G


Harmonic Fields and Maxwell’s Equations

If >> , the fields between the plates are nearly constant over the cross- section,

so that the phasor circuit quantities are and

Substituting these exprsssions in the TL equations for harmonic time dependence,

along with the expressions for , , gives Maxwell's equations

w h

V z hE z I z wH z

G C L

dE z

dzj H z

dH z

dzj E z

x y

xy

yx

( ) ( ) ( ) ( ).

( )( )

( )( )

= − = −

= − = − +( )ωµ ωε σ

w

h

z

y Hy(z)

I(z)x +

V(z)

Ex(z)


Maxwell’s Equations With Medium Loss

With minor manipulation, Maxwell's equations for 1- D propagation of

harmonic waves in a medium with conduction loss can be written

and

The complex equivalent dielectric constant is given by

Let " Then "

In other matierials atomic processes lead to a complex dielectric of the

form

o o o

o r

dE z

dzj H z

dH z

dzj E z

j j

j

xy

yx

r r

o

( )( )

( ) ˆ ( )

ˆ

ˆ

. ˆ

= − = −

= − = −( )= = −( )

ωµ ωε

ε

ε ε ε σ ω ε ε σ ωε

ε σ ωε ε ε ε ε

εε ε ε

εo jr " These processes have a different frequency

dependence for ", but have the same effect on a hamonic wave

−( ).


Constants for Some Common Materials

When conductivity exists, use complex dielectric constant given by

ε = εo(εr - jε") where ε" = σ/ωεo and εo ≈ 10-9/36π

Material* εr σ (mho/m) ε" at 1 GHz

Lime stone wall 7.5 0.03 0.54Dry marble 8.8 0.22Brick wall 4 0.02 0.36Cement 4 - 6 0.3Concrete wall 6.5 0.08 1.2Clear glass 4 - 6 0.005 - 0.1Metalized glass 5.0 2.5 45Lake water 81 0.013 0.23Sea Water 81 3.3 59Dry soil 2.5 -- --Earth 7 - 30 0.001 - 0.03 0.02 - 0.54

* Common materials are not well defined mixtures and often contain water.


Incorporating Material Loss Into Waves

Using the equivalent complex dielectric constant, Maxwell's equations

have the same form as when no loss (conductivity) is present.

The solutions therefore have the same mathematical form with

replaced by

For example, the traveling wave solutions in a material are

and

Here and

are complex quantities.

ε

ε

η

ω µε ω µε ε ε ηµε

µε ε ε

ˆ.

( ) ( )

ˆ "ˆ "

E z V e V e H z V e V e

k jj

xjkz jkz

yjkz jkz

o ro r

= + = −

= = −( ) = =−( )

+ − − + + − − +1


Wave Number and Impedance

The complex wavenumber will have real and imaginary parts

If is less than about we may use the approximations

and

Similarly, for small,

k

k j j

jj

o r

r

o r o rr

o r o r r

≡ − = −( )

≈ ≈

=−( )

≈ +

β α ω µε ε ε

ε ε

β ω µε ε α ω µε εεε

ε ηµ

ε ε εµε ε

εε

"

" ,

"

""

"

10

2

12


Effect of Loss on Traveling Waves

For a wave traveling in the positive direction

(-

The presence of loss (conductivity) results in a finite value of the

attenuation constant . The attenuation (decay) length is 1 .

The magnitude of the field depends on z as given by

z

E z V e V j j z V j z z

E z V z

xjkz

x

( ) exp ( ) exp )exp( )

( ) exp( )

= = − −[ ] = −

= −

+ − + +

+

β α β α

α α

α

|V+|

|V+| /e

1/α z


Attenuation in dB

For a traveling wave, the attenuation in units of deci - Bells is found from

Thus the attenuation rate of the wave in a medium is dB/m

AttnE z

E

V z

V

z e z

x

x

= −

= −

−

= =

+

+20

020

20 8 67

8 67

10 10

10

log( )

( )log

exp( )

log .

.

α

α α

α


Effect of Loss on Traveling Waves, cont.

The instantaneous field of the wave has both sinusoidal variation over a

wavelength and the decay over the attenuation length 1 .

For real amplitude , the spatial variation is given by

Re

or

os( t -

λ π β α

ω β α

ω β α

ω

=

= −[ ] −

−

+

+

+

2

V

E z e V j t z z

V c z z

xj t( ) Re exp ( ) exp( )

)exp( )

1/α

λ=2π/β

V+

V+/e z


Loss Damps Out Reflection in Media

Traveling waveamplitude

z

Reflectingboundary

Incident wave

Reflected wave

E z V z E z V zxIN

x( ) exp( ) ( ) exp( )Re

= − = ++ +α αΓ


Effect of Damping on the |Γ| for a Wall

0 0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0 fGHz

|Γ|2

9/25

1/9

With absorption in the brick wall, the interference minima are

reduced and the reflection coefficient approaches that of the

first air - brick interface or

The fraction of the incident power transmitted through the

wall is 2

Γ

Γ

=−+

= −

≠ −

η ηη η

B o

B o

1 3

1


Summary of Material Loss

• Conductivity is represented in Maxwell’s equationsby a complex equivalent dielectric constant

• The wavenumber k = β − jα and wave impedanceη then have imaginary parts

• The attenuation length = 1/α

• Loss in a medium damps out reflections within amedium

I. Introduction to Wave Propagation - NYU Tandon...

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