July, 2003© 2003 by H.L. Bertoni1 I. Introduction to Wave Propagation Waves on transmission lines Plane waves in one dimension Reflection and transmission

July, 2003 © 2003 by H.L. Bertoni 1

I. Introduction to Wave Propagation

• 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) and current I(z,t) that are functions of distance z along the line and time t

• Lines are characterized by distributed capacitance C 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 l of the wires. Let the constant of proportionality be

C Farads/meter. Then

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 l of the wires. Let the constant

of proportality be L Henrys/meter. Then

V(t) =LldI(t)dt

I(t) +

V(t) -

l

Short circuitB


C and L for an Air Filled Coaxial Line

a

b

€

C =2πεo

ln b a( ) L =

μo

2πln b a( )

Permittivity of vacuum: εo ≈10−9

36π Farads/m

Permeability of vacuum: μo ≡4π ×10−7 Henrys/m

€

Suppose that a=0.5 mm and b=2 mm. Then

C =2πεo

ln4=40.1 pF/m and L =

μoln42π

=0.277 μH/m

Note that

1LC

=1

μoεo

=3×108 m/s and LC

=ln b a( )

2πμo

εo

=ln42π

377=83.2 Ω


C and L for Parallel Plate Line

w

h

z

€

C =εwh

L =μhw

Note that for air between the plates ε=εo and μ=μo so that

1LC

=1

μoεo

=3×108 m/s LC

=hw

μo

εo

=hw

377 Ω


Two-Port Equivalent Circuit of Length zI(z,t) +

V(z,t)

-z z+z z

Lz C z

I(z,t) +

V(z,t)

-

+ I(z +z,t)

V(z+z,t)

-

€

Kirchhoff circuit equations

V(z,t)=LΔz∂I (z,t)

∂t+V(z+Δz,t) I (z,t) =CΔz

∂V(z+Δz,t)∂t

+I (z+Δz,t)

or

V(z+Δz,t)−V(z,t)Δz

=−L∂I (z,t)

∂t

I (z+Δz,t)−I (z,t)Δz

=−C∂V(z+Δz,t)

∂t


Transmission Line Equations

€

Taking the limit as Δz→ 0 gives the Transmission Line Equations

∂V(z,t)

∂z=−L

∂I(z,t)∂t

∂I (z,t)

∂z=−C

∂V(z,t)∂t

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

are in terms of functions F(t- z/v) and G(t+z/v) that are determined by

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

V(z,t)=F(t- z/v)+G(t+z/v) I (z,t)=1Z

F (t-z/v)-G(t+z/v)[ ]

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

−1v

F '(t- z/v)-G'(t+z/v)[ ] =−L1Z

F '(t-z/v)-G'(t+z/v)[ ]

−1vZ

F '(t- z/v)+G'(t+z/v)[ ]=−C F '(t-z/v) +G'(t+z/v)[ ]

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

inside the parentheses of F or G.


Conditions for Existence of TL Solution

€

For the two equations to be satisfied

1v

=LZ

and 1vZ

=C

Multiplying both sides of the two equations gives 1

v2Z=

LCZ

or

v=1LC

m/s

Dividing both sides of the two equations gives vZv

=L

ZC or

Z=LC

Ω

v and Z are interpreted as the wave velocity and wave impedance.


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

€

Assume that G(t+z/v)=0

Then the voltage and current are

V(z,t)=F(t−z/v)=F (−1 v)(z−vt)[ ]

I(z,t)=1Z

F(t−z/v)=1Z

F (−1 v)(z−vt)[ ]

F(t−z/v) represents a wave disturbance

traveling in the positive z direction with

velocity v.

Note that the current in the conductor at

positive potential flows in the direction of

wave propagation.

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 F(t−z/v)=0

Then the voltage and current are

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

I(z,t)=−1Z

G(t+z/v)=−1Z

G (1v)(z+vt)[ ]

G(t+z/v) represents a wave disturbance

traveling in the negative z direction with

velocity v.

Because of the minus sign in I (z,t), the

physical current in the conductor at positive

potential flows in the direction of wave propagation.

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 Excitation

€

Excitation at one end of a semi- infinite length of transmission line.

Source has open circuit voltage VS(t) and internal resistance RS.

Radiation condition requires that excited waves travel away from source.

Terminal conditions at z=0:

VS(t) =RSI (0,t)+V(0,t)

=RS1Z

F(t)+F (t)

or F(t) =Z

Z+RS

VS(t)

VS(t) =−RSI (0,t)+V(0,t)

=−RS

−1Z

G(t)+G(t)

or G(t)=Z

Z+RS

VS(t)

∞

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 l from source.

V(l,t) =F (t−l v) =Z

Z+RS

VS(t−l v)

Delayed version of the source voltage

with the semi- infinite line acting as a

load resisor for the source.

V(−l,t)=G t+(−l v)[ ]=Z

Z+RS

VS(t−l v)

Delayed version of the source voltage

with the semi- infinite line acting as a

load resisor for the source.


Power Carried by Waves

P(z,t)

I(z,t)

V(z,t)

z

€

Instantaneous power P(z,t) carried past plane

perpendicular to z.

P(z,t)=V(z,t)I (z,t)

= F(t−z v)+G(t+z v)[ ]1Z

F(t−z v)−G(t+z v)[ ]

=1Z

F 2(t−z v)−G2(t+z v)[ ]

The two waves carry power independently in the direction of wave

propagation

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

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


Summary of Solutions for TL’s

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

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

• Semi-infinite segment of TL appears at its terminals as a resistance of value Z (even though the wires are assumed to have no resistance)

• The waves carry power independently in the direction of wave propagation


Plane Waves in One Dimension

• Electric and magnetic fields in terms of voltage and 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

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

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

Ex(z,t) =−1h

V(z,t) Volts/m or V(z,t) =−hEx(z,t).

Recall that C =εwh

.

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 w >>h, 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

Hy(z,t)≡1μ

By(z,t)=1μ

−μw

I (z,t)⎡ ⎣

⎤ ⎦ =−

1w

I (z,t) Amps/m or I(z,t)=−wHy(z,t).

Recall that L =μhw

.


Maxwell’s Equations in 1-D

€

Inserting the foregoing expressions for V(z,t), C, I (z,t) and L into the

Transmission Line equations

∂∂z

−hEx(z,t)[ ]=−μhw

⎡ ⎣

⎤ ⎦ ∂∂t

−wHy(z,t)[ ] ∂∂z

−wHy(z,t)[ ] =−εwh

⎡ ⎣

⎤ ⎦ ∂∂t

−hEx(z,t)[ ]

or

∂∂z

Ex(z,t) =−μ∂∂t

Hy(z,t) ∂∂z

Hy(z,t)=−ε∂∂t

Ex(z,t)

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

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

We may think of the plates as being taken to (x,y) → ∞ 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 z .

Ex(z,t) =F (t−z/v)+G(t+z/v) Hy(z,t)=1η

F(t−z/v)−G(t+z/v)[ ]

In air v=1

μoεo

≡c=3×108 m/s is the speed of light and η =μo

εo

=377 Ω

is the wave impedance.

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

constant εr.

For normal media εr >1, but it can be a function of frequency. As and example,

in water at radio frequencies (below 20 GHz) εr =81, but at optical

frequencies εr =1.78.


Power Density Carried by Plane Waves

€

Total instantaneous power carried in parallel plate line

P(z,t)=V(z,t)I (z,t)= −hEx(z,t)[ ] −wHy(z,t)[ ]

=hwEx(z,t)Hy(z,t) watts

Power density crossing any plane perpendicular

to z is

p(z,t)=P(z,t) hw=Ex(z,t)Hy(z,t) watt/m2

=1η

F 2(t−z/v)−G2(t+z/v)[ ]

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

direction from Ex to Hy 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 E must be perpedicular to the direction of

propagation.

Magnetic field vector H must be perpedicular to E and to the

direction of propagation.

The vector cross product p=E ×H watt/m2 is in the direction

of propagation.

The ratio E H is the wave impedance η.


Examples of Polarization

€

Linear polaization of E along x Linear polaization of E along y

E =axcosω(t−z/v)[ ] E =aysinω(t−z/v)[ ]

H =ay1η

cosω(t−z/v)[ ] H =−ax1η

sinω(t−z/v)[ ]

€

Circular polarization

E =axcosω(t−z/v)[ ]+aysinω(t−z/v)[ ]

H =1η

aycosω(t−z/v)[ ]−ax sinω(t−z/v)[ ]{ }

E

x z

H

y

x z

E

y H

€

ax = unit vector along x

ay = unit vector along y


Summary of Plane Waves

• Plane waves are polarized with fields E and H perpendicular to each other and to the direction of propagation

• Wave velocity is the speed of light in the medium• ExH watts/m2 is the power density carried by a

plane wave


Reflection and Transmission at Junctions

• 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

V(0−,t) =V(0+,t)

I(0−,t) =I (0+,t)

Boundary conditions at the

interface of two media

Ex(0−,t)=Ex(0

+,t)

Hy(0−,t)=Hy(0

+,t)

Plane wave propagation and

boundary conditions are analogus

to junctioning of two TL's


Reflection and TransmissionIncident wave

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

HyIn(z,t) Transmitted wave

Reflected wave

v1 and 1 v2 and 2

x

z

€

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

function F1(t- z/v1). Using the boundary conditions we solve for the unknown

functions G1(t+z/v1) and F2(t- z/v2) for the electric fields of the reflected

and transmitted waves: Ex(0−,t)=F1(t)+G1(t)=F2(t)=Ex(0

+,t)

Hy(0−,t) =

1η1

F1(t)-G1(t)[ ]=1η2

F2(t)=Hy(0+,t)


Reflection and Transmission Coefficients

€

Solution of the boundary condition equations for G1(t) and F2(t) in terms of F1(t)

G1(t)=ΓF1(t) F2(t) =ΤF1(t)

The reflection coefficient Γ and transmission coefficient Τ are given by:

Γ =η2 −η1

η2 +η1

Τ =1+Γ =2η2

η2 +η1

Examples:

I. Suppose medium 1 is air so that η1 =η ≡ μo εo =377 and medium 2 has

relative dielectric constant εr =4 so that η2 = μo εrεo =0.5η. Then going

from air- to-dielectric Γad =0.5η −η0.5η +η

=−13

and Τad =1−13

=23


Reflection and Transmission, cont.

€

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

is the dielectric η1 =0.5η ( ) and medium 2 is air η2 =η( ). Then going from

dielectic- to-air, Γda =η −0.5ηη +0.5η

=+13

and Τad =1+13

=43

Note that:

1. Γda =−Γad

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


Reflected and Transmitted Power

€

Instantaneous power carried by the incident wave pIn(z,t), the reflected wave

pRe(z,t), and the transmitted wave pTr(z,t)

pIn(z,t)=ExIn(z,t)Hy

In(z,t) =1η1

F12(t−z v1)

pRe(z,t)=ExRe(z,t)Hy

Re(z,t) =−1η1

G12(t+z v1)

pTr(z,t)=ExTr(z,t)Hy

Tr(z,t) =1

η2

F22(t−z v1)

Just on either side of the interface pIn(0−,t)=1Z1

F12(t) as well as

pRe(0−,t) =1η1

G12(t)=Γ2 1

η1

F12(t) and pTr(0+,t) =

1η2

F22(t) =Τ2 1

η2

F12(t)


Conservation of Power and Reciprocity

€

Conservation of power requires that pIn(0−,t)−pRe(0−,t) =pTr(0+,t) so that

1η1

F12(t)−Γ2 1

η1

F12(t)=Τ2 1

η2

F12(t) or 1−Γ2 =Τ2 η1

η2

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

For waves going from medium 2 to medium 1, the reflection coefficient Γ12 is

the negative of Γ21 going from medium 1 to medium 2. Thus for either

case the ratios pRe(0−,t)

pIn(0−,t)=Γ2 and

pTr(0+,t)pIn(0−,t)

=1−Γ2 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

V(0,t) =RL I (0,t)

F(t)+G(t)=RL

ZF(t)−G(t){ }

Solving for G(t) in terms of F (t),

G(t) =ΓF(t) where the reflection

coefficient is Γ =RL −ZRL +Z

Special cases:

1. Matched termination, RL =Z and Γ =0. Simulates a semi- infinite TL

2. Open circuit, RL → ∞ and Γ =1. Total reflection with V(0,t) =2F (t).

3. Short circuit, RL =0 and Γ =−1. Total reflection with V(0,t)=0.


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 1 v2 and 2 v3 and 3

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

l z

2l/v2

4l/v2

t

€

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 the junction of two transmission lines

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

• The ratio of reflected to incident power =

• Power is conserved so that the ratio of transmitted to incident power = 1 -

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

• At multiple interfaces, delayed multiple interactions complicate the description of the reflected and transmitted fields for arbitrary time dependence


Spatial Variations for Harmonic Time Dependence

• Traveling and standing wave representations of the z dependence

• Period average power• Impedance transformations to account for layered

materials• Frequency dependence of reflection from a layer


Harmonic Time Dependence at z = 0

€

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

dependence exp( jωt) at z=0. Then

V(0,t) =V(0)e jωt =F(t)+G(t)

I(0,t) =I (0)e jωt =1Z

F(t)−G(t)[ ]

where V(0) and I (0) are the complex voltage and current at z=0.

The functions F(t) and G(t) can satisfy these equations only if they too have

harmonic time dependence. Hence

F(t)=V+e jωt and G(t)=V−e jωt

where V+ =12 V(0)+ZI (0)[ ] and V−=1

2 V(0)−ZI (0)[ ] are the complex

voltage amplitudes of the waves traveling in the ±z directions.


Traveling Wave Representation

€

At other locations z≠0

V(z,t)=F(t−z v)+G(t+z v)=V+exp jω(t−z v)[ ]+V−exp jω(t+z v)[ ]

= V+e−jωz v +V−e+jωz v{ }e jωt =V(z)e jωt

I(z,t)=1Z

F(t−z v)−G(t+z v){ }=1Z

V+exp jω(t−z v)[ ]−V−exp jω(t+z v)[ ]{ }

=1Z

V+e−jωz v −V−e+jωz v{ }ejωt =I (z)ejωt

Here V(z) is the phasor voltage and I (z) is the phasor current, which give the

spatial variation for the implied time dependence exp( jωt).

Define the wave number (propagation constant) k ≡ω v m−1. Then

V(z)=V+e−jkz+V−e+jkz and I (z) =1Z

V +e−jkz−V−e+jkz{ }

is the traveling wave representation of phasor voltage and current.


Standing Wave Representation

€

Substituting the expressions for V+ and V− in terms of V(0) and I (0),

and rearranging terms gives the standing wave representation of the phasor

voltage and current:

V(z)=12V(0) e−jkz+e+jkz

[ ]+12 ZI (0) e−jkz−e+jkz

[ ]=V(0)coskz−jZI (0)sinkz

I(z)=12

ZV(0) e−jkz−e+jkz

[ ]+12 I (0) e−jkz+e+jkz

[ ]=I (0)coskz−j1Z

V(0)sinkz

€

The wavenumber is k=ω v =2πf v=2π λ where λ is the

wavelength λ =v f =2π k

For plane waves in a dielectric medium k=ω με


Variation of the Voltage Magnitude

|V+|

z

€

V(z)

€

For I (0)=0 we have a pure standing

wave V(z)=V(0)coskz. Its magnitude

V(z) =V(0) coskz is periodic with

period π k=λ 2.

0 z

€

V(z)

€

V(0)


Standing Wave Before a Conductor

ISC

, 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

E x(0) =0 and H y(0) =I SC

The standing wave field is

E x(z)=12ηISC e−jkz−e+jkz

[ ]

=−jηI SC sinkz

Two waves of equal amplitude and

traveling in opposite directions create

a standing wave.


Standing Wave Before a Conductor, cont.

€

Plot of the magnitude of the standing wave field

E x(z) =ηISC sinkz

ISC

-z


Period Averaged Power

€

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

of the instantaneous power is P(z)=12Re V(z)I

∗(z){ } watts

Using the traveling wave representation

P(z)=12Re V +e−jkz+V−e+jkz

[ ]1Z

V+e−jkz−V−e+jkz[ ]

∗⎧ ⎨ ⎩

⎫ ⎬ ⎭

=1

2ZV+ 2

−V− 2

{ }

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

the incident and reflected waves, and it is independent of z.

For harmonic plane waves p(z) =12 Re E x(z)H y

∗(z){ } watts/m2

In terms of traveling waves p(z) =12η

ExIn 2

−ExRe 2

{ }


Reflection From a Load Impedance

V+

V- ZL

0 z

I(0)

V(0) + ZL

0 z

€

For a complex load impedance ZL

V(0) =V ++V− =ZL I (0) =ZL

ZV +−V−( )

Solving for V− in terms of V + gives

V−=ΓV + where the complex

reflection coefficient Γ is

Γ =ZL −ZZL +Z

Reflected power

PRe =1

2ZV− 2

=1

2ZΓV + 2

=Γ2P In


Summary of Spatial Variation for Harmonic Time Dependence

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

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

• The magnitude of the field for a pure standing wave is periodic in z with period

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

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

• The fraction of the incident power carried by a reflected wave is


Impedance Transformations in Space

• Impedance variation in space• Using impedance for material layers• Frequency dependence of reflection from a brick

wall• Quarter wave matching layer


Defining Impedance Along a TL

I(0)

ZIN V(0) + ZL

-l 0 z

€

At z=0 the ratio of voltage to current

can have some value V(0) I (0)≡ZL

Using the formulas for V(z) and I (z)

we can compute their ratio at z=−l.

Defining this ratio as ZIN(l) we have

ZIN(l) ≡V(−l)I (−l)

=V(0)cos(-kl)−jZI (0)sin(−kl)

I (0)cos(-kl)−j1Z

V(0)sin(-kl)

Dividing numerator and denominator by I (0) and rearranging gives

ZIN(l) =ZZL cos(kl)+jZsin(kl)Zcos(kl)+jZL sin(kl)

=ZZL +jZtan(kl)Z+jZL tan(kl)


Properties of the Impedance Transform

€

The impedance formula

ZIN(l) =ZZL cos(kl)+jZsin(kl)Zcos(kl)+jZL sin(kl)

=ZZL +jZtan(kl)Z+jZL tan(kl)

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 ZL =Z, the imput impedace is matched ZIN =Z

2. The impedance repeats ZIN (l)=ZIN(l+Δl) for kΔl =π or

Δl =π k=λ 2

3. For quarter wave displacement l=λ 4, kl=π 2 and impedance

inverts ZIN (λ 4) =Z2 ZL

4. If ZL =0, then ZIN(l) =jZtan(kl)


Using Transform for Layered Media

Incident wave

ExIn(z) Ex

TR(z) Transmitted

HyIn (z) wave

ExRe(z)

Reflected wave

v1 , 1 v2 , 2 v3 , 3

x

0 l z

ZIN(l) ZL = 3

Z= 2


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 ZL =η3.

Impedance of the finite segment of TL is Z=η2. Wavenumber of this

segment is k2 =ω v2 =ω εr2εoμo =ko εr2

where ko =ω εoμo is the wavenumber of free space.

Input impedance at left surface of the layer is then

ZIN(l) =η2η3 cos(k2l)+jη2sin(k2l)η2 cos(k2l)+jη3sin(k2l)

Reflection coefficient for the wave incident from medium 1 is

Γ =ZIN (l)−η1

ZIN (l)+η1

=η2 η3 −η1( )cos(k2l)+j(η2

2 −η1η3)sin(k2l)η2 η3 +η1( )cos(k2l)+j(η2

2 +η1η3)sin(k2l)


Example 1: Reflection at a Brick Wall

€

HyIN

€

E xIN

w

€

Medium 1 and medium 3 are air

η1 =η3 =ηo ≡μo

εo

Medium 2 is brick with εr2 ≈4

k2 =2ko and η2 =μo

εr2εo

=12ηo

€

Reflection coefficient for the wave incident from air is

Γ =η2 η3 −η1( )cos(k2w)+j(η2

2 −η1η3)sin(k2w)η2 η3 +η1( )cos(k2w)+j(η2

2 +η1η3)sin(k2w)

=j 1

4 ηo2 −ηo

2( )sin(2kow)

2ηo2cos(2kow)+j 1

4 ηo2 +ηo

2( )sin(2kow)

=j 3

4 sin(2kow)2cos(2kow)+j 5

4 sin(2kow)


Example 1: Reflection at a Brick Wall, cont.

€

Let the wall thickness be w=30 cm so that 2kow=4πf

3×108 0.3=4πfGHz

Then pRe pin =Γ2=

9sin2(4πfGHz)64cos2(4πfGHz)+25sin2(4πfGHz)

€

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

power transmitted through the wall is 1−Γ2

0 0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0 fGHz


Example 2: Quarter Wave Layers

Incident wave

ExIn(z) Ex

TR(z) Transmitted

HyIn (z) wave

ExRe(z)

Reflected wave

v1 , 1 v2 , 2 v3 , 3 l=k2)=

x

0 z

€

cos(k2l) =cos(k2 λ2 4) =cos(π /2)=0 and sin(k2l)=sin(π /2) =1

so that ZIN(λ2 /4) =η22 /η3


Example 2: Quarter Wave Layers, cont.

€

For this value of ZIN we have Γ =η2

2 −η1η3

η22 +η1η3

If we choose the layer material such that η22 =η1η3, then Γ =0 and no

reflection takes place.

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

dielectric constant εg

For no reflection: η22 =

μo

εr2εo

=η1η3 =μo

εo

μo

εgεo

or εr2 = εg

Note that the layer thickness is l =λ2 /4=v2

4f=

14 f εr 2εoμo

=vo

4 f εr 2

or l =λo

4 εg4

where λo is the wavelength in air.


Summary of Impedance Transformation

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

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

• Reflection from a layer has periodic frequency dependence with minima (or maxima) separated by f = v2/(2w)

• Quarter wave layers can be used impedance matching 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 G mho/m in addition to the capacitance C farads/m

and inductance L henry/m.

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

Expressions for the circuit quantities are:

G=σwh

C =εwh

L =μhw


Equivalent Circuit for Harmonic Waves

€

In the limit as Δz→ 0 the Kirchhoff circuit equations for the phasor

voltage and current give the TL equations for harmonic time dependence

dV(z)

dz=−jωLI (z)

dI (z)dz

=−G+jωC( )V(z)

+

I(z) V(z)

-z z+z z

I(z) +

V(z)

+ I(z +z)

V(z+z) jLz j C z G


Harmonic Fields and Maxwell’s Equations

€

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

so that the phasor circuit quantities are V(z)=−hE x(z) and I (z) =−wHy(z).

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

along with the expressions for G, C, L gives Maxwell's equations

dE x(z)

dz=−jωμHy(z)

dH y(z)dz

=− jωε+σ( )E x(z)

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

dE x(z)

dz=−jωμHy(z) and

dHy(z)dz

=−jωˆ ε E x(z)

The complex equivalent dielectric constant ˆ ε is given by

ε =εrεo −j σ ω=εo εr −j σ ωεo( )

Let ε"=σ ωεo. Then ε =εo εr −jε"( )

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

form εo εr −jε"( ). These processes have a different frequency

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


Constants for Some Common MaterialsWhen conductivity exists, use complex dielectric constant given by

= o(r - j") where " = o and o 10-9/36

Material* r mho/m)" 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 ε

replaced by ˆ ε .

For example, the traveling wave solutions in a material are

E x(z)=V+e−jkz+V−e+jkz and Hy(z) =1η

V +e−jkz−V−e+jkz{ }

Here k =ω μˆ ε =ω μεo εr −jε"( ) and η =μˆ ε

=μ

εo εr −jε"( )

are complex quantities.


Wave Number and Impedance

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The complex wavenumber k will have real and imaginary parts

k≡β −jα =ω μεo εr −jε"( )

If ε" is less than about εr 10, we may use the approximations

β ≈ω μεoεr and α ≈ω μεoεr ε"2εr

Similarly, for ε" small, η =μ

εo εr −jε"( )≈

μεoεr

1+jε"2εr

⎛

⎝ ⎜ ⎞

⎠ ⎟


Effect of Loss on Traveling Waves

€

For a wave traveling in the positive z direction

E x(z)=V+e−jkz=V+exp−j(β−jα)z[ ]=V+exp(- jβz)exp(−αz)

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

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

The magnitude of the field depends on z as given by

E x(z) =V +exp(−αz)

V+

V+e

z


Attenuation in dB

€

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

Attn=−20log10

E x(z)

E x(0)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

=−20log10

V+exp(−αz)

V+

⎧ ⎨ ⎩

⎫ ⎬ ⎭

=20αzlog10 e{ }=8.67αz

Thus the attenuation rate of the wave in a medium is 8.67α dB/m


Effect of Loss on Traveling Waves, cont.

€

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

wavelength λ =2π β and the decay over the attenuation length 1α.

For real amplitude V +, the spatial variation is given by

Re E x(z)ejωt{ }=V +Re exp j(ωt−βz)[ ]exp(−αz) { }

or

V+cos(ωt-βz)exp(−αz)

V+

V+e z


Loss Damps Out Reflection in Media

Traveling wave amplitude

z

Reflecting boundary

Incident wave

Reflected wave

€

E xIN

(z) =V +exp(−αz) E xRe

(z) =ΓV+ exp(+αz)


Effect of Damping on the || for a Wall

0 0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0 fGHz

€

With absorption in the brick wall, the interference minima are

reduced and the reflection coefficient approaches that of the

first air-brick interface or Γ =ηB −ηo

ηB +ηo

=−13

The fraction of the incident power transmitted through the

wall is ≠ 1−Γ2


Summary of Material Loss

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

• The wavenumber k = j and wave impedance then have imaginary parts

• The attenuation length = 1/ • Loss in a medium damps out reflections within a

medium

Documents

July, 2003© 2003 by H.L. Bertoni1 I. Introduction to Wave Propagation Waves on transmission lines Plane waves in one dimension Reflection and transmission