1

ECE 480

Wireless Systems

Lecture 3

Propagation and Modulation of RF Waves

2

Circular Polarization

Magnitudes of the x – and y – components of are equal

Phase difference is

E z

2

2

Left – Hand Circular (LHC) Polarization

2

Right – Hand Circular (RHC) Polarization

3

Left – Hand Circular (LHC) Polarization

x ya a a 2

j j k z

j k z

j

ˆ ˆE z xa y ae e

ˆ ˆa x j y e

e j

2

2

4

j tE z,t Re E z e

ˆ ˆx a cos t k z y a cos t k z

ˆ ˆx a cos t k z y a sin t k z

2

Convert to polar form

x yE z,t E z,t E z,t

a cos t k z a sin t k z a

12 2 2

12 2 2 2 2

y

x

E z,tz,t tan

E z,t

a sin t k ztan t k z

a cos t k z

1

1

5

Linear polarization

= f (z, t)

f (z, t)

Circular polarization

z,t t k z

f (z, t)

= f (z, t)

E z,t a

x yE z,t a cos t k z a cos t k z

12 2 2

y

x

atan constant

a

1

6

Back to LHC Polarization

Consider the LHC wave at z = 0

t

Inclination angle decreases with time

7

Right – Hand Circular (RHC) Polarization

x ya a a 2

z,t t k z

E z,t a

8

The direction of polarization is defined in terms of the rotation of as a function of time in a fixed plane orthogonal to the direction of propagation

E

9

Example RHC Polarized Wave

An RHC polarized plane wave with electric field modulus of 3 mV/m is traveling in the + y direction in a dielectric medium with

0 04 0

f = 100 MHz

Obtain expressions for E y,t and H y,t

10

Solution

The wave is traveling in the + y direction.

Therefore, the field components are in the x and z directions.

ˆ ˆ ˆz x y

E H

direction of propagation

11

Assign a phase angle of 0 o to the z component of E y

(arbitrary)

E yThe x component of will have a phase shift

2

Both components have a magnitude of a = 3

x z

j j k y j k y

j k y

ˆ ˆE y x E z E

ˆ ˆx ae e z ae

mVˆ ˆx j z e

m

2

3

12

j k y

j k y

ˆH y y E y

ˆ ˆy x j z e

mAˆ ˆz j x e

m

1

13

3

radf

s 82 2 10

r radk

c m

8

8

2 10 4 4

3 10 3

0 12060

4

13

Converting back to the time domain

j t

j k y j t

j t

j k y j t

E Re E y e

ˆ ˆRe x j z e e

mVˆ ˆx sin t k y z cos t k y

m

H Re H y e

ˆ ˆRe z j x e e

mAˆ ˆx cos t k y z sin t k y

m

3

3

3

1

20

14

Elliptical Polarization

Most general case

x ya a 0 0 0

EThe tip of traces an ellipse in the x – y plane

Can be left – handed or right - handed

Major axis:

Minor axis:

Rotation Angle

Ellipticity Angle

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Rotation angle

xacosa

1

2 2

The shape and rotation are defined by the ellipticity angle

atan

a R

1

4 4

aR

a

axial ratio R = 1 Circular

R = Linear

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y

x

o

atan

a

0

00 90

tan tan cos

tan sin sin

0

0

2 22 2

2 24 4

o

o

45

0

Circular

Linear

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Polarization States for Various Combinations of and

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Positive values of corresponding to sin > 0 define left – handed rotation

Negative values of corresponding to sin < 0 define right – handed rotation

How is the type of polarization determined?

a x and a y are, by definition, > 0

y

x

awill vary from

a 0

o 00 90

Two possible values of in this range and

if cos

if cos

0 0

0 0

if sin

if sin

0 0

0 0

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Example: Polarization State

Determine the polarization of a plane wave with an electric field given by

o o mVˆ ˆE z, t x cos t k z y sin t k z

m 3 30 3 45

Solution

Convert the sin term to a cos term by subtracting 90 o

o o mVˆ ˆE z, t x cos t k z y cos t k z

m 3 30 3 45

Convert the – cos term to a + cos term by adding 180 o

o o mVˆ ˆE z, t x cos t k z y cos t k z

m 3 30 3 135

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o oj k z j j k z jˆ ˆE z x e e y e e 30 1353 4Convert to phasor form

o o oy x 135 30 105

y o

x

atan tan .

a

1 10

453 1

3

o o

tan tan cos

tan . cos

. .

.

02 2

106 2 105

3 442 0 2588

0 8900

21

o o o

oo

oo

tan . . , . .

..

..

1

1

2

2 0 89 41 7 221 7 138 3

41 720 85

2138 3

69 152

There are two possible solutions for since the tan function is positive in both the first and third quadrants

Which is correct?

o

cos

.

0

69 15

22

o o

sin sin sin

sin . sin

. .

.

02 2

106 2 105

0 9603 0 9659

0 9276

By a similar analysis,

o. 34 0

The wave is elliptically polarized and the rotation is left - handed

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Plane – Wave Propagation in Lossy Media

E E 2 2 0

C ' j "

' "

2 2 2

can be written as

j

= attenuation constant

= phase constant

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j j

' j "

2 2 2

2 2

2

Equate the real and imaginary parts

'

"

2 2 2

22

Solve for and

' "

'

' "

'

1

22

1

22

1 12

1 12

25

For a uniform plane wave traveling in the + z direction with an electric field

xx

d E zE z

d z

22

20

xˆE x E z

the wave equation becomes

The solution is

z z j zx x xˆ ˆ ˆE z x E z x E e x E e e 0 0

x z j

c c

ˆ Ek EˆH y e e

0

cc

"j

' '

1

21

26

The magnitude of is xE z

z j z zx x xE z E e e E e 0 0

Decreases exponentially with e - z

xy

c

EH

also decreases exponentially with e - z

Define: Skin Depth, s

s 1

Distance that a wave must travel before it is attenuated by

.e

10 3679

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In a perfect dielectrics, , 0 0

In a perfect conductors, , 0

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Expressions are valid for any linear, isotropic, homogeneous medium

"

'"

'"

'

2

2 2

2

10

10 10

10

Low – Loss Dielectric

Quasi – Conductor

(Semiconductor)

Good Conductor

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Low – Loss Dielectric

Consider x1

21

xx , x

1

21 1 12

For

" "j ' j j ' j

' '

1

21 1

2

Divide into real and imaginary parts

" Np

' m

'

2 2

- Same as for lossless medium

30

c

" "j j

' ' '

1 1

2 2

xx , x

1

21 1 12

cc

"j

' '

1

21

c

Same as for the lossless case

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Good Conductor

c

"f

f

fj j j"

2 2

1 1

jj

1

2

32

Semiconductors – Must use exact solution

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Example – Plane Wave in Seawater

A uniform plane wave is traveling downward in the + z direction in seawater, with the x – y plane denoting the sea surface and z = 0 denoting a point just below the surface.

The constitutive parameters of seawater are:

r r

S, ,

m 1 80 4

The magnetic field intensity at z = 0 is given by

o mAˆH ,t y cos t

m 30 100 2 10 15

a. Determine expressions for

b. The depth at which the amplitude of E is 1% of its value at z = 0

E z,t and H z,t

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Solution

a. The general expressions for the phasor fields are

z j zx

x z j z

c

ˆE z x E e e

EˆH z y e e

0

0

r

"

' 3 90

5

480

2 10 1036

9 10

Seawater is a good conductor at 1 KHz

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j j

c

Npf .

m.

.j e . e

3 7

4 4

10 4 10 4 0 126

0 126

0 1261 2 0 044

4

The general expression for E x0 is

jx xE E e 0

0 0

j z j z j tx

. zx

ˆE z,t Re x E e e e e

Vx E e cos t . z

m

0

0

0 126 30 02 10 0 126

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jx z j z j t

j

. z ox

E eˆH z, t Re y e e e

. e

Ay . E e cos t . z

m

0

0

4

0 126 30 0

0 044

22 5 2 10 0 126 45

at z = 0:

ox

AˆH , t y . E cos t

m 3

0 00 22 5 2 10 45

Compare with original expression

o mAˆH ,t y cos t

m 30 100 2 10 15

x

x

. E

mVE .

m

30

0

22 5 100 10

4 44

o o

o

0

0

45 15

60

37

. z o mVˆE z,t x . e cos t . z

m 0 126 34 44 2 10 0 126 60

. z o mAˆH z,t y e cos t . z

m 0 126 3100 2 10 0 126 15

Note that they are no longer in phase. The electric field always leads the magnetic field by 45 o.

b. Set the amplitude to 0.01

. z. e

ln .z m

.

0 1260 01

0 0136

0 126

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Electromagnetic Power Density

Define: Poynting Vector

WS E H

m

2

Direction of S is in the direction of propagation, k

AˆP S ndA S A cos

Power through a surface, A

n unit vector normal to the surface

39

Plane Wave in a Lossless Medium

j k zx

x j k z

ˆE z x E e

EˆH z y e

0

0

Consider a plane wave traveling in the + z direction

Want to find the power density vector, S

40

Time – Domain Approach

x

x

ˆE z ,t x E cos t k z

EˆH z ,t y cos t k z

0

0

x

S z ,t E z ,t H z ,t

Ez cos t k z

2

0 2

41

T

av

x

x

S S z ,t d t TT f

Ez cos t k z d t

Ez

0

2 2

0 2

0

2

0

1 1 2

2

2

�

Time average of S

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Phasor – Domain Approach

av

j k z j k zxx

x

S Re E H *

E *ˆ ˆRe x E e y e

Ez

00

2

0

1

2

2

avS Re E H * is valid for any media

43

Plane Wave in a Lossy Medium

x y

z j zx y

z j zx y

c

ˆ ˆE z x E z y E z

ˆ ˆx E y E e e

ˆ ˆH z x E y E e e

0 0

0 0

1

av

x yz

c

S Re E H *

z E Ee Re

*

2 2

0 02 1

2

jc c e

44

jc c e

zav

c

EˆS z z e cos

2

0 2

2

x yE E E 1

2 2 20 0 0

Note that the average power decays withze 2

zav

c

EˆS z z e cos

2

0 2

2

45

Homework

The electric field of a plane wave is given by

x yˆ ˆE z ,t x a cos t k z y a cos t k z

Identify the polarization state, determine the polarization angles (, ), and sketch the locus of E (0, t) for each of the following cases

ox y

ox y

ox y

ox y

V Va , a ,

m mV V

a , a ,m mV V

a , a ,m mV V

a , a ,m m

3 4 0

3 4 180

3 3 45

3 4 135

46

Homework

In a medium characterized by

r r

S, , .

m 9 1 0 1

Determine the phase angle by which the magnetic field leads the electric field

47

Radiation and Antennas

• An antenna may be considered as a transducer that converts a guided EM wave to a transmitted wave or an incident wave to a guided EM wave

• Antenna dimensions are generally referred to in wavelength units

48

49

Reciprocity

• Antenna radiation pattern: The directional function that characterizes the distribution pattern radiated by an antenna

• Isotropic antenna: A hypothetical antenna that radiates equally in all directions

• Used as a reference radiator to compare antennas

• Reciprocal antennas: Antennas that have the same radiation patterns for transmission as for reception

50

Two aspects of antenna performance

1. Radiation Properties

• Direction of the radiation pattern

• Polarization state of the radiated wave in the TX mode (Antenna Polarization)

• In the RX mode, the antenna can extract only that component of the wave whose E – field is parallel to that of the antennas polarization direction

2. Antenna Impedance

• Pertains to the impedance match between the antenna and the generator

51

Radiation Sources

Two basic types

1. Current sources

• Dipole and loop antennas

2. Aperture fields

• Horn antennas

52

Far – Field Region

The far – field region is at a distance R where the wave may be considered to be a plane wave

ff

DR

22 D = Maximum effective size of the antenna

= Wavelength of the signal

53

Example: Far – Field Distance of an Antenna

A parabolic reflector antenna is 18" in diameter operates at 12.4 GHz. Find the operating wavelength and the far – field distance of this antenna.

Solution

ff

c. m

f .

m" x . m

. "

.DR . m

.

8

9

22

3 100 0242

12 4 10

118 0 457

39 37

2 0 457217 3

0 0242

54

Antenna Arrays

• Can control the phase and magnitude of each antenna individually

• Can steer the direction of the beam electronically

55

Retarded Potentials

Consider a charge distribution as shown

The electric potential V (R) at a point in space specified by the position vector R is given by

v i

'

RV R d '

R'

1

4

i

v

i

R

'

R' R R

= position vector of an elemental volume

= elemental volume

= charge density inside the volume

= distance between the volume and the point

56

If the charge density is time – varying, the obvious solution is

Problem: Does not account for reaction time

Any change in the charge distribution will require a finite amount of time to change the potential

v i

'

R , tV R ,t d '

R'

1

4

57

Retarded Vector Potential

i

'

J R , t t 'A R ,t d '

R'

4

v i

'

R , t t 'V R ,t d '

R'

1

4

p

R't '

u

Retarded Scalar Potential

Delay Time

Valid under both static and dynamic conditions

58

Time – Harmonic Potentials

In a linear system, the parameters all have the same functional dependence on time

Consider a sinusoidal time – varying charge distribution

j tv i v i

v i

R ,t Re R e

R

= phasor representation of v iR ,t

59

j k R'v i

'

R eV R d ' V

R'

1

4

j k R'i

'

J R eA R d '

R'

4

p

ku

60

H A

1In terms of A

H j E

E Hj

1

E j H

H Ej

1

61

The Short (Hertzian) Dipole

Approach: Develop the radiation properties of a differential antenna and use that model to predict other configurations

Characteristics of a Short Dipole

• Current is uniform over the length

50

l

j ti t cos t Re e A

0 0

0

I I

I I

62

At the point Q

j k R'

'

J eA R d '

R'

0

4

ˆk J zc s

02 I

s = cross – sectional area of dipole d ' s d z

limits of integration z 2 2

l l

Assume R' ~ R

63

j k R 2

2

j k R

eˆA z d z

R

ez

R

00

00

4

4

l

l

l

I

I

j k Re

R

Spherical propagation factor

Considers both the magnitude and phase change wrt R

Change to spherical coordinates

64

R Range

Zenith angle

Azimuth angle

65

j k R

R

ˆ ˆz R cos sin

eˆ ˆA R cos sinR

ˆ ˆ ˆR A A A

0 0

4

lI

j k R

R

j k R

eA cos

R

eA sin

R

A

0 0

0 0

4

4

0

l

l

I

I

66

H A

1

E Hj

1

j k R

j k RR

j k R

R

k jH e sin

k R kR

2 k jE e cos

k R kR

k j jE e sin

k R kR kR

H H E

20

2

20

0 2 3

20

0 2 3

1

4

1

4

1

4

0

l

l

l

I

I

I

67

Electric Field lines

68

Far – Field Approximation

R

RkR

, negligiblek R kR

2 3

21

1 1

j k R

R

k eE sin

R

E

EH

0

0

4

0

lj I

Independent of

Proportional to sin

69

Power Density

avS Re E H *

For the short dipole:

av

2

ˆS R S R ,

kS R , sin

32 R

WS sin

m

2 2 20 0 2

2

20 2

lI

70

Define: Normalized Radiation Intensity

max

S R , ,F ,

S

Radiation is maximum when 2

(azimuth plane)

max 2

kS S sin

32 R

R

2 2 20 0 2

0 2

220

2

15

l

l

I

I

71

F , F sin 2

No energy is radiated by the dipole along the direction of the dipole axis and maximum radiation (F = 1) occurs in the broadside direction.