1

ECE 480

Wireless Systems

Lecture 4

Propagation and Modulation of RF Waves

2

Antenna Radiation Characteristics

Antenna pattern:

• Describes the far – field directional properties of an antenna when measured at a fixed distance from the antenna

• 3 – d plot that displays the strength of the radiated field (or power density) as a function of direction (spherical coordinates) specified by the zenith angle and the azimuth angle

• From reciprocity, a receiving antenna has the same directional antenna pattern as the pattern that it exhibits when operated in the transmission mode

3

The differential power through an elemental area dA is

rad av avˆd P S d A S R dA S dA

always in the radial direction in the far – field region

avS

4

radd P S dA

d A R sin d d 2

Define: Solid angle,

d Ad sin d d steradians

R

2

4 for a spherical surface

5

radd P S dAd A

dR

2

d A R d 2

radd P R S R , , d 2

The total power radiated by an antenna is given by

radP R S R , , sin d d

2

2

0 0

6

rad maxP R S F , sin d d

2

2

0 0

rad maxP R S F , d 2

4

F , is the normalized radiation intensity

F , 1

73 – D Pattern of a Narrow – Beam Antenna

8

Antenna Pattern

It is convenient to characterize the variation of F ( , ) in two dimensions

Elevation Plane ( - plane)

Corresponds to a single value of ( = 0 x –z plane) ( = 90 y –z plane)

Azimuth Plane ( - plane)

Corresponds to = 90 o (x – y plane)

Two principle planes of the spherical coordinate system

9

Clearer to express F in db for highly directive patterns

F dB log F10

= 0 plane

10

Side lobes are undesirable

• Wasted energy

• Possible interference

11

Beam Dimensions

Define: Pattern solid angle p

p = Equivalent width of the main lobe

p F , d sr

4For an isotropic antenna with F ( , ) = 1 in all directions:

p sr 4

12

Defines an equivalent cone over which all the radiation of the actual antenna is concentrated with equal intensity signal equal to the maximum of the actual pattern

13

The half – power (3 dB) beamwidth, , is defined as the angular width of the main lobe between the two angles at which the magnitude of F ( , ) is equal to half its peak value

2 1

14

F () is max at = 90 o , 2 = 135 0 , 1 = 45 o , = 135 o – 45 o = 90 o

15

Null Beamwidth, null

Beamwidth between the first nulls on either side of the peak

16

Antenna Directivity

max

av av

p

FD

F F F , d

4

1 114

4

p = Pattern solid angle

For an isotropic antenna, p = 4

D = 1

17

D can also be expressed as

max max

rad av

rad

R S SD

P S

P

R

2

2

4

4

av isoS SS iso = power density radiated by an isotropic antenna

D = ratio of the maximum power density radiated by the antenna to the power density radiated by an isotropic antenna

18

For an antenna with a single main lobe pointing in the z direction:

p xz yz

p xz yz

D

4 4

19

Example – Antenna Radiation Properties

Determine:

a. The direction of maximum radiation

b. Pattern solid angle

c. directivity

d. half – power beamwidth

in the y-z plane for an antenna that radiates into only the upper hemisphere and its normalized radiation intensity is given by

F , cos 2

20

Solution

F , cos 2The statement in the upper hemisphere

can be written mathematically as

F , F cos

elsewhere

2 02

0 2

0

21

a. The function

F cos 2

is maximum when = 0

Polar plot of

F cos 2

b. The pattern solid angle is given by

p F , d

cos sin d d

cosd d sr

4

2 22

0 0

2 23 2

0 00

1 2

3 3 3

22

F cos 2

Polar plot of

c.

p

D

D d B log . d B

4 34 6

2

10 6 7 78

d. The half – power by setting

F cos . 2 0 5

o o

o

1 2

2 1

45 45

90

23

Example – Directivity of a Hertzian Dipole

D

F , sin d d

.

sin d d

4

23

0 0

4

4 41 5

83

For a Hertzian dipole:

F , sin 2

24

Antenna Gain

Define: Radiation Efficiency,

P t = Transmitter power sent to the antenna

P rad = Power radiated into space

P loss = Power loss due to heat in the antenna

= P t – P rad

rad

t

P

P

= 1 for a lossless antenna

25

Define: Antenna Gain, G

max

t

R SG

P

24

max

rad

R SD

P

24

G D

Accounts for the losses in the antenna

26

Radiation Resistance

P loss = Power loss due to heat in the antenna = P t – P rad

I

I

I

t ant loss rad

loss loss

rad rad

P R P P

P R

P R

20

20

20

1

21

21

2

rad rad rad

t rad loss rad loss

P P R

P P P R R

27

To find the radiation resistance:

• Find the far – field power by integrating the far – field power density over a sphere

• Equate to

Irad radP R 20

1

2

28

Example – Radiation Resistance and Efficiency of a Hertzian Dipole

A 4 – cm long center – fed dipole is used as an antenna at 75 MHz. The antenna wire is made of copper and has a radius a = 0.4 mm. The loss resistance of a circular wire is given by

closs

c

fR

a

2

l

Calculate the radiation resistance and the radiation efficiency of the dipole antenna

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Solution

The parameters of copper are

r

c

S.

m

7

1

5 8 10

closs

c

fR

a

.

. .

.

6 7

7

2

0 04 75 10 4 10

2 0 004 5 8 10

0 036

l

30

At 75 MHz:

cm

f

cm.

m

8

6

3 104

75 10

40 01

4

l

This is a short dipole

From before,

maxSR

220

2

15 lI

rad max

RP S

D

24

31

I

I

rad

RP

D R

2220

2

22 2

0

154

40

l

l

I Irad radP R

22 2 20 0

140

2

l

rad

.R

.

2 22 3 0 04

80 804

0 08

l

rad

rad loss

R .. %

R R . .

0 08

0 69 690 08 0 036

32

Half – Wave Dipole Antenna

j ti t cos t cos k z Re cos k z e 0 0I I

z cos k z z

0 4 4I I

In phasor form:

k

2

33

For a short dipole

j k Rk eE sin

R

EH

0 0

0

4

lj I

Expand these expressions to obtain similar expressions for the half – wave dipole

34

j k s

s

j k ed E z z dz sin

s

d E zd H z

0

0

4

I

Consider an infinitesimal dipole segment of length dz excited by a current and located a distance from the observation point

zI

35

The far field due to radiation by the entire antenna is given by

z

E d E

4

4

Two assumptions:

s

s R

1 1

(length factor)

36

j k s

s

j k ed E z z d z sin

s

0

4I

Note that "s" appears in the equation twice – once for the distance away and once for the phase factor

s R is not valid for the length factor

If Q is located at the top of the dipole, the phase

factor is which is not acceptable 2

s R z cos

37

j k R

j k z cosj k ed E z z d z sin e

R

0

4I

j k Rcos cose

E j Isin R

EH

0

0

260

cos cosE

S R ,sin

22

20

22

38

cos cosE

S R , Ssin

22

0 20

22

cos cosE

S R ,R sin

2220

2 20

15 22

I

39

S R , is max when 2

maxS SR

20

0 2

15 I

cos cosS R ,

FS sin

2

0

2

40

Directivity of Half – Wave Dipole

Need P rad and S (R , )

radP R S R , d

cos cossin d d

sin

.

2

4

2

2 2 20

0 0

20

15 2

36 6

I

I

maxSR

20

2

15 I

max

rad

R S RD . . d B

P . R

2 220

2 20

4 1541 64 2 15

36 6

I

I

41

Radiation Resistance of Half – Wave Dipole

radrad

P .R

20

2 20 0

2 2 36 673

I

I I

Recall: for the short dipole (l = 4 cm) at 75 MHz

R rad = 0.08

R loss = 0.036

rad

rad loss

R .%

R R . .

0 08

690 08 0 036

For the half – wave dipole (l = 4 m) at 75 MHz

R loss = 1.8

rad

rad loss

R%

R R .

73

9873 1 8

42

Effective Area of a Receiving Antenna

Assume an incident wave with a power density of S i

The effective area of the antenna, A e , is

inte

i

PA

S

P int = Power intercepted by the antenna

It can be shown:

oc

int Lrad

VP P

R

2

8

ocV = Magnitude of the open – circuit voltage developed across the antenna

43

The power density carried by the wave is

i i

i

ocinte

irad i

E ES

VPA

S R E

2 2

0

2

2

2 240

30

For the short dipole

rad oc iR V E

2280

ll

eA

23

8

44

In terms of D:

e

DA m

2

2

4

Valid for any antenna

Example: Antenna Area

The effective area of an antenna is 9 m 2. What is its directivity in db at 3 GHz?

c.

f

8

9

3 100 1

3 10

eAD . . d B

.

422

4 4 91 13 10 40 53

0 1

45

Friis Transmission Formula

Assumptions:

• Each antenna is in the far – field region of the other

• Peak of the radiation pattern of each antenna is aligned with the other

• Transmission is lossless

46

tiso

PS

R

24For an isotropic antenna: (ideal)

In the practical case, t t tr t iso t t iso

D PS G S D S

R

24

In terms of the effective area A t of the transmitting antenna

t t tr

A PS

R

2 2

47

On the receiving side,t t r t

int r r

A A PP S A

R

2 2

rec r intP P

rec t r t rt r

t

P A AG G

P R R

2

2 2 4

Friis transmission formula

48

rec t r t rt r t t t r r r

t

P A AG G F , F ,

P R R

2

2 2 4

When the antennas are not aligned

(More general expression)

49

Homework

1. Determine the following:

a. The direction of maximum radiation

b. Directivity

c. Beam solid angle

d. Half – power beamwidth in the x – z plane

for an antenna whose normalized radiation intensity is given by:

oF , for ,

elsewhere

1 0 60 0 2

0

Hint: Sketch the pattern first

50

2. An antenna with a pattern solid angle of 1.5 (sr) radiates 30 W of power. At a range of 1 km, what is the maximum power density radiated by the antenna?

3. The radiation pattern of a circular parabolic – reflector antenna consists of a circular major lobe with a half – power beamwidth of 2 o and a few minor lobes. Ignoring the minor lobes, obtain an estimate for the antenna directivity in dB.

51

Analog Modulation

Several basic types

• Amplitude modulation (AM)

• Frequency modulation (FM)

• Pulse code modulation (PCM)

• Pulse width modulation (PWM)

• High frequencies require smaller antennas

• Modulation impresses a lower frequency onto a higher frequency for easier transmission

• The signal is modulated at the transmission end and demodulated at the receiving end

52

Amplitude Modulation

Carrier wave – High frequency signal that transports the intelligence

Signal wave – Low frequency signal that contains the intelligence

53

AM transmitter

• DC shifts the modulating signal

• Multiplies it with the carrier wave using a frequency mixer

• Mixer must be nonlinear

• Output is a signal with the same frequency as the carrier with peaks and valleys that vary in proportion to the strength of the modulating signal

• Signal is amplified and sent to the antenna

54

The mixer is usually a "square law" device, such as a diode or B – E junction of a transistor

Output input 2

Suppose that we apply the following signals to a square law device

f t A cos t

f t A cos t

1 1 1

2 2 2

The output will be

o of t A A cos t A cos t 2

1 1 2 2

55

Homework

o of t A A cos t A cos t 2

1 1 2 2

Determine all possible output frequencies

56

Advantages

• Simplicity

• Cost

Disadvantages

• Susceptible to atmospheric interference (static)

• Narrow bandwidth (550 – 1500 KHz)

57

AM Receiver

• Tunable filter

• Envelope detector (diode)

• Capacitor is used to eliminate the carrier and to undo the DC shift

• Will generally include some form of automatic gain control (AGC)

58

Forms of Amplitude Modulation

In the most basic form, an AM signal in the frequency domain consists of

• The carrier signal

• Information at f c + f m (upper sideband)

• Information at f c - f m (lower sideband)

(US and LS are mirror images)

This wastes transmission power

• Carrier contains no information

• Information is all contained in only one of the sidebands

59

Frequently, in communications systems, the carrier and/or one of the sidebands is suppressed or reduced

• If only the carrier is reduced or suppressed, the process is called "Double – Sideband Suppressed (Reduced) Carrier" (DSSC or DSRC)

• If the carrier and one of the sidebands is suppressed or reduced, the process is called "Single – Sideband Suppressed (Reduced) Carrier" (SSSC or SSRC)

• Often, the carrier and one of the sidebands is totally suppressed. This process is simply called "Single Sideband"

• The carrier must be regenerated at the receiver end

60

Example

Consider a carrier with a frequency c

cc t C sin t

Suppose we want to modulate the carrier with a signal

mm t M sin t

The signal is amplitude – modulated by adding m(t) to C

The expression for this signal is

m cy t C M sin t sin t

Expanding this expression

m c m c

c

cos t cos ty t C sin t M M

2 2

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Convert to frequency domain by taking the Fourier Transform

cy t C m t cos t

c cj t j te e

y t C m t 2

Take Fourier Transform

c c c cY C M C M 1 1

2 2

x = Unit impulse function

62

Eff = 33 % Eff = 100 %Eff = 100 %

63

Modulation Index

peak value of m th

C

Measure of the modulating signal wrt the carrier signal

64

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