polytechnic university© 2002 by h. l. bertoni1 iii. spherical waves and radiation antennas radiate...

Polytechnic University © 2002 by H. L. Bertoni 1

III. Spherical Waves and Radiation

Antennas radiate spherical waves into free space

Receiving antennas, reciprocity, path gain and path loss

Noise as a limit to reception

Ray model for antennas above a plane earth and in a street canyon

Cylindrical waves


Radio Channel Encompasses Cables,

Antennas and Environment Between

• Transmitter impresses information onto the voltage of a high power RF carrier for transmission through the air - called modulation

• Receiver extracts the information from the voltage of a low power received signal - called demodulation

Information

Tx

Information

Rx

CableCable

Transmitting Antenna

Receiving Antenna

Radio Channel


Examples of Different Cellular Antennas

Full wave monopoleabove ground plane

/2

/4/4

Half wave dipole

/2

Dipole in cornerreflector


PCS Antennas


Base Station Antennas


Antennas Radiate Electromagnetic Waves

• EM waves have:

– Electric field E (V/m)

– Magnetic field H (A/m)

• E and H

– Perpendicular to each other and to direction of propagation - Polarization

– Amplitude depends on direction of propagation - Radiation Pattern

Transmitting Antenna

Cable

E

H


Spherical Waves Radiated by Antennas

I terminal Current

Z constant with units of ohms

120

Radial Power Flux

Antenna pattern =

For large r, localized current sources radiate fields in the form of Spherical Waves

( ) 2

,φθf

z

x

y

r

€

E

€

H

€

ar

I


Power Radiation Pattern

• Power density radiated by antenna

P() = ExH* watts/m2

Poynting vector in the radial direction

P()


Omnidirectional Antennas


Parabolic Reflector Antenna


Horn Antennas


Log Periodic Dipole Array


Dual Polarization Patch Antenna


Total Radiated Power

PT is independent of r, as required by

conservation of power.

Normalization for is:

€

PT = P ⋅ardAsphere∫∫ , where dA=r2 sinθdθdφ

PT =12η

ZI2

f θ,φ( )2sinθdθdφ

0

π

∫0

2π

∫

( ) 2

,φθf

€

f θ,φ( )2sinθdθdφ

0

π

∫0

2π

∫ =4π

Then: PT =4π2η

ZI2 and P =arPT

f θ,φ( )2

4πr2 .

dA

€

P

€

ar

r

sphere of area4 2 =r


Antenna Gain and Radiation Resistance for No Resistive Loss

Directive gain = g()= |f ()|2 and

Antenna gain = G = Max. value of g(If isotropic antennas could exist, then |f ()|2 = 1, G = 1

Radiation Resistance Rr = effective resistance seen at antenna terminals

4

4

2

4

2

1

2

22

r

r

Tr

RZ

ZR

ZIPRI

=

=

==


Antenna Directivity, Gain, Efficiency

€

Directivity=Maximum Pointing VectorAverage Pointing Vector

=Pm(r)Pav(r)

=Pm(r)

PT 4πr2( )

Gain=Pm(r)

Pterminal 4πr2( )

includes the effect of antenna resistance

Efficiency =PT

Pterminal

=Gain

Directivity


Short (Hertzian) Dipole Antenna

€

The radiated field can be

written in the desired form

E =aEZIe−jkr

rsinθ

if

f θ( )=32

sinθ

Z=jη23

L2λ

G= f 90°( )2=3/2

Rr =η2π3

Lλ

⎛ ⎝

⎞ ⎠

2

L<<

€

E

€

Hr

z

I (z)

z

I

€

Starting with Maxwell's equations,

it is found that

E =aθ jηLI2λ

e−jkr

rsinθ


Half Wave Dipole Antenna

€

E =aEZIe−jkr

rf θ( )

where

f θ( )=cos

π2

cosθ⎛ ⎝

⎞ ⎠

0.781sinθ

Z=j0.7812π

η

G= f 90°( )2=1.64

10logG =2.2dB

Rr =4πη

j0.7812π

η2

=73Ω

€

E

€

H

rI

z

I (z)

/4

/4

The radiated field can be written:


Summary of Antenna Radiation

• Radiation in free space takes the form of spherical waves

• E, H and r form a right hand system• Field amplitudes vary as 1/r to conserve power• Power varies as 1/r2, and varies with direction

from the antenna• Direction dependence gives the directivity and

gain of the antenna• Radiation resistance is the terminal representation

of the radiated power


Receiving Antennas and Reciprocity

For a linear two-port

V1 =Z11I1 + Z12I2

V2 =Z21I1 + Z22I2

Reciprocity Z12 = Z21

If I2 = 0, V2 = Z12I1 ~ 1/r

For r large,

|Z12| << |Z11|, |Z22 |

+V1

-

+V

2

-r

I1I2

Equivalent Circuit

Z11-Z12 Z22-Z12

Z12

V1

I1

V2

I2


Circuit Relation for Radiation into Free Space

Z11-Z12 Z22-Z12

Z12

+V1

-

I1

+V2

-(open circuit)

V1 = Z11I1V2 = VOC= Z12I1

€

Transmitted power

PT =1/2( )Re V1I1*

( )= 1/2( )Re Z11 I12

( )=(1/2)Rr1 I12

where Rr1 =radiation resistance of antenna 1

Therefore: Z11=Rr1 +jX1

Similarily: Z22=Rr2 +jX2

where Rr2 =radiation resistance of antenna 2


Received Power and Path Loss RatioI2

Z11-Z12 Z22-Z12

Z22*+V1

-

I1

+V2

-Z12

Matched LoadRr2 - jX2

V

€

Current I1 divides between branches: I

2= -I

1

Z12

Z12+ Z22−Z12+Z22∗( )

= -I1

Z12

2Rr 2

Received Power for Matched Load PR

=12 I2

2R

r 2=

12

I1Z12

2Rr 2

2

= I1

2 Z122

8Rr 2

Path Gain PG ≡PR

PT=

I12 Z12

2 8Rr 2

I12Rr1 2

=Z12

2

4Rr1Rr 2

Final expression for PG is the same if antenna 2 radiates and antenna 1 receives.


Effective Area of Receiving Antenna

Effective Area = Ae

€

PR =P ⋅ arAe =PT

gθ,φ( )4πr2 Ae

PT

Z*11

Ae1 Z*22

PT Ae2

€

PG=PR

PT

=g2Ae1

4πr2 and by reciprocity PG =PR

PT

=g1Ae2

4πr2

Therefore g2Ae1 =g1Ae2 or Ae1

g1

=Ae2

g2

= same for all antennas


Effective Area for a Hertzian Dipole

€

E =aθ ZIe−jkr

rf θ( )

gθ( )=(32) sinθ( )

2

Rr =η2π3

Lλ

⎛ ⎝

⎞ ⎠

2

z

L<<r

I I (z)

€

E

sinOC LEV =

For matched termination

orZ11

Z11*

+

Voc

-

RR

RR

+

Voc

-

+ Voc/2-

€

PR =12

Voc 22

RR

=Voc

2

8RR


€

For matched termination:

PR =VOC

2

8RR

=LE sinθ

2

8η2π3

Lλ

⎛ ⎝

⎞ ⎠

2⎡

⎣ ⎢

⎤

⎦ ⎥

=E

2

2η32

sin2θ⎛ ⎝

⎞ ⎠

λ2

4π= P ⋅ar( )g(θ)

λ2

4π

In terms of the effective area PR =P ⋅ arAe.

Comparing expressions, Ae =gθ( )λ2

4π

Effective Area for a Hertzian Dipole - cont.


Path Gain and Path Loss in Free Space

€

For any antenna

Ae1

g1

=Ae2

g2

=Ag

⎛ ⎝ ⎜ ⎞

⎠ Hertz

=λ2

4π or Ae =

λ2

4πg

Path gain in free space

PG≡PR

PT

=g1Ae2

4πr2 =g2Ae1

4πr2 =g1g2

λ4πr

⎛ ⎝

⎞ ⎠

2

For isotropic antennas g1 =g2 =1

PG=λ

4πr⎛ ⎝

⎞ ⎠

2

Path Loss≡PT

PR

=1

PG=

4πrλ

⎛ ⎝

⎞ ⎠

21

g1g2


Path Gain in dB for Antennas in Free Space

Slope=20

-32.4-52.4

-72.4-92.4

r =1 r =10 r =100 r =1000

fGH= 1

PGdB

€

PGdB =−PLdB =10logg1g2

λ4πr

⎛ ⎝

⎞ ⎠

2⎡

⎣ ⎢ ⎤

⎦ ⎥

For isotropic antennas, g1 =g2 =1

For frequency in GHz, λ =c f =0.3 fGH

PGdB =−32.4−20logfGH −20logr


Summary of Antennas as Receivers

• Directive properties of antennas is the same for reception and transmission

• Effective area for reception Ae = g2/4• For matched terminations, same power is received

no matter which antenna is the transmitter

• Path gain PG = PR/PT < 1

• Path loss PL = 1/PG > 1


Noise Limit on Received Power

• Minimum power for reception set by noise and interference

• Noise power set by temperature T, Boltzman’s constant k and bandwidth f of receiver: N = kTf

• For analog system, received power PR must be at least 10N

• For digital systems, the maximum capacity C (bits/s) in presence of white noise is given by the limit

€

C =Δf log2 1+PR

N⎛ ⎝

⎞ ⎠


Sources of Thermal Noise

Sky Temp ~5o -150o K

Ground Temp ~300o K

Physical Temp of Antenna TAP

Physical Temp of Line = TL

Temp of Receiver TR

TA


Thermal Noise Power N

– Boltsman’s constant = k =1.38x10-23 watts/(Hz oK)

– System temperature = TS oK

– Bandwidth = f Hz

– For TS = 300o K and f = 30x103 Hz

• N = 1.24x10-16 watts

• (N)dB = -159.1 dBw = -129.1 dBm

– Noise figure of receiver amplifier F ~ 5 dB

– Effective noise = N + F

• For the example, N + F = -124.1 dBm

€

N =kTsΔf


WalkAbout Phones

Frequency band 450 MHz = 0.667 m

Band width 12.5 kHz

Thermal noise 4x10-18 mW /Hz 5x10-14 mW -133 dBm

Receiver noise figure 5 dB typical

SNR for reception 10 dB for FM

Minimum received power 2x10-12 mW -118 dBm

Transmitted power 500 mW 27 dBm

Maximum allowed path loss (PTr)dB - (PRec)dB 145 dB

Minimum path gain PRec /PTr = 10-14.5 3.2x10-15

Antenna gain / Antenna height Assume 0 dB 1.6 m


Maximum Range WalkAbouts in Free Space

€

PG=G1G2λ

4πR⎛ ⎝

⎞ ⎠

2

=λ

4πR⎛ ⎝

⎞ ⎠

2

>3.2×10−15 =32×10−16

or

R<λ

4π1

32×10−8 =9.4×105 m = 940 km or 563 miles


Summary of Noise

• Noise and interference set the limit on the minimum received power for signal detection

• Thermal noise can be generated in all parts of the communications system

• Miracle of radio is that signals ~ 10-12 mW can be detected


Ground and Buildings Influence Radio Propagation

• Reflection and transmission at ground, walls

• Diffraction at building corners and edges

Diffraction Path

Transmission

Reflection


Two Ray Model for Antennas Over Flat Earth

(Antennas are Assumed to be Isotropic)

€

E1

€

E 2

r1

r2

R

h1h2

Antenna

Image

€

Pr =Pt

λ4π

⎛ ⎝

⎞ ⎠

21r1

exp−jkr1( )+Γ θ( )1r2

exp−jkr2( )2

Γ θ( )=cosθ−a εr −sin2θ

cosθ+a εr −sin2θ

where θ =90°−α and a=1εr for vertical (TM) polarization, or

a=1 horizontal (TE) polarization


Reflection Coefficients at Plane Earth Vertical (TM) and Horizontal (TE) Polarizations

1

Incident Angle , degree0 10 20 30 40 50 60 70 80 90

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Horiz. Pol. r=15-j0.1

Vert. Pol. r=15-j0.1


Path Gain vs. Antenna Separation(h1 = 8.7 m and h2 = 1.8 m)

100

101

102

103

-110

-100

-90

-80

-70

-60

-50

-40

Distance (m)

Pat

h G

ain

(dB

)

Vertical pol.Horizontal pol

Brewster’s angle

f = 900MHz


Sherman Island/Rural


Sherman Island Measurements vs. Theory


Flat Earth Path Loss Dependence for Large R

If R >>h1 and h2 then

r1,2 = R2 + h1 mh2( )2

≈R+1

2Rh1 mh2( )

2=R+

12R

h12 +h2

2( )m

h1h2

Rand Γ(θ) ≈-1

Received power Pr =Pt

λ4π

⎛ ⎝

⎞ ⎠

21r1

exp−jkr1( )+Γ θ( )1r2

exp−jkr2( )2

is approximately Pr =Pt

λ4πR

⎛ ⎝

⎞ ⎠

2

exp jkh1h2

R⎛ ⎝

⎞ ⎠

−exp −jkh1h2

R⎛ ⎝

⎞ ⎠

2

or Pr =Pt

λ4πR

⎛ ⎝

⎞ ⎠

2

2sin kh1h2

R⎛ ⎝

⎞ ⎠

2

=Pt

λ4πR

⎛ ⎝

⎞ ⎠

2

2sin 2πh1h2

λR⎛ ⎝

⎞ ⎠

2


Path Gain of Two Ray Model

€

PG=λ

4πR⎛ ⎝

⎞ ⎠

2

2sin2πh1h2

λR⎛ ⎝

⎞ ⎠

2

At the break point, R=4h1h2

λ the path gain has a local maximum

PG=4λ

4πR⎛ ⎝

⎞ ⎠

2

Past the break point

PG ≈λ

4πR⎛ ⎝

⎞ ⎠

2

22πh1h2

λR⎛ ⎝

⎞ ⎠

2

=h1

2h22

R4

Past the break point, PG is:

Independent of frequency

Varies as 1 R4 instead of 1 R2.


Maximum Range for WalkAbouts on Flat Earth

€

For h1 =h2 =1.6 m, RB =4h1h2

λ=

4(1.6)2

0.667=15.3 m

For R>RB

PG=(h1h2)

2

R4 >3.2×10−15

Solving the inequality for R

R4 <(1.6×1.6)2

3.2×10−15 =0.8×1015

or

R<5.3×103 m = 5.3 km or 3.2 miles


Fresnel Zone Gives Region of Propagation

Fresnel zone is ellipsoid about ray connecting source and receiver and such that r2-r1 =/2– Ray fields propagates within Fresnel zone– Objects placed outside Fresnel zone generate new rays, but

have only small effect on direct ray fields– Objects placed inside Fresnel zone change field of direct ray

r2

r1

r2- r1= /2


Fresnel Zone Interpretation of Break Point

r1

r2

RB

Fresnel zone(r2- r1=

€

Fresnel zone definition: λ 2=r2 −r1Horizontal antenna separation RB for Fresnel zone to touch the ground

λ 2=r2 −r1 = RB2 +(h1 +h2)

2 − RB2 +(h1 −h2)

2 ≈2h1h2

RB

or RB ≅4h1h2

λ


Regression Fits to the 2-Ray Model on Either Side of the Break Point

100 101 102 103

-120

-110

-100

-90

-80

-70

-60

-50

Distance (m)

Pat

h G

ain

(dB

) n1=1.3

n2=3.6

f=1850MHzh1=8.7h2=1.6Model: 2ray, r=15

RB


Six Ray Model to Account for ReflectionsFrom Buildings Along the Street

Each ray seen from above represents two rayswhen viewed from the side:

1. Ray propagating directly from Tx to Rx2. Ray reflected from ground

Ray lengths:

As seen from above

R0 = x2 + zT −zR( )2

Ra = x2 + w+zT +zR( )2

Rb = x2 + w−zT −zR( )2

In 3D

rn1,2 = Rn2 + h1 mh2( )

2

zTzR

p

R0

w

Top view of street canyon showing relevant rays

Rb

Ra


Six Ray Model of the Street Canyon

€

For x>>h1,h2 polarization coupling at walls can be neglected.

Angle of incidence on ground θn =arctanRn

h1 +h2

⎛

⎝ ⎜ ⎞

⎠ ⎟

For each ray pair (vertical polarization)

Vn =e−jkrn1

rn1+ΓH θn( )

e−jkrn2

rn2

Wall reflection angle ψa,b =arctanx

w± zT +zR( )

⎛

⎝ ⎜ ⎞

⎠ ⎟

Path Gain of six rays

PG=λ

4π⎛ ⎝

⎞ ⎠

2

V0 +ΓE ψa( )Va +ΓE ψb( )Vb

2


101 102 103 104-140

-130

-120

-110

-100

-90

-80

-70

-60

-50

-40

Distance (m)

Rec

eive

d P

ower

(dB

W)

2 ray model

6 ray model

Six Ray Model for Street Canyonf = 900 MHz, h1= 10 m, h2= 1.8 m, w = 30 m, zT = zR = 8 m


Received Signal on LOS Route f = 1937 MHz, hBS= 3.2 m, hm = 1.6 m

Telesis Technology Laboratories, Experimental License Progress Report to the FCC, August, 1991.


Summary of Ray Models forLine-of-Sight (LOS) Conditions

• Ray models describes ground reflection for antennas above the earth

• Presence of earth changes the range dependence from 1/R2 to 1/R4

• Propagation in a street canyon causes fluctuations on top of the two ray model

• Fresnel zone identifies the region in space through which fields propagate


Cylindrical Waves Due to Line Source

€

The concept of a cylindrical wave will

be useful for discussing diffraction

Phase is constant over the surface

of a cylinder

For ρ>>λ radiated fields are

E =aEZIe−jkρ

ρf θ( )

H =1η

aρ ×E

Field amplitudes vary as 1/ ρ

to conserve power.

y

z

x

€

E

€

H

LineSource

polytechnic university© 2002 by h. l. bertoni1 iii. spherical waves and radiation antennas radiate...

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