LEC2-PROPAGATION OF RADAR WAVEeee.guc.edu.eg/Courses/Communications/COMM1007 Radar... ·...

RADAR TECHNOLOGY PROF. A.M.ALLAM

3/14/2011 LECTURES �

PROPAGATIONOF RADAR WAVE

RDR .TCH



Abbreviated RF, any frequency within the electromagnetic spectrum associated with radio wave propagation. When an RF current is supplied to an antenna, an

electromagnetic field is created that, then is able to propagate through space.

What is radio frequency?-1

Band designation


3/14/2011 LECTURES � Rx

Matched elementFeeder

Matched element

Rx. antenna

2. Radio linkTx

FeederMatched element

Matched element Tx. antenna

We are here

Propagation of radio waves



Atmospheric layers

Stratosphere

This effect depends on :

1- Frequency.

2- Antenna directivity

3- Proximity of the antenna to the ground

4- The physical nature of the propagation

path, water, land, heavily vegetated

areas,… etc.

3. Factors affecting the wave propagation

ve) because the The signal propagating is modified from straight line (direct wathe following natural environments :effect of

1- Earth: flat or spherical 2- Troposphere

3- Ionosphere 4- Outer free space: homogeneous, isotopic and loss free. 5- Rain, snow and hails



The table shows that spherical segment is always larger than � for UHF, SHF,…,hence the wave at these frequencies will propagate as direct wave

Propagation path, km 1 5 50 500 5000

h, m 0.031 0.78 78 7800 3.75x105

4. Mechanisms of propagation•Direct wave:

hTx Rx

0

Dire

ct ra

y

Diffracted rayThe diffraction is noticeable when the obstacle dimension is comparable to � .

Considers the spherical segment h of the height of the earth is an obstacle due to its spherical shape .


�

Due to steeper decrease in n at lower heights � < 3m (SHF)

Troposheric duct

•Surface wave:The wave propagated close to the earth surface and partly follows the curvature of the globe due to diffraction. For different frequency ranges till UHF you find surface wave specially in short ranges of comm.

•Ground waveSpace direct and reflected wave

•Troposheric waveDue to local irregularities in troposphere � < 10m (VHF,UHF)

•Troposheric duct


Note:Three basic mechanisms that impact propagation in amobile communication system

• Reflection– A propagating electromagnetic wave impinges upon an object

which has very large dimension compared to the wavelength– Reflection occurs from surface of earth and from buildings

and walls

• Diffraction– The radio path between the transmitter and receiver is

obstructed by a surface that has sharp irregularities and itsdimensions are in the order of wavelength

• Scattering– When the medium consists of objects with dimensions that are

small compared to the wavelength


�

The waves propagated over a large ranges and capable of single and multiple

reflections from ionosphere and also scattering from inhomogenities in the

ionosphere � >10m (HF)

Ionospheric wave

(Sky)

Ionosphere

TransmitterReceiver

•Ionospheric wave (sky wave)


Line of sight (LOS) transmission by ground (space) wave and long distance isachieved using MW repeaters rely towers.

Ground (space) wave

ReceiverTransmitter

Direct wave

reflected wave

MW transmission system-5

Advantages: Fewer repeaters compared to coaxial cable

Applications: Telephone service at rate up to 200-300 Mbps

3/14/2011 ��


Consider an isotropic radiator at point 0 fed with power Pt , this power must flow through a spherical surface, hence; the power density W at a point Q is given by:

From Poynting theorem:

(((( ))))

m/vr

P30E

)zonefarin(r4

P120

E

value.s.m.rHxERe

m/wr4

PW

t

2t

2

*

22

t

====∴∴∴∴

ππππ====

ππππ====

====

ππππ====

The isotropic radiator may be considered as a standard reference antenna with power gain G=1. If a practical antenna with a power gain G is placed at 0, then the power received at Q will be increased G times to be (PtG), hence;

s.m.rm/vr

GP30E t====∴∴∴∴

ds [m2]

0

Pt

.Q

•Free space propagation of direct wave


m/ver

GP30E )krt(jt −−−−ωωωω====∴∴∴∴

2

21tr

2

221t

2221t

r

121t

r4GGPPor

)r4(GGP

antennaRxofareaeffectivetheisAr4AGPP

antennaTxofgaintheisGr4

GPW

��

��

��

ππππλλλλ====

ππππλλλλ====

ππππ====

ππππ====Also:

and the instantaneous field is given by:

• The free space experiences a decrease in their power density with a distance by the effect of spreading or divergence, i.e. the energy dispersed on the surface of expanding sphere wherever it advances.

• Losses increase with frequency by the ratio 6 dB/octave or 20 dB/decade.• Higher gain antennas can be used to make up for this loss


•Transmission loss:

2

br4L ��

��

��

λλλλππππ====

i.e the transmission loss for free space propagation is:

λλλλ−−−−ππππ========

−−−−−−−−λλλλ−−−−ππππ========

log20)r4log(20Llog10)dB(L

)dB(G)dB(Glog20)r4log(20Llog10)dB(L

bb

21fsfs

For an ideal isotropic sources G1=G2=121

2 14GG

rLfs ��

��

��====λλλλππππ

It is called the basic transmission loss and can be expressed in dB as:

It is the ratio between the power radiated (transmitted) from the transmitting antenna to the power available at the terminals of the receiving antenna


Find the basic transmission loss in free space for two cases:i) r =10-1 km , λλλλ=200 m ii) r =107 km , λλλλ=3 cm

Example

i) L=39.3 or 15.9 dB ii) L=1.75x1025 or 252 dB

Solution:


��

•Pass gain factor (F)Consider that the earth’s surface is flat, then the reflection point for flat earth given h1, h2, d can be derived as :

2121

22

1221

11

1

21

1

21

1

2

1

2

1

2

2

1

1

/1&

/1

tan

hhd

dhh

hd

hhd

dhh

hd

dd

ddd

hhh

ordd

hh

dh

dh

+=

+=

+=

+=∴

=+

=+

=∴

==ψ

10

141

).( Rjkd e

RfV −≈

πθ

ϕρπ

θ jRjkr ee

RfV .

41

).( 20

2

−≈

The field reaches the receiver produces a voltage proportional to:

and the voltage produced by the reflected wave is proportional to:

h1

h1

h2Reflected wave R2

Direct wave R1

� � Ground

Rx

Tx

d1 d2(d1,d2) is the reflection point

where: f(�) is the radiation field strength pattern ρρρρejϕϕϕϕ is the reflection coefficient at the ground

(1)- Interference zone


��

The total received voltage will be proportional to:

1 2 1( )

1

1( ) [1 ]

4o ojk R jk R Rj

tv f e e eR

ϕθ ρπ

− − −= +

This factor shows how the field at the receiving point differs from the value it would have under free space propagation condition. i.e; it shows the interference effect of the earth on the direct wave propagation.

2 1( )[1 ]ojk R Rjt dv v e eϕρ − −= +

t dv v F=

F is called the path gain factor or ( the interference factor ), or (the attenuation factor), it is the array factor associated with the antenna at height h1 and its image below the surface, with relative excitation of the image antenna being ρρρρe jϕϕϕϕ.

dhhRR

dhhdR

dhhdR

dhh

dd

hhdhhdR

2112

212

2

212

1

2

212

2/12122

122

1

2&

)(21&)(

21

)(2111)(

====−−−−

++++++++====−−−−++++====∴∴∴∴

��

��

�� −−−−++++≈≈≈≈��

��

��

��

�� −−−−++++====−−−−++++====Subs. for (R2-R1) we get the path gain factor as

F= 1+ �ej�e-jko (2 h1 h2 / d)

For the power it will be F2

��


These figures shows ρρρρejϕϕϕϕ as a function of a grazing angle ψψψψ. It is clear that as ψψψψ approaches zero; the reflection coefficient is nearly equal (-1) for both V & H polarization.

σλσλσλσλ====ωεωεωεωε

σσσσ====χχχχ 60,r

ψψψψ Grazing angleϕϕ ϕϕ

Phas

e of

ρρ ρρej ϕϕ ϕϕ

ρρ ρρA

mpl

itude

of ρρ ρρ

ej ϕϕ ϕϕ

Vertical polarization

•Effect of ground reflection (Fresnel expression)

ρρ ρρA

mpl

itude

of ρρ ρρ

ej ϕϕ ϕϕϕϕ ϕϕ

Phas

e of

ρρ ρρej ϕϕ ϕϕ

Horizontal polarization


•Effect of atmospheric refractionAt greater heights the less dense atmosphere results in smaller index of refraction causing the ray to curve or bend in a downward direction in accordance to Snell’s law of refraction:

Since each successive value of nn is smaller than the proceeding value, the angles θθθθn must increase and the ray curves in downward direction. For propagation over a spherical earth, this ray curvature extends the radio horizon beyond the geometrical horizon

n1sinθθθθ1= n2sinθθθθ2=…= nn sin θθθθn

The effect of ray curvature can be taken into account in a simple way for propagation over spherical earth by replacing the earth with an earth having a large radius and considering the rays to propagate along straight lines, providing that nn decreases linearly with height (this is called standard refraction)

a aeae= (4/3)a

ae is the effective radius of the earth


��

In troposphere:

-The water vapor contents are strongly dependent on the weather conditions and sharply decreases with height.

-Pressure decreases with height.

-Temperature decreases with height about 60/km.

Although the troposphere extends out to a relatively low height, it accounts for 4/5 of the entire air mass.

•Detailed analysis of effect of troposphere on ground wave (Troposheric wave)

•8-10km at polar latitude

•10-12km at moderate latitude

•16-18km at equator

Ground

Troposphere

At frequencies above 30MHz there are different mechanisms:

1- Localized fluctuations in refractive index, which can cause scattering.

2- Abrupt change in refractive index as a function of height, which can cause reflection.

3- A more complicated phenomenon known as ducting.

All these mechanisms can carry energy beyond the normal horizon and can cause interferencebetween different radio communication systems. Forward scattering of radio energy may be used as a mechanism for long-distance communication, for frequencies between 300 MHz and 10 GHz.


•It’s a hypothetical troposphere having arbitrary selected set of characteristics reflecting an average condition of the real temperature. It has:- a sea level pressure of 1013 millibars,- a sea level temp. of 150c- a relative humidity of 60%.

•It has a fixed rate of pressure and temperature decrease upward:- 12 millibars / 100 meters of height- 0.550c / 100 meters of height

•It extends out to an altitude of 11 Km.

•Variation of index of refraction:

The refraction index (n) measured in (N) units where (N) is the excess over unity of (n), in millions. i.e.:

•The standard troposphere:

unitsxnN 610)1( −=

Say for n=1.000325 N=325 units

100 200 300

10

8

4

2

6

km

N

•For standard troposphere:

104.0 −−≈ mdhdN

1043.0 −−= mdhdN

•For practical purposes we consider:


The radius of curvature of the ray path is derived as follows:

•Atmospheric refraction:

Atmospheric refraction means; a light or radio ray encounters variation in atmospheric refractive index along its trajectory that cause the ray path to become curved

[ ] ( )

( ) ( )( )( )( )

( ) ( ) 610..sinsin

.sin.

sin..cos

.cos.cos.sinsinsin.cossin

sincoscossinsin)(sin:'

.cos

coscos,

−−=−=

−=∴

−=

+++=++=

++=++=

=∴

≈+

==

dhdN

n

dhdnn

ndn

dhR

ndn

d

dnddndnnn

ddnn

dddnnddnnn

lawssnellfromd

dhR

dhd

dhabm

dab

R

ϕϕϕ

ϕϕϕ

ϕϕϕϕϕϕϕϕϕϕ

ϕϕϕϕϕϕϕ

ϕϕ

ϕϕϕϕ

0

R�

d�

�+ d�

�+ d�

n + dn

n

dh

a

b

For ground waves, the rays are often propagated at small elevation angles for which ��/2

Further approximation is held if: n�1

��

��

��

−−−−====∴∴∴∴

dhdn

nR

��

��

�

−=��

��

�

−=∴

dhdN

dhdn

R6101


The interference equation is usually derived under the assumption that the direct and reflected rays are propagated in straight lines at constant velocity.

In real conditions these paths are curved due to the atmospheric refraction depending on the profile on index of refraction variation.

Replace the earth with another sphere such that the rays propagate in straight lines. Equate the relative curvature between the true rays and the equivalent rays:

( ) 610.11

1111

−+=

−=∴

∞−=−

dhdNa

a

Raa

a

aRa

e

e

for standard troposphere:

With the concept of effective earth’s radius, we replace (a) with (ae) in all equations of the interference zone.

•Effective radius of the earth:

aae 34≈

( ) 610.1 −+=

dhdNa

aae

Ra

ray

Ground

�

ae

ray

Ground


•Forms of atmospheric refraction:

Supper ref.

Standard ref.

Generally; the condition of the troposphere is very close to the standard, and the departure from the normal state will only occur within particular intervals of height.

Supper-refraction : bending of the ray is smaller than that of standard. The radius of curvature of the ray is smaller than the radius of earth, and the rays leaving the transmitter at small angle of elevation will undergo total internal reflection in the troposphere and return to the earth at some distance from the transmitter. On reaching the earth’s surface and reflecting from it again; it skip a very large distance.


3/14/2011 LECTURES ��

The super-refraction affects a volume of the troposphere extending from earth’s surface up to altitude ho (does not exceed 200 m). Within this height the super-refraction results in a surface duct as follows:

•Ray 1,2 starting at large elevation angles > c and undergo partial refraction. They pass through the upper boundary of super-refraction boundary without being trapped there.

•Ray 3; = c at which curvature of the ray equals to that of the earth and thus being horizontal to the earth at an altitude ho.

•Ray 4, 5; within ± c all are trapped by the super-refraction region and they propagate similar to propagation in dielectric wave-guide. The imperfectly conducting surface of the earth acts as the bottom wall of such guide, or duct, and the upper boundary of super-refraction region as the top wall.

hoc

1 23

4

5

Tx


3/14/2011 LECTURES ��

As with the case of dielectric wave-guide, the condition of waves to be supported in a tropospheric duct is that the wavelength shouldn’t exceed �c.

For different duct heights ho , �c is found to be:

Since the height of the tropospheric duct never exceeds 200 m; then the troposphericduct can support UHF, SHF and has no effect on MW, LW. [at ho=200 m, �c =2.4 m].

Thus the conditions of duct (trapping in tropospheric duct) are:

i) dN/dh < -0.157 (existence of super-refracting layer)

ii) within ± c

iii) f > fc

[[[[ ]]]]mhoc42/3 10.5.8 −−−−≈≈≈≈λλλλ

2.4


��

Considering the actual surface of the earth i.e., spherical earth an important question is arising is whether the transmitter and receiver antennas are within the line of sight range LOS of each other ?

eoeoo

o

ahdahda

dah

radina

dah

ah

ah

ahhaa

2211

211

1

12

12

1

11

2&221

.

2

12

1cos

1/1

1cos

==∴��

��

�=∴

=

=∴

−≈−≈

−≈+

=+

=

α

α

αα

α

�

�

h1 h2do1 do2

a

� �

do

aa

Taking that effect into consideration, the path gain factor F is valid only of the link distance is d ≤≤≤≤ 0.8do to be sure that there is LOS between Tx and Rx

•Effect of line of sight range (Optical LOS):

LOS = do1+do2 = do


3/14/2011 LECTURES ��

We can use a flat earth formula, once one gets the reduced heights h1’ and h2

’., for given h1, h2 and d :

LOSfromadhadhhhhhhh ee 2/;2/;';' 222

211222111 =∆=∆∆−=∆−=

Then replace h1,h2 by h1’, h2’ in flat earth formula, hence:

( )

2/1

21

21

21

22

21

11

21

2

21

1

11

''2

1

,

1

''2

'tan

'tan

−

∆−

−−

��

�

�

++=

+=

+=

+=

=∆

��

��

�=��

�

��

�=

hhadd

D

dhh

hdd

hhh

d

eeDF

dhh

R

dh

dh

e

Rjkjϕρ

ψ

D is the divergence factor a = 6371km

•Pass gain factor (F) for spherical earth


Example

Calculate the path gain factor in case of communication between two antennas at heights 30 & 25m, the distance between them is 15Km .The communication is done by vertical polarized wave of � =10 cm propagated over moist soil characterized by (r=10, �=0.01 S/m), use standard troposphere.



-The field at R is calculated using Huygens's principal

-The set of curves from point (R) to (S) at point (l2+�/2) forms a conical surface intersecting the plane of drawing along straight lines RN1 & RN1’

Generally: RNn=l2+n�/2

The intersection between the conical surfaces and the sphere forms a system of concentric circles

The segments bounded by adjacent circles are known as Fresnel zones.

It’s known from optics that the higher order zones cancel one another, i.e. The first Fresnel zone bounds the volume contribution significantly to wave propagation because the fields at reception point are available, [for path difference between two points (rays) is � �/2]

observation

Huygen sphere

sourceT

R

SS

NoN1

N2

•Effect of Fresnel zone ( Radio LOS)

No

N1

N2

l1

l1 l2

N’2

N’1

N2

N1

No


L.. Is the path difference between two rays L = TBR – TR � �/2 (first Fresnel zone) or �/2 � L � � (second Fresnel zone) or � � L � 3�/2 (third Fresnel zone) or (n-1)�/2 � L � n�/2 (defines nth Fresnel zone)

Since TBR =TR + L = constant

It resembles an equation of ellipsoid, its foci at T, R and the region behind both the Tx. and Rx. are excluded from the ellipsoid

T

B

RA

3/14/2011 ��


T Ra b

r1 r2��m

�m

d

��

��

�� ++++====

−−−−−−−−��

��

��++++++++��

��

��

��++++====

−−−−−−−−��

��

��

��

��++++++++��

��

��

��

��++++====

−−−−−−−−++++++++++++====−−−−++++====∆∆∆∆

ba

bab

ba

a

bab

ba

a

basbsadrrL

1121

211

211

11

2

2

2

2

2

2/122/12

222221

ρρρρ

ρρρρρρρρ

ρρρρρρρρ

� �

To derive the minor radius of that ellipsoid:


For fist Fresnel zone L � �/2,

221

2/1121

max

2

dbaford

baab

ba

========≤≤≤≤

++++≤≤≤≤

≤≤≤≤��

��

�� ++++

λλλλρρρρ

λλλλρρρρ

λλλλρρρρ

Measurements shows that to achieve normal MW transmission like free space (LOS propagation) ,the transmission path should pass over all obstacles with a clearance of at least 0.6�m (mid path clearance condition).

T

R

ab

r1

r2��m

�m

d

i.e. the reduced heights of the antenna should be higher than 0.6�m

As f increase, �m decrease. i.e. at LF we can’t use LOS propagation because we need antennas at very high positions

mh ρ6.0'1 ≥


Example

Determine the midpath clearance needed to coincide with the first Fresnel zone for a microwave link cover a distance of 60 km at frequencies: 7, 30, 300 GHz.

�m= 0.5��d

at 7GHz �m =25.354m

at 30GHz �m =12.25m

at 300GHz �m =3.87m


•The effect of a hill can be modeled by a thin plane or a knife edge having the same clearance distance ( height hcfrom the line of sight (LOS) path.

•We assume that there is no significant contribution to the received field from specular reflection, i.e., we shall consider that the received signal is modified from its free space value by the diffraction effect of the knife edge, and no specular reflection.

It’s not always possible to install a communication link such that there are no obstructions such as hills or large buildings that block part of the field from the

transmitting antenna, impeding its arrival at the receiving site.

h1

h2

Direct wave

Ground

Rx

Tx

d1 d2

Hill

Specular reflectionh c

h1

h2

Direct wave

Ground

Rx

Tx

d1 d2

h c Reflect

ed

signal

missing

the Rx

anten

na due

to kn

ife ed

ge

diffra

ction

•Effect of mid-path obstacle (diffraction loss)


Procedure to determine the diffraction loss

( )

( ) hd

hdhdhhhhh

dhdhd

dd

hhhh

pcpc

p

−+=−≈−=

+=−+=

2112

21121121

cosθ

1)The incident field on the surface (S) is expressed by Gaussian-shaped beam

2) The free space field at the receiver site is expressed as the field radiated from the

effective aperture (S) (the integration over y is from –hc to �).

3) The free space field at the receiver site at the absence of the knife edge is calculated as

in step (2) by (y changes from -� to �).

4) The ratio of the field in (2,3) gives the diffraction loss Fd

1

21 dye

aF

ch

ayd

∞

−

−=π

h1

h2

Ground

Rx

Tx

d1 d2

h c

h

h2 - h1�c

�c

P

yz

S


co

c hdd

dH .

2

21λ≈

The diffraction loss can be determined from the graph of Fd versus Hc , where

Notices:

1- at hc = 0, Fd=0.5= -6 dB

i.e., half of the incident radiation is blocked at zero clearance height and this results in 6 dB loss.

2- at Hc = 0.8 or more, the diffraction loss is negligible.


Having an obstacle free 60% of the Fresnel zone gives 0 dB loss otherwise

We have obstacle loss ;diffraction loss or diffraction attenuationbased on knife edge approximation

0 dB

20dB16dB6dB0 dB

First Fresnel zone


At very long, long and medium waves; a terrain with hills up to hundred meters high will be treated as a smooth surface.

ψλπψ sin

4sin2. ir

ir

hhk =

•Effect of rough surfaces:

In cent-metric band; even field covered with grass about 10cm high, will be considered as a rough surface, consequently the reflection coefficient of the earth decreases which decreases the reflected wave.

For simplicity assume that all irregularities are of the same height as shown.

The difference in path length between the upper and lower limits of these irregularities is: 2hir sin� hence the phase difference is:

hir

BA

ψψψψ

Upper limit

ψψψψ

hir

ψψψψ ψψψψ

Lower limit

C

ψψψψψψψψ


A criterion by which the surface may be treated as a smooth one,Rayleigh’s criterion which sets phase difference < ππππ/2 or 0.4ππππ

ψψψψ

λλλλ<<<<ψψψψ

λλλλ<<<<∴∴∴∴

ππππ<<<<ψψψψλλλλππππππππ

<<<<ψψψψλλλλππππ

∴∴∴∴

sin10horsin8h

4.0sinh4

or2

sinh4

1) This is an obligation condition on h of the irregularities to be considered as smooth surface

2) Moreover, due to surface roughness; the reflection coefficient of the earth decreases which consequently decreases the reflected wave


��

•The field strength, due to diffraction, decreases rapidly as the point of absorption moves deeper into the shadow zone, however; the field is still finite and often of sufficient strength to produce useful signal.

•In the near vicinity of the tangent ray, no simple method for calculating the field strength is available (Semi shadow zone), but Fcan be determined with acceptable accuracy by drawing a smooth curve connecting the values of F in the interference zone to those in the shadow zone.

(2)- Diffraction zone (Shadow zone)

Inter

fere

nce z

one

(illu

min

ated

zone

)d �

0.8d o

Sem

i sha

dow

zone

(Inte

rmed

iate

zone

)0.

8 d o �

d �

1.2d

o

Shad

ow zo

ne

(no s

impl

e for

mul

a)d �

1.2d o

•From G.O. point if view, the field strength beyond the LOS or tangent ray is zero, but due diffraction effects, the radiated field penetrates into the shadow zone below the tangent ray.


��

( ) ( ) ( )211 zUzUxVF =

x, z1, z2 are the distance in natural units of length, and heights in natural units of heightHh

z 22 =

( ) [ ]KminLmink

aL

Ld

xwhere ooo

e ,41.284

2, 31

31

λλ=��

��

�==

( ) [ ]minHka

HHh

z oo

e 32

31

21

1 56.472

, λ=��

��

�==

( ) 02.21 2 −= exxV π

V1 is the attenuation function U(z) is the height gain function


��

V1(x) , dB

x

V1(x) , dB

x


��

U1(z) , dB

Z < 1


��

U1(z) , dB

Z > 1

dB


��

Determine F for �=20cm, h1=25m, h2=10m & d=50km. Use the standard troposphere (standard diffraction)

Example

Solution:

( ) ( ) kmhhad eo 648.33102510.8497.22 321 =+=+=

( )zoneshadowzonendiffractiotheinisiteidd

o

..486.1=

( ) ( ) ( )211 .. zUzUxVF =

( ) 994.2,7.162.041.28 3/1 ====Ld

xL

( ) 61.0,538.1,26.162.055.47 22

11

3/2 ======Hh

zHh

zH

0223.0334736 =−=−+−=∴ FordBF

( ) dBxV 36−=∴ ( ) ( ) dBzUdBzU 47 21 −==


3/14/2011 LECTURES ��

Given: h1 , h2 , �o

(3)- Intermediate zone (Semi shadow zone)

Steps:1-Select some points in the interference zone, d� do and determine F at each point

2-Select some points in the diffraction zone, d/do > 1.2 i.e. d=1.25do d= 1.5do and

d=1.75do, and determine F for each point

3-Plot F versus d/do (F in dB=20log dB)

4- Connect the values of F in both zones



In the interference zone:choose d1=0.5do F=1.75=4.86 dB choose d=0.65do F=1.255=1.97 dB

In the shadow zone:choose points: at d=1.25do , d=1.5do and d=1.75do ,

( ) ( ) ( )( )

dBFdBFdBF

dBzU

dBxVdBxVdBxV

pothirdforxpoondforx

pofirstforx

624630

17

968064

int47.6,intsec55.5

int62.4)1.0)(41.28(

)78.48)(25.1(

321

11

321

32

3/11

−=−=−=∴=

−=−=−=∴==

==

kmdo 78.48=

Example:

Solution:

Construct a plot for F versus d/do for a microwave link with antenna heights, h1=h2=35m, �=10cm. Use the plot of V &U


3/14/2011 LECTURES ��

d/do

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