3.Large Scale Propagation-Loss

8/14/2019 3.Large Scale Propagation-Loss

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LargeScale Propagation Loss

1. Wave propagation models

1. LargeScale Path Loss Model (includes shadowing) :It is the decay in signal strength due to distance between Tx and Rx. Remember the formula

Pr(d)=Pr(do)(do/d)

n

. this decay is gradual with distance and slow in nature and can be noticedfor relatively long distances and long time intervals.Shadowing is the random variation in signal strength which is also slow in nature and gradualwith distance.

2. SmallScale fading:It is the fast and random variations of signal strength due to multipath propagation andmovement of mobile station or the surrounding environment. It occurs over small distances(fractions of a wavelength) and is fast in nature.

2. Free Space Propagation Model

By free space we mean that the path between Tx and Rx is clear lineofsight (no obstructions),and that there are no objects in the surrounding of this path.

When this is the case, the received power is given by Friis formula (assuming no loss in antennas orother equipment):

2

2rtt

rd4

GGP)d(P

Ptand Prare transmitted and received powers (Watt), Gtand Grare transmitter and receiver antennagains (dimensionless), d is distance between transmitter and receiver (m), and is the wavelength(m).

Friis formula is valid only for farfield (or Fraunhofer region), that is when d is very largecompared to or to largest dimension (D) of transmitter antennaby largest dimension (D) we mean

Path loss

Path loss+shadowing

Path loss+shadowing+fading


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the largest distance between any two points on the surface of antenna, for rectangular shapedantenna its the diagonal while for elliptical shaped its the major axis.etc.

Farfield is given by:

2D2d

Gain of antenna is given by:

2eA4G

Aeis the effective area of antenna (m2)front area of a dish receiver for example. Aeis then

4

GA

2

e

The quantity PtGtis called the Effective Isotropic Radiated Power (EIRP). Directional transmitterantenna concentrates power in the direction of propagation and reduces power transmitted in otherdirections. When compared to isotropic antenna, the amount of power radiated in the direction of

propagation is therefore more than that radiated by isotropic antenna (of course the power radiatedin other directions is less than that radiated by isotropic antenna, directional antenna onlyconcentrates power and does NOT add power to wave)

WattGPEIRP tt

The power density (in Watt/m2) at the receiver antenna (just before it is collected by antenna) Pdisgiven by

2tt

2d d4

GP

d4

EIRP)d(P

The received power Pr(d) (in Watt) is the product of the power density Pd (Watt/m2) by receiver

antenna effective area Ae(m2)

2

2rtt

2r

2tt

edr

d4

GGP

4

G

d4

GPA)d(P)d(P

which is the Friis formula again.

Power density Pd(d) and electric field intensity E(d) at distance (d) from transmitter antenna arerelated by:

120)]d(E[

R)]d(E[

)d(P2

fs

2

d

The path loss, PL(d), is the ratio of transmitted to received power and is greater than one.


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2rt

2

r

t

GG

)d4()d(P

P)d(PL

In dB the formula becomes

2

2

rt2

rt

2

d4GGlog10

GG)d4(log10dB)d(PL

If we know the received power P(do) at distance do (do in farfield), then we can compute thereceived power Pr(d) at any distance d>do.

2o

orr d

d)d(P)d(P

In dB )dlog(20)dlog(20dB)d(PdB)d(P oorr

Usually dois taken to be 1m for indoor propagation and 100m or 1km for outdoor propagation.

dB, dBW, and dBmFor power calculations:

dB (or dBW) is a unit of power compared to one Watt and is given by

)WattPlog(10dBWPdBP

While dBm is a unit of power compared to one milliWatt

30dBWP)WattP10log(10)mWPlog(10dBmP 3

For power ratios (like gain or loss):Only dB is used (NOT dBW NOR dBm)

dBWPdBWPWattPlog10WattPlog10

WattPWattP

log10dBratioPower 21212

1

The same power ration results if we divide mW over mW or Watt over Watt

dBmPdBmPmWPlog10mWPlog10mWPmWPlog10dBratioPower 2121

2

1

In other words

dBmPdBmPdBWPdBWPdBratioPower 2121

Example 4.1

f=900MHz, largest dimension of transmitter antenna D=1m, find farfield distance?Solution

d > (2D

2

/)

d > (2(1)

2

/(310

8

/90010

6

)

d > 6m


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Example 4.2

Pt=50 Watts, Gt=Gr=1, f=900 MHz find a) Ptin dBm b) Ptin dBW c) Pr(100m) and Pr(1km).Assume free space propagationSolution

Pt[dBm]=10log(501000)=10log(50)+10log(1000)=17+30= 47dBmPt[dBW]=10log(50)=10log(50)=17dBWorPt[dBW]= Pt[dBm]30 = 4730= 17dBW

3. Propagation MechanismsReflection, diffraction, and scattering.

Reflection: when wave falls on a smooth surface which is also large compared to wavelength,ex: earth surface, buildings, walls,etc.

Diffraction: bending of propagating wave when hitting sharp edges, ex: mountain peaks, edgesof wallsetc.

Scattering: when wave falls on rough surface or when dimension of surface is small compared towavelength, ex: street signs, tree leafs, walking peopleetc.

4. Reflection

When a wave travels from one medium to another, part of it is reflected back into the first mediumand the other part is refracted (or transmitted) through the second medium. The amount ofreflected or transmitted parts depend on polarization (parallel or perpendicular) of wave,

permittivity () and permeability () of both media, and angle of incident i. In conductors, itdepends on conductivity () and frequency (f) as well.

We will consider transverse EM plane waves. In such waves E and H are orthogonal to each otherand direction of propagation is orthogonal to plane containing E an H. We define plane ofincidence as the plane containing the incident, the reflected, and the transmitted waves. In thefigure shown it is the plane of the paper.A wave is said to be parallel polarized if the electric field E is contained in the plane of incidence,figure (a).A wave is said to be perpendicular polarized if the electric field E is perpendicular to the

plane of incidence, figure (b).


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We are interested in two cases, when first medium is air or vacuum (1=o, 1=o, 1=0) andsecond medium is perfect dielectric (2=roand 2=o2=0). The other case is when first mediumis air or vacuum and the other medium is a perfect conductor (2= )

In both cases,

ri

And in the case of reflection from perfect dielectric

)90sin()90sin( t22i11

Or, since we assumed 1=2=o, 1=o, and 2=ro

)90sin()90sin( tri

Or

)cos()cos( tri

Here i, r, and t are measured with respect to surface of boundary (NOT the normal ofboundary).

What about the amount of the reflected wave?At the surface of boundary we have the following relation

ir EE

is called Fresnel reflection coefficient, and its value depends on polarization of incident wave.

For parallel polarized wave (again we are assuming 1=2=o, 1=o, and 2=ro)

)(cos)sin(

)(cos)sin(

i2

rir

i2

rir||

While for perpendicular polarized wave we have

)(cos)sin(

)(cos)sin(

i2

ri

i2

ri

Note that r1 for any medium

What if E field is neither contained in, nor perpendicular to surface of incidence?In this case we decompose the wave into two components, parallel and perpendicular and dealwith each component separately.


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If we plot the magnitudes of ||and as function of ifor fixed values of rwe get the followinggraphs

Note that there is a value of ithat makes ||equal to zero, this value of iis called the Brewsterangle B. (for there is no such value of ithat makes it zero)

Solving for ||= 0, we get

11

1

1

r2r

rB

Example 4.4

Find ||and when iapproaches zero and medium 1 is free space and medium 2 is perfectdielectric.Solution:

1)0(cos)0sin(

)0(cos)0sin(2

rr

2rr

||

1)0(cos)0sin(

)0(cos)0sin(2

r

2

r

This is an important result since for wireless communications usually the distance between Txand Rx is very large compared to antenna heights and the angle of incidence iis very small and can

be approximated as zero.

Reflection from perfect conductors

When medium 2 is a perfect conductor ( = ), then ||=1 and = 1. Both are constants and areindependent of i.


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5. Ground Reflection (2ray) Model

In the previous chapter we used n=4 for path loss exponent, in this section well see why. We willderive the 2ray formula from free space formula.First we need to know how the electric filed decays with distance (d) in free space. So

120

)d(Ed4

EIRP)d(P2

o2o

od

120

)d(E

d4

EIRP)d(P

2

2d

Dividing the two equations we get

d

d)d(E)d(E oo

Remember that E(d) and E(do) are the amplitudes of a sinusoidal electric component of an EM waveand are always positive.Now consider the situation in which Rx antenna receives a lineofsight (LOS) and a groundreflected versions of the same transmitted signal as shown.

Each one of the two paths alone is represented by freespace model. Heights of Tx and Rx antennasare ht and hr. The ground distance between Tx and Rx towers is (d), while the LOS distance

between Tx and Rx antennas is (d) which can be more than (d) because htand hrare not necessarilyequal. The distance traveled by the reflected path is (d) which is clearly larger than (d). We define=dd.

2rt

22rt

2 )hh(d)hh(ddd

can be approximated by (this is the first approximation)

dhh2rt


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If the magnitude of a freespace model electric field is given at some reference point (do) then themagnitudes of LOS and reflected paths are

d

dE

d

d)d(EE ooooLOS

ddE

dd)d(EE oooog

What about phaseshift? Phaseshift between LOS path and reflected path is given by

2

can be calculated using the exact or the approximate values of .

So if the electric field received by LOS is considered to be at zero phase as tcosd

dEcoo then the

electric field received by reflected path is tcos

d

dEc

oo

We will use phasor diagram, but before that, if (d) is very large compared to h tand hrthen (d), (d)and (d) are close to each other and hence

d

dE

d

dE

d

dE oooooo

In other words the amplitudes of the two received signals are almost equal. Still though, there is asignificant difference in the phases.

The total electric field at the receiver is the phasor sum of the LOS and reflected waves. We willassume that (d) is large and so i 0 and = 1.

2sind

dE2

)cos(22d

dE

)cos()1)(1(211d

dE

)(101d

dE

101ddE

d

dE0

d

dE

d

dE0

d

dEEE)d(E

oo

oo

22oo

oo

oo

oooo

oooogLOSTOT


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When /2


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6. DiffractionAny wave (including EM waves) that isobstructed by an obstacle can propagatearound this obstacle and reach into theshadow region, this phenomenon is

called diffraction.Diffraction can be explained byHuygens principle which states that:we can think of the propagating EMwave as small point sources located atthe wavefront, these small pointsources radiate small EM waves. If wesum the vectors of these small EMwaves, we get the original EM wave.Huygens principle is a simplified

model to deal with EM waves.

Knifeedge diffraction model

To study diffraction mathematically, we will use a simplified method in which we representmountains, buildings, etc by a knifeedge. The knife edge is supposed to be very thin and extendsinfinitely in and out of the page plane.

The question is: what is the value of the electric field with the presence of the knifeedge obstacle?

The answer is in terms of the freespace electric field Eowhich is the electric field in absence of theknifeedge obstacle or any other obstacle including earth surface. So we will use the free spacemodel to find Eo.

120

E

d4

GP)d(P

2o

2tt

d

d GP30Etto

Now we find Ed, the electric field with the presence of the knifeedge.

od E)v(FE

F(v) is called the complex Fresnel integral.

How to find F(v)?First we define (v), the FresnelKirchoff diffraction parameter. In the figure above we assume h r


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between receiver and knife edge which is also very near to (d2), (here we are assuming that d1andd2are much larger than htand hr). Important too is the height of the knifeedge above the lineofsight (h). Note that (h) can be zero if knifeedge is on lineofsight, or it can be negative if knifeedge is under lineofsight.

Now we define (v) as

21

21

dddd2

hv

At the end we calculate F(v), the Fresnel complex integral, as

hr

hththr

hobs

small sources(Huygens)

Diffracted waves (vector summationof small waves from small sources)

hthr

d1 d2

d1 d2

d1d2

h=hobshr

hthr

d1 d2

h=0

hthr

d1 d2

(h) negative

d

d1

d1

d2

d2


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2

dt)2

tjexp(

2j1

vF

Dont worry! We will not evaluate this difficult integral. Instead, we will use a graph that representthe dB value of F(v) versus (v). The dB value of F(v) is called the diffraction gain G d(v).

)v(Flog20vGd

Here we use 20log[] and not 10log[] since it is not a power ratio.

Example:4.7

=1/3 m, d1=1km, d2=1km, find diffraction loss for a) h=25m, b) h=0, c) h=25m.

Solution:

a) 74.2)1000)(1000)(3/1(

10001000225dddd2hv21

21

Using the graph we find diffraction gain Gdto be approximately 22 dB (or diffraction loss is 22dB)

b) h=0 so v=0, using graph the diffraction loss is 6dB

c) 74.2)1000)(1000)(3/1(

10001000225

dd

dd2hv

21

21

Using the graph we find diffraction loss to be approximately 0 dB


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Example:4.8

f=900 MHza) find diffraction loss

b) find obstacle height required toinduce 6dB diffraction loss.Solution:

a) h=75m25m2km/12km=70.8m

25.4

)2000)(10000)(3/1(2000100002

8.70v

Using graph we find that diffractiongain Gd26dB or diffraction loss is26dB.

b) From graph we find that to get diffraction loss of 6dB (Gd6dB) we need h=0, which means thatthe obstacle edge is exactly on the LOS between Tx and Rx. This means thathobs=25+25m2km/12km=29.17m.

Fresnel ZoneThe path length difference betweenlineofsight and diffracted signal ()is given by

21

212

21

22

2

22

1

dd

dd

2h

ddhdhd

The phase difference between lineofsight signal and diffracted signal isgiven by

21

212

ddddh2

We are looking for values of (h) that makes the phase difference between LOS and diffracted pathequal to multiples of , that is n , which correspond to constructive and destructiveinterference cases. These values of (h) are called Fresnel distances of order (n), or simply rn.Solving for n we get

2121

n dd

ddnr

50m

25m

100m

h

10km 2km

75m

ht

hr

h

d1 d2

d1

d2

221 hd 22

2 hd

LOS


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If we find rnfor all possible d1, d2(remember d1+d2=d), we end up with an ellipsoid for each (n).These ellipsoids have the transmitter and receiver antennas at their foci as shown.

Recall that

21

21

dd

dd2hv

then2v

2

Now, what are the values of v (or h) after which F(v) keep decreasing below 0dB? (in other words,the diffracted signal strength keeps falling below LOS strength?)From the graph we find that after v = 0.8, F(v) keeps decreasing below 0dB. This value of (v)corresponds to

12121

21

21 r57.0dd

dd

2

8.0dd2

dd8.0h

In other words, in order to guarantee diffracted signal to be equal in strength to LOS signal, wehave to keep obstacles outside 57% of the first Fresnel zone, or approximately 55% (sometimes itstaken as 60%).

7. Scattering

Scattering is the reflection of EM wave from a rough surface, in this case the EM wave does notreflect in one direction like in smooth reflection, but scatters in different directions. Question is,when is a surface considered to be rough or smooth?

Rayleigh criterion answers this question. When the maximum separation between top and bottomparts of the surface, called (h) is greater than the critical height (hc), the surface is rough, otherwiseit is smooth. Critical height (hc) is given by

ic sin8

h

n=1, or =n=2, or =2n=3, or =3


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In this case, the reflected electric field intensity is less than that in the case of smooth (specular)reflection. So the reflection coefficient must be multiplied by scattering loss factor ( s)

srough s


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The probability that the received signal, P r(d)[dB], is greater than a specific value, [dB] (in dBtoo), is given by

dx)x(f])d(PPr[ Xr

Again, we do not need to evaluate this integral; instead, we will use Qfunction tables.

)d(P

Q])d(PPr[ rr

Of course, the probability that Pr(d) is less than is

)d(P

Q)d(P

Q1])d(PPr[1])d(PPr[ rrrr

Always remember: Pr(d), , and all are in dB.Example:4.9(c,d) modified!The received power at some reference do=100m was found to be 0dBm, n=4.4, =6.875dB. c) findthe average received power at d=2km d) find Pr[Pr(2km)>60dBm)

Solution:

dB25.57m100

km2log)4.4(10]dB)[m100(P]dB[)km2(P rr

65542.034458.01)4.0(Q1)4.0(QdB875.6

)dBm25.57(dBm60Q]dBm60)d(PPr[ r

Note that the term (60dBm(57.25dBm)) = 2.75dB (not dBm!) which is the same if we convertthe 60dBm to 90dB and the 57.25dBm to 87.25dB then subtract them!

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3.Large Scale Propagation-Loss