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Propagation over a Plane Earth:

In many practical situations, the free space model gives a very

bad estimate of the actual path loss. Especially in situations when the

received signal is a superposition of many reflected waves, the averagereceived power is in most cases much lower than predicted by the

free space model. This is the case in many mobile radio applications,

where the receiver antenna is typically very close to the ground. To

give some insight, let us consider a very simplified situation,

with one direct line-of-sight transmission path and one indirect

path that is reflected by the ground, as shown in Fig. 7.1. The

received signal is now equal to the sum of two signals, the direct signal

and the indirect signal. Both are subject to a path loss given by the free

space model, but the indirect signal has traversed a slightly longer

distance. The indirect signal has a phase difference relative to

the direct signal, corresponding to its somewhat longer

distance and to the phase shift at the ground reflection.To simplifies

the path loss derivation, it is common practice to assume that the ground

reflection is lossless and causes a phase change of 180 degrees. Simple

geometry then shows that the phase difference between the direct

and indirect signal is

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where d is the distance between antennas and and are the heights of transmit

and receive antenna heights respectively.

when the distance between the two antennas is assumed much larger that

the antenna heights, which is true in most practical situations, this phasedifference is approximately,

The path loss may now be approximated as

This result can be further simplified by the approximation sin(x)~x when

dis much larger than the antenna heights, as

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This is an inverse fourth-power loss formula in distance, as compared to an

inversesecond-order loss formula for the free space case. A more detailed

derivation of this result may be found in [3]. In decibel scale it becomes

and here it is clearly seen that a doubling of the distance now leads to a 12 dB loss

in average received power instead of 6 dB as in free space. Moreover, we see that

increasing one of the antenna heights by a factor of 2 leads to a 6 dBpower gain.With the approximations made, there is no carrier frequency

effect in this formula.The reason for the increased path loss is that the two

signals add destructively.

therefore,pathloss is given as,

L=32.5+20 logf +20 logd

plane-Earth propagation, which differs from the freespace in three ways,

l. As a consequence of the assumption that d >>,andthe angle is small, and

wavelength cancels out of the equation, leaving it to be essentially frequency

independent.

2. It shows an inverse fourth-power law, rather than the inverse-square law of free

space propagation. These points to a far more rapid attenuation of the power

received.

3. It shows the effect of the transmit and receive antenna heights on propagation

losses. The dependence on antenna heightmakes intuitive sense.

Program to find the variation of pathloss with distance:f1=50*10^6;

f2=100*10^6;

f3=500*10^6;

f4=1*10^9;

f5=5*10^9;

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f6=10*10^9;

x1=20*log(f1);

x2=20*log(f2);

x3=20*log(f3);

x4=20*log(f4);x5=20*log(f5);

x6=20*log(f6);

d=0:10:500;

y=20*log(d);

lp1=-(32.5+x1+y);

lp2=-(32.5+x2+y);

lp3=-(32.5+x3+y);

lp4=-(32.5+x4+y);lp5=-(32.5+x5+y);

lp6=-(32.5+x6+y);

plot(d,lp1,d,lp2,d,lp3,d,lp4,d,lp5,d,lp6)

axis('square')

xlabel('distance in km')

ylabel('pathloss lp in dB')

Output:

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0 100 200 300 400 500-390

-380

-370

-360

-350

-340

-330

-320

-310

-300

distance in km

pathloss

lp

in

dB