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Problem2 3

1) Using the free space path loss model, the maximum acceptable in-building penetration loss for a base station with 5km coverage is 95.18 dB. 2) Using the two-ray path loss model, the maximum acceptable in-building penetration loss is 95.737 dB. 3) Both path loss models provide similar results for the maximum acceptable in-building penetration loss, being approximately 95 dB.

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64 views2 pages

Problem2 3

1) Using the free space path loss model, the maximum acceptable in-building penetration loss for a base station with 5km coverage is 95.18 dB. 2) Using the two-ray path loss model, the maximum acceptable in-building penetration loss is 95.737 dB. 3) Both path loss models provide similar results for the maximum acceptable in-building penetration loss, being approximately 95 dB.

Uploaded by

kanny
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Problem 2.

3:

In a mobile communication network, the minimum required signal to noise ratio is 12 dB. The
background noise at the frequency of operation is -115 dBm. If the transmit power is 10W,
transmitter antenna gain is 3dBi, the receiver antenna gain is 2dBi, the frequency of operation is
800 MHz. and the base station and mobile antenna heights are 100m and 1.4m respectively.
Determine the maximum in building penetration loss that is acceptable for a base station with a
coverage of 5km if the following path loss models are used:

1) Free space path loss model


2) Two-ray path loss model

Solution:

Transmit Power Pt = 10W

Convert to dBm:

10logPt = 10log(10W / 1 mW) = 40 dBm.

Frequency, f = 800 MHz

Distance : d =5 km = 5000 meters

Transmitter antenna Gain , Gt = 3 dBi

Receiver antenna Gain , Gr= 2 dBi

Convert antenna gains to dBm :

Transmitter Antenna Gain Gr = 5.15dBm

Receiver Antenna Gain Gr = 4.15dBm

1) Free Space Path loss model.


Pr = Pt * Gt * Gr * (λ / 4* ∏ * d)2
Received power at distance one meter i.e. d = 1,
Po = Pt * Gt * Gr * ( λ / 4* ∏ )2
10logPo = 10logPt + 10logGt + 10logGr + 20log( λ / 4* ∏ )
= 40 + 5.15 + 4.15 + 20 * log (0.375/(4 * 3.14)) (because λ= c/f = 0.375)
= 40 + 5.15 + 4.15 + 20 * log (0.02985)
= 40 + 5.15 + 4.15 + 20 *(-1.5250)
= 49.30 – 30.50
= 18.80 dBm
Now,
Pr = Po / d2
10logPr = 10logPo - 20log( d )
= 18.80 - 20 log(5000)
= 18.80 – 73.98
= -55.18 dBm
Therefore,
Path Loss :
L = 10logPt - 10logPr
= 40 – (-55.18)
= 95.18 dBm……………….Ans of Path Loss

2) Two Ray Path loss model.

Pr = Pt * Gt * Gr * ((hb2 * hm2)/d4)
Received power at distance one meter i.e. d = 1,
10logPr = 10logPt + 10logGt + 10logGr + 20log(hb) + 20log(hm) – 40 log(d)
= 40 + 5.15 + 4.15 + 20*log (100) + 20*log (1.4) - 40*log (5000)
= 40 + 5.15 + 4.15 + 40 + 2.922 – 147.959
= -55.737 dBm
Therefore,
Total Path Loss :
L = 10logPt - 10logPr
= 40 – (-55.737)
= 95.737 dBm……………….Ans of Path Loss

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