Frequency reuse factor | Assignment

PART: (A)
1. What is the requirement on cluster size to support the GoS requirement?
Given Data:
Average call request rate = 3 calls/day
Average holding time = 15min
Gos = 2%
User density = 200 users
Traffic intensity per user = 3 × 15 = 45 Erlangs
Number of channels or cluster size = (200 × 45) ×2% = 180 channels`
2. What is the requirement on pathloss exponent to satisfy the SIR requirement?
Given:
S/I = 12Db;
N= 180 channels
Frequency reuse factor = 1/N
Frequency reuse factor = 1/180 = 0.0055
Frequency reuse factor = (N + S/I)1/ γ
Where,
γ = pathloss exponent
0.0055 = (180 + 120) 1/γ
γ = -1.0104
3. What is the minimum base station antenna height to satisfy the SIR requirement?
Given Data:
Fc = 1800MHz,
C= 3× 108 m/sec
λ = c /f = 3× 108 / 1800 × 103 =166.67 Hz.
Height of antenna = 0.25× λ = 0.25× 166.67 = 41.67 m.
4. Compute the worst case SIR experience.
Given:
R= 2 km
H= 6400km= Height of earth
D=√ 2hR
D = √ 2× 2× 6.4 ×106 = 5059.64
Suppose, n=4;

The worst case can be determined in SIR:
SIR = R−4
2 ( D−R )−4 + 2 ( D+ R )−4 +2 ( D )−4
SIR = 2−4
2 ( 5059.64−2 )−4 +2 (5059.64 +2 )−4 + 2 ( 5059.64 )−4
SIR = 7.0285
5. Assume that base stations transmit at Pt = 10W. Both the interference signals and the
desired signal are subject to shadowing with spread = 4 dB. Compute the worst-case
outage probability.
Pr = Pt /4 πd2 = 10/ 4 π× (5059.64)2 = 3.1085
Outage probability can be determined in the form of reception failure probability for
given ratio so the notation;
SIR = Z =X/Y = 10 / 3.1085 = 3.2169
1. Describe a 1200 sectoring scheme to increase the SIR for the system. Recompute the
outage probability with the deployed sectoring.
Sectoring is the method which is used for cellular system to improve the SIR
performance; cells are divided into the radial sectors in this method. 1200 cell sectoring
reduces the first tier co- channel interferes from 6 to 2. The distance for first tier or the
outage probability will be at D and D+ 0.7R.

PART: B

1. Consider a multipath fading channel with the following impulse response
>> h=[0 0 0 0];
while sum(h~=0)<4
h= 2*randn(1, 4);
[~, k]= max(abs(h));
h= h/h(k);
end
>> plot(h)
plot(h)
h= -0.238027264631479 -0.811867792480743 1 -0.381687371221368
k= 3