# Date due: Essay Example

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MOBILE COMMUNICATION FUNDAMENTAL 11

Mobile Communication Fundamental

**Mobile Communication Fundamental**

**Question 1**

SIR= 1/ ((2 × (Q^{-α} + (Q-1)^{-α} + (Q+1)^{-α})

SIR = 1/ ((2Q^{-α} + (2Q-2)^{-α} + (2Q+2)^{-α})

Therefore: SIR = 2Q^{α} + (2Q-2)^{ α} + (2Q+2)^{ α}

Where Q= √3N; N= cluster size; α = path loss exponent which is 3.5 in this case

SIR is given as 12dB

Hence, 12 = 2(√3N)^{ 3.5}+ 2 (√3N)^{ 3.5}— 2^{3.5} + 2(√3N)^{ 3.5} + 2^{3.5}

12 = 6 (√3N)^{ 3.5}

2 = √3^{3.5} × N^{3.5}

The cluster size is therefore 6

**Worst case SIR for the 6 cluster size**

Q =D/R=√3N

D = √18 × R

D = 4.2426R

SIR = R^{-3.5}/ ((2 × (D^{-3.5} + (D-R) ^{-3.5}+ (D+R) ^{-3.5})

SIR = R^{-3.5}/ 6D^{-3.5}

SIR = 6D^{3.5}/ R^{3.5}

But D = 4.2426R

Therefore SIR = (6 × 4.2426R^{3.5}) / R^{3.5}

Worst case SIR = 6 × 4.2426 = 25.4556

LmediandB (urban) = 69.55+ 26.16 log fc −13.82 log ht −a(hr) + (44.9−6.55 log ht) log d

Fc = 900MHz

Ht = 40m; a(hr) is negligible

D = 0.04 km

Therefore SIR = 69.55 + 26.16 log 900 – 13.82 log 40 + (44.9 -6.55 log 40) log 0.04

SIR = 69.55 + 77.28 – 22.14 + 34.41log 0.04

= 124.69 – 48.10

SIR = 76.59

Considering this worst case SIR, it can be deduced that it meets the SIR requirements. This is because of the fact that the interference signal is much higher than the received desired signal power as a result of the small distance of 0.04 kilometers. This in turn has had effect on the co-channel reuse factor Q making it larger because of the increased cluster size.

In this case, we shall assume that P_{0} and P are average signals received at mobile stations located respectively at distances d_{0} = 40m and d = 2000m, it follows that;

P_{0} = P (d_{0}/d)^{-α}

Expressed in dB, the equation can be written as

P_{0dB} = P_{dB} -10α log_{10} (d_{0}/d)

However, since there is no mention of deterministic path loss in the question, shadowing effect is of paramount focus.

The probability density function of the attenuation in dB is;

*P* (X_{dB}) = 1/ (√2πσ_{dB}) exp (-*x*^{2}_{dB})/ (2σ^{2}_{dB})

Therefore the probability that the attenuation exceeds 145dB is:

*Pr* (X_{dB} > 145dB) = ƒ_{145}^{α} 1/ (√2πσ_{dB}) exp ((-*x*^{2}_{dB})/ (2σ^{2}_{dB})) d*x*_{dB}

Remember X is the path loss due to shadowing.

Path loss dB (d = 2000m) = path loss dB (d_{0} = 40m) – 40αlog 40 (40/2000)

= 6 – 140 ×0.03204

= 6 – 4.4856 = 1.5144dB

Hence; *Pr* (1.5144dB > 145dB) = ƒ_{145}^{α} 1/ (√2πσ_{dB}) exp ((-*x*^{2}_{dB})/ (2σ^{2}_{dB})) d*x*_{dB}

Let λ = *X*dB/√2σdB; then:

*Pr* (1.5144dB > 145dB) = ƒ^{α}_{145}/ (√2σdB) * 1/√π exp (-λ^{2}) dλ

= ½ erfc (145/√2σdB)

In this case, the erfc is the universally known complementary error function:

This function however cannot be calculated analytically. Numerical solutions for this are found through the table available in the Rappaport book (table D.2 that provides all solutions to the erfc for factor *X =1- erf (x)*).

Using this table, it is found out that;

*Pr* (1.5144dB > 145dB) = 0.098479

In the case of 1.5GHz

Number 2;

LmediandB (urban) = 69.55+ 26.16 log fc −13.82 log ht −a(hr) + (44.9−6.55 log ht) logd

Fc = 1.5GHz= 1500MHz

Ht = 40m; a(hr) is negligible

D = 0.04 km

Hence, SIR = 69.55 + 26.16 log 1500 – 13.82 log 40 + (44.9 -6.55 log 40) log 0.04

= 69.55 + (26.16* 3.17609) — 22.14 + 34.41log 0.04

= 152.63651 — 48.10

SIR = 104.53651

Number 3;

In this case, we shall assume that P_{0} and P are average signals received at mobile stations located respectively at distances d_{0} = 40m and d = 2000m, it follows that;

P_{0} = P (d_{0}/d)^{-α}

Expressed in dB, the equation can be written as

P_{0dB} = P_{dB} -10α log_{10} (d_{0}/d)

However, since there is no mention of deterministic path loss in the question, shadowing effect is of paramount focus.

The probability density function of the attenuation in dB is;

*P* (X_{dB}) = 1/ (√2πσ_{dB}) exp (-*x*^{2}_{dB})/ (2σ^{2}_{dB})

Therefore the probability that the attenuation exceeds 145dB is:

*Pr* (X_{dB} > 1500dB) = ƒ_{1500}^{α} 1/ (√2πσ_{dB}) exp ((-*x*^{2}_{dB})/ (2σ^{2}_{dB})) d*x*_{dB}

Remember X is the path loss due to shadowing.

Path loss dB (d = 2000m) = path loss dB (d_{0} = 40m) – 40αlog 40 (40/2000)

= 6 – 140 ×0.03204

= 6 – 4.4856 = 1.5144dB

Hence; *Pr* (1.5144dB > 1500dB) = ƒ_{1500}^{α} 1/ (√2πσ_{dB}) exp ((-*x*^{2}_{dB})/ (2σ^{2}_{dB})) d*x*_{dB}

Let λ = *X*dB/√2σdB; then:

*Pr* (1.5144dB > 1500dB) = ƒ^{α}_{1500}/ (√2σdB) * 1/√π exp (-λ^{2}) dλ

= ½ erfc (1500/√2σdB)

In this case, the erfc is the universally known complementary error function:

This function however cannot be calculated analytically. Numerical solutions for this are found through the table available in the Rappaport book (table D.2 that provides all solutions to the erfc for factor *X =1- erf (x)*).

Using this table, it is found out that;

*Pr* (1.5144dB > 1500dB) = 1.01875

At the moment the base station antenna height decreases, the Sir decreases following larger capacity but potentially high interference resulting from decreased distance. This on the contrary would increase the probability because of the high interference from the path loss.

**Question 2**

In this case, the formula of calculating traffic intensity of a cell is applied.

Therefore Erlang B formula used is;

A = *U×*A_{u} where A is the traffic intensity of a cell, U is the number of users and A_{u} is the traffic intensity per user.

But A_{u} = λ × H in which λ is the average number of call requests per unit time per user; H is the average duration of a call.

Hence, A_{u} = (1 × 3)/60 = 0.05 Erlangs

A = U × 0.05

22.4 = 0.05U

U = 22.4/0.05 = 448 users.

However, P_{blocking} (A, N) = (A^{N}/N)/ (Σ^{N}_{i=0} A^{i}/i)

A= total offered traffic density

N = number of trunked channels

Users still remain 448.

From the online calculator, 5% of blocking probability and 22.4 channels available, the corresponding traffic intensity A is 896. A_{u }was initially found to be 0.05 Erlangs.

Therefore, U = 896/ 0.05 = 17,920 users.

Note that K = G_{p} (I_{0}/E_{b})

But K = number of users

G_{p} = processing gain.

Given E_{b}/I_{0} = 6dB; it means that I_{0}/E_{b} = 1/6dB

And therefore 448 = G_{p}/6dB

G_{p} = 448 × 6 = 2688dB

In this case, interference realized is 50/10 = 5. Therefore power at the station can be computed as follows:

I = (K-1) Ps

P_{s} = I/ (K-1) = 5/ 447 = 0.01119watts

Hence the total number of users acceptable K = (W/R)/ (E_{b}/I_{0})

447 = (W/R)/ 6

W/R = G_{p} = 447 × 6 = 2682 which is the ultimate processing gain.

First, the formula applied in this case is;

K = G_{α}G_{v }((W/R)/ (E_{b}/I_{0})

K = maximum number of users in a single cell

G_{α} = sectoring factor = 2.4

G_{v} = stop transmission rate = 2.67

W = 22.4 MHz

R = 9600b/s

E_{b}/I_{0} = 6dB

Hence K = 2.4 × 2.67 × (22.4/9600)/6

= 9600/ (0.6 * 2.67* 22.4)

= 272 users

**References:**

Dahlman, E. Parkvall, S. & Skold, J. (2011). *4G LTE/LTE-Advanced for Mobile Broadband*. London: Academic Press.

Dahlman, E. (2008). *3G evolution: HSPA and LTE for mobile broadband*. London: Academic Press.

Schwartz, M. (2005). *Mobile wireless communications*. Cambridge: Cambridge University Press.

Kuhn, V. (2006). Wireless communications over MIMO channels: applications to CDMA and multiple antenna systems. California: John Wiley and Sons.

Gow, G. & Smith, R. (2006). *Mobile and wireless communications: an introduction*. New York: McGraw-Hill International.

Vaughan, R. et al. (2003). *Channels, propagation and antennas for mobile communications*. Sidney: IET.

Rhee, M. (2009). *Mobile communication systems and security*. California: John Wiley and Sons.

Singal. (2011). *Wireless Communications*. New York: Tata McGraw-Hill Education