Tutorial (4) Problem (1):

Transcription

1 Problem (1): Tutorial (4) Consider the WLAN supported within the campus of the GUC. Assume that an access point is located in the second floor. The transmitter power at the access point is 5 dbm. Free space pathloss may be assumed up to a distance of 1.5 m away from the transmitter such that power received at a distance of 1.5m could be evaluated as dbm. Afterwards a path-loss exponent of nsf (same floor) could be assumed. i. We would like to estimate the path-loss exponent nsf (same floor) if the signal strength received at a laptop in an office within the same floor that is 17 m away is -90 dbm. Assume from the access point, the LOS path is obstructed by three walls. (Assume PAF of 5 db for the first two walls and 3 db for the third wall) P R (d) = P R (d o ) 10n SF log ( d d o ) PAF 90 = n SF log ( 17 ) ( ) 1.5 n SF = 3.33 ii. Estimate the signal strength at a laptop in the cafeteria area (C1, ground floor) 6 m away from the access point and the LOS path is obstructed by two walls. (Assume PAF of 4 db for each wall and FAF of 12 db for two floors) P R (d) = P R (d 0 ) 10n SF log ( d d 0 ) PAF FAF P R (d) = log( 6 ) 12 = dbm 1.5 iii. What is the multi-floor pathloss exponent that would achieve the same received power as in part (ii) P R (d) = P R (d 0 ) 10n MF log(d/d 0 ) PAF = n MF log(4) n MF = 5.3 iv. What is the carrier frequency for the WLAN system used P L (d 0 ) follows free space = P T P R (d 0 ) = = 41.5dB

2 P L (d 0 ) = 20log( 4Πd 0 λ ) = 41.5 (Assume G T = G R = 1) 4Πd 0 λ = λ = fc = 1.91 GHz N.B.: P L (d 0 ) = 10log (( 4Πd 0 λ )2 ), We can use this formula because free space pathloss may be assumed up to a distance of 1.5 m. Problem (2): If the received power at a reference distance do = 1km is equal to 1mW, find the received power at a distance of 2km from the same transmitter for the following path loss models: (Assume f = 100MHz, ht = 40m, hr = 3m, Gt = Gr = 0dB) i. n=4 P R (d) = P R (d o ) 10nlog ( d ) = 10 log(1) 10(4) log( 2 ) = dbm d o 1 ii. Hata model for a large city environment P R (d = 2 km) = P T L 50 (d = 2 km) P T = P R (d o ) + L 50 (d = 1 km) P R (d o ) = 10 log(1) = 0 dbm f c > 1500 MHz Extended Hata will be used L 50 (Urban) = logf c 13.2logh te a(h re ) + ( logh te ) log d + C M For Large City: a(h re ) = 3.2(log11.75h re ) = 3.2(log ) = 2.69 db C M = 3dB L 50 (d = 1 km) = log log ( log40) log = db

3 L 50 (d = 2 km) = log log ( log40) log = db P T = = dbm P R (d = 2 km) = = dbm Problem (3): If the average received power (P R (d) = PT PL(d) ) is -57dBm at d=2km taking into account the shadowing effect having a Gaussian Distribution with zero mean and standard deviation = 6dB. Find the percentage of area within a 2km radius cell that receives signals greater than a value -60dBm. P r (P R (d = 2km) > 60) = P r (P R (d) X > 60) = P r ( 57 X > 60) = P r (X < 3) = 1 P r (X > 3) = 1 Q ( 3 ) = 1 Q (1 ) = = Assume n=2 n = 6 2 = 3 Percentage of Area=9 %

4 Problem (4): Consider four-cell frequency reuse. Cell B1 is the desired cell and B2 is a co-channel cell as shown in the figure below. For a mobile located in cell B1, find the cell radius R to give coverage percentage of 95% inside the cell. Coverage is considered to be satisfied when the Signal to Interference ratio (SIR) at the mobile is greater than 1 db. Assume the following: - Co-channel interference is due to base B2 only - Carrier frequency, fc=90 MHz - Reference distance d0=1 Km (assume free space propagation from the transmitter to d0) - Transmitter power available at any base station, Pt=10 W PL(dB) between the mobile and base B1 is given as P L (db) = P d L (d o ) + 25log ( ) X d = db o PL(dB) between the mobile and base B2 is given as P L (db) = P d L (d o ) + 40log ( ) X d = 0dB o n = 2.5 = 3.2 N.B.: We used this n from the first equation because we are concerned with the coverage inside cell B1, so we must use the path-loss exponent of this cell (B1). U(γ) = 0.95

5 P r (SIR > 1) = 0. SIR = P S P I > 1 db We will design on the worst SIR, which occurs when the mobile is on the edge of B1 (Furthest point from B1 and nearest point to the interferer B2) P S = P R (R) = P T P L (R) = P T (P R L (d o ) + 25log ( ) X d ) o P I = P R (D) = P T P L (D) = P T (P D L (d o ) + 40log ( ) X d ) o P I = P T (P D L (d o ) + 40log ( )) (N. B. = 0) d o SIR = P S P I = P T (P R L (d o ) + 25log ( ) X d ) (P T (P D L (d o ) + 10(4)log ( ))) o d o = 25 log(r) + 25 log(d o ) + 40 log(d) 40 log(d o ) + X = 25 log(r) 15 log(d o ) + 40 log(d) + X BS1 2.5 R D BS2 3 2 R D = (2.5R) 2 + ( R) = R

6 SIR = 25 log(r) + 40 log(2.646 R) 15 log(d o ) + X = 15 log(r) X P r (SIR > 1) = P r (15 log(r) X > 1) = P r (X > 15 log(r) ) 15 log(r) = Q ( ) = 0. Since, the value of Q(z) is greater than 0.5, therefore, z is a negative number, because Q(+ve number) < 0.5, Q(0) = 0.5 and Q( ve number) > 0.5 Therefore, Q(the positive number) = 1 0. = log(r) Q ( ) = 0.2 From the Q-Table Q(0.) = log(r) R = 3.16 Km = 0. Problem (5): Suppose a mobile station is located in network comprised of three base stations BS1, BS2 and BS3, as shown in the next figure. The distance between any two base stations is D=1600m. The received power in dbm at base station i, from the mobile station, is modeled as: Pr,i(d)dBm = Po 10nlog (di/do) + χi where i=1,2,3 and di is the distance between the mobile and base station i in meters. Po is the received power at distance do from the mobile antenna and n is the path loss exponent. χi is a zero mean Gaussian random variable with standard deviation in db, that models the variation of the received signals due to shadowing. Assume that the random components χi of the signals received at different base stations are independent of each other. The minimum usable signal for acceptable voice quality at the base station receiver

7 is Pr,min and the threshold level for handoff initiation is Pr,HOin dbm. Assume that the mobile is currently connected to BS1. A handoff occurs when the received signal at the base station BS1, from the mobile, drops below threshold Pr,HO, and the signal received at some other base station is greater than the minimum acceptable level Pr,min. Given the parameters in the following table, determine the probability that a handoff occurs (Pr[handoff]) Parameter Value BS 1 MT BS 2 BS 3 n 3 db Po 0dBm do 1m Pr,min -11 dbm Pr,ho -112 dbm Handoff occurs when: Pr,BS1<Pr,HO And (Pr,BS2>Pr,min OR Pr,BS3>Pr,min) P r (handoff) = P r (P r,bs1 < P r,ho ) (P r (P r,bs2 > P r,min ) + P r (P r,bs3 > P r,min )) = P r (P o 10nlog ( d 1 d o ) + χ 1 < 112) (P r (P o 10nlog ( d 2 d o ) + χ 2 > 11) + P r (P o 10nlog ( d 3 d o ) + χ 3 > 11)) = P r (0 10(3)log ( d 1 1 ) + χ 1 < 112) (P r (0 10(3)log ( d 2 1 ) + χ 2 > 11) + P r (0 10(3)log ( d 3 1 ) + χ 3 > 11))

8 d 1 = d 2 = = 00 m d 3 = = m 00 BS MT d3 BS3 BS2 P r (handoff) = P r (0 10(3)log(00) + χ 1 < 112) (P r (0 10(3)log(00) + χ 2 > 11) + P r (0 10(3)log(135.65) + χ 3 > 11)) = P r (χ 1 < ) (P r (χ 2 > ) + P r (χ 3 > 23.75)) = (Q ( = (Q ( )) (1 Q ( ) + 1 Q ( )) )) (2 Q ( ) Q ( )) = (Q(3.113)) (2 Q(3.63) Q(2.969)) (Q(3.1)) (2 Q(3.9) Q(3.0)) = ( ) =