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(ELEC343)01Fall_Mid1sol.pdf
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ELEC 343 Wireless Communications
Midterm #1

Solution

November 6, 2001
1. (25%) Base Station (BS) A has six nearest neighbors, B1,,B6, using the same frequency channels. Each of these BSs is 5km away from BS A. The cell radius of BS A is 800m and the cell radii of BSs B1 to B6 are all 1km. Assume that all the BSs transmit at the same power equal to 20dBW and the channel has a PL(100m) = 80dB and a path loss exponent n=3. a) (5%) What is the received power in dBm (not dBW) from the BS A as predicted by the path
loss model for a mobile standing at the edge of the cell of BS A?
Tx Power = 20dBW = 50dBm;
PL(500m) = 80dB + 10*n*log10(800/100) = 80 + 27.1 = 107.1dB
Rx Power = 50dBm C 107.1dB = -57.1dBm

b) (5%) Assuming that this mobile is equal distance (5km) away from all 6 interferers and taking into account interference from these 6 interferers only, what is the total interference power?
PL(3km) = 80dB + 10*n*log10(5000/100) = 80 + 51.0 = 131.0dB
Interference power from 1 interferer = 50dBm C 131.0dB = -81.0dB
Total interference power = 10*log10(6) + (-81.0dB) = -73.2dBm

c) (5%) What are the signal-power-to-interference-power ratio (SIR) and the signal-power-to-noise-and-interference-power ratio (SNIR), assuming a 1MHz transmission at around 27.C (or 300K) and a noise figure of 10dB?
SIR = -57.1dBm C (-72.2dBm) = 15.1dB Total Ambient Noise Power = 10*log10(1.38*10-23*300*106) = -113.8dBm Total Effective Noise Power = -113.8dBm +10dB = -103.8dBm Total Noise and Interference Power = 10*log10(10(-72.2/10)+10(-103.8/10)) = -72.2dBm SNIR = 15.1dB
d) (5%) What is the predicted Eb/N0, if we treat the total interference and noise power spectral density level as N0 and the data rate is 500kbps?
15.1dB =10*log10( EbRb/(N0B)) = Eb/N0 - 3dB
Eb/N0 = 18.1dB
e) (5%) Let the Log-normal Shadowing effect is modeled by XA and XB1 to XB6. If these shadowing effects are independent and each has a standard deviation of 4dB, express the random Eb/N0 , denoted by gb, in dB, taking into account the Shadow fading effect. What is the standard deviation of the SNIR?
gb = Eb/N0 + XA C 10*log10(10^(XB1/10)+ + 10^(XB6/10))
The standard deviation cannot be computed easily, but it will be larger than 4dB.
2. (17 %) A system design engineer from HK went oversea to help a mobile phone operator to deploy a cellular phone network in Chicago. Based on his experience in HK, he estimated the path loss exponent to be 4 and then designed a system using a cluster size N=4 and omni-directional antenna to satisfy a required SIR at 14dB. Unfortunately, after the measurement has completed in Chicago, the path loss exponent turns out to be 3.2.
a) (5%) What is the predicted SIR, based on the assumption that all BSs transmit at the same power level?
n
3.2

(
3N )(
3*4 )
SIR == = 8.88 = 9.5dB.
k 6
b) (8%) The engineer plans to use either a 120.-sector antenna or increase the cluster size to 9 to satisfy the 14dB SIR requireme