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(ELEC343)02Fall_MidTermsol.pdf
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ELEC 343 Wireless Communications
Midterm #1
Fall 2002

Solution
October 29, 2002
1. (28%) Base Station (BS) A has six nearest neighbors, B1,,B6, using the same frequency channels. There is a mobile user (user A) in Cell A connected to the network via BS A. Moreover, user A is 1km away from BS A and 5 km away from B1,, B5, and B6. Considering the forward (or downlink) communication only, we find that the transmit power level of BS A is 20dBm, but due to the size difference of the cells, the transmit power levels of B1,,B6, are 20dBm, 20dBm, 30dBm, 30dBm , 30dBm, and 30dBm, respectively. Assume that the wireless channel has a PL(100m) = 75dB and a path loss exponent n=3. Moreover, assume that the bandwidth is 1MHz, the temperature is 27.C (or 300K) and the mobile user A is using a very good receiver with a noise figure of 0.5dB. For your information, the Boltzman Constant is 1.38*10-23 Watt per Kevin per Hz.
a) (3%) What is the received power level at mobile user A from BS A in dBm as predicted by the path loss model?
d
PrA = Pt -PL(d ) -n10log( )
0 d0

1000
= 20 -75 -30log( )
100
-
=-85 dBm (3.16210 9 mW )
b) (6%) What is the received power level at mobile user A from BS B1,, B5, and B6 in dBm as predicted by the path loss model?
PrB1 =PrB2
5000

= 20 -75 -30log( )
100
-
=-105.97 dBm (2.52910 11 mW )
PrB3 =PrB4 =PrB5 =PrB6
5000

= 30 -75 -30log( )
100
-
=-95.97 dBm (2.52910 10 mW )
c) (9%) What is the signal-power-to-interference-power ratio (SIR)? What is the signal-power-to-noise-and-interference-power ratio (SNIR)? What is the Eb/N0, treating the total interference and noise power spectral density level as N0 if the data rate is 1Mbps?
-23 6
KTB = (1.3810 )(300)(110 )
=-113.83dBm
PNoise = -113.83 + 0.5

-12
= -113.33dBm (4.645 10 mW )
3.16210-9
SIR =
-11 -10
2(2.52910 ) + 4(2.52910 ) = 2.977 = 4.738dB
3.16210-9
SNIR =
-11 -10 -12
2(2.52910 ) + 4(2.52910 ) + 4.64510 = 2.964 = 4.72dB
. P .. B .
Eb r
= .. .... ..
N0 PR
. N +I .. b .
1106

= 4.72 +10log( )
1106
= 4.72dB
d) (5%) What is SNIR if the mobile receiver uses a cheaper low-noise amplifier (LNA) with noise figure of 9dB?
PNoise = -113.83 + 9
-
= -104.83dBm (3.28910 11 mW )
3.16210-9
SNIR =
-11 -10 -11
2(2.52910 ) + 4(2.52910 ) + 3.28910 = 2.887 = 4.61dB
e) (5%) In your opinion, can the mobile user use a cheaper LNA if it can tolerate about 0.3dB degradation in the SNIR? Why?
No. In the above calculation, we are just calculating the mean value of SNIR, without considering shadowing effect. The extra 0.3dB tolerance on SNIR should be used to prevent degradation due to shadowing effect, and hence a cheaper LNA should not be used.
2. (22%) A cellular network operating in Hong Kong uses a cluster size equal to 3 and omni-directional antennas. The SIR performance is found to be acceptable and measurement data shows that the path loss exponent