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(ELEC343)Fall03_Mid-term1_solutions.pdf
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ELEC 343 fall 03
Mid-term 1
Solution

Question 1
a)

where Pt is the transmit power,
is the PL(d0) in linear scale, d0 = 100m & di = 5km for i = 1,.,6 , dA = 1km




b) No. As the transmit power Pt will be cancelled out in the equation, it does not depend on Pt. Also, Transmit power is the same for the signal part and the interference part. Hence increasing it will not affect the ratio.

c) SNIR will always be less than SIR. Since the SIR is only 2.27 or 3.57dB, SNIR cannot be 10dB regardless of the transmit power level.
If the target is 2dB instead of 10dB, then

SNIR =

where PN = 1.38x10-23x300x200x103x100.25 Watt = 1.472 x 10-15 Watt

We then let PN =
and we have





Then by PN =
,
Pt =

=

= 7.76 x 10-4 Watt
= -31.1dBW



d) SNIR will always be less than SIR. Since the SIR is only 2.27 or 3.57dB, SNIR cannot be 10dB regardless of the transmit power level.
If the target is 2dB instead of 10dB, then make use of the noise figure which equals to 0.5dB instead of 2.5dB.

PN = 1.38x10-23x300x200x103x100.05 Watt = 9.29 x 10-16 Watt
by PN =

Pt =

=

= 4.89 x 10-4 Watt
= -33.1 dBW




Question 2
a) Since , we can easily get when N=9, i=3, j=0.

b) The figure is draw as follows:


c) In the formula , the k indicate the nearest cells from the center cell that use the same frequency channels. Since there is no sectoring or cell splitting involved, the k will equal to 6.


d) Since the GSM system uses 200kHz frequency channels, in either uplink or downlink we will have
channels
and the 9 cells in a cluster will get

channels
since the channel must be assigned in such as way as to minimize the co-channel interferences, we divide the 9MHz band into 45 channels, i.e. f1,f2,f45, and mark the cells in a cluster as G1, G2, G9, then we will assign the channels in such a way:
G1={ f1, f10, f19, f28, f37 }
G2={ f2, f11, f20, f29, f38 }
G3={ f3, f12, f21, f30, f39 }
G4={ f4, f13, f22, f31, f40 }
G5={ f5, f14, f23, f32, f41 }
G6={ f6, f15, f24, f33, f42 }
G7={ f7, f16, f25, f34, f43 }
G8={ f8, f17, f26, f35, f44 }
G9={ f9, f18, f27, f36, f45 }
The same scheme applies to both uplink and downlink channels.



Question 3
a) For the forward link, using high TX power will
1) Increase the SNIR, allows better performance when the user is in heavy shadowing.

2) If the current SNIR is much lower than SIR, increasing the TX power will increase SNIR toward SIR. If the increase is significant, it may allow the redesign of N to increase the capacity.



b) For the reverse link, using more sensitive receiver will allow heavy shadowed mobile to reach the BS if there is very little interference. However, if the interference is the determining factor, it will not increase performance nor capacity.

Question 4




= 48.19
= 16.83 dB

Question 5
a)