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Unformatted text preview: 34 Problems and Solutions (Chapter 5) 1. An octagon shaped cell is closer to a circle as compared to a hexagon.
Explain why such a cell is not used as an ideal shape of the cell? [Solution] Hexagonal modeling of cells achieves complete packing of space without
any overlap or gaps between cells. Though octagons (or any polygon of
sides 7 or more) are closer in shape to circles than hexagons, they cannot
be packed without causing either overlap or gaps or both. 2. A new wireless service provider decided to employ a cluster of 19 cells as
the basic module for frequency reuse. (a) Can you identify one such cluster structure? (b) Repeat (a) for N = 28. (c) Can you get an alternative cluster structure for part (a)? (d) What is the reuse distance for the system of part (c)? (e) Can you ﬁnd the worst case cochannel interference in such a. system? [Solution] (a)19cellclusteri=3,j=2=>Nm4+9+6=19 (b) 28cellcluster,i=4,j=2=>N=16+4+8=28 (c) An alternative cluster will turn out to be the same, but seen from a
diﬁ'erent angle, i.e., with the shift parameters (i, j) interchanged. It
will have the same reuse distance as the 19cell cluster pattern shown
above. (d) Reuse distance D = «3m: = Jim = 7.5512 (e) Let the reuse distance D and the radius of the cell R. T = (2(9— W + 2e)": +2(q+ 1)~'r)’
where q = 5% = JET? = 7.55 (for N = 19). 3. Two adjacent B83 3' and j are 30 Inns apart. The signal strength received
by the MS is given by the following expressions. GIGIR
L(I) P(z) = where L(a:) = 69.55 + 26.1610gm f¢(MHz) — 13.8210gm hb(m) — a[hm (m)]
+ [44.9 —_6.55 Iogm hi, (m)] log10(x), and z is the distance of the MS from BS i. Assumelunity gain for G, and
G", given that P.(t) = 10 Watts, = 100 Watts, f, = 300 MHz, hr = 40 m,
hm = 4 m, a = 3.5, a: = 1 km, and Jig(1013 the transmission power ofBS
j. (a) What is the power transmitted by BS 1', so that the MS receives
signals of equal strength at z? (b) If the threshold value E = 1 dB and the distance where handoﬁ is likely to occur is 2 km from BS j, then what is the power transmitted
by BS 3‘? ' [Solution] (5) Assurne that the distance of the MS from base station i is 1:. Since
Gt: ,=1. SincengGT=1, Pi P5
m7 = M30 — z) (I) 36 P. = 10 10800) = 10 dB
By substituting data given in L (x), we get L(z) = 69.55 + 64.801 — 22.140  14 + 34.40610g(:1:)
= 98.211 L(30 — z) = 98.211 + 3440610380 — x)
= 98.211 + 50.315
= 148.526 Substituting in Equation (1), we have 10 _ P;
98.211 ‘ 148.586' Solving for 135, we get
P} = 15.123 dB = 32.512 Watts (b) Assume that the distance of the MS from the BS 8' is z. and if E = 1
(113 then the point of hand off will be when Pm(i) + 11133 = Pm(j)
Pmﬁ) = 10 logm 10 = 10 dB
Similar to (a), we get L(:c) = 69.55 + 64.801 — 22.140 — 14 + 34.40610gm(1:)
= 98.211 + 34.40610gm(x)
= 98.211 + 34.40610g10 2
= 108.5682 L(30  x) = 98.211 + 34.406 logm(30 — 1:)
= 98.211 + 34.406 logm (28)
= 148.0019. Using Equation (1), We get (10 + 1) 4: 148.0019 = 108.5682 11 P...(j) Thus,
' ' Pm(j) = 14.995 dB = 31.589 Watts. 37 4. If each user keeps a traﬁc channel busy for an average of 5% time and an
average of 60 requests per hourm or generated, what is the Erleng value? ' [Solution] The request rate r = 60/3600 = 1/60 requests/sec
Holding time = 0.05 x 3600 = 180 sec
Therefore the offered traﬂic load in Erlangs is =request rate x holding
time = (1/60) x 180 = 3 Erlangs. — 5. Prove that D = RV 3N.
[Solution] To prove that D = R‘/(3N)
Let R be the radius of the cell and D the reuse distance. C1
D Ad
I (21")
1 (2le C2 where i and j are the number of cells in the corresponding directions. In
the ﬁgure as we traverse from ClCZCI, we ﬁrst move 1 cell in the i
direction, from CI to CZ and then 2 cells in the 3' direction from C2 to
Cl. _ ' We have R1 =Rcos30=‘/?§R (2) The distance D can be found using the casino law, 132 = (312121))2 + (312125)? — 2i(2R1)j(2R1) cos 120
= (i2 + j2)(2R1)2 + £31231)?
= (2R1)2(i2+j2+ij)
= (2R1)2(N).
where N is the number of cells in the cluster.
Using Equation (2) described above, we get
DZ=3N122=>D=J§NR 38 (a) _ 6. Prove that N = i2 +_ji2 + ij.
[Solution]
Using the basic analysis in Problem 5.5. and the result obtained for D,
we get
D” = («21:39)2 + (j(2R1))2 — 259353931) cos 120
= (i2 +j2)(2R1)2 + £21231)” = (2R1)2(£2 + 1'2 + £1”). Therefore, 31% 122 = (mir’a2 +j2 + £1)
However, (23‘)2 = 3122 Hence, mam? =3*R’(i2+j2 +£j)
Therefore, N=W+ﬂ+m 7. The size and shape of each cluster in a cellular, need to be designed care
fully so as to cover adiacent spoke in a nonoverlapped manner. Deﬁne
such pettens for the following eluster sizes: (a) 4cell
(b) 90811
(c) 13oell
(d) 37cell [Solution] (b) 39 (d) 8. A cellular scheme employed a. cluster of size 16 cells. Later on, it was
decided to use two diﬁemnt clusters of size 7 and 9 cells. Is it possible to
replace each original cluster by two new clusters? Explain clearly. [Solution] A 16oell cluster cannot be substituted by two clusters of 7 and 9 cells
each, as the resultant conﬁguration of cells will not be as closely packed as
the original 16oell cluster and therefore will have gaps. This means that
the original area covered by the 16cell cluster, may not be fully covered
in the new conﬁguration. It is also pessible that certain areas not covered
earlier will be covered now, which is not acceptable. 9. For the following cell pattern, Figure for problem 5.9 41 (a) Find the reuse distance if radius of each cell is 2 lrms. (b) If each channel is multiplexed among 8 users, how many calls can be
simultaneously processed by each cell if only 10 channels per cell are
reserved for control, assuming a total bandwidth of 30 MHz available
and each simplex channel of 25 kHz ? (c) If each user keeps a traﬂic channel busy for an average of 5% time
and an average of 60 requests per hour are generated, what is the
Erlang value “.7 [Solution] (9.) D = V3NR=> D = 2 *(31: 12)”
' The reuse distance = 12 kms
(13) One duplex channel = 2 (BW of one simplex channel ) = 2 s 25 = 50
kHz
Number of channels = (10%)— 10* 12 = 600— 120 =_4so channels
Numberofchannelspercell=5§=40lcell Total numberofcallspercell=8=r40=320 calls/cell (c) The request rate a = 5% = 6—10 requats/second Holding time = 5% = .05 * 3600 = 5 * 36 = 180 seconds
Therefore the oﬁ‘ered trafﬁc load in Erlangs is
a. = request rate a holding time = (316) as 180 = 3 Erlangs. 10. A TDMAbased system shown in the Figure, has a total bandwidth of
12.5 MHz and contains 20 control channels with equal channel Spacing of
30 kHz. Here, the area of each cell is equal to 8 km” and cells required to
cover a total area of 3600 km”. Calculate the following: (a) Number of trafﬁc dunnels/cell
(b) Reuse distance Figure of Problem 5.10.
I [Solution] (a) 6
12.5)(10 20 £133; A“. 44 trafﬁc channels/cell 42 (b) D=_\/3WR =v3x9x4
=9.12Km 11. During a busy hour, the number of calls per hour for sari of the 12 cells of
a cellular cluster is 2220, 1900, 4000, 1100, 1000, 1200, 1800, 2100, 2000,
1580, 1800 and 900. Assume that 75% of the car phones in this cluster
are used during this period and that one call is made per phone. (3.) Find the number of customers in the system. (b) Assuming the average hold time of 60 seconds, What is the total
Erlang value of the system ? (c) Find the reuse distance D if R = 5 kms.
[Solution] (9.) Number of customers in the system = (2220 + 1900 + 4000 + 1100 +
1000 + 1200 + 1300 + 2100 + 2000 + 1580 + 1800 + 900)/0.75 = 16200 (b) Avasge number of calls in each cell = (2220 +1900 +4000 + 1100 +
1000 + 1200 + 1800 + 2100 + 2000 + 1580 + 1800 + 900) / 12 = 1800 Holding time = 60 seconds Request rate a = :15—g%] = 0.5 calls/second Therefore the offered traﬂic load in Erlangs is a = request rate us holding time = 0.5 at 60 = 30 Erlangs (c) R=5km
Reuse distance=¢3NR=v3=k12s5=301an 12. Given a bandwidth of 25 MHz and a. frequency reuse factor of 1 and RF
channel size of 1.25 MHz and 38 calls per RF channel, ﬁnd: (a) The number of RF channels for CDMA.
(b) The number of permissible calls per cell (CDMA). [Solution] (:1) The number of RF channels = 1%; = 20 channels (b) The number of permissible calls per cell = 20 * 38 = 760 calls per
cell. 13. _ 14. 15. 43 If a wireless service provider has 20 cells to cover the whole service area,
with each cell having 40 channels, how many users can the provider sup port if a. blocking probability p of 2 % is required? Assume that each user
makes an average of three calls/hour and each call duration is an average
of three minutes. (Erlang B values are given in the Appendix A) [Solution]
For 40 channels and a blocking probability of 2% (i.e., 0.02), we get an A = 30.997 Erlangs (from the Erlang B table)
number of calls per hour A=( timeinseconds )akholdmg timemseconds
number of customers It 3 n In 3
= (——————)*3*60— 20 — 30.997 solving for n, we get, 1: = 206.67 customers per cell Total number of users in the system = 206.674: 20 = 4133 users (approx) The following ﬁgure shows a cellular architecture. Is there some speciﬁc
reason why it could have been designed this way ? Figure 5. problem 5.14 [Solution] The cellular architecture is designed in a way to cater to increased traﬂic
in some areas. Smaller cells can handle a larger number of users per
square kilometer as the number of allocated channels to each cell is the
same. In addition, interference among cells using the same frequencies is
minimized. This way there is a. greater frequency reuse and hence greater
capacity within the limited bandwidth provided. The following ﬁgure shows the cell structure of a. metro area. Can you , explain Why this might have been designed so? 16. 4 4
4 s 4
s 4 a 4
4 ‘ 4
a 4 12 4 a
4 12 1i 4 4 4 12 ‘l
" 4 ‘12‘2ii‘21‘2‘ 2 4
4 20 20 4 4
5 2 12 20 12 12 a
1 4 2a 29 4
4 412122012124 4 4
a ‘ 12 12 4 a
4 12 4
4 ‘ 4
4 a , a
4
a
4 4
'4 Figure for problem 5.15 [Solution] The smaller cells and the greater number of channels assigned to the cells
at the center of the cluster indicate regions of very high traﬂic (like the
business district of a large metropolis). This provides better service in
terms of coverage and capacity, in an area where the greatest traﬂic is
expected to be generated. Moving away from the business areas, out
wards towards the suburbs, the number of channels available per cell also
decreases, in expectation with the lesser number of calls made from the
residential areas of a city. It is also possible that the inner cells may be
activated/deactivated based on dynamic conditions like traﬂic or time of
the day. This is a way of using limited resources frequency. Prove the following for a hexagonal cellular system with radius R, reuse
distance D and given the value of N: (a) N = 3, prove D = 3R.
(b) N = 4, prove D = mil.
(c) N = 7, prove D = \f2_1R. [Solution]
Using reuse distance D: R4/3N (a) N=3, D=\/3*3R=3R
(b) N24, D=¢3*4R=¢ER
(c) N=7, D=v3s7R=x/2_1R _ 45 17. In Figure 5.14, calculate the cochannel interference ratio in the worst case
for the forward channel, given N = 7, R = 3 ions, and 7 = 2.  [Solution] D = V3NR
q=%=—3RN'—Ii =V3N=V3$7=4.58
2'— 1
I _ 2(q 1f" +2(9)“’ + 2(q+ 1)‘"’
=3.17 18. What is meant by handoﬂ' interval and handoﬁ' region? Explain their
usefulness with appropriate diagams. [Solution]
Handoﬂ" interval is the time duration required for completing a handofi’. Handoﬂ" region is region where the mobile station requests the base station
for handcifs. 19. What are the diﬂ'erences between adjacent channel interference and cochan—
nel interference? Explain with suitable diagrams.
[Solution]
Adjacent channel interference is caused when adjacent channels affect the
signal in the current channel due to side bands. Clochannel interference is the interference caused due to ongoing transmissions in different cells
which are using the same channel. ' 20. What are the advantages of cellsectoring? Explain with suitable dia grams.
[Solution] Cell sectoring helps in reducing cochannel interference and improve spa
tial reuse of the frequencies. ...
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 Cellular network, Erlangs, reuse distance

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