1. Cellular and Wireless Networks System Design Fundamentals
Fundamentals of
Cellular and Wireless Networks
Lecture ID: ET- IDA-113/114
, v13
Prof. W. Adi
Tutorial-1
The Cellular Concept
System Design Fundamentals

2. Cellular Design Fundamentals Summary1 2 3 4 5 6 7 R
Hexagon area ≈ 2,6 R2 D R 7 6 2 7
Cluster size N = i2 + ij + j2
Where i, j are integers 1 6 2 7 1 5 3 6 2 1
Single cluster
Cluster size N=7 4 5 3 7 4 5 3
R: Cell radius 4 6 2 7 D
Frequency reuse factor is 1/N 1 6 7 5 3
Frequency reuse distance D = R  3N 1 6 2 7 4 5 3 1
Co-Channel reuse ratio Q = D/R = 3N 4 5 3 6 2 7 7 4 1 6 2
Co-channel interference ratio S/I:
S/I = R-n /  Di–n
Simplified S/I = Qn / 6 = 1/6 (3N)n/2
=> N = 1/3 (6 S/I)2/n 5 3 1 6 2 4 5 3 1 4 5 3
First interference tier 4
Network cell layout
Approx. W.c. S/I

3. Problem 1.1:
If a total of 33MHz of bandwidth is allocated to a particular FDD cellular telephone system which
uses 25 kHz simplex channels to provide full duplex voice and control channels.
1. Compute the number of channels available per cell for a system cluster size N= 4, 7 and 12 cells
2. Compute the total channel capacity of the network if only 50 base stations are to be installed.
3. Finde the frequency reuse distance D if the if an area of 250 square kilometer is to be covered assuming the use of hexagonal cell shape.
Solution 1.1:
1. Channel bandwidth = 25 kHz x 2 simplex channels= 50 kHz/duplex channel
The total available channels= /50 = 660 channels
For N=4  number of channels/cell = 660/4  165 channels
For N=7  number of channels/cell = 660/7  95 channels
For N=12  number of channels/cell = 660/12  55 channels
For N=4, 165 channels/cell x 50 = 8250 channels frequency reuse factor 1/N= ¼=0.25
For N=7, channels/cell x 50 = 4750 channels frequency reuse factor 1/N= 1/7=0.14
For N=12, 55 channels/cell x 50 = 2750 channels frequency reuse factor 1/N= 1/12=0.08
Every base station covers 250/50 = 5 square kilometers  2.6 R2 = 5 => R = 1.39 km
Frequency resuse distance for N= 4  D = R  3N = 1.39  3x4 = 4.78 km
Frequency resuse distance for N= 7  D = R  3N = 1.39  3x7 = 6.32 km
Frequency resuse distance for N= 12  D = R  3N = 1.39  3x12 = 8.28 km

4. Problem 1.2:
What is the number of RF channels per cell in the GSM network for the frequencyspectrum allocation as shown below
Compute the cluster size N for GSM system if the required signal to interference ratio S/I is 14 dB and the path loss exponent n= 4.
What is the number of channels per cell in GSM under the given S/I condition?
What is the number of omnidirectional base stations having 2KM coverage radius for an area of 312 square kilometer. Use the hexagonal cell layout model.
What is the maximum number of simultanious users in the network
What would be the maximum number of simultanious users in the network if the signal to interference ratio S/I is reduced to 10 dB ?
Comput the frequency reuse distance for both S/I cases.
MHz for uplink (reverse) -> 25 MHz bandwidth
MHz for downlink (forward) -> 25 MHz bandwidth
200 KHz per digital channel
8 voice channels per digital channel

5. Solution 1.2: Co-channel interference ratio S/I: S/I = R-n /  Di–n
As shown in the above Figure, we have KHz/200 KHz=125 channels, one of which is reserved for guard band and 124 channels remain for users. On each channel 8 voice channels are carried by using 8 time-multiplexing. Total number of voice channels=124 x 8 = 992
Cluster size for a given S/I is : N = 1/3 (6 S/I)2/n
14 dB  S/I=25.11
N = 1/3 (6 x 25.11)2/4  => N=4
Maximum number of simultaneous users per cell is:
124 (RF channels) x 8 users/channel = 248 voice channels per cell
4 Cells/cluster
3 Cell area is = 2.6 R2 = 2.6 x 22 = 10.4 KM2
Number of required base stations = 312/10.4 = 30 base stations
Maximum number of simultaneous users (channels) = 248 user/cell x 30 cells = 7440 channels
5. 10 dB  S/I=10, N = 1/3 (6 x 10)2/4  2.58 cells/cluster => N=3
124 (RF channels) x 8 users/channel = 330 channels per cell
3 Cells/cluster
Maximum number of system channels = 330 user/cell x 30 cells = 9900 channels
Co-channel distance D = R  3N = 2 KM  3x4 = 6.93 KM for N=4
Co-channel distance D = R  3N = 2 KM  3x3 = KM for N=3
Co-channel interference ratio S/I:
S/I = R-n /  Di–n
Simplified S/I = Qn / 6 = 1/6 (3N)n/2
=> N = 1/3 (6 S/I)2/n

6. Problem 1.3:
What is the number of channels per cell in the AMPS network for the frequencyspectrum allocation as shown below
Compute the cluster size N for AMPS system if the required signal to interference ratio S/I is 18 dB and the path loss exponent n= 4.
What is the number of channels per cell in AMPS under the given S/I condition?
What is the number of omnidirectional base stations having 2KM coverage radius for an area of 312 square kilometer. Use the hexagonal cell layout model.
What is the maximum number of simultanious users in the network
What would be the maximum number of simultanious users in the network if the signal to interference ratio S/I is reduced to 14 dB ?
Comput the frequency reuse distance for both S/I cases.
AMPS Advanced Mobile Phone System Frequency specrum allocation

7. Solution 1.3: Co-channel interference ratio S/I: S/I = R-n /  Di–n
As shown in the Figure, we have 416 channels: 21 of which are control channels and 395 channels remain for users
Cluster size for a given S/I is : N = 1/3 (6 S/I)2/n
18 dB  S/I=63
N = 1/3 (6 x 63)2/4  => N=7
Maximum number of simultaneous users per cell (channels per cell) is:
395 (voice channels)  = 56 channels per cell
7 Cells/cluster
3 Cell area is = 2.6 R2 = 2.6 x 22 = 10.4 KM2
Number of required base stations = 312/10.4 = 30 base stations
Maximum number of simultaneous users = 56 user/cell x 30 cells = 1,680 channels
5. 14 dB <=> S/I=25.11, N = 1/3 (6 x 25.11)2/4  4.09 cells/cluster => N=4
395 (voice channels)  = 99 channels per cell
4 Cells/cluster
Maximum number of system channels = 99 user/cell x 30 cells = 2,970 channels
Co-channel distance D = R  3N = 2 KM  3x7 = 9.16 KM for N=7
Co-channel distance D = R  3N = 2 KM  3x4 = 6.93 KM for N=4
Co-channel interference ratio S/I:
S/I = R-n /  Di–n
Simplified S/I = Qn / 6 = 1/6 (3N)n/2
=> N = (1/3) (6 S/I)2/n

8. Problem 1.4:
find the S/I ratio for the cellular model shown below taking into account first and second tier
interfering cells. Use i=2, j=1 and N=7 cluster size. D2 D1

9. Solution 1.4:
30° D2
D1 cos 30°+ D1
D1 cos 30°+
30°

10. D1 = R  3N =>
D2 = 2 D1 cos 30° = 2 D1 ½  3 = D1  3 =3R  N
D2 =3R  N

11. Problem 1.5: Solution 1.5: Pt R=1390 m P0 = Pt - 45 d0 = 10 m Pr Thus:
Referring to problem 1.1. If the cell radius R in a cellular network is 1390 meter and the mobile device sensitivity is –80 dBm (Pt (dBm)= 10 log10 Pt ) . Find the required transmitter power of a base station assuming that the received power at the distance of 10 meter from the base station antenna is 45 dB less than the transmitted power. (use the simplified propagation model with n=4).
Solution 1.5: Pt
R=1390 m
P0 = Pt - 45
Thus:
Pt = P0 + 45
Pt =
Pt = dBm
Pt (dBm)= 10 log10 Pt (mW)
50.72(dBm) = 10 log10 Pt
Pt= = mw
Pt= watt
d0 = 10 m
P0 =
? dBm Pr
45 dB loss
-80dBm
Pr = P0 ( dr/d0) –n
for n=4
10 log Pr = 10 log P log dr + 40 log d0
Pr(dBm) = P0(dBm) log log 10
-80 dBm = P0(dBm) log
P0(dBm) = - 80 dBm + 40 x
P0(dBm) = - 80 dBm – 40 = 5.72 dBm