1. Mobile Computing COE 446 Wireless Multiple Access
Tarek Sheltami
KFUPM
CCSE
COE
Principles of Wireless Networks
K. Pahlavan and P. Krishnamurth
3/31/2017

2. Outline Comparison between FDMA, TDMA and CDMA Format complexity
System capacity
Handoff
Power control
Traffic Engineering
Erlang B (blockage probability)
Eralng C (delay probability)
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3. Comparison of CDMA, TDMA and FDMA
Format Flexibility
The chief advantage of CDMA relative to TDMA is its flexibility in timing and quality of transmission
CDMA users are separated by their codes, unaffected by the transmission time relative to other users
Each user is far more liberated from other users
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4. Comparison of CDMA, TDMA and FDMA..
Performance of Multipath Fading:
Multipath in wireless channels causes frequency selective fading
If BW is narrowband system coincides with the location of fade, no useful signal is received
As we increase the transmission BW, fading occupy only portion of the transmission band, providing an opportunity for a wideband receiver to take advantage of the portion of the transmission band not under fade and a more reliable communication link
The wider the BW, the better is the opportunity for averaging out the faded frequency
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6. Comparison of CDMA, TDMA and FDMA..
Performance of Multipath Fading:
Sectored antenna and spread spectrum were not used in 1G FDMA systems, because they were analog systems
System Capacity:
Depends on a number issues:
The frequency reuse factor
Speech coding rate
Type of antenna
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9. FDMA
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11. TDMA
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13. 16.7 users per cell
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14. Comparison of CDMA, TDMA and FDMA..
Handoff:
Occurs when a received signal in an MT becomes weak and another BS can provide a stronger signal to the MT
The 1G FDMA cellular systems often used the so-called hard decision handoff in which BS controller monitors the received signal from BS and at the appropriate time switches the connection from BS to another
TDMA systems used the so-called mobile-assisted handoff in which the MT monitors the received signal from available BSs and reports it to the BS controller, which then makes the decision on the handoff
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15. Comparison of CDMA, TDMA and FDMA..
Handoff:
Because adjacent cells in both FDMA and TDMA are using different frequencies, the MT has to disconnect from and reconnect to the network, which will appear as a click to the user
Signals fluctuate anyway because they are arriving over radio channels
At the edge of the cells, users experience a period of poor signal quality and possibly several clicks during the completion of the handoff process
Because adjacent cells in a CDMA network use the same frequency, an MT moving from one cell to another can make seamless handoff by the use of the signal combining
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16. Comparison of CDMA, TDMA and FDMA..
Handoff:
When an MT approaches the boundary between cells, it communicates with both
A controller combines the signals from both links to form a better communication link
When a reliable link has been established with new BS, the mobile stops communicating with the previous BS and communication is fully established with the new BS (soft handoff)
Soft handoff provides a dual diversity for the received signal from two links, which improves the quality of reception and eliminates clicking as well as the ping-pong problem
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17. Comparison of CDMA, TDMA and FDMA..
Power Control:
Power control is necessary for FDMA and TDMA systems to control adjacent channel interference and mitigate the unexpected interference caused by the near-far-problem
In FDMA and TDMA some sort of power control is needed to improve the quality of voice delivered to the user
In CDMA, the capacity of the system depends directly on power control and an accurate power control mechanism is needed for proper operation of the network
CDMA systems adjust the transmitted power more often and with smaller adjustment steps to support a more refined control of power
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18. Comparison of CDMA, TDMA and FDMA..
Power Control:
Better power control saves on transmitted power of the MT, which increases the life of the battery
The more refined power control in CDMA system, the more power management of the MTs, which is important practical issue for users of the MTs
Implementation Complexity:
Spread spectrum requires greater circuit complexity than conventional modulation
This leads to higher electronic power consumption and larger weight and cost MTs
Gradual improvement in battery and IC technologies have made it transparent to the user
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19. Performance of Fix-assignment Access Methods
In wired network two types of blockage:
When the calling number is not available
When the telephone company out of resources
In most cellular systems:
Blockage results in a response that is busy tone
A message “All circuits are busy….”
We will refer to # 2 in the rest of the course
Telephone service providers design their network so that the blockage rate at the peak traffic is always below a certain percentage
Cellular operators try to keep this average below 2%
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20. Performance of Fix-assignment Access Methods..
The blockage rate is a function of the number of subscribers, number of initiated calls and length of conversation
In cellular networks, the number of subscribers operation is a cell is a function of time
In downtown areas, everyone uses their cell phones during the day and in the evenings they use them in their residential area which is covered by a different cell
Traffic fluctuations in cellular telephone networks
To cope with the limitations, cellular operators use complex frequency assignment strategies to share the available resources in the optimal manner
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21. Traffic Engineering Using the Erlang Equations
Erlang EQ. is the core in traffic engineering for telephone application
Two basic equations are used for traffic engineering (Erlang B and Erlang C)
Erlang B relates to the probability of blockage B(N, ρ) to the number of channels N, and the normalized call density in unit channels ρ
Where ρ = λ/μ, λ is Poisson arrival rate and μ is the service rate of the calls
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22. Example 1 (Capacity using Erlang B formula)
We want to provide a wireless public phone service with five lines to a ferry crossing between islands carrying 100 passengers where on average each user makes a three-minute telephone call every two hours. What is the probability of a passenger approaching the telephones and none of the five lines are available? Is it acceptable for cellular network?
Solution:
From the above problem we have the following data:
Number of channels = 5 channels
Number of users = 100 passengers
Average number of calls/user = 1 call
The total duration = 120 minutes
The average duration/call = 3 minutes
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23. 55
B(5, 2.5)=0.07
0.02
2.5 3
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24. Example 2 (Capacity using Erlang B formula)
An IS-136 cellular phone owns 50 cell sites and 19 traffic per cell each with bandwidth of 30 kHz. Assuming each user makes three calls per hour and the average holding time per call is 5 minutes, determine the total number of subscribers that the service provider can support with a blocking probability of 2%.
Solution:
From the above problem we have the following data:
Number of channels = 19 * 3 = 57 channels/cell
Average number of calls/user = 3 call
The total duration = 60 minutes
The average duration/call = 5 minutes
From the chart:
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25. 3/31/2017

26. 55
N=57
0.02
Back
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27. Traffic Engineering Using the Erlang Equations..
The Erlang C is formula results in waiting time in a queue if a cell does not get through, but it is buffered until the channel is available
The probability that a call gets delayed:
Because of the complexity of the calculation, tables and graphs are used to provide values for the probability based on normalized values of ρ
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28. Traffic Engineering Using the Erlang Equations..
The probability of having a delay that is more than a time t is given by:
This indicates the exponential of the delay time
The average delay is give by the average of the exponential distribution:
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29. Example (Call Delay using Erlang C formula)
For the ferry described in example#1, solve the following questions:
What is the average delay for a passenger to get access to the telephone?
What is the probability of having a passenger waiting more than a minute to access the telephone?
Solution:
From the above problem we have the following data:
Number of channels = 5 channels
Normalized traffic density ρ = 2.5
The average duration/call = 3 minutes  μ = 1/3 min-1
From the chart:
P[Delay > 0] = 0.13
a) Using equation (2):
b) Using equation (1):
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30. 0.13
2.5
Back
4/29/2008