Wireless Multimedia System
Professor Hsiao-kuang Wu (hsiao@csie.ncu.edu.tw)
Homework 3: (Due April 27, 2005)
1. In the last decades, progress in microelectronics and VLSI technology has fostered the widespread
use of computing and communication applications in portable electronic devices. A new universal
radio interface called Bluetooth has been developed enabling electronic devices to connect and
communicate via short range radio connections. What particular group of companies proposed the
Bluetooth technology? How does Bluetooth radio system coexist with cellular system and Wireless
LAN system? Why dynamic slot allocation scheme could reach better performance? What is the slot
time for Bluetooth system? What is the maximum user rate for Bluetooth? Does CDMA scheme be
applied in one particular piconet? Does CDMA scheme be applied in one scatternet?
(1) Ericsson, Nokia, IBM, Toshiba, and Intel publish the Version 1.0of
the specification in July 1999. Over 1000 companies have signed as
adopters of the technology.
(2) Bluetooth and the 802.11 standards can peacefully coexist because each
was designed for a distinct purpose. The 802.11 standards are
specifically designed for running WLANs that create a network of
computers and connect users to the Internet. Bluetooth, on the other
hand, is designed to allow portable devices to communicate with each
other. However, many agree that Bluetooth can be used in an office or
classroom to connect mobile devices in the same space that a WLAN is
running to create a computer network.
Both WLAN and bluetooth technologies are popular wireless connections
for voice and data communications. the two technologies use the same
frequency spectrum, which must be shared effectively for both
technologies to be used at the same time. An advanced Packet Traffic
Arbitration (PTA) algorithm with voice packet prioritization ensures
seamless collaboration between Bluetooth and WLAN and excellent
voice/audio quality. Adaptive Frequency Hopping (AFH) capability to
actively avoid WLAN frequency channels that are in use.
(3) DSA maximizes the bandwidth utilization and minimizes latency within
a piconet. DSA scheme enables for not only slave-to-slave
communication but also serves as multi-device communication (MDC),
hence emulating a multicast-like communication within the piconet.
(4) A single time slot in the Bluetooth system lasts 625us. It can be though
of as the time it takes to send one packet from one Bluetooth device
to another.
(5) The communication channel can support both data (asynchronous) and
voice (synchronous) communications with a total bandwidth of 1 Mb/sec.
The supported channel configurations are as follows:
Configuration
Max. Data Rate Upstream
Max. Data Rate Downstream
3 Simultaneous Voice
64 kb/sec X 3 channels
64 kb/sec X 3 channels
Symmetric Data
433.9 kb/sec
433.9 kb/sec
Asymmetric Data
723.2 kb/sec or 57.6 kb/sec
57.6 kb/sec or 723.2 kb/sec
Channels
(6) Bluetooth radios communicate using a time division duplex discipline (TDD). Because
more than two devices share the piconet medium, the access technique is TDMA. Thus,
piconet access can be characterized as FH-TDD-TDMA.
(7) Occasionally, two piconets will use the same physical channel during the same
time slot, causing a collision and lost data. However, because this will happen
infrequently, it is readily accommodated with forward error correction and error
detection/ARQ techniques. Thus, a form of CDMA is achieved between devices on different
piconets in the same scatternet; this is referred to as FH-CDMA.
2. (A) If a total of 33 MHz of bandwidth is allocated total particular FDD cellular telephone system
which uses two 25 kHz simplex channels to provide full duplex voice and control channels, compute
the number of channels available per cell if a system uses (1) 3-cell reuse (2) 7-cell reuse. (3) 12-cell
reuse. (B) If a signal to interference ratio of 15 dB is required for satisfactory forward channel
performance of a cellular system, what is the frequency reuse factor (D/R) and cluster size (N) that
should be used for maximum capacity if the path loss exponent is (a) n = 4, (b) n = 3 ? Assume that
there are 6 co-channels cells in the first tier, and all of them are at the same distance from the mobile.
Use suitable approximations. (Candidate Cluster sizes could be 3, 7, 12, 13).
(A)(1)
33MHz
1
  220 (channels)
25kHz  2 3
(2)
33MHz
1
  94.29 (channels)
25kHz  2 7
(3)
33MHz
1
  55 (channels)
25kHz  2 12
R 4
1 D
15dB  10 log(
)  10 log[  ( ) 4 ]
4
6 R
6 D
D
 3.71
(B)(a) R
D
(3.71) 2
 3N , N 
 4.59
R
3
N  7
R 3
1 D
15dB  10 log(
)  10 log[  ( ) 3 ]
3
6 R
6 D
D
 5.7
(b) R
D
(5.7) 2
 3N , N 
 10.83
R
3
 N  12
3. (A) A given radio spectrum (or bandwidth) can be divided into a set of disjoint or non-interfering
radio channel. All such channels can be used simultaneously while maintaining an acceptable received
radio signal. Channel allocation schemes can be divided into a number of different categories
depending on the comparison basis. Describe FCA (Fixed Channel Allocation), DCA (Dynamic
Channel Allocation) and HCA (Hybrid Channel Allocation). (B) Describe what is the meaning of
Handoffs, intra-cell handoff, inter-cell handoff? In order to determine handoff initiations, several
strategies have been proposed: Relative signal strength, Relative signal strength with threshold,
Relative signal strength with hysteresis (plus optionally dwell timer). Explain the approaches.
(A) FCA: assign a number of channels to each cell permanently according
to some reuse pattern.
DCA: all channels are placed in a pool and dynamically assigned to a
new call to satisfy minimum CIR.
HCA: a combination of fixed and dynamic channel allocation schemes,
with the channels divided into fixed and dynamic sets. This means that
each cell is given a fixed number of channels that is exclusively used
by the cell. Request for a channel from the dynamic set is initiated
only when a cell has exhausted using all its channels in the fixed set.
(B) Handoff: change the radio channel due to the radio link degradation
or channel reorder.
Intra-cell handoff: change channel in the same cell.
Inter-cell handoff: user moves to a new cell, so the channel must be
reallocated.
Relative signal strength: always choose the channel with the strongest
signal.
Relative signal strength with threshold: when current signal is below
the receiver threshold and signal from another station is higher than
handoff threshold, handoff.
Relative signal strength with hysteresis: handoff when the signal is
lower than a near station for a period of time.
4. .(a) What are the difference between Active Mode and Power Saving Mode?
(b) How does Power Saving Mode Station know when to wait up to get the TIM (Traffic Indication
Map)?
(c) What is the purpose of TIM (Traffic Indication Map)?
(d) How does Power Saving Mode Station to inform AP that he is ready to accept the buffered data?
(e) For BSS multicast services, how does AP transmit multicast or broadcast to all of the Stations if
at least one of the stations are in power saving mode? (hint DTIM)
(a) active mode : STA 可以在任何時間接收 frame。在 active mode 中，STA
必須隨時在 awake state。
power save mode: STA 每隔一段時間去聽 selected beacons，若 beacon 中
的 TIM element 指示有 MSDU 要傳來，則必須回 PS-Poll frame，在 PS mode
中，STA 大部分時間是在 Doze state, 而在接收 selected beacons、
broadcast or multicast transmission 與傳送時，進入 awake state。
(b) 在 PS mode 的 STA 應該依據 STA 的 ListenInterval 與 ReceiveDTIMs
parameters periodically 去聽 beacon(TIM 包含在 beacon 中)。
(c) TIM 可以告訴 STA，AP 是否有要傳給它的 buffered data。若有，STA 就必
須保持在 awake state，否則它可以回到 dozen state 進入 power save mode
節省 power，直到下一次聽 beacon 的時間再 wake up。
(d) power saving mode station 由 beacon 中知道 AP 有自己的 buffered data，
則由 response AP 一個 PS-Poll frame 告知 AP 他已準備好要接收 buffered
data。
(e) 若有任何一 STA 在 PS mode, AP 會 buffer 所有 broadcast 與 multicast
MSDUs, 直到下一個包含 delivery TIM(DTIM)的 beacon 發送時，緊接著
beacon 把這些 MSDUs 發送給所有的 STA.
5.
(A) Direct Sequence Cellular system adopts universal frequency reuse policy. Explain what is
universal frequency reuse policy? What is the advantage and disadvantage of this policy? (B) If
W=1.25 MHz, R=9600 bps, and a minimum acceptable E/N is found to be 10 dB, determine the
maximum number of users that can be supported in a single-cell CDMA system using
omni-directional base station antennas and voice activity factor α = 3/8, K = 1/2, suppose the
background noise is approximately neglected, where W/R is called the processing gain, (C)
Explain why reverse link is vulnerable to the “near far” problem, and what is solution to maximize
the system capacity? (d) Explain soft handoff and softer handoff?
(A) universal frequency reuse policy: 一個 cell 內所使用的 frequency，
在隔壁 cell 就可以馬上被 reuse，不需要區格多個 cell 之後才再利用。
advantage:

If traffic at the location increases, the introduction of new
cell will be less restricted compared with FDMA or TDMA.

Universal frequency reuse turns out to be beneficially affects
or improves the capacity of the system.

Avoids the complicated issue of frequency-management planning
when additional cells are introduced.

Applies to all users in the assigned spectrum or channel, to all
cell sites, and to all media, terrestrial and satellite-both
geostationary and low earth orbit.

With CDMA, seamless handover between media becomes possible.
 enables the new function soft handover
disadvantage:

Creates additional interference when increase in the effective
receiving angle enlarges overlapping region and causes more
interference to the user.

Near-far problem.
(B)
(C)(i) The goal in CDMA systems is to assure that all mobiles achieve the
same received power levels at the Base Station. The reverse link is
vulnerable to the near-far problem because of the strongest (nearby)
reverse link can capture the intended receiver but weaker (far away)
reverse link get ignored by the receiver. In order to minimize the
Reverse Link Near-Far problem, mobiles that are closer to the Base
Station must transmit less power than mobiles that are further away
from the Base Station. For Example, DS-CDMA is mitigating the problem
by using a combination of open-loop and closed-loop power control.
(ii) CDMA is interference limited. Therefore reducing interference is
also most important in increasing the capacity. The power control
reduced the average mobile transmitted powers as well.
(D) Soft handoff: A Mobile Station is communicating simultaneously with
the prior Base Station as well as the new BS. Such a scheme, just for
some short duration of time, is called soft handoff or make before
break.
Softer handoff: Similar to soft handoff, softer handoff is the soft
handoff between two sectors of the same cell. For a perfect sectorized
antenna pattern, softer handoff cannot be applied. In practice, the
antenna patterns do not fit the sector area perfectly, and there is
an overlapping of the two antenna patterns between the sectors. The
overlapping of the sector antenna patterns generates additional
interference on both the reverse and forward links.
Trunking and Grade of Service:
Example: a radio channel is occupied for thirty minutes during an hour carries 0.5 Erlangs of traffic.
Capacity of an Erlang B System
AC
C! =GOS (Grade of Service)
Pr[blocking]=
c
Ak

k  0 k!
Number of
Channels C
Capacity (Erlangs) for GOS
= 0.01
= 0.005
= 0.002
= 0.001
2
0.153
0.105
0.065
0.046
4
0.869
0.701
0.535
0.439
5
1.36
1.13
0.900
0.762
10
4.46
3.96
3.43
3.09
20
12.0
11.1
10.1
9.41
24
15.3
14.2
13.0
12.2
40
29.0
27.3
25.7
24.5
70
56.1
53.7
51.0
49.2
100
84.1
80.9
77.4
75.2
Problem 6
How many users can be supported for 0.5% blocking probability for the following number of trunked
channels in a blocked calls cleared system? (A) 2, (b) 5, (c) 10, (d) 20, (e) 100. Assume each user
generate 0.1 Erlangs of traffic.
ANS:
(A) 0.105/0.1=1.05 也就是 1 個 user
(b)1.13/0.1=11.3 也就是 11 個 user
(c)3.96/0.1=39.6 也就是 39 個 user
(d)11.1/0.1=111 也就是 111 個 user
(e)80.9/0.1=809 也就是 809 個 user
Problem 7
An urban area has a population of two million residents. Three competing trunked mobile networks
(systems A, B, and C) provide cellular service in this area. System A has 394 cells with 19 channels
each, system B has 98 cells with 57 channels each, and system C has 49 cells, each with 100 channels.
Find the number of users that can be supported at 2% blocking if each user averages two calls per hour
at an average call duration of three minutes. Assuming that all three trunked systems are operated at
maximum capacity, compute the percentage market penetration of each cellular provider.
ANS:
System A
Traffic intensity per user, Au = H = 2 (3/60) = 0.1 Erlangs
from the Erlang B chart, the total carried traffic, A, is obtained as 12 Erlangs.
Therefore, the number of users that can be supported per cell is
U = A/ Au = 12/0.1 = 120
total number of subscribers that can be supported by System A = 120 394 = 47280
System B
Given:
Traffic intensity per user, Au = H = 2 (3/60) = 0.1 Erlangs
from the Erlang B chart, the total carried traffic, A, is obtained as 45 Erlangs.
Therefore, the number of users that can be supported per cell is U = A/ Au = 45/0.1 = 450
total number of subscribers that can be supported by System B = 450 98 = 44100
System C
Traffic intensity per user, Au = H = 2 (3/60) = 0.1 Erlangs
from the Erlang B chart, the total carried traffic, A, is obtained as 88 Erlangs.
Therefore, the number of users that can be supported per cell is U = A/ Au = 88/0.1 = 880
total number of subscribers that can be supported by System C is equal to 880 49 = 43120
Therefore, total number of cellular subscribers that can be supported by these three systems are 47280+
44100+43120 =134500 users.
Percentage market penetration:
System A: 47,280/2,000,000 = 2.36%
System B: 44,100/2,000,000 = 2.205%
System C: 43,120/2,000,000 = 2.156%