MCN 7104 Lecture Notes

Mobile Communications

2011

Dr. Bulega Tonny Eddie

Department of Networks

College of Computing and IT

Makerere University

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1 Cellular Systems Intro
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1.1 Generation Zero
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1.2 Cellular
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1.3 Key Terms
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1.4 The Cellular Radio
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2.0 Frequency Reuse

2.1 Cellular Geometry

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2.1.1 Channel Assignment within Group
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2.2 Handoff
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2.3 Co-Channel Interference
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2.4 Downtilt
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2.5 Adjacent Channel Interference
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3 Trunking
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3.1 Blocked calls cleared
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3.3 Discussion
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4 Increasing Capacity and Coverage
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4.1 Sectoring
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4.2 Microcells
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4.3 Repeaters

1 Cellular Systems Intro

Reading: Ch 1, Ch 3.1 from Rappaport.

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1.1 Generation Zero

The study of the history of cellular systems can help us understand the need for the system design
concepts we have today.
One of the major developments in WWII was the miniaturization of FM radio components to a
backpack or handheld device (the walkie-talkie), a half-duplex (either transmit or receive, not
both) push-to-talk communication device. After returning from war, veterans had the expectation
that wireless communications should be available in their civilian jobs [19]. But the phone system,
the Public Switched Telephone Network (PTSN) was: wired, and manually switched at telephone
exchanges. In 1952, the mobile telephone system (MTS) was designed to serve 25 cities in the US
[6] (Including one in Salt Lake City [5]). In each city, an additional telephone exchange office
was created for purpose of connection with the mobile telephones [19]. The MTS and later the
improved mobile telephone system (IMTS), introduced in 1964, were not particularly spectrally
efficient.
• They were allocated a total bandwidth of about 2 MHz. Frequency modulation (FM) was
the bet technology, and was narrowband, so they chose to operate as frequency division
multiple access (FDMA), in which each channel was allocated a non-overlapping
frequency band within the 2 MHz.
• The PTSN is full duplex (transmit and receive simultaneously) in IMTS, so it required
two channels for each call, one uplink (to the base station) and one downlink (to the
mobile receiver). Note MTS had been half duplex, i.e., only one party could talk at once.
• The FCC required them to operate over an entire city (25 mile radius). Since the coverage
was city wide and coverage did not exist outside of the cities, there was no need for
handoff.
Control was manual, and the control channel was open for anyone to hear. In fact, users were
required to listen to the control channel. When the switching operator wanted to connect to any
mobile user, they would announce the call on the control channel. If the user responded, they
would tell the user which voice channel to turn to. Any other curious user could listen as well. A
mobile user could also use the control channel to request a call to be connected. The system was
congested, so there was always activity.

The demand was very high, even at the high cost of about $400 per month (in 2009 dollars).
There were a few hundred subscribers in a city [6] but up to 20,000 on the waiting list [19]. The
only way to increase the capacity was to allocate more bandwidth, but satisfying the need would
have required more bandwidth than was available.
The downsides to MTS took a significant amount of technological development to address, and
the business case was not clear (AT&T developed the technologies over 35 years, but then largely
ignored it during the 1980s when it was deployed [6]).

1.2 Cellular

The cellular concept is to partition a geographical area into “cells”, each covering a small fraction
of a city. Each cell is allocated a “channel group”, i.e., a subset of the total list of channels. A
second cell, distant from a first cell using a particular channel group, can reuse the same channel
group. This is called “frequency reuse”. This is depicted in Figure 3.1 in Rappaport. This assumes

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that at a long distance, the signals transmitted in the first cell are too low by the time they reach
the second cell to significantly interfere with the use of those channels in the second cell.
There are dramatic technical implications of the cellular concept. First, rather than one base
station, you need dozens or hundreds, deployed across a city. You need automatic and robust
mobility management (handoff) to allow users to cross cell lines and continue a phone call. Both
of these are actually enabled by semiconductor technology advancement, which made the base
stations and the automated wired PSTN cheaper [19].
What is frequency reuse? What is required for a handoff? These are to be covered in the next
lecture.

1.3 Key Terms

Communication between two parties (a “link”), in general, can be one of the following:
• Simplex : Data/Voice is transferred in only one direction (e.g., Paging). Not even an
acknowledgement of receipt is returned.
• Half Duplex : Data/Voice is transferred in one direction at a time. One can’t talk and listen at
the same time. One channel is required.
• Full Duplex : Data/Voice can be transferred in both directions between two parties at the same
time. This requires two channels.
In a cellular system, there is full duplex communication, between a base station and a mobile. The
two directions are called either uplink (from mobile to base station) or downlink (from BS to
mobile). The downlink channel is synonymous with “forward channel”; the uplink channel is
synonymous with the “reverse channel”.
Simultaneous communication on the many channels needed for many pairs of radios to
communicate can be accomplished by one of the following:
• Frequency division multiple access (FDMA): Each channel occupies a different band of the
frequency spectrum. Each signal can be up-converted to a frequency band by multiplying it by a
sinusoid at the center frequency of that band, and then filtering out any out-of-band content, as
you have learned LECETRICAL.
• Time division multiple access (TDMA): Every period of time can be divided into short
segments, and each channel can be carried only during its segment. This requires each device to
be synchronized to have the same time clock.

See Figures 9.2 and 9.3, pages 450-453, in the Rappaport book.
Physical “parts” of a cellular system:

1. Public switched telephone network (PSTN): Wired telephone network, connecting homes,
businesses, switching centers.

2. Mobile switching center (MSC), a.k.a. mobile telephone switching office (MTSO): Controls
connection of wireless phone calls through the base stations to the PSTN. Connected either by
wire or by wireless (microwave relay) to the base stations.

3. Base station (BS): Maintains direct wireless connection to cell phones in its cell. Typically
maintains many connections simultaneously. Has multiple antennas, some for downlink and some
for uplink. See Figure 1.5, page 14, in the Rappaport book.

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In cellular systems, there are actually two types of channels: (1) Control, and (2) Communication.
The control channel is needed to tell the mobile device what to do, or for the mobile to tell the BS
or MSC what to do. The communication channel is the “voice channel” or data channel, the
actual information that the user / system needs to convey in order to operate. Since we also have a
forward and reverse channel (for full duplex comms), we have

1. FCC: Forward control channel

2. FVC: Forward voice channel
3. RCC: Reverse control channel
4. RVC: Reverse voice channel

Information needed for/from the mobile user: (1) mobile identification number (MIN), basically
the telephone number; (2) Electronic serial number (ESN), information about who will get billed
for this call, and needed to find the home location register (HLN), where the information about
this mobile is stored; (3) station class mark, which indicates how much power a mobile is able to
transmit.
We will review “How a call is made from PSTN to mobile” handout from Prof. Cynthia Furse.

1.4 The Cellular Radio

A cell is the area in which a mobile is served by a single BS. What is the power transmitted by the
radios in a cell system? Limits differ by country.
1. Basestation maximum = 100 W maximum Effective Radiated Power (ERP), or up to 500 W in
rural areas [20]
2. Cell phone: typically 0.5 W; but limited by power absorbed by human tissue in test
measurements.
The measurement is called Specific Absorption Rate (SAR). For TDMA, transmitter is only on a
fraction of the time. For CDMA, transmit power is lowered when close to a BS.
Cell phone exposure limits are typically set to meet both US and European standards. Note that
cellular community uses the term “mobile station” (MS) to describe the cell phone or mobile,
even though it is an oxymoron.

2 .0 Frequency Reuse
2.1 Cellular Geometry
When the signal from the base station is weak, the mobile will not be able to be served by the BS.
What shape is a cell? See Figure 1. These are in order from best to worst:
1. A random shape dependent on the environment.
2. Circular (theoretical): If path loss was a strictly decreasing function of distance, say, 1/dn,
where d is the distance from BS to mobile, then the cell will be a perfect circle. This is never true
in reality.
3. An approximation to the theoretical shape: required for a tessellation (non-overlapping
repetitive placement of a shape that achieves full coverage. Think floor tiles.) Possible “tile”
shapes include triangles, squares, hexagons. Hexagons are closest to reality.
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Figure 1: Theoretical coverage area, and measured coverage area.
In (b), from measurements, with red, blue, green, and yellow indicating signal strength, in
decreasing order. From Newport et. l.[12].

What is the size of the coverage area of a cell? (From C. Furse)

• Macrocell : Diameter typically greater than 2000 feet, up to 25 miles
• Microcell : Diameter typically 400 to 2000 feet
• Picocell : Diameter typically 100 feet

There are also cell sites in trucks that can be driven to replace downed cell towers after natural
disasters, or to create additional capacity for large gatherings (football games, rock concerts),
called “cell on wheels” (COW) or “cell on light truck sites” (COLTS).
As we mentioned in lecture 1, a cellular system assigns subsets, “channel groups”, of the total set
of channels to each cell. Call the total number of channels S, and the number of channel groups N.
Then there are on average k = S/N channels per cell. (In reality, k may vary between groups.)
Then with N channel groups, how do we assign them? We want cells that reuse group A, for
example, to be as far apart as possible.

2.1.1 Channel Assignment within Group

See Section 3.3. Which channels should be assigned to a cell? First, it is best to separate channels
in the group in frequency as much as possible to reduce adjacent channel interference (studied
later). But which channels are assigned? Two ways:

1. Fixed assignment: Each basestation has a fixed set of channels to use. Simple, but a busy cell
will run out of channels before a neighboring cell. System performance will be limited by the
most crowded cell.

2. Dynamic allocation: Each base station can change the channels it uses. Channels in
neighboring cells must still be different. This requires more careful control, but increases the
capacity.
For example, a typical city needs more channels in its business districts during the day and in its
residential areas at night and on weekends.
For general shapes, this can be seen as a graph coloring problem, and is typically covered in a
graph theory course.
For hexagons, we have simple channel group assignment. Consider N = 3, 4, 7, or 12 as seen in
Figure 2. A tesselation of these channel groupings would be a “cut and paste” tiling of the figure.
The tiling of the N = 4 example is shown in Figure 3.
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Figure 2: Hexagonal tesselation and channel groupings for N = 3, 4, 7, and 12.

Example: Call capacity of N = 4 system

Assume that 50 MHz is available for forward channels, and you will deploy GSM. Each channel
is 200 kHz, but using TDMA, 8 simultaneous calls can be made on each channel. How large is k?
How many forward calls can be made simultaneously for the cellular system depicted in Figure 3?
Solution: There are 50 MHz / 0.2 MHz or 250 total channels. With N = 4, then k = 250/4 = 62.5,
and with (about) 62.5 channels, 8(62.5) = 500 calls can be made simultaneously in each cell.
There are 28 cells on the cell map in Figure 3, so the total forward calls is 28(500) = 14 × 103
calls can be made simultaneously.
Why wouldn’t you choose N as low as possible? There are interference limits, which will be
discussed in more detail later.
How do you generally “move” from one cell to the co-channel cell (a second cell assigned the
same channel group)? All cellular tiling patterns can be represented using two non-negative
integers, I and j. The integer i is the number of cells to move from one cell in one direction.
Then, turn 60 degrees counter-clockwise and move j cells in the new direction. For Figure 3, this
is i = 2, j = 0.
In this notation, the number of cells can be shown to be: What is the distance
between two co-channel cell BSes? If the distance between the BS and a vertex in its cell is called
R, its “radius”, then you can show this co-channel reuse distance D is:

The ratio of is called Q, the co-channel reuse ratio.

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2.2 Handoff
See Section 3.4. As a mobile travels beyond the coverage region of its serving BS, it must be
transferred to better BS. If the received power drops too low prior to handoff, the call is
“dropped”. Rappaport denotes this minimum received power, below which a call cannot be
received, as Pr,minimumuseable. We want to initiate a handoff much prior to this point, so we set a
higher threshold Pr,handoff at which the MSC initiates the handoff procedure. Note the signal
strength varies quickly due to multipath fading, but we are most interested an short-term averaged
received power.
Because power may go down quickly, particularly at high mobile speeds, this handoff needs to
happen quickly. In GSM, handoff is typically within 1-2 seconds. In AMPS, this was 10 seconds
(higher potential for dropped calls!)
Define handoff margin as ∆
∆ = P r,handoff – P r,minimum usable

How much margin is needed to handle a mobile at driving speeds?

Example: Handoff Margin

Let the speed of a mobile be v = 35 meters/sec. For n = 4, a cell radius of 500 meters (the
distance at which the power is at the threshold), and a 2 second handoff, what ∆ is needed?
Solution: Assume the mobile is driving directly away from the BS, so distance d changes by 70
meters in two seconds. Consider the received power at the two times:

Taking the difference of the two equations (the 2nd minus the 1st),

Note that in this simple example, the propagation equation used is for “large scale path loss” only,
which changes slowly. Typically, shadowing (caused by large geographical features and
buildings blocking the signal) will play a more important role in quick changes in received power.
Mobile handoff strategies:

1. MSC controlled: all BSes measure all RVC, using a spare “locator receiver”. The MSC
decides when to handoff.
2. Mobile-assisted hand-off (MAHO): mobile measures the FCC from neighboring BSes,
and reports them to the MSC. This ends up leading to faster handoffs. MAHO is used in
2G systems.
This assumes that there is a channel in the new BS to offer the entering mobile! But there may not
be, and the call may be dropped for this reason. Users complain about dropped calls. So BSes
may reserve “guard channels” purely for handoff purposes, which then are not offered to mobiles
making new calls.

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CDMA (Verizon, e.g.) phones do not require handoff as we’ve described above (here called
“hard handoff”). In CDMA, a user does not need to switch “channel”, so handoff changes only
which BS is receiving the signal and sending the replies. In fact, if multiple BSes receive the
signal from the same mobile, the MSC can combine / choose the best among the three

Discussion: What are some of the problems with handoff vs. what the Rappaport book has
presented?

2.3 Co-Channel Interference

What is the ratio of signal power to interference power?

This is the critical question regarding the limits on how low we may set N. This ratio is
abbreviated S/I. Signal power is the desired signal, from the base station which is serving the
mobile. The interference is the sum of the signals sent by co-channel base stations, which is not
intended to be heard by mobiles in this cell. The S/I ratio is defined as:

where Ii is the power received by the mobile from a co-channel BS, of which there are i0, and S is
the power received by the mobile from the serving BS.
As a first order, before we get more complicated, we model the received power as inversely
proportional to distance to the n power, for some constant path loss exponent n

for some real value constant c.

We typically look at the worst case, when the S/I is the lowest. This happens when the mobile is
at the vertex of the hexagonal cell, i.e., at the radius R from the serving BS. So we know

Figure 4: Desired, and interfering signal for a mobile (M) from a serving and co-channel base
station

What are the distances to the neighboring cells from the mobile at the vertex? This
requires some trigonometry work. The easiest approximation is (1) that only the first “tier” of co-
channel BSes matter; (2) all mobile-to-co-channel-BS distances are approximately equal to D, the
distance between the two co-channel BSes. In this case, where i0 is the number of co-channel
cells in the first tier. For all N, we typically have i0 = 6 (try it out!); this will change when using
sector antennas, so it can be useful to leave i0 in the denominator. It is useful to report the S/I in
dB, because S/I requirements are typically given in dB.
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Example: AMPS design

Assume that 18 dB of S/I is required for acceptable system operation. What minimum N is
required? Test for n = 3 and n = 4.

For n = 3, N = 17.4; for n = 4, N = 6.5. Clearly, a high path loss exponent is important for
frequency reuse.

2.4 Downtilt

The Rappaport does not cover antenna downtilt, but it is an important practical concept. Compare
the elevation angles from the BS to mobile (Q1 in Figure 4) and co-channel BS to the mobile (Q2
in Figure 4). Note Q2 is lower (closer to the horizon) than from the serving BS. The great thing is,
we can provide less gain at angle Q2 than at Q1, by pointing the antenna main lobe downwards.
This is called downtilt. For example, if the gain at Q1 is 5 dB more than the gain at Q2, then the
we have added 5 dB to the S/I ratio. Having a narrow beam in the vertical plane is also useful to
reduce the delay spread and thus inter-symbol interference (ISI) [2], which we will introduce I
the 2nd part of this course. This narrow vertical beam is pointed downwards, typically in the
range of 5-10 degrees. The effect is to decrease received power more quickly as distance
increases; effectively increasing n. This is shown in Figure 5. How do you calculate the elevation
angle from a BS to a mobile? This angle is the inverse tangent of the ratio between BS height ht
and horizontal distance from the mobile to BS, d. But, at very low ratios, we can
approximate . So the angle is ht/d.

Figure 5: A diagram of a BS antenna employing downtilt to effectively increase the path loss at
large distances. From [9]

Ever wonder why base station antennas are tall and narrow? The length of an antenna in any
dimension is inversely proportional to the beamwidth in that dimension. The vertical beamwidth
needs to be low (5-10 degrees), so the antenna height is tall. The horizonal pattern beamwidths
are typically wide (120 degrees or more) so the antenna does not need to be very wide. For more
information, perhaps for a project, please consult [7].
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Discussion: What are some of the problems with coverage and frequency reuse vs. what the
Rappaport book has presented?

2.5 Adjacent Channel Interference

Our standard, non-ideal radios do not perfectly filter out any out-of-band signals. Thus any signal
that a mobile sends in another channel (besides its assigned channel) is interference at the BS
w.r.t. the desired signal sent by another mobile in that channel. Each mobile’s receiver also must
filter out out-of-band signals from the BS, which does send signals on all channels. One standard
way of making this problem less difficult is to assign non-adjacent channels within each cell’s
channel group.

Example: Assign S = 50 channels into groups for N = 7

Solution: We should have about k = 50/7 ≈ 7 channels per group. For group 1, use forward
channels {1, 8, 15, 22, 29, 36, 43, and 50}. For group i, i = 2 . . . 7, let the channels
for group i consist of {i, i + 7, i + 14, i + 21, i + 28, i + 35, i + 42}.
Because of the spacing in channels used in one group, the specifications for transmit and receive
filters will be less strict, and can be more effective at reducing adjacent channel interference.
There is still the near-far effect. If a TX near the BS is producing just a little bit of out-of-band
noise, it might swamp out the desired signal transmitted by a TX far away to the same BS.
One solution is power control, i.e., reducing the power level TXed by mobiles close to the BS,
since a high TX power is not necessary in that case. This reduces their out-of-band noise as well.
This means that antennas far away must transmit larger power than those nearby. Compared to a
mobile transmitting full power all the time, power control extends battery life, and generally
reduces even co-channel interference on the reverse channel. However, power control requires,
well, control. Controlling a mobile means communication from the BS to the mobile to inform it
whether to increase or decrease its power, which then requires data overhead.
Tight power control is particularly required in CDMA systems, in which the “near-far problem”
is even more of a problem.

3 Trunking

Trunking refers to sharing few channels among many users. Let U be the number of users, and C
be the number of channels. Each user requires a channel infrequently, so a dedicated channel for
each user is not required. But, the request for a channel happens at random times, and so for any
C < U, it is possible that there will be more requests than channels.
• Erlang: A “unit” of measure of usage or traffic intensity. One Erlang is the traffic
intensity carried by one channel that is occupied all of the time. 0.1 Erlang is the same
channel occupied only 10% of the time.
• Average holding time: Average call duration, denoted H.
• Call rate: Average number of calls per unit time, denoted λ. Typically taken to be at the
busiest time of day.

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 • Total offered traffic intensity: The total amount of traffic users request of the system,
denoted A.
• Grade of Service (GOS): The probability an offered call will be blocked (and thus not
served, or carried by the system).

Rappaport presents that an average user will request (offer) this much traffic, Au = λH. For
example, if a user makes on average, two calls per hour, and that call lasts an average of 3
minutes,

Then, to compute the total offered traffic intensity, and the total offered traffic intensity per
channel (denoted Ac),
A = UAu, Ac = A/C
For the above example, assume that there are 1000 users and 200 channels. Then A = 1000(0.1)
= 100, and Ac = 100/200 = 0.5.
Note that Ac is a measure of the efficiency of the utilization of the channels.
How should we design our system? Obviously, Ac should be less than one (A < C); or we’ll never
satisfy our call demand. But how should we set U, Au, C to satisfy our customers?
First choice: what do we do when a call is offered (requested) but all channels are full?
• Blocked calls cleared: Ignore it!
• Blocked calls delayed: Postpone it!

3.1 Blocked calls cleared

1. Call requests are a Poisson process. That is, the times between calls are exponentially
distributed, and memoryless.
2. Call durations are also exponentially distributed.
3. Rather than a finite number U of users each requesting Au traffic, we set the total offered
traffic as a constant A, and then let U → ∞ and Au → 0 in a way that preserves UAu = A.
This is the “infinite number of users” assumption that simplifies things considerably.
These assumptions, along with the blocked calls cleared setup of the system, leads to the Erlang
B formula:

Since C is very high, it’s typically easier to use Figure 3.6 on page 81. By setting the desired
GOS, we can derive what number of channels we need; or the maximum number of users we can
support (remember A = UAu); or the maximum Au we can support (and set the number of minutes
on our calling plans accordingly).

3.2 Blocked calls delayed

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Instead of clearing a call; put it in a queue (a first-in, first-out line). Have it wait its turn for a
channel. (“Calls will be processed in the order received”). There are now two things to determine

1. The probability a call will be delayed (enter the queue), and

2. The probability that the delay will be longer than t seconds.

The first is no longer the same as in (2); it goes up, because blocked calls aren’t cleared, they
“stick around” and wait for the first open channel.
Here, we clarify the meaning of GOS for a blocked calls delayed system. Here it means the
probability that a call will be forced into the queue AND it will wait longer than t seconds before
being served (for some given t). We need a couple additional assumptions:

1. The queue is infinitely long. In a computer system, this translates to infinite memory.
2. No one who is queued gives up / hangs up (rather than wait).

With these assumptions, we can derive the Erlang C formula, for the probability that a call will be
delayed:

It is typically easiest to find a result from Figure 3.7, on page 82. Once it enters the queue, the
probability that the delay is greater than t (for t > 0) is given as

The two combined are needed to find the marginal (overall) probability that a call will be delayed
AND experience a delay greater than t, the event that we are quantifying in GOS.
GOS = P [delay > t] = P [delay > t|delay > 0] P [delay > 0]

Example: N = 7 cell cluster

A 7 cell cluster (with N = 7) has 28 MHz allocated to it for forward channels and each channel is
200 kHz. Assume blocked-called-delayed and a probability of delay of 1%, and each user makes
one 10 minute call every 3 hours. (a) What is the number of users that can be supported? (b) What
is P [delay > 10] seconds? (c) What if it was a blocked-calls-cleared system with QOS of 1%?
Solution: 30 MHz / 200 kHz = 150 channels, divided among 7 cells, so about 20 channels per cell
(after 1 control channel per cell). (a) With 20 channels, and probability of delay of 1%, looking at
figure 3.7, we see A is about 11. With 11 = AuU and Au = 10/(3 × 60) = 1/18, we have that U =
198 ≈ 200. But this is per cell. So there can be 7(200) = 1400 users in the 7 cells. (b) in each cell,
C = 20, A = 11, H = 600 sec. So P [delay > t] = (0.01) exp[−(20 − 11)(10)/600] = 0.01 exp(−
0.15) = 0.0086. (c) From Figure 3.6, A ≈ 12, so U = 216, for a total of 1512 total users.

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3.3 Discussion

What are the problems or benefits we see from the assumptions we’ve made? Are call requests
“memoryless”? Is the exponential interarrival time assumption accurate? When catastrophic
events occur, or major news breaks, what happens? How should a communications system be
designed to handle these cases

4 Increasing Capacity and Coverage

4.1 Sectoring
In sectoring, we divide each cell into three or six “sectors” which are then served by three or six
separate directional antennas, each with beamwidth of about 120 or 60 degrees.

We showed in Lecture 2 that the S/I ratio is given by where N is the reuse ratio, and
i0 is the number of first-tier co-channel base stations. When we used omnidirectional antennas at
each BS, we saw that i0 = 6 regardless of N. By using sector antennas at the BSes, we will show
that i0 will be lower. By reducing the S/I ratio for a given N, we allow a system to be deployed for
a lower N, and therefore a higher capacity system.
However, each cell’s channel group must be divided into three sub-groups. These new groups
have 1/3 or 1/6 the number of channels, and thus the trunking efficiency will be lower.

Example: Decrease in trunking efficiency for constant N

Let N = 7, each cell has C = 100 channels, and users who make calls with λ = 0.01 per minute
with average holding time 3 minutes. For blocked-calls-cleared and a GOS of 2%, what is the
number of users which can be supported in this cell? Next, using 120 degree sectoring, and
otherwise identical system parameters, what is the number of users which can be supported in this
cell?

What percentage reduction in capacity does sectoring with constant N cause?

Solution: For C = 100 and GOS= 0.02, from Figure 3.6, I read A ≈ 99. Thus with Au = 0.01(3) =
0.03, we could support U = 99/0.03 = 3300. For the sectoring case, C = 33.3 in each sector, and
from Figure 3.6, A = 24. So we could support U = 24/0.03 ≈ 800 per sector, or 2400 total in the
cell. The number of users has reduced by 28%.

Example: Reducing N with sector antennas

For the same system, now assume that with 120 degree sectoring, that N can be reduced from 7 to
4. What number of users can be supported?
Solution: Now, the number of channels in each cell goes up to 100(7/4) = 175. So each sector has

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C = 58 channels. With GOS = 2%, from Figure 3.6, A ≈ 48, so U ≈ 1600, for a total of 4800
users per cell. This is a 45% increase upon the N = 7 non-sectored cell.
Why does i0 reduce? Consider again a mobile at the edge of a cell. We need to determine which
of the first tier BSes contribute significantly to the interference signal. Refer to Figures 3.10, 3.11,
for N = 7, P3.28(b) for N = 3, and to Figure 6 for N = 4.

Figure 6: 120 degree sectoring for cellular system with N = 4. Only two first tier BSes
significantly interfere with the middle BS.

So how much does sectoring improve the S/I ratio for particular values of N?

Example: Sectoring improvement compared to Omni

Compared to when i0 = 6, how much does S/I improve with sectoring?

Solution: Recall that . In dB terms,

So with i0 = 6, the latter term is 7.8 dB. If i0 = 1, 2, and 3, the same term is 0, 3.0, or 4.8 dB.
So, the improvement is 3, 4.8, or 7.8 dB. The particular value of i0 that can be obtained is a
function of N and whether 60 or 120 degree sectoring is used.

4.2 Microcells

When we introduced “cells” we said the radius was a variable R. The idea of using microcells is
that for a densely populated area, we cut the size of the cell by half. In this microcell-covered area,
the concept of frequency reuse occurs described earlier, only with smaller R. The smaller R also
has the benefit that transmit powers would be cut by a factor of 2n (see Rappaport 3.7.1 for
details). The other main benefit is that by reducing the area of a cell by a factor of four (forced by
cutting R by two) the capacity in the microcell area is increased by four. For example, consider
Figure 7, which shows an original macrocell grid, next to an “inserted” microcell area.
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However, at the edges of the microcell area, there is a conflict. Cells that were separated by
distance R √ 3N for the initial R are no longer separated by that much. Conflicts in channel
assignments at the edges are solved by splitting the channel group into two sub-groups. These
subgroups can have different sizes, e.g., the sub-group used for the microcell might have fewer
channels assigned to it compared to the macrocell.
Another problem is that the number of handoffs is increased, since users travel through microcells
more quickly. This can be addressed using umbrella cells (page 66) or microcell zones (Section
3.7.4).

4.3 Repeaters

This is Section 3.7.3 in Rappaport. Repeaters can be used to increase the coverage area,
particularly into buildings, tunnels, and canyons. They are bidirectional (they amplify forward
and reverse channels). However, repeaters don’t add any capacity to the system, they just increase
the reach of a BS or MS into “shadowed” areas.

4.4 Discussion

What are some of the problems with the assumptions made in this analysis?

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