Cellular Mobile Communication Concepts

Both techniques increase capacity, but in different ways:

• Cell splitting divides a congested cell into smaller cells, each with its own base station and lower power. More cells means more reuse of the same channels, so capacity rises roughly in proportion to the number of new cells (e.g. halving the radius gives about 4 times the capacity in that area).
• Sectoring replaces the omni antenna with directional antennas (3 × 120° or 6 × 60°). It reduces the number of co-channel interferers (6 → 2 for 120°), raising S/I, so a smaller cluster size N can be used. Capacity increases only through the smaller N; channels are also divided among sectors, reducing trunking efficiency.

Recommendation

For a city where demand keeps growing, cell splitting is the best long-term choice because it multiplies capacity in proportion to the number of new cells and can be repeated (macro → micro → pico) only where traffic is high. Sectoring alone gives a limited, one-time gain and lowers trunking efficiency.

In practice the engineer should first apply sectoring at existing sites (quick, cheap, improves S/I) and then split cells in hot spots (city centre, markets) as traffic grows, using lower power and down-tilted antennas. Combined, they give the maximum capacity at reasonable cost.

Given: population = 2 000 000; cells = 394; channels per cell C = 19; GOS = 2 % blocking; λ = 2 calls/hour; H = 3 min.

Step 1: Traffic per user

Au = λ × H = 2 × (3/60) = 0.1 Erlang

Step 2: Traffic per cell from Erlang B table For N = 19 channels and B = 2 %:

A (per cell) = 12.33 Erlangs

Step 3: Users per cell

U = A / Au = 12.33 / 0.1 = 123.3 ≈ 123 users per cell

Step 4: Total users supported

Total = 123 × 394 = 48 462 users
(Without rounding: 123.3 × 394 = 48 580)

Step 5: Market penetration

Penetration = 48 462 / 2 000 000 × 100 = 2.42 %

Answer: About 48 462 users (≈ 48 500) can be supported at 2 % blocking, giving a market penetration of about 2.42 %.

Why the microcell zone concept is needed

When sectoring is used to increase capacity, the cell is split into sectors, which increases the number of handoffs and the load on the switching centre (MSC), and also reduces trunking efficiency. The microcell zone concept, proposed by Lee, improves capacity and S/I without extra handoffs.

Microcell zone concept

• A cell is divided into several zones (e.g. 3), each with a low-power, directional zone site (transmitter/receiver) located at the cell edge pointing inward.
• All zone sites are connected by fibre or microwave to a single base station, and they share the same radio equipment and channels of the cell.
• As the mobile moves from one zone to another within the cell, the base station simply switches the channel to the new zone site; the mobile keeps the same channel, so no handoff is processed by the MSC.
• A channel is active only in the zone where the mobile is, so radiation is localised and co-channel interference is reduced. This improves S/I (Lee reports about 18 dB with N = 3 instead of N = 7), giving about 2.33 times capacity of a 7-cell omni system.

          Zone 1
          (Tx/Rx)
            \  /
     ┌────── BS ──────┐
  Zone 2 (Tx/Rx)   Zone 3 (Tx/Rx)
   (links by fibre/microwave to BS)

Advantages: higher capacity, fewer handoffs, better S/I, useful along highways and in urban corridors.

In practical cellular systems, handoff design must handle users with very different speeds and the limited availability of cell sites.

1. High-speed and low-speed users

   • Fast vehicles cross small cells in seconds; handoffs must be quick, while pedestrians may never need one.
   • Using many microcells for fast users overloads the MSC with handoffs.

2. Umbrella cell approach

   • Large, high-power umbrella (macro) cells serve high-speed users; small microcells under them serve slow users.
   • The speed of a user is estimated from signal variations; fast users are moved to the umbrella cell, reducing the number of handoffs.

   ┌───────── Umbrella (macro) cell ─────────┐
   │  (◯)   (◯)   (◯)   (◯)  ← microcells    │
   └─────────────────────────────────────────┘

3. Difficulty of new cell sites – in cities it is hard to obtain new tower sites, so existing sites and power levels must be used carefully; mixed macro/micro layouts must be planned.

4. Cell dragging – a slow user with a strong line-of-sight signal can travel far into a neighbouring cell without handoff because the signal stays above threshold, causing interference. Thresholds and radio coverage must be carefully set.

5. Handoff threshold margin Δ = Pr,handoff − Pr,min – if too large, unnecessary handoffs; if too small, calls drop before handoff completes. The dwell time and handoff processing time (1G ~10 s, 2G ~1–2 s) are considered.

6. Hard and soft handoff – CDMA systems use soft handoff (connected to several base stations at once), giving fewer dropped calls.

7. Intersystem handoff – when a mobile moves between different operators' MSCs or systems, extra signalling and compatibility are needed.

8. Mobile-assisted handoff (MAHO) – in 2G, the mobile measures nearby base stations' power and reports it, making handoffs faster.

Two schemes for prioritising handoff

Dropping an ongoing call is worse than blocking a new call, so handoffs are given priority:

1. Guard channel concept

   • A fraction of the channels in each cell is reserved exclusively for handoff requests; new calls can use only the remaining channels.
   • Reduces dropped calls, but lowers total carried traffic because guard channels may be idle. Dynamic channel assignment makes it more efficient.

2. Queuing of handoff requests

   • There is a finite time interval between the signal reaching the handoff threshold and falling to the minimum usable level. During this time the handoff request is placed in a queue and served as soon as a channel becomes free.
   • Reduces forced termination, but cannot guarantee zero drops because the waiting time is limited.

 Signal level
   │ \
   │  \ ← handoff threshold: request queued
   │   \
   │    \ ← minimum usable level: call drops
   └──────────── time
       |<-->| queuing time available

Trunking terms

• Grade of service (GOS): a measure of the ability of a user to access the trunked system during the busiest hour; usually given as the probability that a call is blocked (Erlang B) or delayed beyond a given time (Erlang C). Example: GOS = 2 %.
• Traffic intensity: the measure of channel time utilisation, i.e. the average channel occupancy, measured in Erlangs. For one user, Au = λH; total offered traffic A = U·λ·H. One Erlang = one channel occupied continuously.
• Average holding time (H): the average duration of a typical call. Example: H = 3 minutes.
• Blocked call (lost call): a call that cannot be completed at the time of request because all channels are busy (congestion). In Erlang B systems blocked calls are cleared; in Erlang C they are queued.

Given: number of cells = 32, R = 1.6 km, N = 7, total traffic channels S = 336.

Area of one hexagonal cell:

A_cell = (3√3/2) R² = 2.598 × 1.6² = 2.598 × 2.56
       = 6.651 km²

Total area covered:

A_total = 32 × 6.651 = 212.83 km²

Traffic channels per cell:

k = S / N = 336 / 7 = 48 channels per cell

Total simultaneous calls:

C = 48 × 32 = 1536 calls

(Number of clusters = 32/7 ≈ 4.57, and C = M × S gives the same value 4.57 × 336 = 1536.)

Answer: Area ≈ 212.8 km², 48 channels per cell, 1536 simultaneous calls.

Frequency reuse

Frequency reuse is the use of the same radio channels in different cells that are far enough apart that their co-channel interference stays within limits. The total channels S are divided among N cells of a cluster, and the cluster is repeated M times over the service area, so capacity C = M × S grows with the number of clusters without extra spectrum.

   ___     ___
  / 2 \___/ 7 \
  \___/ 3 \___/     Cells with the same number
  / 1 \___/ 1 \     use the same channels
  \___/ 4 \___/     (N = 7 cluster)

Channel assignment strategies

1. Fixed channel assignment (FCA)

   • Each cell is given a fixed set of channels in advance.
   • A call is blocked if all channels of that cell are busy.
   • Borrowing strategy: a cell may borrow a channel from a neighbouring cell if none of its own is free; the MSC ensures the borrowed channel does not cause interference.
   • Simple and needs little control, but cannot adapt to changing traffic.

2. Dynamic channel assignment (DCA)

   • Channels are not permanently assigned; on each call request the MSC allocates a channel from the common pool using an algorithm that considers future blocking, reuse distance, channel usage and cost.
   • Uses channels efficiently and lowers blocking, but needs real-time data collection and heavy MSC computation.

3. Hybrid channel assignment – some channels fixed in each cell, the rest in a common pool assigned dynamically.

Given: single high-power transmitter: 40 voice channels over 140 km². Area divided into 7 equal cells; each cell supports 30 % of total channels.

i) Coverage area of each cell

A_cell = 140 / 7 = 20 km²

ii) Total voice channels: cellular vs non-cellular

Channels per cell:

k = 30 % of 40 = 0.3 × 40 = 12 channels per cell

Total channels in the cellular system (each cell uses its own 12 channels; channels are reused in non-adjacent cells):

C_cellular = 7 × 12 = 84 channels

Non-cellular system:

C_non-cellular = 40 channels

Ratio:

84 / 40 = 2.1

Answer: (i) 20 km² per cell; (ii) the cellular system provides 84 channels versus 40, i.e. 44 more channels or 2.1 times the capacity, because frequencies are reused in different cells.

Co-channel interference (CCI)

• Interference between signals from cells that use the same set of frequencies (co-channel cells) under frequency reuse.
• Cannot be removed by increasing transmitter power, because the interfering signal grows equally. It is reduced only by physically separating co-channel cells by a minimum distance D, i.e. a larger reuse ratio Q = D/R = √(3N), or by sectoring.
• Measured by S/I = (√(3N))ⁿ / i₀ for the first tier of i₀ interferers. AMPS needs S/I ≥ 18 dB, which gives N = 7 with n = 4.

Adjacent channel interference (ACI)

• Interference from signals at frequencies adjacent (next) to the wanted signal, caused by imperfect receiver filters letting nearby frequencies leak into the passband.
• Becomes serious through the near–far effect: a nearby transmitter on an adjacent channel overpowers a weak, distant wanted signal.
• Reduced by careful filtering and channel assignment – not assigning adjacent channels in the same cell, keeping a large frequency separation between channels in a cell (e.g. channels 1, 22, 43, … in AMPS), and power control.

Proof of Q = √(3N)

Assumptions: hexagonal cells of radius R (centre to vertex); distance between centres of two adjacent cells = √3 R.

Locating a co-channel cell: in a hexagonal layout, the nearest co-channel cell is found by moving i cells along a chain of hexagons, turning 60° anticlockwise, and moving j cells.

 Co-channel cell
        *
       /        From cell A move i cells
      / j cells    along one axis, turn 60°,
     /  (60° turn) move j cells.
 A *-----------*
    i cells

The two paths have lengths i√3R and j√3R, and the angle between them (inside the triangle) is 120°. By the cosine rule:

D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R) cos 120°
   = 3R²i² + 3R²j² + 3R²ij        (cos 120° = −1/2)
   = 3R² (i² + ij + j²)

Relating to cluster size: the area of a hexagon is proportional to the square of the distance between centres of neighbouring hexagons. A cluster of N cells repeats with centre spacing D, while single cells repeat with spacing √3R, so

N = (D / √3R)² = i² + ij + j²

Substituting:

D² = 3R² N
D  = R √(3N)
Q  = D / R = √(3N)

A larger Q means more separation between co-channel cells, so less co-channel interference but fewer channels per cell (less capacity).

Given: N = 7, B = 1 %, H = 2 min, λ = 1 call/hour, 57 channels per cell, blocked calls cleared (Erlang B).

Traffic per user:

Au = λH = 1 × 2/60 = 0.0333 Erlang

Omnidirectional antennas

All 57 channels form one trunk group. From the table (57 channels, 1 %):

A = 44.2 Erlangs per cell
Users per cell = 44.2 / 0.0333 = 1326 users

60° sectoring

Each cell has 6 sectors, channels per sector = 57/6 = 9.5 ≈ 9 channels. From the table (9 channels, 1 %):

A per sector = 3.783 Erlangs
A per cell   = 6 × 3.783 = 22.70 Erlangs
Users per cell = 22.70 / 0.0333 ≈ 681 users

Trunking capacity loss

Loss = 44.2 − 22.70 = 21.50 Erlangs per cell
     = 21.50 / 44.2 × 100 = 48.6 %

Answer: Trunking efficiency loss ≈ 21.5 Erlangs per cell (≈ 48.6 %), about 645 fewer users per cell. Sectoring breaks a large trunk into small ones, which are less efficient; the gain from sectoring must come from the smaller cluster size it permits.

Handoff (handover) is the transfer of an ongoing call from one base station (or channel) to another without interrupting it, when the mobile moves out of the coverage of the serving cell.

A handoff is started when the received signal falls to a handoff threshold Pr,handoff, which is set above the minimum usable signal Pr,min. The difference is the handoff margin Δ = Pr,handoff − Pr,min.

(a) Improper handoff

 Signal      BS1                    BS2
   │  \                            /
   │    \_ Pr,handoff  (A)        /
   │       \                    /
   │ ........\ Pr,min (B) ... /..
   │        call dropped here
   └──────────────────────────────── distance

The signal at point A reaches the handoff threshold, but the handoff is not completed before the signal falls below Pr,min at B (because of a too-small Δ, slow processing at the MSC, or no free channel in BS2). The call is dropped.

(b) Proper handoff

 Signal      BS1                    BS2
   │  \                            /
   │    \_ Pr,handoff (A)  ______/
   │       \         ____/  handoff to BS2 (C)
   │ ........\ Pr,min ..........
   └──────────────────────────────── distance

At A the handoff threshold is reached, the MSC finds a channel in BS2 and the call is transferred at C, before the signal from BS1 drops below Pr,min. The call continues without interruption.

The co-channel reuse ratio Q = D/R, where D is the distance between centres of the nearest co-channel cells and R is the cell radius.

Assumptions: hexagonal cells of radius R (centre to vertex); distance between centres of two adjacent cells = √3 R.

Locating a co-channel cell: in a hexagonal layout, the nearest co-channel cell is found by moving i cells along a chain of hexagons, turning 60° anticlockwise, and moving j cells.

 Co-channel cell
        *
       /        From cell A move i cells
      / j cells    along one axis, turn 60°,
     /  (60° turn) move j cells.
 A *-----------*
    i cells

The two paths have lengths i√3R and j√3R, and the angle between them (inside the triangle) is 120°. By the cosine rule:

D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R) cos 120°
   = 3R²i² + 3R²j² + 3R²ij        (cos 120° = −1/2)
   = 3R² (i² + ij + j²)

Relating to cluster size: the area of a hexagon is proportional to the square of the distance between centres of neighbouring hexagons. A cluster of N cells repeats with centre spacing D, while single cells repeat with spacing √3R, so

N = (D / √3R)² = i² + ij + j²

Substituting:

D² = 3R² N
D  = R √(3N)
Q  = D / R = √(3N)

A larger Q means more separation between co-channel cells, so less co-channel interference but fewer channels per cell (less capacity).

Interference is the major factor limiting the capacity and quality of cellular systems. It causes cross-talk in voice calls, missed and dropped calls, and errors in data. Sources include other mobiles in the same cell, calls in neighbouring cells, other base stations on the same frequency, and non-cellular systems leaking into the band.

1. Co-channel interference (CCI)

• Caused by cells that use the same frequency set under frequency reuse.
• Cannot be reduced by increasing power; reduced by increasing the co-channel reuse ratio Q = D/R = √(3N).
• S/I = (√(3N))ⁿ / i₀ for i₀ first-tier interferers; e.g. N = 7, n = 4 gives about 18.7 dB.

2. Adjacent channel interference (ACI)

• Caused by signals on neighbouring frequencies leaking through imperfect receiver filters; worsened by the near–far effect.
• Reduced by good filtering, keeping channels in a cell widely separated in frequency, and power control.

3. Other types

• Intersymbol interference from multipath delay spread (digital systems).
• Multiple access interference in CDMA from other users' codes.
• Out-of-band and intermodulation interference from other radio systems.

Methods to reduce interference

• Proper cluster size and frequency planning
• Sectoring with directional antennas and down-tilt
• Power control (keep each mobile at the minimum needed power)
• Good filters and channel assignment
• Frequency hopping, equalisers and diversity

Interference-limited design trades capacity (small N) against quality (high S/I).

Grade of Service (GoS)

GoS is a measure of the ability of a user to access a trunked system during the busiest hour. It is usually expressed as the probability that a call is blocked (or delayed beyond a set time). Example: GoS = 2 % means 2 calls in 100 are blocked during the busy hour.

Measurement in a blocked-calls-cleared system: a blocked call is not queued; the user must try again later. Assuming Poisson call arrivals, exponential holding times and infinite users, the blocking probability is given by the Erlang B formula:

            Aᶜ / C!
Pr[block] = ─────────────
            Σ (k=0..C) Aᵏ / k!

where A = offered traffic (Erlangs) = U·λ·H and C = number of channels. In practice GoS is measured as the ratio of blocked call attempts to total call attempts in the busy hour, and the Erlang B chart/table gives A for a given C and GoS.

Optimal N for S/I ≥ 15 dB, n = 3

Using S/I = (√(3N))ⁿ / i₀, with i₀ = 6 (omni), 2 (120°), 1 (60°) first-tier interferers. Valid cluster sizes are N = i² + ij + j² = 3, 4, 7, 9, 12, …

(a) Omnidirectional (i₀ = 6)

N = 7 : (√21)³ / 6 = 96.23 / 6 = 16.04 → 12.05 dB  < 15 ✗
N = 9 : (√27)³ / 6 = 140.30 / 6 = 23.38 → 13.69 dB < 15 ✗
N = 12: (√36)³ / 6 = 216 / 6    = 36.00 → 15.56 dB ≥ 15 ✓

Optimal N = 12.

(b) 120° sectoring (i₀ = 2)

N = 4 : (√12)³ / 2 = 41.57 / 2 = 20.78 → 13.18 dB < 15 ✗
N = 7 : 96.23 / 2  = 48.12      → 16.82 dB ≥ 15 ✓

Optimal N = 7.

(c) 60° sectoring (i₀ = 1)

N = 3 : (√9)³  = 27.00 → 14.31 dB < 15 ✗
N = 4 : (√12)³ = 41.57 → 16.19 dB ≥ 15 ✓

Optimal N = 4.

Comment

Sectoring reduces the number of first-tier co-channel interferers, so the required S/I is met with a smaller cluster. A smaller N gives more channels per cell (S/N) and hence more capacity: going from N = 12 to N = 4 gives three times as many channels per cell. The cost is more antennas, more handoffs between sectors and lower trunking efficiency. With n = 3 (less path loss than n = 4), interference falls off more slowly, so larger clusters are needed than in the usual n = 4 case.

Co-channel interference (CCI)

• Interference between signals from cells that use the same set of frequencies (co-channel cells) under frequency reuse.
• Cannot be removed by increasing transmitter power, because the interfering signal grows equally. It is reduced only by physically separating co-channel cells by a minimum distance D, i.e. a larger reuse ratio Q = D/R = √(3N), or by sectoring.
• Measured by S/I = (√(3N))ⁿ / i₀ for the first tier of i₀ interferers. AMPS needs S/I ≥ 18 dB, which gives N = 7 with n = 4.

Adjacent channel interference (ACI)

• Interference from signals at frequencies adjacent (next) to the wanted signal, caused by imperfect receiver filters letting nearby frequencies leak into the passband.
• Becomes serious through the near–far effect: a nearby transmitter on an adjacent channel overpowers a weak, distant wanted signal.
• Reduced by careful filtering and channel assignment – not assigning adjacent channels in the same cell, keeping a large frequency separation between channels in a cell (e.g. channels 1, 22, 43, … in AMPS), and power control.

Cell footprint

The cell footprint is the actual geographical area over which a base station's signal is strong enough for reliable service (above the minimum usable level). The real footprint is irregular because of terrain, buildings and antenna patterns; it is modelled as a hexagon for planning, since hexagons tile the area without gaps or overlaps and approximate a circle closely.

Interference tier

Co-channel cells around a given cell are arranged in rings called tiers:

• First tier: the 6 nearest co-channel cells, each at distance D = R√(3N) from the centre cell.
• Second tier: the next ring of 12 co-channel cells at about 2D, and so on (third tier 18 cells at about 3D).

              (I)          (I)
                  \       /
         (I) ──── [ Serving ] ──── (I)     I = first-tier
                  /   cell  \           co-channel cell
              (I)          (I)          at distance D

Because received power falls as d⁻ⁿ (n = 2 to 4), the second-tier cells at 2D contribute only about (1/2)ⁿ of the first tier's interference each (1/16 for n = 4). Therefore, in practice, only the first tier is considered:

S/I = R⁻ⁿ / Σ(i=1..i₀) Dᵢ⁻ⁿ ≈ (D/R)ⁿ / i₀ = (√(3N))ⁿ / i₀

with i₀ = 6 for omni antennas, 2 for 120° sectoring and 1 for 60° sectoring (roughly). Example: N = 7, n = 4 gives S/I = 4.583⁴/6 = 73.5 ≈ 18.66 dB.

Handoff (handover) is the process of transferring an ongoing call from one channel or base station to another without interrupting the call, when the mobile moves into a different cell or the signal quality falls.

Handoff process

1. The base station (or mobile) continuously monitors received signal strength (RSSI) and quality.
2. When the signal from the serving cell falls to a handoff threshold (Pr,handoff), a handoff is initiated.
3. Neighbouring base stations measure the mobile's signal (1G locator receiver) or the mobile measures neighbours' signals and reports them (MAHO in 2G).
4. The MSC selects the best target cell, finds a free channel there and instructs the mobile to retune.
5. The call is switched to the new base station before the signal reaches the minimum usable level (Pr,min); otherwise the call drops.

 Power   BS1 signal             BS2 signal
   │ \                         /
   │   \ ← Pr,handoff          /
   │     \  Δ               /
   │       \ ← Pr,min     /
   └──────────────────────────── distance
   BS1          handoff         BS2

The margin Δ = Pr,handoff − Pr,min must be chosen carefully: too large causes unnecessary handoffs, too small causes dropped calls.

Practical handover considerations

1. High- and low-speed users – vehicles pass through small cells quickly, causing many handoffs; pedestrians may never need one. Handoff strategy must handle both.
2. Umbrella cell approach – large high-power umbrella cells (on tall towers) serve fast users, while microcells beneath serve slow users. Speed is estimated from signal fading rate; fast users are moved to the umbrella cell, reducing handoffs and MSC load.

 ┌──────── Umbrella cell (fast users) ────────┐
 │   (◯)    (◯)    (◯)    (◯)  microcells     │
 └────────────────────────────────────────────┘

3. Difficulty in obtaining new cell sites – in cities, operators often add capacity by placing new cells at existing sites, so mixed macro/micro planning and careful power levels are needed.
4. Cell dragging – a slow user with a strong LOS signal moves far into the neighbouring cell without handoff because the signal stays above threshold, causing interference. Careful threshold and coverage setting reduces it.
5. Handoff time and margin – 1G took about 10 s for handoff, needing Δ of 6–12 dB; 2G with MAHO needs only 1–2 s and Δ of 0–6 dB.
6. Soft handoff (CDMA) – the mobile connects to several base stations at once and the best signal is chosen, reducing dropped calls.
7. Intersystem handoff – when moving between MSCs or networks, extra signalling and billing (roaming) issues arise.
8. Prioritising handoffs – guard channels and queuing of handoff requests reduce forced termination.

Given: λ = 3 calls/hour, H = 5 min, blocked calls cleared (Erlang B).

a) Traffic intensity per user

Au = λH = 3 × 5/60 = 0.25 Erlang

b) One channel, 1 % blocking

From the table, N = 1, B = 1 %: A = 0.010 Erlang.

U = A / Au = 0.010 / 0.25 = 0.04 user

Less than one user can be supported; i.e. even a single user (0.25 E) would give far more than 1 % blocking. So effectively no user (0.04 user) can be served at 1 % GOS with one channel.

c) Five trunked channels, 1 % blocking

From the table, N = 5, B = 1 %: A = 1.36 Erlangs.

U = 1.36 / 0.25 = 5.44 ≈ 5 users

d) Number of users doubled

U = 2 × 5 = 10 users
A = 10 × 0.25 = 2.5 Erlangs

From the table, N = 5 and A = 2.50 corresponds to B = 7 %. (Erlang B formula gives 6.97 %.)

If 5.44 users are doubled (10.88 users, A = 2.72 E), blocking is about 8.7 %.

Is it acceptable? No. The blocking probability rises from 1 % to about 7 %, i.e. 7 of every 100 call attempts fail in the busy hour. This is well above the usual design GOS of 1–2 %, so service quality is unacceptable. More channels are needed (from the Erlang B table, about 7 channels would be needed for 2.5 E at 1 %).

Note: trunking 5 channels supports 5.44 users compared with 5 × 0.04 = 0.2 users on 5 separate single channels, showing the trunking gain.

Improper vs proper handoff

A handoff starts when the signal from the serving base station falls to the handoff threshold Pr,handoff, which is kept above the minimum usable level Pr,min by a margin Δ.

 Signal  BS1                          BS2
   │ \                                /
   │   \ A: Pr,handoff reached     /
   │     \                  ____/
   │.......\ B: Pr,min ....../...........
   │  (improper: call drops at B)
   │  (proper: transferred to BS2 at C, before B)
   └───────────────────────────────── distance

Too large a Δ is also bad: it gives unnecessary handoffs and loads the MSC.

Cellular concept for N = 7

In the cellular concept, the service area is divided into small hexagonal cells, each served by a low-power base station. The total S channels are divided among N = 7 cells forming a cluster, so each cell gets S/7 channels. The cluster is repeated across the area (frequency reuse).

        ___
    ___/ 2 \___
   / 7 \___/ 3 \
   \___/ 1 \___/
   / 6 \___/ 4 \
   \___/ 5 \___/
       \___/

• N = 7 is obtained with i = 2, j = 1 (N = i² + ij + j² = 4 + 2 + 1).
• Reuse ratio Q = √(3×7) = 4.58, so co-channel cells are 4.58R apart.
• With n = 4 and six interferers, S/I = 4.58⁴/6 ≈ 73.5 = 18.7 dB, which meets the 18 dB needed by AMPS. That is why N = 7 is the classic choice.
• Frequency reuse factor = 1/7.

Handoff margin Δ is the difference between the signal level at which a handoff is initiated and the minimum usable signal level for acceptable voice quality:

Δ = Pr,handoff − Pr,minimum usable

 Received
 power (dBm)  BS1                     BS2
    │ \                               /
 Pr,handoff ─ ─\─ ─ ─ ─ ─ ─ ─ ─ ─ ─ /─ ─
    │  ↑ Δ       \                /
 Pr,min ─ ─ ─ ─ ─ ─\─ ─ ─ ─ ─ ─ /─ ─ ─ ─
    │                \         /
    └──────────────────────────────── distance
     BS1         handoff region    BS2

• Typical minimum usable level: −90 to −100 dBm.
• If Δ is too large, unnecessary handoffs occur, loading the MSC.
• If Δ is too small, there may not be enough time to complete the handoff before the signal falls below Pr,min, and the call drops.
• Δ depends on handoff processing time: 1G systems needed about 6–12 dB; 2G systems with mobile-assisted handoff need only about 0–6 dB.

Spectral efficiency of a cellular system is the traffic (or number of channels) carried per unit bandwidth per unit area, e.g. channels/MHz/km². It depends directly on how often the same frequency is reused in space.

Relation between reuse distance and cluster size:

D = R √(3N)     →     N = (D/R)² / 3

A smaller reuse distance D (for a given R) means a smaller cluster size N.

Effect on capacity: with S total duplex channels,

Channels per cell  k = S / N
Total capacity     C = M × S = (number of cells / N) × S

So C ∝ 1/N ∝ 1/D². If D is minimised:

1. Each cell gets more channels (S/N larger).
2. The same frequency is reused more times over the same area (more clusters M).
3. More users are served per MHz per km², so spectral efficiency is maximised.

Example: with S = 420 channels and 70 cells:

Limit: D cannot be reduced indefinitely, because co-channel interference rises:

S/I = (√(3N))ⁿ / i₀ = (D/R)ⁿ / i₀

Therefore the minimum reuse distance that still meets the required S/I (e.g. 18 dB for AMPS, giving N = 7 for n = 4) gives the maximum spectral efficiency. Sectoring, power control and better modulation allow a smaller D and hence higher efficiency.

Given: N = 7, area of each hexagonal cell = 23 km².

Step 1: Cell radius from hexagon area

A = (3√3/2) R² = 2.598 R²
R² = 23 / 2.598 = 8.853
R  = 2.975 km

Step 2: Co-channel reuse ratio

Q = √(3N) = √21 = 4.583

Step 3: Minimum co-channel distance

D = Q × R = 4.583 × 2.975 = 13.63 km

Answer: The minimum distance between centres of co-channel cells is about 13.63 km.

Handoff (handover) is the process of transferring an ongoing call from one channel or base station to another without interrupting the call, when the mobile moves into a different cell or the signal quality falls.

Handoff process

1. The base station (or mobile) continuously monitors received signal strength (RSSI) and quality.
2. When the signal from the serving cell falls to a handoff threshold (Pr,handoff), a handoff is initiated.
3. Neighbouring base stations measure the mobile's signal (1G locator receiver) or the mobile measures neighbours' signals and reports them (MAHO in 2G).
4. The MSC selects the best target cell, finds a free channel there and instructs the mobile to retune.
5. The call is switched to the new base station before the signal reaches the minimum usable level (Pr,min); otherwise the call drops.

 Power   BS1 signal             BS2 signal
   │ \                         /
   │   \ ← Pr,handoff          /
   │     \  Δ               /
   │       \ ← Pr,min     /
   └──────────────────────────── distance
   BS1          handoff         BS2

The margin Δ = Pr,handoff − Pr,min must be chosen carefully: too large causes unnecessary handoffs, too small causes dropped calls.

Types of handover with applications

1. Hard handover ("break before make") – the old link is released before the new one is set up; the mobile changes frequency or time slot. Used in FDMA/TDMA systems: AMPS, GSM, LTE.
2. Soft handover ("make before break") – the mobile is connected to two or more base stations at the same time on the same frequency, and the MSC selects or combines the best signal. Used in CDMA systems: IS-95, CDMA2000, WCDMA. Fewer dropped calls.
3. Softer handover – between two sectors of the same base station in CDMA; signals combined at the base station.
4. Intra-cell handover – change of channel within the same cell due to interference; used in GSM.
5. Inter-cell (intra-MSC) handover – between cells under the same MSC.
6. Inter-MSC / intersystem handover – when the mobile moves to a cell controlled by another MSC or system (e.g. 4G to 3G when LTE coverage ends, or roaming).
7. Mobile-controlled, network-controlled and mobile-assisted (MAHO) handover – based on who measures and decides; MAHO is used in GSM, where the mobile reports neighbour cell power levels and the network decides.

Assumptions: all base stations transmit equal power, path loss exponent n is the same everywhere, the mobile is at the cell edge (distance R from its own base station), and only the first tier of i₀ co-channel cells at distance D is considered.

Signal-to-interference ratio:

S / I = S / Σ(i=1..i₀) Iᵢ

Received power falls with distance as d⁻ⁿ (log-distance path loss):

Pr = P₀ (d/d₀)⁻ⁿ
S  ∝ R⁻ⁿ
Iᵢ ∝ Dᵢ⁻ⁿ

Substituting:

S / I = R⁻ⁿ / Σ(i=1..i₀) Dᵢ⁻ⁿ

All first-tier interferers at equal distance Dᵢ = D:

S / I = R⁻ⁿ / (i₀ D⁻ⁿ)
      = (D/R)ⁿ / i₀

Using the co-channel reuse ratio D/R = Q = √(3N):

S / I = (√(3N))ⁿ / i₀

      I       I
        \   /
   I ── [BS]·Mobile ── I     i₀ = 6 first-tier
        /   \              co-channel cells at D
      I       I

Example: N = 7, n = 4, i₀ = 6:

S/I = (√21)⁴ / 6 = 441 / 6 = 73.5 = 18.66 dB

which meets the 18 dB requirement of AMPS. For sectoring, i₀ = 2 (120°) or 1 (60°).

The paper as printed does not give the data for this question, so a typical set of values is assumed below to show the full method:

• Service area = 1765 km², cell radius R = 1.5 km (hexagonal cells)
• Required S/I = 18 dB, path loss exponent n = 4, omnidirectional antennas (i₀ = 6)
• Total available duplex channels S = 660 (e.g. 33 MHz with 2 × 25 kHz per duplex channel)

a) Cell cluster size

Find the smallest valid N (N = i² + ij + j²) with S/I = (√(3N))ⁿ / i₀ ≥ 18 dB:

N = 3 : (√9)⁴ / 6  = 13.5  → 11.30 dB ✗
N = 4 : (√12)⁴ / 6 = 24.0  → 13.80 dB ✗
N = 7 : (√21)⁴ / 6 = 73.5  → 18.66 dB ✓

N = 7.

b) Number of cell clusters in the service area

Area of one cell = 2.598 R² = 2.598 × 1.5² = 5.846 km²
Number of cells  = 1765 / 5.846 = 301.9 ≈ 302 cells
Number of clusters = 302 / 7 = 43.1 ≈ 43 clusters

c) Maximum number of users at any instant

Channels per cell = 660 / 7 = 94.3 ≈ 94
Max users = 94 × 302 = 28 388
(or by clusters: 43 × 660 = 28 380)

About 28 400 users can be served simultaneously (one user per channel).

General method: (1) choose N from the S/I requirement, (2) number of cells = area / (2.598 R²), clusters = cells / N, (3) maximum simultaneous users = (S/N) × number of cells = M × S.

Handover (handoff) is the process of transferring an ongoing call or data session from one channel or base station to another without interruption, when the mobile moves into another cell or the signal quality falls.

Process

• The serving base station monitors signal strength. When it falls to the handoff threshold Pr,handoff, the MSC looks for a better neighbouring cell and a free channel.
• The call must be transferred before the signal falls to the minimum usable level Pr,min. The difference Δ = Pr,handoff − Pr,min is the handoff margin; too large Δ causes unnecessary handoffs, too small Δ causes dropped calls.
• In 1G, base stations measured the mobile's signal; in 2G (GSM) the mobile measures neighbours and reports them (mobile-assisted handoff, MAHO).

 Power  BS1                BS2
  │ \                     /
  │   \ Pr,handoff      /
  │     \  Δ         /
  │.......\ Pr,min /........
  └──────────────────────── distance

Types

• Hard handover (break before make) – GSM, LTE.
• Soft handover (make before break) – CDMA, WCDMA; connected to several base stations at once.
• Intra-cell, inter-cell, inter-MSC (intersystem) handover.

Prioritising handover: guard channels reserved for handoffs and queuing of handoff requests, because dropping a call is worse than blocking a new one.

Practical issues: umbrella cells for high-speed users, cell dragging, and dwell time variation due to fading.

Given: city area = 1500 km²; 12-cell reuse pattern (N = 12); R = 1.387 km; total spectrum = 28.5 MHz; full duplex channel bandwidth = 25 kHz (the "25 MHz" in the paper must be 25 kHz); GOS = 2 % (Erlang B); Au = 0.03 Erlang per user; traffic per cell = 84 Erlangs.

(a) Number of cells in the service area

Area of one cell = 2.598 R² = 2.598 × 1.387²
                 = 2.598 × 1.9238 = 4.998 ≈ 5 km²
Number of cells  = 1500 / 4.998 = 300.1 ≈ 300 cells

(b) Number of channels per cell

Total channels = 28.5 MHz / 25 kHz = 1140 channels
Channels/cell  = 1140 / 12 = 95 channels

(c) Maximum carried traffic

For 95 channels at 2 % GOS, the Erlang B chart gives about 84 Erlangs per cell (as given).

Max carried traffic = 300 × 84 = 25 200 Erlangs

(d) Total number of users for 2 % GOS

Users = Total traffic / Au = 25 200 / 0.03 = 840 000 users

(e) Number of mobiles per unique channel

= 840 000 / 1140 = 736.8 ≈ 737 mobiles per channel

(Each unique channel is reused in 300/12 = 25 cells.)

(f) Theoretical maximum users served at one time

All channels in all cells busy at once:

= 95 × 300 = 28 500 users

This is about 28 500 / 840 000 = 3.4 % of the subscribers.

Frequency reuse is the technique of using the same set of radio channels in different cells that are far enough apart, so that co-channel interference stays within a tolerable limit. It lets a cellular system with a limited spectrum serve a very large number of users.

Frequency reuse concept

• The total S duplex channels are divided into N groups. Each cell in a group of N neighbouring cells (a cluster) gets k = S/N channels, so a cluster uses the full spectrum once.
• The cluster is then repeated across the service area. Cells with the same letter (same channel set) are co-channel cells.
• Each base station uses low power and antennas placed so that coverage stays roughly inside its own cell.
• For hexagonal cells, cluster size is N = i² + ij + j² (i, j = 0, 1, 2, …), so N = 1, 3, 4, 7, 9, 12, …
• If the cluster is repeated M times, total capacity is C = M·k·N = M·S. Smaller N means more repetitions and more capacity, but co-channel cells come closer and interference increases.
• The frequency reuse factor is 1/N, since each cell gets 1/N of the channels.

      B   C          B   C
    G   A   D  ...  G   A   D
      F   E          F   E
   cluster 1 (N=7)  cluster 2
  Cell A in both clusters uses the same channel set

Co-channel reuse ratio

The co-channel reuse ratio Q is the ratio of the distance D between the centres of the nearest co-channel cells to the cell radius R:

Q = D / R = √(3N)

• D = R√(3N) for hexagonal geometry, so Q depends only on cluster size.
• Large Q (large N): co-channel cells are far apart, so co-channel interference is low and transmission quality is better, but each cell has fewer channels and capacity is lower.
• Small Q (small N): more channels per cell and higher capacity, but more co-channel interference.
• Q links directly to signal-to-interference ratio. With path loss exponent n and i₀ first-tier interferers: S/I ≈ (√(3N))ⁿ / i₀ = Qⁿ / i₀.

Example: for an S/I requirement of 18 dB with n = 4 and i₀ = 6, N = 7 (Q = 4.58) gives S/I = 4.58⁴/6 ≈ 73.5 = 18.66 dB, which is acceptable.

Proof that Q = √(3N)

In a hexagonal layout, to reach the nearest co-channel cell you move i cells along one chain of hexagons, turn 60°, and move j cells more (this is how N = i² + ij + j² is defined).

• Distance between centres of two adjacent hexagons = √3·R (R = centre-to-vertex radius).
• Using axes 60° apart and the cosine rule on the triangle with sides i·√3R and j·√3R and the angle between them 120°:

D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R)cos120°
   = 3R²(i² + j² + ij)
   = 3R²·N
D  = R√(3N)
Q  = D/R = √(3N)

start --(i cells)--> turn 60° --(j cells)--> co-ch cell
        \___________________ D ______________________/

Optimal N for S/I = 15 dB, n = 4

Required S/I = 15 dB = 10^1.5 = 31.62. Using S/I = Qⁿ / i₀ = (√(3N))⁴ / i₀ = (3N)²/i₀. Valid N: 1, 3, 4, 7, 9, 12, …

a) Omnidirectional antennas (i₀ = 6)

(3N)²/6 ≥ 31.62  →  3N ≥ 13.77  →  N ≥ 4.59
N = 4: Q = 3.464, S/I = 144/6 = 24
       = 13.80 dB < 15 (fails)
N = 7: Q = 4.583, S/I = 441/6 = 73.5
       = 18.66 dB > 15 (ok)

N = 7

b) 120° sectoring (i₀ = 2)

(3N)²/2 ≥ 31.62  →  3N ≥ 7.95  →  N ≥ 2.65
Try N = 3: Q = 3, S/I = 81/2 = 40.5 = 16.07 dB > 15 (ok)

(The exact worst-case geometry, S/I = 1/[Q⁻⁴ + (Q+0.7)⁻⁴], gives 17.52 dB for N = 3, also acceptable.) N = 3

c) 60° sectoring (i₀ = 1)

(3N)² ≥ 31.62  →  3N ≥ 5.62  →  N ≥ 1.87
Try N = 3: S/I = 81 = 19.08 dB > 15 (ok)
(N = 1 gives 9 = 9.54 dB, fails)

N = 3

Effect of trunking efficiency

Sectoring splits each cell's channels among the sectors, so each sector is a smaller trunk and trunking efficiency drops. 60° sectoring gives the same N = 3 as 120° sectoring, but splits channels into 6 groups instead of 3, losing more trunking efficiency for no gain in cluster size.

Answer: N = 7 (omni), N = 3 (120°), N = 3 (60°). Considering trunking efficiency, 120° sectoring with N = 3 is the best choice.

As demand grows, the channels per cell become insufficient. Capacity and coverage are improved mainly by cell splitting, sectoring, the microcell zone concept, and repeaters for coverage.

1. Cell splitting

• A congested cell is divided into smaller cells (microcells), each with its own base station, lower antenna height and lower transmit power.
• Since more cells fit in the same area, the cluster is repeated more times and channel reuse increases.
• If the radius is halved (R → R/2), the number of cells in the area becomes about 4 times, so capacity is about 4 times.
• Power of the new cell: to keep the same received power at the edge, Pt2 = Pt1/2ⁿ; for n = 4, Pt2 = Pt1/16 (12 dB less).
• Drawbacks: more base stations, more handoffs, and old and new cells must coexist during the transition.

2. Sectoring

• A single omnidirectional antenna is replaced by 3 directional antennas (120° sectors) or 6 antennas (60° sectors). Each sector uses a subset of the cell's channels.
• A sector receives interference from only part of the first-tier co-channel cells: i₀ drops from 6 to 2 (120°) or 1 (60°).
• Higher S/I allows a smaller cluster size (e.g. N = 7 → N = 4 or 3), which raises capacity.
• Drawbacks: more antennas, loss of trunking efficiency, and extra handoffs between sectors.

     120° sectoring          Interferers seen
        /\                   Omni : 6
       /α \                  120° : 2
      /----\                 60°  : 1
     | β  γ |

3. Microcell zone concept

• A cell is divided into zones (usually 3). Each zone site is connected to a single base station by fibre or microwave.
• The mobile is served by the zone with the strongest signal; when it moves between zones in the same cell, the base station just switches zones and no handoff at the MSC is needed. The channel stays the same.
• Since only one zone transmits on a channel at a time, interference is local, D/R improves and capacity can rise (about 2.33 times in the textbook example) without extra handoffs.

4. Coverage enhancement: repeaters

• Repeaters (range extenders) receive, amplify and retransmit signals to cover holes, tunnels, buildings and remote roads.
• Distributed antenna systems and in-building systems improve indoor coverage.

Co-channel vs adjacent channel interference

Co-channel interference (CCI) is caused by signals from other cells that use the same frequency. Adjacent channel interference (ACI) is caused by signals on neighbouring frequencies that leak into the passband of the wanted channel.

Proof that Q = √(3N)

To reach the nearest co-channel cell, move i cells along a chain of hexagons, turn 60° and move j cells. Centre-to-centre distance of adjacent hexagons = √3R. Applying the cosine rule (angle between the two moves = 120°):

D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R)cos120°
   = 3R²(i² + ij + j²) = 3R²N
D  = R√(3N)
Q  = D/R = √(3N)

Channel calculation

Given: total spectrum = 20 MHz, simplex channel = 25 kHz, so a duplex channel needs 2 × 25 = 50 kHz.

a) Duplex channels = 20 MHz / 50 kHz
                   = 20 000 kHz / 50 kHz = 400

b) Channels per cell (N = 4) = 400 / 4 = 100

Answer: a) 400 duplex channels; b) 100 duplex channels per cell.

Handoff (handover) is the process of transferring an ongoing call from one channel or base station to another without interrupting the call, as the mobile moves from one cell into another or when signal quality falls.

Basic handoff idea

• Each BS watches signal level. A handoff threshold is set a little above the minimum usable level:

Δ = Pr,handoff − Pr,minimum usable

• If Δ is too large, there are unnecessary handoffs; if too small, the call may drop before handoff completes.
• The system checks that the drop in signal is not just momentary fading, so it averages the signal over time (dwell time).

 Signal
   |  BS1 signal          BS2 signal
   |\                        /
   | \______ handoff ______/
   |   \    threshold    /
   |----\--------------/----- minimum usable level
   +------------------------------ distance
   BS1           A   B          BS2
   (handoff made between A and B)

Handoff strategy in GSM: mobile assisted handoff (MAHO)

GSM uses Mobile Assisted Handoff and hard handoff (break-before-make).

1. The BSC sends the MS a list of neighbour cells (BCCH allocation list).
2. In idle TDMA time slots, the MS measures the received level (RXLEV) of up to 32 neighbours, and RXLEV and RXQUAL (bit error quality) of the serving cell.
3. The MS reports the six strongest neighbours plus serving cell values to the BTS on the SACCH about every 480 ms.
4. The BTS adds its own uplink measurements and timing advance (distance).
5. The BSC decides when to hand off, based on level, quality, distance or traffic load, and selects a target cell.
6. The target channel is reserved, the MS is commanded to switch, it sends access bursts on the new channel, and the old channel is released.

Because the MS does the measuring, handoff decisions are faster (about 1–2 s) than in network-controlled handoff of first-generation systems.

Types of GSM handoff

Prioritizing handoffs

• Guard channels: some channels are reserved only for handoff requests.
• Queuing of handoff requests: a request is queued while the mobile is still in the overlap area.
• Umbrella cells are used for fast-moving users to reduce the number of handoffs.