Give step-by-step solution with explanation and final answer:
total of
3
4
.
6
5
MHz of bandwidth is allocated to a particular FDD cellular telephone system that uses two
2
5
kHz simplex channel

Question

Give step-by-step solution with explanation and final answer:
total of 
3
4
.
6
5
 ﻿MHz of bandwidth is allocated to a particular FDD cellular telephone system that uses two 
2
5
 ﻿kHz simplex channels to provide full duplex voice and control channels. Assume Erlang B with blocking probability 
1
%
 ﻿and a seven
-
cell reuse pattern for the system.
(
5
 ﻿pts
)
a
.
 ﻿Determine the number of channels allocated in each cell.
b
.
 ﻿What is the traffic capacity, 
A
,
 ﻿in Erlangs per cell?
c
.
 ﻿Determine the number of users that can be supported per cell if an average call for a user takes three minutes and the average call rate per user, 
λ
=
1
 ﻿calls per hour.
d
.
 ﻿If each cell covers five square kilometers, then how many subscribers could be supported in an urban market that is 
5
0
k
m
×
5
0
k
m
 ﻿for the case of omnidirectional base station antennas?
e
.
 ﻿If each cell covers five square kilometers, then how many subscribers could be supported in an urban market that is 
5
0
k
m
×
5
0
k
m
 ﻿for the case of 
1
2
0
°
 ﻿sectored antennas? What is the percentage degradation in trunking efficiency when going from the omnidirectional unsectored system to 
1
2
0
°
 ﻿sectoring
?

VIPSolutions · Accepted Answer

# Step-by-Step Solution

Given:

**Total bandwidth:** $ .46\,\text{MHz} $
- **Channel bandwidth (simplex):** $ 25\,\text{kHz} $ (for full duplex, use two simplex channels)
- **Blocking probability:** $ 1\% $ ($ B = .01 $), use Erlang B
- **Reuse pattern:** 7 cells ($ N = 7 $)
- **Average call duration:** 3 minutes
- **Average call rate per user:** $ \lambda = 1 $ call/hour
- **Cell area:** $ 5\,\text{km}2 $
- **Market area:** $ 50\,\text{km} \times 50\,\text{km} = 250\,\text{km}^2 $
- **Antenna sectorization:** Omnidirectional (part d), 120° sectors (part e)

---

## a. Number of Channels Allocated per Cell

### **Step 1: Find Total Number of Full Duplex Channels**

- Each full duplex channel needs $ 2 \times 25\,\text{kHz} = 50\,\text{kHz} $
- Total number of full duplex channels:
  $$
  \text{Total channels} = \frac{\text{Total BW}}{\text{BW per full duplex channel}} = \frac{3,460\,\text{kHz}}{50\,\text{kHz}} = 69.2
  $$
  - So, **69 full duplex channels** (since you can't have a fraction of a channel).

### **Step 2: Channels per Cell (with 7-cell reuse)**

- Each cell gets $ \frac{\text{Total channels}}{N} $:
  $$
  \text{Channels per cell} = \frac{69}{7} \approx 9.857
  $$
- **Final Answer:**
  - **Each cell is allocated 9 channels** (rounded down to the nearest whole channel).

---

## b. Traffic Capacity $ A $ in Erlangs per Cell

- Use Erlang B formula with:
  - $ C = 9 $ channels
  - $ B = .01 $ (1% blocking)

Find the maximum offered traffic ($ A $) such that $ B(9, A) = .01 $.

### **Step 1: Find $ A $ using Erlang B table**

From Erlang B tables (or online calculators):

- For $ C = 9 $, $ B = .01 $ → $ A \approx 5. $ Erlangs

### **Final Answer:**
- **Traffic capacity per cell is $\boxed{5.}$ Erlangs**

---

## c. Number of Users Supported per Cell

Given:
- Average call duration: 3 min ($ = .05 $ hours)
- Average call rate per user: $ \lambda = 1 $ call/hour
- Traffic per user ($ a $):
  $$
  a = \lambda \times \text{average call duration (in hours)} = 1 \times .05 = .05\,\text{Erlangs/user}
  $$
- Total traffic per cell: $ 5. $ Erlangs

### **Step 1: Users Supported**
$$
\text{Number of users} = \frac{\text{Total traffic}}{\text{Traffic per user}} = \frac{5.}{.05} = 100
$$

### **Final Answer:**
- **Each cell can support 100 users**.

---

## d. Subscribers Supported in a $ 50 \,\text{km} \times 50\,\text{km} $ Market (Omnidirectional)

- Market area: $ 250\,\text{km}^2 $
- Area per cell: $ 5\,\text{km}^2 $
$$
\text{Number of cells} = \frac{250}{5} = 500
$$
$$
\text{Total users} = 500 \text{ cells} \times 100 \text{ users/cell} = 50,000
$$

### **Final Answer:**
- **50,000 subscribers** can be supported in the market.

---

## e. Subscribers Supported with $ 120^\circ $ Sectoring

### **Step 1: Effect of Sectoring**

- $ 120^\circ $ sectoring divides each cell into 3 sectors.
- Each sector gets its own set of channels (i.e., frequency reuse factor is improved, interference is reduced).
- Typically, the reuse factor can be reduced (reuse pattern can be smaller).

Assuming sectoring allows the reuse factor to drop from $ N = 7 $ to $ N = 4 $:

$$
\text{Channels per cell with sectoring} = \frac{69}{4} = 17.25 \to 17 \text{ channels per cell}
$$

### **Step 2: New Traffic Capacity per Cell**

For $ C = 17 $ channels, $ B = .01 $:

Look up Erlang B table:

- $ C = 17 $, $ B = .01 $ → $ A \approx 11. $ Erlangs

$$
\text{Users per cell} = \frac{11.}{.05} = 220
$$

$$
\text{Total users} = 500 \text{ cells} \times 220 = 110,000
$$

### **Step 3: Trunking Efficiency Degradation**

Trunking efficiency is related to the number of channels per trunk; as channels per sector decrease (since each cell's channels are split into sectors), the trunking efficiency drops.

But here, with sectoring, total channels/cell increased, so trunking efficiency **improves**, not degrades.

#### **Percentage Degradation Calculation**

- Users per cell (unsectored): $ 100 $
- Users per cell (sectored): $ 220 $
- **Trunking efficiency improves by:**

$$
\text{Percentage improvement} = \frac{220 - 100}{100} \times 100\% = 120\%
$$

So, **there is no degradation; there is a 120% improvement in trunking efficiency**.

### **Final Answers:**

- **110,000 subscribers** can be supported with $ 120^\circ $ sectoring in the market.
- **No degradation** in trunking efficiency; instead, **trunking efficiency improves by 120%** when going from an unsectored to a $ 120^\circ $ sectored system.

---

# **Summary Table**

| Part | Parameter                     | Value / Calculation           | Answer                                      |
|------|-------------------------------|------------------------------|---------------------------------------------|
| a    | Channels per cell             | $ \lfloor 69/7 \rfloor $   | 9                                           |
| b    | Traffic per cell              | Erlang B, $ C=9, B=.01 $  | 5. Erlangs

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