Overview of Wireless Networks: Introduction

Overview of Wireless Networks: Introduction

Overview of Wireless Networks: Existing Network Infrastructure
Public Switched Telephone Network (PSTN): Voice Internet: Data Hybrid Fiber Coax (HFC): Cable TV

Overview of Wireless Networks: Market Sectors for Applications
Four segments divided into two classes: voice-oriented and data-oriented, further divided into local and wide-area markets Voice: Local: low-power, low-mobility devices with higher QoS – cordless phones, Personal Communication Services (PCS) Wide area: high-power, comprehensive coverage, low QoS - cellular mobile telephone service Data: Broadband Local and ad hoc: WLANs and WPANs (WPAN-Wireless Personal Area Network) Wide area: Internet access for mobile users

Overview of Wireless Networks: Evolution of Voice-Oriented Services
Year Event Early 1970s First generation of mobile radio at Bell Labs Late 1970s First generation of cordless phones 1982 First generation Nordic analog NMT 1983 Deployment of US AMPS 1988 Initiation of GSM development (new digital TDMA) 1991 Deployment of GSM 1993 Initiation of IS-95 standard for CDMA 1995 PCS band auction by FCC 1998 3G standardization started FDMA – Frequency Division Multiple Access NMT – Nordic Mobile Telephony AMPS – Advanced Mobile Phone System GSM – Global System for Mobile Communication TDMA – Time Division Multiple Access IS-95 – Interim Standard 95 CDMA – Code Division Multiple Access PCS – Personal Communication System FCC – Federal Communication Commission

Overview of Wireless Networks: Evolution of Data-Oriented Services
Year Event 1979 Diffused infrared (IBM Rueschlikon Lab - Switzerland Early 1980s Wireless modem (Data Radio) 1990 IEEE for Wireless LANs standards Announcement of Wireless LAN products 1992 HIPERLAN in Europe 1993 CDPD (IBM and 9 operating companies) 1996 Wireless ATM Forum started 1997 U-NII bands released, IEEE completed, GPRS started 1998 IEEE b and Bluetooth announcement 1999 IEEE a/HIPERLAN-2 started HIPERLAN – High Performance Radio LAN CDPD – Cellular Digital Packet Data U-NII – Unlicensed National Information Infrastructure GPRS – General Packet Radio Service

Overview of Wireless Networks: Different Generations
1G Wireless Systems: Analog systems Use two separate frequency bands for forward (base station to mobile) and reverse (mobile to base station) links: Frequency Division Duplex (FDD) AMPS: United States (also Australia, southeast Asia, Africa) TACS: EU (later, bands were allocated to GSM) NMT-900: EU (also in Africa and southeast Asia) Typical allocated overall band was 25 MHz in each direction; dominant spectra of operation was 800 and 900 MHz bands. AMPS – Advanced Mobile Phone System TACS – Total Access Communication System NMT – Nordic Mobile Telephony

2G Wireless Systems: Four sectors Digital cellular GSM (EU/Asia): TDMA IS-54 (US): TDMA IS-95 (US/Asia): CDMA PCS – residential applications CT-2 (EU,CA): TDMA/TDD DECT(EU):TDMA/TDD PACS (US): TDMA/FDD CT-2 – Cordless Telephone 2 DECT – Digital Enhanced Cordless Telephone PACS – Personal Access Communication System

2G Wireless Systems: Four sectors (cont’d) Mobile data CDPD shares AMPS bands and site infrastructure; GPRS shares GSM’s radio system - data rates suitable for Internet WLAN – Unlicensed bands, free of charge and rigorous regulations: very attractive! IEEE and IEEE b use DSSS physical layer; HIPERLAN/1 uses GMSK; IEEE a and HIPERLAN/2 use OFDM: next generation CDPD – Cellular Digital Packet Data GPRS – General Packet Radio Service DSSS – Direct Sequence Spread Spectrum GMSK – Gaussian Minimum Shift Keying OFDM – Orthogonal Frequency Division Multiplexing

3G and Beyond Purpose: develop an international standard that combines and gradually replaces 2G digital cellular, PCS, and mobile data services, at the same time increasing the quality of voice, capacity of the network, and data rate of the mobile data services. Radio transmission technology of choice: W-CDMA 3G was envisioned to provide multimedia services to users everywhere

3G and Beyond WLANs provide broadband services in hot spots WPANs connect personal devices together: laptop, cellular phone, headset,speakers, printers WLAN and WPAN are the future of broadband and ad hoc wireless networks WPAN’s first standard: bluetooth – lower rates than WLAN but uses a voice-oriented wireless access method for integration of voice and data services

Relative coverage, mobility, and data rates of generations of cellular systems and local broadband and ad hoc networks.

Operation of Wireless Networks
Getting familiar with terms: MS/MT: Mobile Station/Mobile Terminal BS: Base Station MSC: Mobile Switching Center HLR: Home Location Register (database) VLR: Visitor Location Register (database) Cellular Network Architecture Mobility Management

Cellular Network Architecture
Location Register (Database) Mobile Switching Center Radio Network Base Station Controller MSC Backbone Wireline Network Mobile Terminal Base Station Cell

BASIC ARCHITECTURE Home Location Register (HLR)
BACKBONE TELEPHONE NETWORK Visitor Location Register (VLR) Mobile Switching Center MSC (MSC) VLR Mobile Terminal (MT) Local Signaling Long Distance Signaling

Cellular Concept A CELL is the radio coverage area by a Base Station (BS). The most important factor is the SIZE and the SHAPE of a CELL. Ideally, the area covered a by a cell could be represented by a circular cell with a radius R from the center of a BS. Many factors may cause reflections and refractions of the signals, e.g., elevation of the terrain, presence of a hill or a valley or a tall building and presence in the surrounding area. The actual shape of the cell is determined by the received signal strength. Thus, the coverage area may be a little distorted. We need an appropriate model of a cell for the analysis and evaluation. Many posible models: HEXAGON, SQUARE, EQUILATERAL TRIANGLE.

Cell Shape (a) Ideal Cell (b) Actual Cell (c) Different Cell Models R

Size and Capacity of a Cell per Unit of Area and Impact of the Cell Shape on System Characteristics

Cellular Concept - Example
Consider a high-power transmitter that can support 35 voice channels over an area of 100 km2 with the available spectrum If 7 lower power transmitters are used so that they support 30% of the channels over an area of 14.3 km2 each. Then a total 7*30% * 35 = 80 channels are available instead of 35. 2 3 1 7 4 6 5

Cellular Concept If two cells are far away from enough that the same set of frequencies can be used in both cells, it is called frequency reuse. With frequency reuse, a large area can be divided into small areas, each uses a subset of frequencies and covers a small area. With frequency reuse, the system capacity can be expanded without employing high power transmitters.

Capacity Expansion by Frequency Reuse
Same frequency band or channel used in a cell can be “REUSED’ in another cell as long as the cells are far apart and the signal strength do not interfere with each other. This enhances the available bandwidth of each cell. A group of cells that use a different set of frequencies in each cell is called a cell cluster.

NUMBER OF CELLS IN A CLUSTER

CELL CLUSTER

FREQUENCY REUSE Example: A typical cluster of 7 such cells and 4 such clusters with no overlapping area F1 F2 F3 F4 F5 F6 F7 F1 F2 F3 F4 F5 F6 F7 |------ | |D F1 F2 F3 F4 F5 F6 F7 F1 F2 F3 F4 F5 F6 F7  FREQUENCY REUSE DISTANCE D

RULE to Determine the Nearest Co-Channel Neighbors Determining the Cluster Size
j To find nearest co-channel neighbors of a particular cell Step 1: Move i cells along any chain of hexagons; Step 2: Turn 60 degrees counterclockwise and move j cells. i and j measure the number of nearest neighbors between co-channel cells The cluster size, N, N = i2+ij+j2 i 3 2 1 4 If i =2 and j = 0, then N = 4 If i = 2 and j = 1, then N =7 2 1 3 1 2

RULE to Determine the Nearest
Co-Channel Neighbors Determining the Cluster Size

Frequency Reuse The distance between 2 cells using the same channel is known as the REUSE DISTANCE D. There is a close relationship between D, R (radius of each cell) and N (the number of cells in a cluster) D = (sqrt 3N) . R The REUSE FACTOR is then D/R = sqrt (3N)

Frequency Reuse Let N be the cluster size in terms of number of cells within it and K be the total number of available channels without frequency reuse. N cells in the cluster would then utilize all K available channels. Each cell in the cluster then uses 1/N-th of the total available channels. N is also referred as the frequency reuse factor of the cellular system.

Capacity Expansion by Frequency Reuse
Assume each cell is allocated J channels (J<=K). If the K channels are divided among the N cells into unique and disjoint channel groups, each with J channels, then K = J N The N cells in a cluster use the complete set of available frequencies. The cluster can be replicated many times. Let M be the number of replicated clusters and C be the total number of channels in the entire system with frequency reuse, then C is the system capacity and computed by C = M J N

Cellular System Capacity - Example
Suppose there are 1001 radio channels, and each cell is Acell = 6 km2 and the entire system covers an area of Asys = 2100km2. Calculate the system capacity if the cluster size is 7. How many times would the cluster of size 4 have to be replicated in order to approximately cover the entire cellular area? Calculate the system capacity if the cluster size is 4. Does decreasing the cluster size increase the system capacity? Solution: 1. J=K/N=143, Acluster=N*6=42km2, M=2100/42=50, C=MJN=50,050 chs. 2. N=4, Ac=4*6=24km2, M=2100/24=87. 3. N=4, J = 1001/4 = 250 chs/cell. C = 87 * 250 * 4 = 87,000 chs. Decrease in N from 7 to 4 increase in C from 50,050 to 87,000.  Decreasing the cluster size increases system capacity. So the answer is YES!

Geometry of Hexagonal Cells (1) How to determine the DISTANCE between the nearest co-channel cells ?
Planning for Co-channel cells D is the distance to the center of the nearest co-channel cell R is the radius of a cell j D i 30o R

Geometry of Hexagonal Cells (2)
Let D be the actual distance between two centers of adjacent co-channel cells where D= Let Dnorm be the distance from the center of a candidate cell to the center of a nearest co-channel cell, normalized with respect to the distance between the centers of two adjacent cells, . Note that the normalized distance between two adjacent cells either with (i=1,j=0) or (i=0,j=1) is unity.

Let D be the actual distance between the centers of two adjacent co-channel cells. D is a function of Dnorm and R. From the geometry we have From N and Dnorm equations

With the actual distance between the centers of two adjacent hexagonal cells, the actual distance between the center of the candidate cell and the center of a nearest co-channel is then For hexagonal cells there are 6 nearest co-channel neighbors to each cell. Co-channel cells are located in tiers. In general, a candidate cell is surrounded by 6k cells in tier k. For cells with the same size the co-channel cells in each tier lie on the boundary of the hexagon that chains all the co-channel cells in that tier.

As D is the radius between two nearest co-channel cells, the radius of the hexagon chaining the co-channel cells in the k-th tier is given by k.D. For the frequency reuse pattern with i=2 and j=1 so that N=7, the first two tiers of co-channel cells are given in Figure. It can be readily observed from Figure that the radius of the first tier is D and the radius of the second tier is 2.D.

Number of Cells in A Cluster
A candidate cell has 6 nearest co-channel cells. Each of them in turn has 6 neighboring co-channel cells. So we can have a large hexagon. This large hexagon has radius equal to D which is also the co-channel cell separation. The area of a hexagon is proportional to the square of its radius, (let =2.598), R D

Number of Cells in A Cluster
The number of cells in the large hexagon is then In general the large hexagon encloses the center cluster of N cells plus 1/3 the number of the cells associated with 6 other peripheral large hexagons. Hence, the total number of cells enclosed by the large hexagon is

We assume the size of all the cells is roughly the same, as long as the cell size is fixed co-channel interference will be independent of transmitted power of each cell. The co-channel interference will become a function of q where q = D/R = sqrt (3N). q is the CO-CHANNEL REUSE RATIO and is related to the cluster size. A small value of q provides larger capacity since N is small. For large q, the transmission quality is better, smaller level of co-channel interference. By increasing the ratio of D/R spatial separation between co-channel cells relative to the coverage distance of a cell is increased.Thus, interference is reduced from improved isolation of RF energy from the nmber of cells per cluster N co-channel cells.

Furthermore, D (distance to the center of the nearest cochannel cell) is a function of NI and S/I in which NI is the number of co-channel interfering cells in the first tier and S/I = received signal to interference ratio at the desired mobile receiver.

Frequency Reuse Ratio The frequency reuse ratio, q, is defined as
q = D/R which is also referred to as the co-channel reuse ratio. Also  q = sqrt(3N) Tradeoff q increases with N. A smaller value of N has the effect of increasing the capacity of the cellular system and increasing co-channel interference Tradeoff between q and N

Interference MAJOR LIMITING FACTOR for Cellular System performance is the INTERFERENCE Implications:  CROSS TALK  Missed and Blocked Calls. SOURCES OF INTERFERENCE? Another mobile in the same cell A call in progress in neighboring cell. Other base stations operating in the same frequency band Non-cellular systems leaking energy into cellular frequency band

Interference 1. CO-CHANNEL INTERFERENCE
2. ADJACENT CHANNEL INTERFERENCE

CO-CHANNEL INTERFERENCE
Frequency Reuse  Given coverage area cells using the same set of frequencies  co-channel cell!!! Interference between these cells is called CO-CHANNEL INTERFERENCE. (Thermal noise  increase SNR and combat it). However, co-channel interference  cannot be overcome just by increasing the carrier power of a transmitter. Because increase in carrier transmit power increases the interference. Reduce co-channel interference Co-channel cells must be physically separated by a minimum distance to provide sufficient isolation.

Co-Channel Interference
Intracell Interference: interferences from other mobile terminals in the same cell. Duplex systems Background white noise Intercell interference: interferences from other cells. More evident in the downlink than uplink for reception Can be reduced by using different set of frequencies Design issue Frequency reuse Interference System capacity Bottomline: It determines link performance which in turn dictates the frequency reuse plan and overall capacity of the system.

Cell Site-to-Mobile Interference (Downlink) Mobile-to Cell-Site Interferences (Uplink)

Base  Mobile  DOWNLINK Mobile Base  UPLINK UPLINK All mobiles in 6 cells + central cell assigned to the same frequency channel DOWNLINK All base stations in 6 cells and central cell have the same frequency channel. DOTTED LINES show the interference of all 6 mobiles (all co-channel) received at central base station (interference) Actual signal is from the mobile in the center cell to its own base station. (Uplink Signal Interference ratio)

Base  Mobile  DOWNLINK CASE From the base stations (from co-channel cells) interference received by the mobile in the center cell. Desired signal is from the base to mobile in the center cell. Alarge is the area of the hexagonal cells of the large one. Asmall is the area of each cell. Alarge/Asmall  A number of cells in this each repetitous pattern (3N).

For simplicity, we consider only the average channel quality as a function of the distance dependent path loss. Signal-to-Co-channel interference ratio, (S/I), at the desired mobile receiver which monitors the forward channel is defined by S is the desired signal power from desired base station Ii interference power caused by the i-th interfering co-channel cell base station. NI is the number of co-channel interfering cells

The desired signal power S from desired base station is proportional to r-, where r is the distance between the mobile and the serving base station.  is the path loss component. The received interference, Ii, between the ith interferer and the mobile is proportional to (Di)-. The white background noise is neglected in the interference-dominant environment. Assume the transmisson powers from all base stations are equal, then we have

Consider only the first tier of interfering cells, if all interfering base stations are equidistant from the desired base station and if this distance is equal to the distance D between cell centers, then the above equation can be simplified to: (i.e., r=R and assume Di=D and use q=D/R):

Frequency reuse ratio, e.g., NI = 6 

Example: In AMPS systems; for =4, S/I = 18dB (i.e., 63.1), [20 log (S/I)  dB] are acceptable; then (assume N=6) q = (6 63.1)1/4  4.41. Thus, the cluster size N should be (from eq. q=sqrt(3N) N = q2/3 = 6.79  7. i.e.,A 7-cell reuse pattern is needed for an S/I ratio of 18dB. Based on q=D/R, we can select D by choosing the cell radius R.

An S/I of 18 dB is the measured value for the accepted voice quality from the present day cellular mobile receivers. Sufficient voice quality is provided when S/I is greater than or equal to 18dB.

Example: Co-Channel Interference
If S/I = 15 dB required for satisfactory performance for forward channel performance of a cellular system. What is the Frequency Reuse Factor? What Cluster Size should be used for maximum capacity? (Use path loss component of =3 and =4) . Assume 6 co-channels all of them (same distance from the mobile)

N= 7 and =4 The co-channel reuse ratio is q=D/R=sqrt(3N)=4.583 Or dB  greater than the minimum required level  ACCEPT IT!!! b) N= 7 and =3 Or dB  less than the minimum required level  REJECT IT!!!

We need a larger N. Use eq. N =i2+ij+j2 for i=j=2  next possible value is N=12. q=D/R=sqrt(3.N) = 6 and =3 Or dB  N=12 can be used for minimum requirement, but it decreases the capacity since 12 cell reuse offers a spectrum utilization of 1/12 within each cell.

Worst Case Co-Channel Interference i. e
Worst Case Co-Channel Interference i.e., mobile terminal is located at the cell boundary where it receives the weakest signal from its own cell but is subjected to strong interference from all all the interfering cells. We need to modify our assumption, i.e., assume Di=D. The S/I ratio can be expressed as R D-R D+R D Used D/R=q and =4. Where q=4.6 for normal seven cell reuse pattern.

Example: Worst Case Cochannel Interference (2)
A cellular system that requires an S/I ratio of 18dB. (a) If cluster size is 7, what is the worst-case S/I? (b) Is a frequency reuse factor of 7 acceptable in terms of co-channel interference? c) If not, what would be a better choice of frequency reuse ratio?

Example: Worst Case Cochannel Interference (2)
Solution (a) N=7  q = If a path loss component of =4, the worst-case signal- to-interference ratio is S/I = 54.3 or 17.3 dB. (b) The value of S/I is below the acceptable level of 18dB. To ncrease S/I  we need to decrease I, I.e., Increase the frequency reuse factor, q=5.20 by using N =9. The S/I becomes then or 19.8dB. Acceptable…

ADJACENT CHANNEL INTERFERENCE
Interference resulting from signals which are adjacent in frequency to the desired signal is called ADJACENT CHANNEL INTERFERENCE. WHY? From imperfect receiver filters (which allow nearby frequencies) to leak into the pass-band. NEAR FAR EFFECT: Adjacent channel user is transmitting in very close range to a subscriber’s receiver, while the receiver attempts to receive a base station on the desired channel. NEAR FAR EFFECT also occurs, when a mobile close to a base station transmits on a channel close to one being used by a weak mobile. Base station may have difficulty in discriminating the desired mobile user from the “bleedover” caused by the close adjacent channel mobile.

ADJACENT CHANNEL INTERFERENCE
How to reduce? Careful filtering Channel assignment no channel assignment which are all adjacent in frequency. Keeping frequency separation between each channel in a given cell as large as possible. e.g., in AMPS System there are 395 voice channels which are divided into 21 subsets each with 19 channels. In each subset, the closest adjacent channel is 21 channels away. 7-cell reuse -> each cell uses 3 subsets of channels. 3 subsets are assigned such that every channel in the cell is assured of being separated from every other channel by at least 7 channel spacings.

Cell Splitting A method to increase the capacity of a cellular system by dividing one cell into more smaller cells. Each smaller cell has its own base station and accordingly antenna height and transmission power can be reduced. Cell splitting reduces the call blocking probability because the number of channels is increased. But it increases the handoff rate, i.e., more frequent crossing of borders between the cells. We have the formula in calculating path loss: Pr(dBW) = P0(dBW) - 10 log10(d/d0) where d0 is the distance from the reference point to the transmitter, and P0 is the power received at the reference point.

Cell Splitting (2) Let Pt1 and Pt2 be the transmit power of the large cell BS and medium cell BS, respectively. The received power at the edge of large cell is Pr1 = P  log10(R/d0) The received power at the edge of large cell, Pr1 is proportional to Pt1 (R)- . The received power at the edge of R/2 cell, Pr2 is proportional to Pt2 (R/2)- . With the equal received power, we have Pt1 (R)- = Pt2 (R/2)- , i.e., Pt1/Pt2= 2  R R/2

Example – Cell Splitting
Suppose each BS is allocated 60 channels regardless of the cell size. Find the number of channels contained in a 3x3 km2 area without cell splitting, i.e., R= 1km and with cell splitting, R/2 = 0.5km. The number of cells for R=1km. Each large cell can cover 3.14km2, for 9 km2 approximately need 9/3.14 => 3 cells. However, 3 hexagon cannot cover a square of 3x3. A better approximation is 4 cells. So the number of channels is 4x60=240. With small cells, the number of cells is approximately (1/0.5)2x4 = 16. Then the number of channels is 16x60=960.

Cell Sectoring (Directional Antennas)
Omni-directional antennas allow transmission of radio signals with equal power strength in all directions. Reality is an antenna covers an area of 60 degrees or 120 degrees  DIRECTIONAL ANTENNAS!!!! Cells served by these antennas are called SECTORED CELLS!!! Many sectored antennas are mounted a BS tower located at the center of the cell and an adequate number of antennas is placed to cover the entire 360 degrees of the cell. .

CELL SECTORING Directional Antennas (Sectoring)
1 1 2 2 4 3 3 120 DEGREE SECTOR 90 DEGREE SECTOR OMNIDIRECTIONAL 6 5 1 4 2 3 60 DEGREE SECTOR

Cell Sectoring (Directional Antennas)
Advantages of Cell Sectoring: Borrowing of channels Coverage of smaller area by each antenna and hence lower power is required in transmitting radio signals. Helps to decrease interference between co-channels. Also the spectrum efficiency of the overall system is enhanced. .

Co-Channel Interference Reduction with the Use of Directional Antennas (Cell Sectoring)
The basic form of antennas are omnidirectional. Directional antennas can increase the system capacity. If we sectorize the cell with 120o in each sector, the S/I becomes: (The number of interferers is reduced from 6 to 2.) The capacity increase in (S/I)120 vs (S/I)omni is 3. 1 2 1 3 2 3

Worst-Case Scenario in 120o Sectoring
Let D be the distance between the adjacent co-channel cells. With the distance approximation and use path loss component , the signal-to-interference ratio is D+0.7R 1 2 3 D

Fixed Channel Assignment (FCA)
Each cell is allocated a predetermined set of voice channels. The BS is the entity that allocates channels to the requests. If all channels are used in one cell, it may borrow a channel from its neighbors through MSC. Fast allocation, but may result high call blocking probabilities.

Dynamic Channel Assignment (DCA)
Voice channels are not allocated to each cell permanently. When a request is received at the BS, this BS request a channel from MSC. DCA can reduce the call blocking probability, but it needs real-time data collection and signaling transmission between BS and MSC.

Call Admission Control
CAC is used to avoid congestions over the radio links and to ensure the QoS requirements of ongoing services. Quality of service (QoS) Packet-level factors Packet loss rate, packet delay, packet delay variation, and throughput rate. Grade of service (GoS) Call-level factors New call blocking probability, handoff call dropping probability, connection forced termination probability.

CAC Procedure Determine the amount of available channels, i.e., the number of channels for accepting new and handoff requests. When the N-th request arrives, i.e., there are (N-1) ongoing services. If there are enough resources to admit the N-th request, then the new request is admitted. Otherwise, it will be denied. In order to maintain the continuity of a handoff call, handoff calls are given higher priority than the new call requests. The prioritized call admission is implemented by reserving channels for handoff calls. This method is referred to as guard channels. Fixed reservation and dynamic reservation.

Cell Capacity The average number of mobiles requesting service (average call arrival rate):  The average length of time a mobile requires service (the average holding time): T The offered traffic load: A = T e.g., in a cell with 100 mobiles, on an average, if 30 requests are generated during an hour, with average holding time T=360 seconds, then the arrival rate =30/3600 requests/sec. A servicing channel that is kept busy for an hour is quantitatively defined as one Erlang. Hence, the offered traffic load (A) by Erlang is then

Call Blocking How likely a new user can get a connection established successfully? Admission control of new calls. It is measured by the probability of call blocking, which is a quality of service (QoS) factor, a.k.a., (GoS) factor. Assume we have a total number of C channels in a radio cell. If the number of active users during any period of time is C, then the call blocking probability is 1. If and only if the number of ongoing calls is less than C, the probability of call blocking will be less than 1.

Erlang B Probability of an arriving call being blocked Prob that (Blocked Calls Cleared; Lost)) is where C is the number of channels in a cell. Prob(A,C) is also called blocking probability, probability of loss, or probability of rejection.

Erlang C Probability of an arriving call being delayed, i.e.,
probability that no trunk (server=channel) is available for an arriving call in a system with C channels and the call is delayed, is where Prob(A, C) is the probability of an arriving call being delayed with load A and C channels.

SEMANTICS Prob that calls are lost..
GOS for telephone calls (realstic values 10^{-2} – 10^{-3} Physical Interpretation-> Ratio of calls rejected to the total number of calls. What can we do with these Erlang formulas? Given a fixed offered traffic A in Erlangs and a fixed device capacity C, then find the prob of blocking… Determine the offered traffic in Erlangs that produces a given blocking probability for a fixed device capacity C. Determine the required device capacity given the blocking probability and the offered traffic in Erlangs.

EXAMPLE An urban PC area has a population of 2 Million residents. A cellular company serves this area. System has 394 cells with 19 channels each. Find the number of users that can be supported at 2% blocking if each user averages 2 calls/per hour at an average call duration of 3 minutes!!

EXAMPLE Prob of Blocking 0.02 (GOS) Number of Channels  C=19
Traffic Intensity per User  A/mu = 2* 3/60 = 0.1 Erlangs From Erlang B chart, total carried traffic obtained as 12 ERLANGS. The number of users which can be supported per cell is 12/0.1 = 120. There are 39 cells  total number of subscribers supported is 120 * 394 =

Efficiency (Utilization)
Example: for previous example, if C=2, A= 3 then B(2, 3) = 0.6, Blocking probability, i.e., 60% calls are blocked. Total number of rerouted calls = 30 x 0.6 = 18 Efficiency = 3(1-0.6)/2 = 0.6

Summary The advantage of cellular communications
Capacity extension by frequency reuse Cell cluster and cochannel cells Number of cells in a cluster Frequency reuse ratio Co-Channel interference Impact of cluster size Worst-case cochannel interference Traffic load and call blocking probability Average delay Probability of queuing delay Cell splitting and sectoring Fixed channel allocation and dynamic channel allocation

Overview of Wireless Networks: Introduction

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