Embed Size (px)
Citation preview
Mobile Cellular Mobile Cellular System System (Continued…)
Channels Reuse[[Remaining Part of 1G 1G Cellular Cellular Systems]Systems]
ALAK ROY.Assistant ProfessorDept. of CSENIT [email protected]
SLIDE #2.2
64
Channels Reuse• Cell structure can reuse frequency only when
certain distance is maintained between cells that use the same channels.
Channel Assignment Strategies: (to achieve the objectives – increasing capacity & min. interference)
• Fixed frequency assignment:– certain channels are assigned to a certain cell– problem: different traffic load in different cells– Borrowing strategy allow a cell to borrow channels
from a neighboring cell (MSC supervises the procedure, without interfering donor cell)
65
Channels Reuse (Contd.)• Dynamic frequency assignment:
– base station chooses frequencies depending on the frequencies already used in neighbor cells
– more capacity in cells with more traffic– assignment can also be based on interference
measurements
66
Interference• Co-channel interference
– Signals from cells that share a channel cause co-channel interference
– Can’t remove it by increasing power.• Adjacent channel interference
– Signals (which are adjacent in frequency to the desired signal) from adjacent cells and within same cell cause this.
– Near-far problem (An adjacent channel user is transmitting in very close range to a subscriber’s receiver, while receiver attempts to receive signal from BS on the desired channel).
– Use careful filtering and channel assignments to reduce it.• But, available channels decrease for incoming calls.
67
Frequency reuse factor
• Total available channels = S• N “adjacent” cells (called a cluster) share S
channels (Note: N is called Cluster size)• System has M clusters• Each cell gets k channels
– S = k N
• Capacity of the system is C = MkN • Frequency reuse factor is N
68
Problem 1
• Consider a cellular system in which there are a total of 1001 radio channels available for handling traffic. Suppose, tha area of a cell is 6 km2 and the area of the entire system is 2100 km2.
i. Calculate the system capacity if the cluster size is 7ii. How many times would the cluster of size 4 have to be
replicated in order to approximately cover the entire cellular area?
iii. Calculate the system capacity if the cluster size is 4iv. Does decreasing the cluster size increase the system
capacity? Justify your claim.S= 1001, N=?, k=? , M=?, C= MkN
69
ALAK ROY.Assistant ProfessorDept. of CSENIT [email protected]
UCS-805 MOBILE COMPUTING
NIT Agartala, Dept of CSEJan-May,2011
Mobile Cellular SystemsMobile Cellular Systems
GSM : GSM : 2G Cellular Systems2G Cellular Systems
SLIDE #2.3
71
2G Cellular Systems
• Four Major Standards:– GSM (European)– IS-54 (later becomes IS-136, US)– JDC (Japanese Digital Cellular)– IS-95 (CDMA, US)
72
Example: GSM
• Frequency Band– 935-960, 890-915 MHz– Two pieces of 25 MHz band
(same as AMPS)• AMPS has 833 user channels• How about GSM?
73
Time Division Multiple Access (TDMA)
• The mobile users access the channel in round-robin fashion.
• Each station gets one slot in each round.
Slots 2, 5 and 6 are idle
74
FDMA/TDMA, example GSM
1 2 3 7 8
f
t
124
1
124
1
20 MHz
200 kHz
890.2 MHz
935.2 MHz
915 MHz
960 MHz
Each freq. carrier is divided into 8 time slots.
75
Number of channels in GSM
• Freq. Carrier: 200 kHz• TDMA: 8 time slots per freq carrier
• No. of carriers = 25 MHz / 200 kHz= 125
• No. of user channels = 125 * 8 = 1000
76
Capacity Comparison• Reuse factor
– 7 for AMPS– 3 for GSM (why smaller reuse factor?)
• What’s the capacity of GSM relative to AMPS?
A. one half of AMPS B. the sameC. 3 times larger D. 10 times larger
77
Answer• AMPS
– reuse factor = 7– no. of users / cell = 833 / 7 = 119
• GSM– reuse factor = 3– no. of users / cell = 1000 / 3 = 333– almost 3 times larger than AMPS!
78
Multiple Access Methods
Three major types:– Frequency Division Multiple Access (FDMA)– Time Division Multiple Access (TDMA)– Code Division Multiple Access (CDMA)
• Frequency hopping (FH-CDMA)• Direct sequence (DS-CDMA)
79
Frequency-Time Plane
Time
Frequency
Partition of signal space into time slotsand frequency bands
80
FDMA
Time
Frequency
Different users transmit at different frequency bandssimultaneously.
81
TDMA
Time
Frequency
Different users transmit at different time slots.
Each user occupy the whole freq. spectrum.
82
Frequency Hopping CDMA
Frequency
Time
At each successive time slot, the frequency band assignments are reordered.
Each user employs a code that dictates the frequency hopping pattern.
83
Synchronization
• The previous figure implies that each signal synchronizes with each of the other signals.
• In practice, this is not the case.• Frequency hops may collide, but it does not
occur frequently. – How often collisions occur depends on the choice
of codes.
84
Direct Sequence CDMA
Time
Frequency
All users occupy the whole bandwidth all the time.
Signals of different users overlap with one other.
How can it be done?
85
CDMA Encoding
• Each user is assigned a unique signature sequence (or code), denoted by (c1,c2,…,cM). Its component is called a chip.
• Each bit, di, is encoded by multiplying the bit by the signature sequence:
Zi,m = di cm
86
Encoding Example
• Data bit d1 = –1
• Signature sequence (c1,c2,…,c8) = (+1,+1,+1,–1,+1,–1,–1,–1)
• Encoder Output(Z1,1,Z1,2,…,Z1,8) = (–1,–1,–1,+1,–1,+1,+1,+1)
87
Bandwidth
• Note that the chip rate is much higher than the data rate.
• Consider our previous example.– Suppose the original data signal occupies a
bandwidth of W. – What is the bandwidth of the encoded signal?
88
Spread Spectrum Technique
Time
Frequency
Time
Frequency
Encoding
The bandwidth expands by a factor of M.
M is called spreading factor or processing gain.
89
CDMA Decoding
• Without interfering users, the receiver would receive the encoded bits, Zi,m , and recover the original data bit, di, by computing:
M
mmmii cZ
Md
1,
1
90
CDMA Decoding Example
(c1,c2,…,c8) = (+1,+1,+1,–1,+1,–1,–1,–1)(Z1,1,Z1,2,…,Z1,8) = (–1,–1,–1,+1,–1,+1,+1,+1)
(–1,–1,–1,–1,–1,–1,–1,–1)
di = –1
multiply
add and divide by M
91
92
Multiuser Scenario
• If there are N users, the signal at the receiver becomes:
• How can a CDMA receiver recover a user’s original data bit?
N
n
nmimi ZZ
1,
*,
93
Multiplied by the signature sequence of user 1
2-user example
94
Signature Sequences
• In order for the receiver to be able to extract out a particular sender’s signal, the CDMA codes must be of low correlation.
• Correlation of two codes, (cj,1,…, cj,M) and (ck,1,…, ck,M) , are defined by inner product:
M
mmkmj cc
M 1,,
1
95
The Meaning of Correlation
• What is correlation?– It determines how much similarity one sequence has with
another.– It is defined with a range between –1 and 1.
Correlation Value Interpretation
1 The two sequences match each other exactly.
0 No relation between the two sequences
–1 The two sequences are mirror images of each other.
Other values indicate a partial degree of correlation.
96
