WCDMA Fundamentals by Dr. Hatem MOKHTARI

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Transcript of WCDMA Fundamentals by Dr. Hatem MOKHTARI
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© Cirta Consulting LLC 19992004
WCDMA Fundamentals
Dr. Hatem MOKHTARICirta Consulting LLC
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© Cirta Consulting LLC 19992004
Spread Spectrum Modulation
Sn ε( )SW
Radio PropagationChannel
ε−1( ) S’n
Transmitter Receiver
NarrowBand Signal
Narrowband Signal
n(t)
i(t)
SPREAD SPECTRUM SYSTEM BLOCK DIAGRAM
Sn : Narrowband Modulated Binary Sequence (information : Speech or Data)ε( ) : Spreading function using high chiprate modulationε1( ) : Despreading function using the same sequence as ε( )n(t) : Gaussian WhiteNoise ; i(t) : Interference
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Digital Modulation in WCDMA : QPSK for DL and BPSK in UL
PSK or Phase Shift Keying is advantageous compared to other modulations :
Constant AmplitudeData information is “hidden” in the Phase componentRobustness to Noise and Interference because Noise in general affects the Amplitude and not the Phase Component
QPSK is a 4state Modulation scheme :Used in DL because of High Data Rate demandSame Properties as the PSK : Constant Amplitude and Data in the Phase component
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BPSK ModulationBinary Phase Shift Keying Modulation is a twostate Modulation schemeIn BPSK the signal can take two states :
A binary digit is mapped to high frequency carrier sinusoidal waveform of a given phase as given below :For a 1 Transmitted symbol :
for a 0 Transmitted symbol :
Where the Amplitude, , Eb is the transmitted Energy per bitThe duration of this sinusoidal waveform is Tb
( ) ( )tfTEtfAts cb
bc ππ 2cos22cos)(1 ==
( ) ( )tfTEtfAts cb
bc πππ 2cos22cos)(2 −=+=
b
b
TEA 2
=bTt ≤≤0
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To ensure that each transmitted binary digit contains an integer number of cycles, the carrier frequency fc has to fulfill the condition : where nc is a fixed integer number. For the above example nc = 2
1 0 0 1
Tcarrier = 1/fc
Tb = nc*Tcarrier
b
cc T
nf =
BPSK Modulation Example
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Modulation Example : BPSKBPSK Modulation shifts the PHASE of the DATA modulated carrier by 180 degrees.
Mathematically this can be represented as a multiplication of the carrier by a function c(t) which takes the values +1 and 1
Assume data modulated carrier of power P and frequency ω0 and phase modulated θd(t)
PhaseModulator
Binary Data
( )tP 0cos2 ω
( ))(cos2 0 ttP dθω +
)(tc
( ))(cos)(2 0 tttcP dθω +
BPSK DS SS Transmitter
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Modulation Example : BPSKThe Wideband Signal is transmitted througha channel having a delay td
The received signal is mixed with interference and Gaussian Noise
Despreading is done by remodulating the wideband signal with adhoc delayed spreadingcode as shown bellow
BandpassFilter[ ] )()(cos)(2 0 tntttcP ddd ++−+− φτθωτ
)( 'dtc τ−
BPSK DS Receiver
Data PhaseDemodulator Estimated Data
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Modulation Example : BPSKThe remodulation or correlation of the received signal with the delayed spreading codeis a critical function in all SS Systems
The signal component of the Output of the despreading Mixer is given by :
[ ]φτθωττ +−+−−= )(cos)'()(2)( 0'
ddddn tttctcPtS
τ’d is the receiver’s best estimate of the Transmission delay
Since c(t) equals +1 or 1 the product will be +1 if the delaysτd = τ’d that is, if the spreading code and the despreading code are SYNCHRONIZED.
)'()( dd tctc ττ −−
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BPSK Spectral DensityBefore Spreading and for Tb as a bit duration the twosided power spectral Density is given by :
After the Spreading we use the Chip duration Tc instead of Tb because of the involvement of the Chip (code) :
Where sinc function (Also known as the Cardinal Sinus) is given by :
( )[ ] ( )[ ]{ }bbbn TffcTffcPTfS 02
02 sinsin
21)( ++−=
( )[ ] ( )[ ]{ }ccc TffcTffcPTfSn 0
20
2' sinsin21)( ++−=
xxxc )sin()(sin =
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BPSK Power Spectral Density
0.4
0.2
0
0.2
0.4
0.6
0.8
19
00
800
700
600
500
400
300
200
100 0
100
200
300
400
500
600
700
800
900
x (DEGREES)
SIN
(x)/x
00.10.20.30.40.50.60.70.80.9
1
900
800
700
600
500
400
300
200
100 0
100
200
300
400
500
600
700
800
900
x (DEGREES)
(sin
x/x)
^2
xxxc )sin()(sin =
22 )sin()(sin
=xxxc
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QPSK Modulation* QPSK is very similar to BPSK, except that now one of four possible waveforms is transmitted through the channel.
* Each Waveform can represent two binary digits
* Es is the Transmitted Signal Energy per Symbol
* Ts is the Symbol Duration. Note that each symbol can represent 2 binary digits unlike BPSK in which each symbol was just a sungle binary digit
* fc is the carrier frequency equal to nc/Tc as in BPSK description
( ) 4,3,2,14
122cos2)( =
−+= iwhereitf
TEts cs
si
ππ
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BASIS FunctionsIn this case we use ORTHONORMAL basis functions as follows :
The Coordinates on the signal Constellation (or Space) diagram are given by :
r(t) = si(t) + n(t) is the received signal : Either signal s1 or s2 + Random Noise
QPSK Modulation
( )tfT
t cs
πφ 2cos2)(1 = ( )tfT
t cs
πφ 2sin2)(2 =
∫=sT
dtttrx0
11 )()( φ ∫=sT
dtttrx0
22 )()( φ
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The signal Constellation diagram for QPSK is shown bellow.The signal points are mapped to a pair of binary digits as shownThe Decision boundaries are shown as solid Horizontal Vertical linesNotice how this mapping has been chosen so that neighboring signal points differ in only a SINGLE BINARY DIGIT
QPSK Modulation : Constellation
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QPSK ModulationFor example, signal B(11) and D(10) differ in only binary digit positionIf for example, the signal point A is transmitted and a Symbol Error occurs, it is very likely that the received symbol will be either C or D.This Type of Mapping is called GRAY ENCODINGThe gray encoding scheme used will mean that on average, we can expect the probability of an error in a binary digit to half of the probability of an error in a symbol
ExampleUnder the conditions of no noise, the coordinates of a signal point are given by :
and
( )
−=
412cos1
πiEx s
( )
−−=
412sin2
πiEx s
Test : Verify the above formulas for i=1
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QPSK Modulation : Answer
dttftftfTE
dtttrx cc
T T
cs
ss s
−== ∫ ∫ )
4sin()2sin()
4cos()2cos()2cos(
2)()(
0 011
πππππφ
( )dttftftfTE
dtttrxs sT T
cccs
s∫ ∫ −×==0 0
211 )2cos()2sin()2(cos
222
)()( πππφ
( ))2cos(121cos2 θθ += )2sin(
21cossin θθθ =
21sEx = 22
sEx −=
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QPSK Coherent Receiver
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CDMA Multiple Access : Principal of Spread Spectrum (SS)Each User encodes its signalCode Signal Bandwidth (W) > Information Bandwidth
The Receiver knows the code sequence
Transmission
Spread Spectrum
f f
f
P
f
Reception
Despreading
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CDMA Multiple Access Advantages : Multiple Access Features
1. All Users’ Signals overlap in TIME and FREQUENCY2. Correlating the Received Signal despreads ONLY the WANTED SIGNAL
p
f f
S1 p
S1xC1
p
f f
S2 p
S2xC2 f
p
S2 X C2 X C1
f
p S1 = S1 X C1 X C1
RECEIVER of USER 1
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CDMA Multiple Access Advantages : Interference Rejection
p
f f
S1 p
S1xC1
p
f
I
f
p
f
p
IxC1 I
S1
Correlation Narrowband Interference Spread the power
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CDMA Principles : Multiplexing
Radio PropagationChannel
D/AA
A
B1
B2
m1(t)
m2(t)
c1(t)
c2(t)
c1(t)
c2(t)
c1(t) and c2(t) are Orthogonal Codes : 0)()(0
21 =∫T
dttctc
∫
∫
m’1(t)
D/Am’2(t)
Transmitter Receiver
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Walsh Codes (1/6)Since all the WCDMA users use the same RF Band in the DL, to avoid mutual interference Walsh codes are used.Hadamard Matrix is a recursive Matrix :
Where
Example : N=2
=
NN
NNN HH
HHH 2
=
1000
2H
=
=
=
3
2
1
0
22
224
0111
1000
1000
1000
WWWW
HHHH
H
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Walsh Codes (2/6)Walsh codes are thus given by :
Afterwards, replacing 0 by 1 we obtain the real Walsh codes used in WCDMA
Note : Except W0 all the codes satisfy orthogonality and dot product conditions to be used in WCDMA.
[ ][ ][ ][ ]1111
11111111
1111
3
2
1
0
−−=−−=
−−=−−−−=
WWWW
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Walsh Codes : Example (3/6)Assume three different users with three different data sequences :
Assume we have a Spreading factor of 4 then :The Spread Spectrum Signal would be for user1 :
[ ][ ][ ]111)(
111)(111)(
3
2
1
++−=−++=+−+=
tmtmtm
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 11m1(t)
W1(t)
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1S1(t)
×
=
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Walsh Codes : Example (4/6)The Resulting SS Signal for the three users can be written as follows :
And the bit sequence would be :
If no error is encountered each user decodes its signal using the despreading function, for user j for example :
m1(t) = w1(t)C(t) = [4 4 4] and using an integrator m1(t) is fully recovered which yields the original signal : [+1 1 +1]
m2(t) = w2(t)C(t) = [4 4 4] and using an integrator m2(t) is fully recovered which yields the original signal : [+1 +1 1]
m3(t) = w3(t)C(t) = [4 4 4] and using an integrator m3(t) is fully recovered which yields the original signal : [1 +1 +1]
)()()()()()()( 332211 tmtwtmtwtmtwtC ++=
[ ]113113113111)( −−−−−−−−−=tC
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C(t)
W1(t)
C(t) W1(t)∫bT
dttWtC0
1 )()(1 or 1
[ ]113113113111)()( 1 −−−−−=tWtC
Walsh Codes : Example (4/6))('
1tm
)('1tm is computed over the information period Tb using a summation
]444[)('1
−=tm
4 4 4
The forward link of the CDMA system modeled uses orthogonal Walsh codes to separate the users. Each user is randomly allocated a Walsh code to spread the data to be transmitted. The transmitted signals from all the users are combined together, then passed through a radio channel model. This allows for clipping of the signal, adding multipathinterference, and adding Guassian noise to the signal. The receiver uses the same Walsh code that was used by the transmitter to demodulate the signal and recover the data. After the received signal has beendespread using the Walsh code, it is subsampled back down to the original data rate. This is done by using an integrate and dump filter, followed by a comparator to decide whether the data was a 1 or a 0. The received data is then compared with the original data transmitted to calculate the bit error rate (BER). The RMS amplitude error is also worked out. The signal level after it has been demodulated and filtered, is compared with the expected amplitude of the signal based on the transmitted data. The RMS amplitude error directly relates to the bit error rate, so is a useful measurement to make.
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Walsh Codes : SelfTestCompute the Hadamard Matrix for N=4
What are the Possible Orthogonal Walsh Codes ?
Given a signal m1(t) and m2(t) as follows :m1(t) = [1 1]; m2(t) = [1 1]; m3(t) = [1 1]; Compute the composite spread spectrum signal Verify that m1(t), m2(t) and m3(t) are fully recovered
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Walsh Codes SelfTest : Answer
=
=
1001001101011111
0110110010100000
0110110010100000
0110110010100000
44
448 HH
HHH
First row is not considered as an orthogonal code, all the remaining rows are Walsh codes by replacing each 0 by 1. Basically 7 Walsh codes are thus generated.
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Walsh Codes SelfTest : Answerm1=[1 1]m2 = [1 1]m3 = [1 1]
m1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1W1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1S1 = W1*m1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
m2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1W2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1S2 = W2*m2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
m3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1W3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1S3 = W3*m3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
C=W1*m1+W2*m2+W3*m3 1 1 3 1 1 1 3 1 1 3 1 1 1 3 1 1
W1*C 1 1 3 1 1 1 3 1 1 3 1 1 1 3 1 1After Integrator 8 8After Threshold Decision 1 1
W2*C 1 1 3 1 1 1 3 1 1 3 1 1 1 3 1 1After Integrator 8 8After Threshold Decision 1 1
W3*C 1 1 3 1 1 1 3 1 1 3 1 1 1 3 1 1After Integrator 8 8After Threshold Decision 1 1
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CDMA Principles
CrossCorrelation Rxy(τ) :
Crosscorrelation if τ=0 :
If x and y are discrete sequence (binary signals):
Example of orthogonal codes :
∫ −=T
xy dttytxR0
)()()( ττ
∫=T
xy dttytxR0
)()()0(
iIii
T yxYXRxy ∑≤≤
==1
.)0(
−−
=
1111
X
−
−
=
1111
Y [ ] 01111
1111
.1111)0( =−+−=
−
−
−−=xyR
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CDMA Principles
To be used in DSSS CDMA Codes must satisfy the following conditions :
Zero CrosscorrelationNumber of +1s and 1s must be the sameDot Product must be equal to 1
Example :Dot product of the previous example is :
14/)1111(4/. =+++=XX T
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CDMA PrinciplesTc : Chip Rate of the PN CodeTb : Information rate (voice/data)
m1(t)
Tb 2Tb
1 1 1
3Tb
f
M1(f)
1/Tb
Tc 4Tc
f
C1(f)
1/Tb
c1(t)
1/Tc
f
C1(f)* M1(f)
1/Tb 1/Tc
m1(t).c1(t)
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CDMA FundamentalsW/R : Defined as the system processing gain
In CDMA, the Reverse Link Capacity is often the limiting link in terms of capacityIn CDMA : Uplink Receive Power is equal from all MSs.Per user : S/N = 1/(M1)
M : Total Number of users in the cellS = S (The wanted signal)N = (M1)S => S/N = 1/(M1)
Example : If M=7 then S/N = 1/6
if M>>1 then
M : The Number of simultaneous users a CDMA cell can support
o
b
NERW
M ≈
S
S
S
S
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Eb = Signal Power / Bit Rate = S/Rb
No = Noise Power / Bandwidth = N/W bo
b
RW
NS
NE
×=
Signal to Noise Ratio
Processing Gain
Example :Given a Demodulator Performance
Bit rate Rb = 8 kpbsBandwidth W = 1.2 Mcps => G = W/Rb = 150 = 21 dB
dBNE
o
b 6>
dBdBdBRW
NE
NS
dBbdBo
b
dB
15216 −=−>
−
=
CDMA Multiple Access : Principal of Spread Spectrum (SS)
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η+−=
11..
11
RW
MNE
o
b
CDMA Fundamentals : CapacityIf other users from other cells are considered, the actual cell becomes loaded and :
where η is the loading factor (0 < η < 1)We define F as the Frequency reuse :
η+=
11F Where η is given by :
ceInterferencellOwn
ceInterferenCellOther
II
__
__=η
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CDMA Fundamentals : Capacity
Cell B
Cell A
Cell C
B1
B2
C1
C2
Interference Introduced by Users in the Neighboring Cells
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CDMA Fundamentals : Capacity
Cell B
Cell A
Cell C
Sectorization Reduces Interference and addsa Gain to the system : Sectorization Gain
Unwanted interferersrejected by antennapattern of Cell A
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Sectorization Gain :
TriSectors : λ = 3 (2.5 in practice)
6Sectors : λ= 6 (5 in practice)
Sectorization Gain = λ = Total Interfering Power from all Directions/ Perceived Interference Power by the sector antenna. G is the antenna pattern in given direction
CDMA Fundamentals : Capacity
∫
∫= π
π
θθθ
θθλ 2
0
2
0
)()0()(
)(
dIGG
dI
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Voice Activity Factor : Interference is reduced when the user is not transmitting
The final value for M :
CDMA Fundamentals : Capacity
vRW
MNE
o
b 1..1
1..1
1 λη+−
=
vNERW
M
o
b
λη
.1
1.+
≈
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ShannonHartley TheoremGeneric Model
Radio Propagation Channel +
Bandlimited White Noise
Received SignalNoise+Signal
1. In Theory, without presence of Noise the channel capacity is infinite
2. For a given Signal Level the capacity tends to a constant value as thebandwidth increases.
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CDMA Multiple Access PrincipleShannon Theorem
+×=
NSWC 1log 2
Channel Capacity C (Bit/s) given by Shannon Theorem :
W : System Bandwidth (Hz)S/N : Signal to Noise Ratio (numerical value)C : System Capacity (bit/s)
Same Capacity
Wide W and Low S/N (such as in WCDMA)
Narrow W and Large S/N (such as in GSM)
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C can also be written as :
where
assuming W to be infinite :
Thus for any degrading S/N we can transfer data with low BER if we increase the bandwidth used to transfer the data. Numerical Example :
For a data rate of 10kbps, and SNR of 20 dB then W = 694.4 kHz
ShannonHartley Theorem
+×
×=
+
×=x
xNSWN
S
WCo
o 11ln44.12ln
1ln
SWNx o=
××=
++−
×=
+
×=
→→∞→ NSW
t
ttt
NS
tt
NSC
ttx44.1
...!3!2lim44.1)1ln(lim44.1lim
32
00
00
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Bandwidth EfficiencyBandwidth Efficiency is a measure of how well a particular modulation and errorcontrol coding scheme is making use of the available bandwidth Example :
If a system requires 4 kHz of bandwidth to continuously send 8000 binary digits/sec, the bandwidth efficiency = 8000/4000 = 2 bits/s/HzNotice that to double the rate at which binary digits are sent over the given communication channel, we would require a bandwidth efficiency = 16000/4000 = 4 bits/s/HzIn this case of an ideal system, rb = 1/Tb = C, where C is given by the ShannonHartley theorem :
Or equivalently :
+×=
+×=
WC
NEW
NSWC b .1log1log
022
−
=
⇒
−=
WCN
E
WCN
E WC
dB
bWC
b 12log101210
00
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Bandwidth Efficiency Versus Eb/No
0.1
1
107
.1
5.9
4.5
2.7
0.7
1.76
4.49
7.48
10.7 14
17.5
Eb/No (dB)
Ban
dwid
th E
ffic
ienc
y, R
/W
Bandwidth Efficiency
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Bandwidth EfficiencySelfTest
* A Digital Cellular Phone System is required to work at a bandwidth efficiency of4 bits/s/Hz to ensure sufficient users to make it profitable.
** What is the minimum Eb/No in dB required to ensure that users on theedge of the coverage area receive errorfree communication ?
** To Double the number of users on the existing communication system,by what amount should the basestation and handsets transmitted powers be
increased to maintain coverage and errorfree communication ?
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CDMA Principles
λ/2Mobile
distance
Am
plitu
de
The MS crosses 2 fades in : v2λ
Example : @ 900 MHz and v = 90 km/h (25 m/s)MS crosses fades every 6.67 ms
@ 1800 MHz MS crosses fades every 3.335 ms
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CDMA Principles : Delay SpreadTb
τ
BTS
Propagation Time T1
Propagation Time T2
Delay Spread = τ = T2  T1
Here τ < Tb implies Interference
Example :D1 = 150 m and D2 = 200 m => T1 = 150/3*108 = 0.5 µsand T2 = 0.66 µs => τ = 0.16 µs Assume we have a UMTS Service R = 3.84 Mchps => Tb = 0.26 µs τ < Tb => INTERSYMBOL INTERFERENCE PRESENT
Multipath Immunity
WCDMA is inherently tolerant to multipath delay spread signals as any signal which is delayed by more than one chip time becomes uncorrelated to the PN code used to decode the signal. This results in the multipath simply appearing as noise. This noise leads to an increase in the amount of interference seen by each user subjected to themultipath and thus increases the received BER.
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CDMA Principles : Delay Spread
t
Received Power
Time (µs)
τ1= 3µsτ2= 4µs
τ3
Inter Symbol Interference can occur if the delay spread τn is greater than
one symbol period : The higher the bit rate, the more ISI occur
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Example 1:Let us consider a Mobile Communications System that uses Rb = 270.83 kbpsThe bit period is thus Tb = 1/270830 = 3.69 µsConclusion : bit period almost equal to 4 µs as shown on the delay spread power profile => ISI would normally exist ! Without use of Viterbibased EQUALIZER such as in GSM
Example 2:Let us consider a Mobile Communications System that uses Rb = 1.2288 Mbps = 1228800 bpsThe bit period is thus Tb = 1/ 1228800 = 1 µsConclusion : bit period is much LESS than 4 µs as shown on the delay spread power profile => ISI would not normally exist !Important note : CDMA Rake Receive uses a special form of Time Diversity to recover the signal. CDMA Rake receiver combines multipath components and suppresses phase differences provided that delays are not very small
CDMA Principles : Delay Spread
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The Principal of Maximum Ratio Combining in CDMA Rake Receiver
Transmitted Symbol Amplitude Phase
Received Signalat each time delay
Modified SignalUsing Channel Estimator
CombinedSymbol
θ
Figure #1
Figure #2
Figure #3
θ
θ
θ
θ
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Block Diagram of CDMA Rake Receiver
Correlator Phase Rotator
DelayEqualizer
CodeGenerator
ChannelEstimator Finger # 1
Matched Filter
∑ I
∑Q
I
Q
Correlator Phase Rotator
DelayEqualizer
CodeGenerator
ChannelEstimator Finger # 2
Correlator Phase Rotator
DelayEqualizer
CodeGenerator
ChannelEstimator Finger # 3
Combiner
Timing (finger allocation)
Input RF SignalQI
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Digitized input samples are received from RF Frontend in the form of I and Q components
Code Generator and Correlator : Perform despreading and integration to user data symbol
Channel estimator : Uses the Pilot symbols to estimate the channel state
Phase Rotator : aligns the symbols to the initial phase (phase cancellation)
Delay Equalizer : Compensates the Delay in the arrival times of the symbols in each finger
Rake Combiner : Sums up the channelcompensated symbols, thereby providing MULTIPATH DIVERSITY against Fading.
Matched Filter : Determines and Updates the Current Multipath Delay Spread. This is used to assign the Rake fingers to the largest Peaks (Maximum Combining)
CDMA Rake Receiver : Components
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CDMA Principles: Delay Spread In Multipath Environment :
Received power can be written as :
Fourier Transfer Function :
∑ ∑= =
−=→−=N
n
N
nnnnn fjafSfRtsatr
1 1).2exp()()()()( τπτ
∑=
−==N
n
fjn
neafSfRfH
1
.2
)()()( τπ
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CDMA Principles : Delay Spread
f
H(f
)
Example with twoequal amplitude paths : a1=a2=A
2A
τ21
τ1
τ23
τ2
)cos(2)( τπffH =
1. FrequencySelective Fading is evident in the nulls of the Magnitude Spectrum
2. WCDMA is more advantageous than CDMA when the delays are small such as 0.4 ms (Dense Urban and Urban Environments)
3. WCDMA using 5 Mbps (bit period of 0.2 ms) better than IS95 CDMA using only 1.2288 Mbsp (bit period 1 ms) when ISI are to be considered in Dense Urban areas
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CDMA Fundamentals : Power Control
MS2
Pr,2Pr,2
MS1
P = 21 dBmP = 21 dBm
NearFar Problem
PL1 = 100 dB
PL2 = 90 dBPr,1 = EIRP(MS1)  PL1 = 21  100 = 79 dBmPr,2 = EIRP(MS2)  PL2 = 21  90 = 69 dBm
(S/N)1 = Pr,1  Pr,2 = 10 dB(S/N)2 = Pr,2  Pr,1 = +10 dB
MS2 must be Power Controlled by 10 dB to have the same S/N for both users MS1 and MS2
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What is the initial MS power to be used ?
Maximum MS Power : Advantage : High likelihood to reach the BTSDrawback : Uplink interference
Minimum MS Power :Advantage : Low Uplink interference Drawback : Low probability to reach the BTS
Solution (IS95 used in WCDMA also) :Use of Access Probes : MS Power Rise portions done graduallyAdvantage : Avoid Uplink interference and Reach the BTS with sufficient Transmit power
CDMA Fundamentals : Power Control
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OPENLOOP Power Control:MS Transmits its first access probes at relatively low powerMS waits for a response back from the BTSIf after a Random time the MS has no acknowledgment from the BTS, a second Access Probe is performed (at a slightly higher power)Process is repeated until the MS receives a response from the BTS
CDMA Fundamentals : Power Control
MSTransmit Power
Initial Transmit Power
FirstAccesProbe Correction
SecondAccesProbe Correction
Random Time Intervals
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OPENLOOP Power Control: Purely MobileControlled Operation
Step Size for a Single Access Probe is Specified by the System Parameter PWR_STEP (equivalent to BSS RF Parameters in GSM)
The Standard Specifies that the MS uses initially the Received power from the BTS :If the MS receives strong DL level => MS assumes BTS close and then transmit at low levelAlternatively MS transmits at high level if the initial DL Level is low
Initial MS Transmit Power (dBm) :
NOM_PWR and INIT_PWR : Broadcast by the BS in the “access parameters message”These can be set by the Operators for further fine tuning
CDMA Fundamentals : Power Control
PWRINITPWRNOMpp rinitialt __73, ++−−=
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During a call :MS continues to calculate PrPath Loss changes as the MS moves and Pr alsoOpenLoop PC adjusts the MS Transmit power using :
VERY IMPORTANT NOTES :UL and DL frequencies are differentFast Rayleigh Fading is FrequencySelectiveOpenLoop PC too slow to compensate Rayleigh FadingHowever, OpenLoop PC suitable for LogNormal (slow) FadingCorrelation between UL and DL concerning SlowFading but not Fast Rayleigh Fading
ClosedLoop PC compensates Fast Rayleigh Fading
CDMA Fundamentals : Power Control
∑+++−−= ..Pr.__73 CorrectAccPWRINITPWRNOMpp rt
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CLOSEDLOOP Power Control : Involves BOTH the MS and the BS
During the call :BS monitors the UL and measures the link QualityIf the UL quality becomes bad : BS commands MS to powerupIf the If the UL quality becomes too good : BS commands MS to powerdown
CDMA Fundamentals : Power Control
BS Monitors UL Eb/No
Eb/No > Threshold
BS Commands MS to
PowerDown
BS Commands MS to
PowerUp
Yes
No
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CLOSEDLOOP PCBS Sends PC commands to the MS using the DLPC Commands are in the form of PowerControl Bits (PCBs)Each Powerup or Powerdown amount is +1 dB or 1 dBMS response must be fast (Fast Rayleigh Fading)PCBs are sent through the Traffic channelBits are robbed (stolen) from the Traffic Channel in order to send the PCBs
CDMA Fundamentals : Power Control
Conv. Encoder
R=1/2MUX SpreadingVocoder
12.2 kbps 24.4 kbps 24.4 kbps 3.84 Mchip/s
PCB @ 1500 bps (1500 Hz)
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CDMA Fundamentals : Power Control
o
b
NE Threshold
INNER LOOP1. Measure Eb/No2. Compare Eb/Noto Threshold3. Decide which PCBto send
OUTER LOOP
Adjusting Eb/NoThreshold
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SOFTHANDOVERDuring the communication, the MS simultaneously maintains connection with two or three Base Stations,A Traffic Channel is maintained with both cells,
CDMA Fundamentals : Handover
BS1 (Home Cell)
BS2 (Target Cell)
RNC
MS Combines the two signalsusing the Rake Receiver
DOWNLINK
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SOFTHANDOVER
CDMA Fundamentals : Handover
BS1 (Home Cell)
BS2 (Target Cell)
UPLINK
Demodulated Frame 1Demodulated Frame 2
SELECTORRNC
Selects the Best Frame withthe best FER value !
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CDMA Fundamentals : HandoverSOFTERHANDOVER
Softer Handover is considered when two cells (or sectors) of the same site are involved
on the DL the same process happens at the MS : demodulation and Maximum Combining using the Rake Receiver features
on the UL, two sectors of the same site simultaneously receive two signals from the mobile
The signals are demodulated and combined inside the Site, but only ONE frame is sent back to the RNC
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CDMA Fundamentals : Handover
SOFTER HANDOVER ILLUSTRATION
RNC
Only 1 Frame sentback to the RNC
BTS
Home Cell Target Cell
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CDMA Fundamentals : HandoverHARDHANDOVER
HardHandover occurs from CDMA carrier to another CDMA Carrier or CDMA to Analog Carrier. CDMA to CDMA HO is sometimes called DtoD Handover.
HardHandover in UMTS may concern WCDMA to GSM or WCDMA to GSM1800
f1
f2
f2
f1
BS1 BS2
MS
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HANDOVER PROCESS in WCDMAEach Sector is distinguished from one another by its PILOT channel : 4 Logical Channels on the DL are Defined : Pilot, Paging, Synch., and Traffic Channels
Pow
er
Frequency
PILOT CHANNEL
PAGING CHANNELSYNCHRONIZATION CHANNEL
TRAFFIC CHANNEL 1 : USER 1
TRAFFIC CHANNEL 2 : USER 2
TRAFFIC CHANNEL K : USER K
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HANDOVER PROCESS in WCDMA
6Sector Site
3Sector Site
Each Sector uses a different PILOT and is assigned a different PN Code with an offset to distinguish it from other Sectors
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For Traffic(Voice and Data) we use Eb/No, but for PILOT a special term to describe the SNR is the Ec/Io
Ec/Io : Energy per chip per interference densityEc : related to the spreading code, hence the term “chip”
Since no Baseband data is transmitted through the Pilot, there is no Despread and bits are not recovered, unlike for Traffic channels (baseband)
Ec/Io : defines the signal strength of a PILOT channel
The PILOT is not Despread : No despreading gain !
HANDOVER PROCESS in WCDMA
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MS is an intimate in the SHO in WCDMA : MS Measures the Ec/Io and report it the the BTSBTS transmits a PN Sequence using different offsets for each sectorEc/Io gives a good indication on whether or not a sector should the serving cell
HANDOVER PROCESS in WCDMA
PN1PN2
PN3Example with 3 PILOTS
Highway or Road
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During HO Management Process :MS maintains in its Memory four exclusive lists of SectorsSectors stored in the form of PN offsets
Four Exclusive List of sectors called : SETSActive SetCandidate SetNeighbor SetRemaining Set
HANDOVER PROCESS in WCDMA
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Active Set (A) :Contains all Sectors communicating with the MS on traffic channelsIf the Active Set contains only one Pilot : NO SHOIf the Active Set contains more than one Pilot : the MS is maintaining communication with all the Active Set Sectors with Different Traffic ChannelsActive can contain 6 Pilots : UL Diversity maximizedA Pilot enters the (A) only if the RNC decides and then sends a “handover Direction Message” to the message to add that pilot into (A)
Candidate Set (C) :Contains Pilots whose Ec/Io are sufficient to make them HO CandidatesIf Ec/Io > Threshold (Pilot Detection Threshold or T_ADD in Database), the Pilot will be added to the (C)A Pilot is deleted from the (C) and moved to the Neighbor Set (N) if Ec/Eo< Threshold (“Pilot Drop Threshold” or T_DROP in Database)
HANDOVER PROCESS in WCDMA
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HANDOVER PROCESS in WCDMAA Pilot is removed from (C) if its Ec/Io drops below T_DROP threshold for morethan duration specified by T_TDROP : „Handover Drop Timer ExpirationValue“ and put in (N) setThe (C) Set can contain at least 6 Pilots
T_DROP
Counter = 0 Counter = T_DROP
Ec/Io
time
Pilot put in (N) setPilot still in (C) set
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The Neighbor Set (N) : contains those Pilots that are in the Neighbor List of the MS´s Serving CellInitially :
(N) contains Pilots sent to the MS in the ´Neighbor List Message´ by theServing Cell
To keep current Pilots in the (N), MS keeps an aging counter for each Pilot in (N) : NGHBR_MAX_AGE
Counter is reset (counter = 0) when Pilot is moved from (A)/(C) to (N)
Counter incremented for each Pilot in (N) whenever a Neighbor List update message is received
The (N) Set can contain up to 20 Pilots
HANDOVER PROCESS in WCDMA
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The Remaing Set (R) : Contains all possible Pilots in the System for this UMTS carrier Frequency, except pilots in (A), (C) or (N)
In (R) Pilot PN offsets are defined by the Parameter PILOT_INCExample :
If PILOT_INC = 4 ; each sector in the Network can only transmit 0, 4, 8 , 12, etc.
PILOT_INC is sent to the MS inthe „Neighbor List Message“ and „Neighbor List Update Message“
HANDOVER PROCESS in WCDMA
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HANDOVER PROCESS in WCDMA
Ec/Io
Distance
T_ADDT_DROP
SourceCell A
TargetCell B
1 Pilot (A) in Active Set
1 Pilot (B) in Active Set
2 Pilots (A&B)In Active Set : MS is in SHO
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HARD HANDOVER in WCDMAHard HO includes interfrequency HO :
Cells with multiple Carriers with high loadWCDMA GSM
f1, f2f1
f1f1
f1f1
f1GSM
WCDMA
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HARD HANDOVER in UMTS
A
B
CHO
Qua
lity
Mea
s ure
(dB
)
TimeReplace Aby C
Replace Cby A
Replace Aby B
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Cell Breathing Concept in UMTS
LOAD
The Cell Size Shrinks due to Loading
1. UL NoiseRise due to owncell or othercell UL interference translated into LowerUL MAPL : MAPL2 = MAPL1  NoiseRise (dB)
2. NoiseRise computed Using : 10log10(1η) ; η being the Loading due to interference
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Cell Breathing : RF Propagation Environment
Basic Concept for Loading and Propagation Environment :Dense Urban : Lp = 127 + 45*log10(r)MAPL = 126.7 dB using η = 0.5 (50 % loading)η = 0.5 => owncell interference is twice the othercell interferencer = 0.984 km but reduces to 0.501 km using η = 0.75 loading49 % Range Loss in DENSE URBAN
50 %Loading
75 % Loading
HIGHER LOADING
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Suburban : Lp = 123 + 33*log10(r)MAPL = 133.7 dB using η = 0.5 (50 % loading)r = 2.1 km and reduces to r = 1.7 km @ η = 0.7519 % Range Loss in SUBURBAN
Cell Breathing : RF Propagation Environment
MAPL (η = 0) Ideal Case
MAPL (η = 50%) or Noise Rise = 3 dB
MAPL (η = 75 %) or Noise Rise = 6 dB
Log(d)
Dense Urban
Suburban
Path
Los
s (dB
)
3 dB6 dB
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Soft CapacityWCDMA System is limited by interference sharing between neighbor cellsCapacity is dynamically changing due to different traffic variationHard Capacity is limited by Hardware Limitation : TRX namely
Equally Loaded Cells Less Interference in the Neighboring Cells=> Higher Capacity in the Middle Cell
Soft Capacity = (Erlang Capacity with Soft Blocking / Erlang Capacity with Hard Blocking)  1
Number of Channels N Number of Channel Pool = N*(1+i)
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i is defined by :i = Other Cell Interference / Own Cell Interference
i = 0.55 for Omni Celli = 0.55 for 2sector celli = 0.65 for 3sector celli = 0.75 for 4sector celli = 0.85 for 6sector cell
Soft Capacity Calculation Procedure :1. Compute the Number of Channels N in equally loaded case based on Uplink Load Factor :
2. Npool = N*(1+i)3. Use Erlang B table to compute the Offered Capacity
Offered Capacity (Erlang) = ErlangB(Npool, 2%)4. Soft Capacity (Erlang) = Offered Capacity(Erlang) / (1+i)5. Soft Capacity (%) = (Soft Capacity (Erlang)/Offered Capacity (Erlang)) 1
Soft Capacity
+
=
o
b
NEvi
RW
N)1(
η
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ExampleAssumptions
Computation of Number of Channels :W/R = 3.84*1000/12.2 = 314.75Eb/No = 4 dB = 2.51v = 0.67 (67 % Voice Activity Factor)N = 60.4 channelsNpool = 60.4*(1+0.55) = 93.6 ChannelsOffered Capacity = ErlangB (94 channels, 2% GoS) = 82.17 ErlangSoft Capacity = 82.17/(1+0.55) = 53 ErlangHard Capacity = ErlangB (61 channels, 2% GoS) = 50.6 ErlangSoft Capacity (%) = 53.6/50.6 1 = 4.74 approximately 5%
Soft Capacity
P A R A M E T E R V A L U EB it R a te S p ee ch @ 1 2 .2 k b p sV o ic e A c tiv ity F ac to r 6 7 %E b /N o S p p ech : 4 d Bi 0 .5 5 a ssu m in g a n O m n i C e llN o ise R ise 3 d B (5 0 % L o ad F ac to r)G o S 2 %
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CAPACITY in UMTS : Uplink Capacity  Sensitivity is affected by interference (Loading)
Maximum Capacity is related to the amount of UL interference a system can tolerate
Mmax : Maximum number of simultaneous users supported by a single cellcarrier @ 100% Loading assuming that all users are using the same service (e.g. speech or Data). Also called the Pole Capacity.
For a Multiservice system :
Mn : The Number of simultaneous users for the n th service
( )ULULI η−×−= 1log10
max
10101MMULI
UL =−=−
η
...101max,3
3
max,2
2
max,1
110 +++=−=−
MM
MM
MMULI
ULη
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Maximum UL Capacity figures
F Values are : 0.67 for Omni,
0.93 for 3sector, 0.4 for Microcells (without the presence of Fast Fading)
numericalICF
M
×+
+=)1(
11max
1010dBdBo
bRW
NE
IC
−
=
CAPACITY in UMTS : Uplink Capacity
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CAPACITY in UMTS : Uplink Capacity
SiteC onfiguration
Pedestrian A3 km /h w ithout
Fast Fading
Pedestrian B3 km /h w ithFast Fading
V ehicular A 120km /h w ith Fast
FadingO m ni 0.84 / 0 .67 0.67 0.673Sector 1.16 / 0 .93 0.93 0.93M icrocell 0 .5 / 0 .4 0.4 0.4
F Values on the UL
Service Pedestrian AM m ax
w ithout FastFading
Pedestrian BM m ax
w ith FastFading
Vehicular AM m ax
W ith FastFading
Speech 12.2 kbps 94.6 / 105.9 84.4 56.3Circuit 128 kbps 8.5 / 9.3 7.9 6.8Circuit 384 kbps 4.2 / 4.6 3.7 3.2Packet 128 kbps 11.1 / 12.3 9.9 8.1Packet 384 kbps 4.3 / 4.8 4.0 3.4
Typical Uplink Mmax Values for a 3 Sector Site
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Example of Computation of Mmax
1. Assume we use Packet Data @ 384 kbps and a trisectorial Site Configuration; What is the Mmax ?
Answer : W/R = 3840/384 = 10 = 10 dB; Eb/No = 1.5 dBF = 0.93 and C/I = 0.14125 => Mmax = 4.66 users
2. Assume we use Packet Data @ 144 kbps and a 3sector site configuration; What is the Mmax?
CAPACITY in UMTS : Uplink Capacity
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In the DL, each user is subjected to different interference levelsTherefore, no single interference level is valid in the DLFor dimensioning purposes, the current solution is to use simulation. Mostly, researchers do use Monte Carlo Random Number Generation Algorithms
CAPACITY in UMTS : Downlink Capacity
Urban Environment  Pedestrian A 3 km/hAssumptions : 18 dBi Antennas, PL = 134.7 + 35.2 log(R),
Antenna Height 30 m, No Body Loss Considered
0102030405060708090
0 0.5 1 1.5 2 2.5 3 3.5 4
Cell Range (km)
Rel
ativ
e Lo
adin
g (%
)
0 dB5 dB10 dB15 dB20 dB25 dB
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CAPACITY in UMTS : Downlink Capacity
Service Dense Urban &Suburban
Rural
Speech 12.2 kbps 86.4 64.1Circuit 128 kbps 7.3 4.1Circuit 384 kbps 2.7 1.6Packet 128 kbps 8.4 5.2Packet 384 kbps 2.8 1.7
Recommended Mmax Values for 3sector site Configuration
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CAPACITY in UMTS : Downlink Capacity
Open Area  Vehicular A 120 km/hAssumptions : 18 dBi Antennas, PL = 105 + 33.8 log(R), Antenna Height 50
m, 3 dB Body Loss
0
20
40
60
80
100
0 5 10 15 20 25 30 35 40 45 50
Cell Range (km)
Rel
ativ
e Lo
adin
g (%
)
0 dB5dB10dB15dB20dB25dB
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CAPACITY in UMTS : Downlink Capacity
ExampleEstimate the Number of Speech Users supported in an Urban Environment :
Cell Range : 1 km18 dB Building Penetration Loss4 dB Feeder Loss18 dBi Antenna Gain
Answer1. Take the Mmax value for Speech 12.2 kbps
It would be 86.4 usersFind at which relative load the 25 dB curve crosses the 1 km range
It would be 40%Calculate the supported relative Load :
It would be 86.4 X 0.4 = 34 simultaneous users
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Total Loading :Loading = M1/M1max + M2/M2max + M3/M3max + …
Within a Cell, one service may have higher loading than another which leads to :
1 >= M1/(L1*M1max) + M2/(L2*M2max) + M3/(L3*M3max) +…Li : is the Maximum Load of Service i that a cell is able to support at a given range
Example of Mixed ServicesAssume we are interested in Speech @ 7.95 kbps mixed with Data @ 32 kbps in an Open Area (Vehicular A) at a range of 20 km. A 6 dB incar Penetration loss and 2 dB Feeder Loss are assumed :M1max = 64.1 users for Speech @ 7.95 kbpsM2max = 16.4 users for Packet Data @ 32 kbps
CAPACITY in UMTS : Mixed Services (1/2)
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A) For Speech, find the Relative Load the 8 dB (BL + BPL + Lf+j) Curve Crosses the 20 km Cell Range
Ans. 65%B) For Packet Data, find at which Relative Load the 5 dB (BPL + Lf+j ) Curve Crosses the 20 km Cell Range :
Ans. 75%Note 1: This examples confirms that some Services may have higher relative load than othersNote 2 : The above Loads are the Maximum Allowed Loads for Each Service, when offered as a Single ServiceC) Determine M1 and M2 that fulfill the following Equations :
1 >= M1/(0.65*64.1) + M2/(0.75*16.4) (Cell Range Limitation)0.65 >= M1/64.1 + M2/16.4 (Total Load Limitation)At this stage it is necessary to determine the capacity distribution between Service Types : We assume Speech Traffic approximately 10 times Packet Data Traffic, hence : M1 = 10*M2This Leads to M1 = 29 and M2 = 3 ; we verify that Loading = 63 % < 65%
CAPACITY in UMTS : Mixed Services (1/2)
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Concluding Remarks : CDMA System Capacity
Capacity, or number of simultaneous users M, is directly proportional to the Processing Gain of the System
Capacity is inversely proportional to the required Eb/No of the system. The lower the required threshold Eb/No , the higher the capacity
Capacity can be increased if one can decrease the amount of loading from users in adjacent cells
Spatial filtering, such as sectorization, increases system capacity. For example, a 6sector cell would have a higher capacity than a 3sector cell.