Wcdma Fundamentals By Dr. Hatem Mokhtari

WCDMA Fundamentals Dr. Hatem MOKHTARI Cirta Consulting LLC

© Cirta Consulting LLC 1999-2004

1

Spread Spectrum Modulation n(t) Narrow-band Signal

Narrow-Band Signal Sn

ε( )

Transmitter

SW Radio Propagation Channel

i(t)

ε−1( )

S’n

Receiver

SPREAD SPECTRUM SYSTEM BLOCK DIAGRAM Sn : Narrow-band Modulated Binary Sequence (information : Speech or Data) ε( ) : Spreading function using high chip-rate modulation ε-1( ) : Despreading function using the same sequence as ε( ) n(t) : Gaussian White-Noise ; i(t) : Interference © Cirta Consulting LLC 1999-2004

2

Digital Modulation in WCDMA : QPSK for DL and BPSK in UL „

„

PSK or Phase Shift Keying is advantageous compared to other modulations : ‹ Constant Amplitude ‹ Data information is “hidden” in the Phase component ‹ Robustness to Noise and Interference because Noise in general affects the Amplitude and not the Phase Component

QPSK is a 4-state Modulation scheme : ‹ Used

in DL because of High Data Rate demand ‹ Same Properties as the PSK : Constant Amplitude and Data in the Phase component © Cirta Consulting LLC 1999-2004

3

BPSK Modulation „ „

Binary Phase Shift Keying Modulation is a two-state Modulation scheme In 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 :

s1 (t ) = A cos(2πf c t ) = ‹

2 Eb cos(2πf c t ) Tb

for a 0 Transmitted symbol :

s2 (t ) = A cos(2πf c t + π ) = − ‹ ‹ ‹

2 Eb

2 Eb cos(2πf c t ) Tb

Where A = T the Amplitude, 0 ≤ t ≤ Tb , b Eb is the transmitted Energy per bit The duration of this sinusoidal waveform is Tb

© Cirta Consulting LLC 1999-2004

4

BPSK Modulation Example Tb = nc*Tcarrier

1

0

0

1

Tcarrier = 1/fc

To ensure that each transmitted binary digit contains an integer number of cycles, the n c carrier frequency fc has to fulfill the condition : f c = where nc is a fixed T b integer number. For the above example nc = 2

© Cirta Consulting LLC 1999-2004

5

Modulation Example : BPSK BPSK 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)

Binary Data

Phase Modulator

2 P cos(ω 0t )

2 P cos(ω 0t + θ d (t ) ) 2 P c(t ) cos(ω 0t + θ d (t ) )

c(t )

BPSK DS SS Transmitter © Cirta Consulting LLC 1999-2004

6

Modulation Example : BPSK The 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 ad-hoc delayed spreading code as shown bellow

2Pc(t −τ d ) cos[ω0t +θd (t −τ d ) + φ ] + n(t)

Bandpass Filter

Data Phase Demodulator

Estimated Data

c(t − τ d' ) BPSK DS Receiver © Cirta Consulting LLC 1999-2004

7

Modulation Example : BPSK The re-modulation or correlation of the received signal with the delayed spreading code is a critical function in all SS Systems The signal component of the Output of the despreading Mixer is given by :

Sn' (t) = 2Pc(t −τ d )c(t −τ 'd ) cos[ω0t +θd (t −τ d ) + φ ] τ’d is the receiver’s best estimate of the Transmission delay Since c(t) equals +1 or -1 the product c(t −τ d )c(t −τ 'd ) will be +1 if the delays τd = τ’d that is, if the spreading code and the despreading code are SYNCHRONIZED.

© Cirta Consulting LLC 1999-2004

8

BPSK Spectral Density „

Before Spreading and for Tb as a bit duration the two-sided power spectral Density is given by :

Sn ( f ) = „

}

After the Spreading we use the Chip duration Tc instead of Tb because of the involvement of the Chip (code) :

S n' ( f ) = „

{

1 PTb sin c 2 [( f − f 0 )Tb ] + sin c 2 [( f + f 0 )Tb ] 2

{

}

1 PTc sin c 2 [( f − f 0 )Tc ] + sin c 2 [( f + f 0 )Tc ] 2

Where sinc function (Also known as the Cardinal Sinus) is given by :

sin c( x) =

sin( x) x

© Cirta Consulting LLC 1999-2004

9

BPSK Power Spectral Density 1 0.8

SIN(x)/x

0.6

sin c( x) =

0.4

sin( x) x

0.2

900

800

700

600

500

400

300

200

0

100

-100

-200

-300

-400

-500

-600

-700

-800

-0.2

-900

0

-0.4 x (DEGREES)

1 0.9 0.8 0.6 0.5 0.4 0.3 0.2 0.1 900

800

700

600

500

400

300

200

0

100

-100

-200

-300

-400

-500

-600

-700

-800

0 -900

 sin( x)  sin c 2 ( x) =    x 

2

(sinx/x)^2

0.7

x (DEGREES) © Cirta Consulting LLC 1999-2004

10

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

si (t ) =

2 Es π  cos 2πf c t + (2i − 1)  4 Ts 

where i = 1,2,3,4

* 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

© Cirta Consulting LLC 1999-2004

11

QPSK Modulation „

BASIS Functions ‹ In this case we use ORTHONORMAL basis functions as follows :

2 φ1 (t ) = cos(2πf c t ) Ts ‹

The Coordinates on the signal Constellation (or Space) diagram are given by : Ts

x1 = ∫ r (t )φ1 (t )dt 0

‹

2 φ2 (t ) = sin (2πf c t ) Ts

Ts

x2 = ∫ r (t )φ 2 (t )dt 0

r(t) = si(t) + n(t) is the received signal : Either signal s1 or s2 + Random Noise

© Cirta Consulting LLC 1999-2004

12

QPSK Modulation : Constellation ‹

‹

‹

‹

The signal Constellation diagram for QPSK is shown bellow. The signal points are mapped to a pair of binary digits as shown The Decision boundaries are shown as solid Horizontal Vertical lines Notice how this mapping has been chosen so that neighboring signal points differ in only a SINGLE BINARY DIGIT

© Cirta Consulting LLC 1999-2004

13

QPSK Modulation ‹ ‹ ‹ ‹

‹

‹

For example, signal B(11) and D(10) differ in only binary digit position If 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 ENCODING The 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 ‹ Example Under the conditions of no noise, the coordinates of a signal point are given by :

and

π  x1 = Es cos (2i − 1)  4 

Test : Verify the above formulas for i=1

π  x2 = − Es sin (2i − 1)  4 

© Cirta Consulting LLC 1999-2004

14

QPSK Modulation : Answer Ts

2 Es x1 = ∫ r (t )φ1 (t )dt = Ts 0

Ts

π



π 

∫ cos(2πf t )cos(2πf t ) cos( 4 ) − sin(2πf t ) sin( 4 ) dt c

c

c

0

Ts

T 2 Es 2 s x1 = ∫ r (t )φ1 (t )dt = × cos 2 (2πf c t ) − sin( 2πf c t ) cos( 2πf c t ) dt ∫ Ts 2 0 0

(

cos 2 θ =

x1 =

1 (1 + cos(2θ ) ) 2

Es 2

)

sin θ cos θ =

1 sin( 2θ ) 2

Es x2 = − 2

© Cirta Consulting LLC 1999-2004

15

QPSK Coherent Receiver

© Cirta Consulting LLC 1999-2004

16

CDMA Multiple Access : Principal of Spread Spectrum (SS) „ „

Each User encodes its signal Code Signal Bandwidth (W) > Information Bandwidth

Transmission

Spread Spectrum

f

f „

The Receiver knows the code sequence

Reception

P Despreading

f

f

© Cirta Consulting LLC 1999-2004

17

CDMA Multiple Access Advantages : Multiple Access Features 1. All Users’ Signals overlap in TIME and FREQUENCY 2. Correlating the Received Signal despreads ONLY the WANTED SIGNAL p

p

S1

RECEIVER of USER 1

S1xC1 f

p

f

S1 = S1 X C1 X C1

p

S2 X C2 X C1 p

p

S2

f

f

S2xC2 f

f

© Cirta Consulting LLC 1999-2004

18

CDMA Multiple Access Advantages : Interference Rejection p

p

S1

S1xC1 f

f p

p

p

I

I

S1 IxC1 f

f

f

Correlation Narrowband Interference Spread the power © Cirta Consulting LLC 1999-2004

19

CDMA Principles : Multiplexing A

B1

m’1(t) ∫

D/A

m1(t) c1(t)

c1(t)

Radio Propagation Channel

m’2(t)

B2

D/A

∫

m2(t)

A c2(t)

c2(t)

Receiver

Transmitter c1(t) and c2(t) are Orthogonal Codes

:

T

∫ c (t )c (t )dt = 0 1

2

0

© Cirta Consulting LLC 1999-2004

20

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 :

H H 2N =  N H N „

„

Where

HN  H N 

0 0  H2 =   0 1  

Example : N=2

H 2 H4 =  H 2

0 H 2  0 =  H 2  0 0

0 0 0 W0    1 0 1 W1  = 0 1 1  W2  1 1 0  W3 

© Cirta Consulting LLC 1999-2004

21

Walsh Codes (2/6) „

Walsh codes are thus given by :

W0 = [− 1 − 1 − 1 − 1] W1 = [− 1 1 − 1 1]

W2 = [− 1 − 1 1 1] W3 = [− 1 1 1 − 1] „

„

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.

© Cirta Consulting LLC 1999-2004

22

Walsh Codes : Example (3/6) „

Assume three different users with three different data sequences :

m1 (t ) = [+ 1 − 1 + 1]

m2 (t ) = [+ 1 + 1 − 1] m3 (t ) = [− 1 + 1 + 1] „ „

W1(t) m1(t)

Assume we have a Spreading factor of 4 then : The Spread Spectrum Signal would be for user1 : -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

1

-1

= S1(t)

-1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1 1 -1 -1 1 -1 1 1 -1 1 -1

© Cirta Consulting LLC 1999-2004

23

Walsh Codes : Example (4/6) „

The Resulting SS Signal for the three users can be written as follows :

C (t ) = w1 (t ) m1 (t ) + w2 (t ) m 2 (t ) + w3 (t ) m 3 (t ) „

And the bit sequence would be :

C (t ) = [− 1 − 1 − 1 3 „

−1 −1 3

−1 −1 3

− 1 − 1]

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]

© Cirta Consulting LLC 1999-2004

24

Walsh Codes : Example (4/6) C(t) W1(t)

C(t)

m1' (t )

1 or -1

Tb

∫ C (t )W (t )dt 1

0

W1(t)

C (t )W1 (t ) = [1 − 1 1 3 1 − 1 − 3

4

− 1 1 3 1 − 1]

-4

4

m1' (t ) = [4 − 4 4] m1' (t ) is computed over the information period Tb using a summation © Cirta Consulting LLC 1999-2004

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 multipath interference, 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 been despread using the Walsh code, it is sub-sampled 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.

25

Walsh Codes : Self-Test „

Compute 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

© Cirta Consulting LLC 1999-2004

26

Walsh Codes Self-Test : Answer

H H8 =  4 H 4

0 0  0 H 4  0 = H 4  0 0  0 0 

0 1 0 1 0 1 0 1

0 0 1 1 0 0 1 1

0 1 1 0 0 1 1 0

0 0 0 0 1 1 1 1

0 1 0 1 1 0 1 0

0 0 1 1 1 1 0 0

0 1 1  0 1 0  0 1 

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. © Cirta Consulting LLC 1999-2004

27

Walsh Codes Self-Test : Answer m1=[-1 1] m2 = [1 -1] m3 = [1 1] m1 W1 S1 = W1*m1

-1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1

m2 W2 S2 = W2*m2

1 -1 -1

1 1 -1 1 -1 1

1 1 1

m3 W3 S3 = W3*m3

1 -1 -1

1 1 1

C=W1*m1+W2*m2+W3*m3

-1

-1 3

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 -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 -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 1 -1 -1 1 -1 -1

-1 -1 -1

3

-1

-1

3

-1 -1 -1

1

3

1 -1

1 -1 -1

3 -1

-1

3

1

-1 8 1 -1 -8 -1

1 -1 -3 -1

1

-1

-3 -1 -8 -1

W2*C After Integrator After Threshold Decision

1

3 -1

1

1

3 -1 8 1

1 -3

-1 -1

1 -3 -1

W3*C After Integrator After Threshold Decision

1 -1

3

1

-1

3

1

-1

1

1

1 8 1

3

1

-1 1 -1

1 1 1

W1*C After Integrator After Threshold Decision

1

1

1 1 1

1 1 1

3 -1

1 8 1

© Cirta Consulting LLC 1999-2004

28

CDMA Principles T

„

„

Cross-Correlation Rxy(τ) : Cross-correlation if τ=0 :

R xy (τ ) = ∫ x (t ) y (t − τ ) dt 0 T

Rxy (0) = ∫ x(t ) y (t )dt 0

„

If x and y are discrete sequence (binary signals):

Rxy ( 0 ) = X T .Y =

∑x y

1≤ i ≤ I „

i

i

Example of orthogonal codes : − 1 − 1 − 1 1 1 − 1 Rxy (0) = [− 1 − 1 1 1].  = 1 − 1 + 1 − 1 = 0 Y =  X =  1 1 1       − 1 1 − 1

© Cirta Consulting LLC 1999-2004

29

CDMA Principles „

To be used in DS-SS CDMA Codes must satisfy the following conditions : ‹ Zero

Cross-correlation ‹ Number of +1s and -1s must be the same ‹ Dot Product must be equal to 1 „

Example : ‹ Dot

product of the previous example is :

X T . X / 4 = (1 + 1 + 1 + 1) / 4 = 1 © Cirta Consulting LLC 1999-2004

30

CDMA Principles m1(t)

1

-1

Tb c1(t)

M1(f)

1

2Tb

3Tb

Tc : Chip Rate of the PN Code Tb : Information rate (voice/data)

1/Tb

f

C1(f) f

Tc m1(t).c1(t)

4Tc

1/Tb

1/Tc

C1(f)* M1(f) f 1/Tb

1/Tc

© Cirta Consulting LLC 1999-2004

31

CDMA Fundamentals „

„

„

„

„

„

„

W/R : Defined as the system processing gain In CDMA, the Reverse Link Capacity is often the limiting link in terms of capacity In CDMA : Uplink Receive Power is equal from all MSs. Per user : S/N = 1/(M-1) ‹ M : Total Number of users in the cell ‹ S = S (The wanted signal) ‹ N = (M-1)S => S/N = 1/(M-1) Example : If M=7 then S/N = 1/6

S S S

S

W if M>>1 then M ≈ R Eb No M : The Number of simultaneous users a CDMA cell can support

© Cirta Consulting LLC 1999-2004

32

CDMA Multiple Access : Principal of Spread Spectrum (SS) Eb = Signal Power / Bit Rate = S/Rb No = Noise Power / Bandwidth = N/W

Eb S W = × N o N Rb

Signal to Noise Ratio Processing Gain Example : Given a Demodulator Performance Eb > 6dB

No

Bit rate Rb = 8 kpbs Bandwidth W = 1.2 Mcps => G = W/Rb = 150 = 21 dB

E  W  S   > 6 dB − 21dB = − 15 dB   =  b  −   N  dB  N o  dB  Rb  dB © Cirta Consulting LLC 1999-2004

33

CDMA Fundamentals : Capacity If other users from other cells are considered, the actual cell becomes loaded and :

Eb 1 1 W = . . No M −1 R 1+η where η is the loading factor (0 < η < 1) We define F as the Frequency reuse :

1 F= 1+η

Where η is given by : η =

I Other _ Cell _ Interference I Own _ cell _ Interference

© Cirta Consulting LLC 1999-2004

34

CDMA Fundamentals : Capacity

Cell A

B1

Cell B

C2

B2 C1

Cell C

Interference Introduced by Users in the Neighboring Cells © Cirta Consulting LLC 1999-2004

35

CDMA Fundamentals : Capacity

Cell A

Cell B

Unwanted interferers rejected by antenna pattern of Cell A

Cell C

Sectorization Reduces Interference and adds a Gain to the system : Sectorization Gain © Cirta Consulting LLC 1999-2004

36

CDMA Fundamentals : Capacity „

Sectorization Gain :

„

Tri-Sectors : λ = 3 (2.5 in practice)

„

6-Sectors : 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 2π

λ=

∫ I (θ )dθ

2π

0

G (θ ) ∫0 G(0) I (θ )dθ

© Cirta Consulting LLC 1999-2004

37

CDMA Fundamentals : Capacity „

Voice Activity Factor : Interference is reduced when the user is not transmitting

Eb 1 W 1 1 . . .λ. = No M −1 R 1 + η v „

The final value for M : W    R  . 1 .λ M ≈   Eb  η + 1 v   N  o

© Cirta Consulting LLC 1999-2004

38

Shannon-Hartley Theorem „

Generic Model Band-limited White Noise

Radio Propagation Channel

+

Noise+Signal

Received 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 the bandwidth increases. © Cirta Consulting LLC 1999-2004

39

CDMA Multiple Access Principle Shannon Theorem Channel Capacity C (Bit/s) given by Shannon Theorem :

S   C = W × log 2  1 +  N  W : System Bandwidth (Hz) S/N : Signal to Noise Ratio (numerical value) C : System Capacity (bit/s) Wide W and Low S/N (such as in WCDMA) Same Capacity Narrow W and Large S/N (such as in GSM)

© Cirta Consulting LLC 1999-2004

40

Shannon-Hartley Theorem „

C can also be written as :

„

where

„

assuming W to be infinite :

x=

 S   ln1 + N oW   S  1   × x ln1 +  = 1.44 ×  C =W × ln 2  x  No 

WN o S

 S   S  ln( 1 + t )  lim  lim lim C = 1 . 44 ×  = 1 . 44 ×  x→ ∞ t→0 N t N  0  t→0  0 „

„

t−

t2 t3 + + ...  S  2! 3! = 1 . 44 × W ×   t N 

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

© Cirta Consulting LLC 1999-2004

41

Bandwidth Efficiency „

„

Bandwidth Efficiency is a measure of how well a particular modulation and error-control 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/Hz ‹ Notice 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/Hz ‹ In this case of an ideal system, rb = 1/Tb = C, where C is given by the Shannon-Hartley theorem :  E C S  C = W × log 2 1 +  = W × log 2 1 + b .   N  N0 W  ‹

Or equivalently :

C W

 E Eb 2 −1 = ⇒  b C N0  N0 W

  = 10 log  dB

10

  C  2W −1   C       W

© Cirta Consulting LLC 1999-2004

42

Bandwidth Efficiency

Bandwidth Efficiency Versus Eb/No

17.5

14

10.7

7.48

4.49

1.76

-0.7

-2.7

-4.5

-5.9

1 -7.1

Bandwidth Efficiency, R/W

10

0.1 Eb/No (dB)

© Cirta Consulting LLC 1999-2004

43

Bandwidth Efficiency Self-Test * A Digital Cellular Phone System is required to work at a bandwidth efficiency of 4 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 the edge of the coverage area receive error-free communication ? ** To Double the number of users on the existing communication system, by what amount should the base-station and handsets transmitted powers be increased to maintain coverage and error-free communication ?

© Cirta Consulting LLC 1999-2004

44

Amplitude

CDMA Principles

distance

λ/2

Mobile The MS crosses 2 fades in :

λ 2 v

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 © Cirta Consulting LLC 1999-2004

45

CDMA Principles : Delay Spread Tb Propagation Time T1

Delay Spread = τ = |T2 - T1| Propagation Time T2

τ Here τ < Tb implies Interference

BTS

Example : D1 = 150 m and D2 = 200 m => T1 = 150/3*108 = 0.5 µs and T2 = 0.66 µs => τ = 0.16 µs Assume we have a UMTS Service R = 3.84 Mchps => Tb = 0.26 µs τ < Tb => INTER-SYMBOL INTERFERENCE PRESENT

© Cirta Consulting LLC 1999-2004

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 the multipath and thus increases the received BER.

46

CDMA Principles : Delay Spread Received Power

τ1= 3µs

τ2= 4µs τ3

Time (µs)

t

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

© Cirta Consulting LLC 1999-2004

47

CDMA Principles : Delay Spread „

Example 1: ‹ ‹ ‹

„

Let us consider a Mobile Communications System that uses Rb = 270.83 kbps The bit period is thus Tb = 1/270830 = 3.69 µs Conclusion : bit period almost equal to 4 µs as shown on the delay spread power profile => ISI would normally exist ! Without use of Viterbi-based EQUALIZER such as in GSM

Example 2: ‹ ‹ ‹ ‹

Let us consider a Mobile Communications System that uses Rb = 1.2288 Mbps = 1228800 bps The bit period is thus Tb = 1/ 1228800 = 1 µs Conclusion : 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

© Cirta Consulting LLC 1999-2004

48

The Principal of Maximum Ratio Combining in CDMA Rake Receiver Transmitted Symbol - Amplitude - Phase

Received Signal at each time delay

Modified Signal Using Channel Estimator

Combined Symbol

θ θ Figure #1 Figure #2 Figure #3

θ θ θ

© Cirta Consulting LLC 1999-2004

49

Block Diagram of CDMA Rake Receiver

Correlator

Input RF Signal Code Generator

I Q Channel Estimator

Channel Estimator

Code Generator

Channel Estimator

Delay Equalizer

Finger # 2 Phase Rotator

Correlator

Delay Equalizer

Finger # 1 Phase Rotator

Correlator Code Generator

Phase Rotator

Delay Equalizer

I

∑I

∑Q

Q

Combiner

Finger # 3

Timing (finger allocation)

Matched Filter

© Cirta Consulting LLC 1999-2004

50

CDMA Rake Receiver : Components „

Digitized input samples are received from RF Front-end 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 channel-compensated 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)

© Cirta Consulting LLC 1999-2004

51

CDMA Principles: Delay Spread „

In Multipath Environment : ‹ Received

power can be written as :

N

N

n =1

n =1

r (t ) = ∑ a n s (t − τ n ) → R ( f ) = S ( f ) ∑ a n exp( − j 2π f .τ n )

‹ Fourier

Transfer Function :

N R( f ) H(f ) = = ∑ a n e − j 2πf .τ n S ( f ) n =1

© Cirta Consulting LLC 1999-2004

52

CDMA Principles : Delay Spread H(f)

Example with two-equal amplitude paths : a1=a2=A H ( f ) = 2 cos( π fτ )

2A

1 2τ

1

τ

3 2τ

2

τ

f

1. Frequency-Selective 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 IS-95 CDMA using only 1.2288 Mbsp (bit period 1 ms) when ISI are to be considered in Dense Urban areas © Cirta Consulting LLC 1999-2004

53

CDMA Fundamentals : Power Control Near-Far Problem Pr,1 = EIRP(MS1) - PL1 = 21 - 100 = -79 dBm Pr,2 = EIRP(MS2) - PL2 = 21 - 90 = -69 dBm

PL2 = 90 dB Pr,2

Pr,2 P = 21 dBm

P = 21 dBm PL1 = 100 dB

MS1 (S/N)1 = Pr,1 - Pr,2 = -10 dB (S/N)2 = Pr,2 - Pr,1 = +10 dB

MS2 MS2 must be Power Controlled by -10 dB to have the same S/N for both users MS1 and MS2

© Cirta Consulting LLC 1999-2004

54

CDMA Fundamentals : Power Control „

What is the initial MS power to be used ? Maximum MS Power : ) Advantage : High likelihood to reach the BTS ) Drawback : Uplink interference ‹ Minimum MS Power : ) Advantage : Low Uplink interference ) Drawback : Low probability to reach the BTS ‹

„

Solution (IS-95 used in WCDMA also) : ‹ Use of Access Probes : MS Power Rise portions done gradually ‹ Advantage : Avoid Uplink interference and Reach the BTS with sufficient Transmit power

© Cirta Consulting LLC 1999-2004

55

CDMA Fundamentals : Power Control „

‹ ‹ ‹ ‹

OPEN-LOOP Power Control:

MS Transmits its first access probes at relatively low power MS waits for a response back from the BTS If 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

MS Transmit Power

First Acces Probe Correction

Second Acces Probe Correction

Random Time Intervals

Initial Transmit Power

© Cirta Consulting LLC 1999-2004

56

CDMA Fundamentals : Power Control „

„

„

„

„ „

OPEN-LOOP Power Control: Purely Mobile-Controlled 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 level ‹ Alternatively MS transmits at high level if the initial DL Level is low Initial MS Transmit Power (dBm) :

p t ,initial = − p r − 73 + NOM _ PWR + INIT _ PWR

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

© Cirta Consulting LLC 1999-2004

57

CDMA Fundamentals : Power Control „

During a call : ‹ MS continues to calculate Pr ‹ Path Loss changes as the MS moves and Pr also ‹ Open-Loop PC adjusts the MS Transmit power using :

p t = − p r − 73 + NOM _ PWR + INIT _ PWR + ∑ Acc . Pr .Correct . „

VERY IMPORTANT NOTES : ‹ ‹ ‹ ‹ ‹

„

UL and DL frequencies are different Fast Rayleigh Fading is Frequency-Selective Open-Loop PC too slow to compensate Rayleigh Fading However, Open-Loop PC suitable for Log-Normal (slow) Fading Correlation between UL and DL concerning Slow-Fading but not Fast Rayleigh Fading

Closed-Loop PC compensates Fast Rayleigh Fading

© Cirta Consulting LLC 1999-2004

58

CDMA Fundamentals : Power Control „

„

CLOSED-LOOP Power Control : Involves BOTH the MS and the BS

During the call : ‹ BS monitors the UL and measures the link Quality ‹ If the UL quality becomes bad : BS commands MS to power-up ‹ If the If the UL quality becomes too good : BS commands MS to powerdown BS Commands MS Yes to Power-Down

BS Monitors UL Eb/No

Eb/No > Threshold

No

BS Commands MS to Power-Up

© Cirta Consulting LLC 1999-2004

59

CDMA Fundamentals : Power Control CLOSED-LOOP PC BS Sends PC commands to the MS using the DL PC Commands are in the form of Power-Control Bits (PCBs) Each Power-up or Power-down amount is +1 dB or -1 dB MS response must be fast (Fast Rayleigh Fading) PCBs are sent through the Traffic channel Bits are robbed (stolen) from the Traffic Channel in order to send the PCBs „

„ „ „ „ „ „

PCB @ 1500 bps (1500 Hz)

Vocoder

12.2 kbps

24.4 kbps Conv. Encoder R=1/2

3.84 Mchip/s

24.4 kbps

MUX

Spreading

© Cirta Consulting LLC 1999-2004

60

CDMA Fundamentals : Power Control

OUTER LOOP Adjusting Eb/No Threshold

Eb No

Threshold

INNER LOOP 1. Measure Eb/No 2. Compare Eb/No to Threshold 3. Decide which PCB to send

© Cirta Consulting LLC 1999-2004

61

CDMA Fundamentals : Handover „ „

„

SOFT-HANDOVER

During the communication, the MS simultaneously maintains connection with two or three Base Stations, A Traffic Channel is maintained with both cells, BS2 (Target Cell) RNC

BS1 (Home Cell)

DOWNLINK

MS Combines the two signals using the Rake Receiver © Cirta Consulting LLC 1999-2004

62

CDMA Fundamentals : Handover „

SOFT-HANDOVER RNC

Demodulated Frame 1

Demodulated Frame 2

SELECTOR Selects the Best Frame with the best FER value !

UPLINK BS2 (Target Cell) BS1 (Home Cell) © Cirta Consulting LLC 1999-2004

63

CDMA Fundamentals : Handover „ „

„

„

„

SOFTER-HANDOVER

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

© Cirta Consulting LLC 1999-2004

64

CDMA Fundamentals : Handover Home Cell

Only 1 Frame sent back to the RNC RNC

Target Cell

BTS

SOFTER HANDOVER ILLUSTRATION © Cirta Consulting LLC 1999-2004

65

CDMA Fundamentals : Handover „ „

„

HARD-HANDOVER

Hard-Handover occurs from CDMA carrier to another CDMA Carrier or CDMA to Analog Carrier. CDMA to CDMA HO is sometimes called D-to-D Handover. Hard-Handover in UMTS may concern WCDMA to GSM or WCDMA to GSM1800

BS2

BS1 f1

f1

f2

MS

f2

© Cirta Consulting LLC 1999-2004

66

HANDOVER PROCESS in WCDMA Each Sector is distinguished from one another by its PILOT channel : 4 Logical Channels on the DL are Defined : Pilot, Paging, Synch., and Traffic Channels

Power

„

TRAFFIC CHANNEL K : USER K

TRAFFIC CHANNEL 2 : USER 2 TRAFFIC CHANNEL 1 : USER 1 SYNCHRONIZATION CHANNEL PAGING CHANNEL PILOT CHANNEL Frequency © Cirta Consulting LLC 1999-2004

67

HANDOVER PROCESS in WCDMA Each Sector uses a different PILOT and is assigned a different PN Code with an offset to distinguish it from other Sectors

3-Sector Site 6-Sector Site © Cirta Consulting LLC 1999-2004

68

HANDOVER PROCESS in WCDMA „

„

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 density ‹ Ec : 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 !

© Cirta Consulting LLC 1999-2004

69

HANDOVER PROCESS in WCDMA „

MS is an intimate in the SHO in WCDMA : ‹

MS Measures the Ec/Io and report it the the BTS

‹

BTS transmits a PN Sequence using different offsets for each sector

‹

Ec/Io gives a good indication on whether or not a sector should the serving cell

PN2

Highway or Road

PN1

PN3

Example with 3 PILOTS

© Cirta Consulting LLC 1999-2004

70

HANDOVER PROCESS in WCDMA „

„

During HO Management Process : ‹ MS maintains in its Memory four exclusive lists of Sectors ‹ Sectors stored in the form of PN offsets

Four Exclusive List of sectors called : SETS ‹Active Set ‹Candidate Set ‹Neighbor Set ‹Remaining Set

© Cirta Consulting LLC 1999-2004

71

HANDOVER PROCESS in WCDMA „

Active Set (A) : ‹ ‹ ‹

‹ ‹

„

Contains all Sectors communicating with the MS on traffic channels If the Active Set contains only one Pilot : NO SHO If the Active Set contains more than one Pilot : the MS is maintaining communication with all the Active Set Sectors with Different Traffic Channels Active can contain 6 Pilots : UL Diversity maximized A 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 Candidates If 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)

© Cirta Consulting LLC 1999-2004

72

HANDOVER PROCESS in WCDMA „

„

A Pilot is removed from (C) if its Ec/Io drops below T_DROP threshold for more than duration specified by T_TDROP : „Handover Drop Timer Expiration Value“ and put in (N) set The (C) Set can contain at least 6 Pilots

Ec/Io

Pilot still in (C) set

Pilot put in (N) set

T_DROP

Counter = 0

Counter = T_DROP

time

© Cirta Consulting LLC 1999-2004

73

HANDOVER PROCESS in WCDMA „

„

„

„

„

„

The Neighbor Set (N) : contains those Pilots that are in the Neighbor List of the MS´s Serving Cell Initially : ‹ (N) contains Pilots sent to the MS in the ´Neighbor List Message´ by the Serving 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

© Cirta Consulting LLC 1999-2004

74

HANDOVER PROCESS in WCDMA „

„

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_INC ‹ Example : )

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“ ‹

© Cirta Consulting LLC 1999-2004

75

HANDOVER PROCESS in WCDMA Source Cell A

Ec/Io

Target Cell B

2 Pilots (A&B) In Active Set : MS is in SHO 1 Pilot (B) in Active Set

T_ADD T_DROP

1 Pilot (A) in Active Set

Distance

© Cirta Consulting LLC 1999-2004

76

HARD HANDOVER in WCDMA „

Hard HO includes inter-frequency HO : ) )

Cells with multiple Carriers with high load WCDMA GSM

f1 f1

f1 f1, f2 f1

f1

GSM

f1

WCDMA

© Cirta Consulting LLC 1999-2004

77

HO Quality Measure (dB)

HARD HANDOVER in UMTS A

B C Replace A by C

Replace C by A

Replace A by B

Time

© Cirta Consulting LLC 1999-2004

78

Cell Breathing Concept in UMTS The Cell Size Shrinks due to Loading LOAD

1. UL Noise-Rise due to own-cell or other-cell UL interference translated into Lower UL MAPL : MAPL2 = MAPL1 - Noise-Rise (dB) 2. Noise-Rise computed Using : -10log10(1-η) ; η being the Loading due to interference © Cirta Consulting LLC 1999-2004

79

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 => own-cell interference is twice the other-cell interference ‹ r = 0.984 km but reduces to 0.501 km using η = 0.75 loading ‹ 49 % Range Loss in DENSE URBAN

HIGHER LOADING 50 % Loading

75 % Loading

© Cirta Consulting LLC 1999-2004

80

Cell Breathing : RF Propagation Environment ‹ ‹ ‹

Path Loss (dB)

‹

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 = 0.75 19 % Range Loss in SUBURBAN

Dense Urban Suburban

MAPL (η = 0) Ideal Case

3 dB 6 dB

MAPL (η = 50%) or Noise Rise = 3 dB MAPL (η = 75 %) or Noise Rise = 6 dB

Log(d) © Cirta Consulting LLC 1999-2004

81

Soft Capacity „ „ „

WCDMA System is limited by interference sharing between neighbor cells Capacity is dynamically changing due to different traffic variation Hard Capacity is limited by Hardware Limitation : TRX namely

Number of Channels N

Equally Loaded Cells

Number of Channel Pool = N*(1+i)

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 © Cirta Consulting LLC 1999-2004

82

Soft Capacity „

i is defined by : ‹ i = Other Cell Interference / Own Cell Interference ) ) ) ) )

„

i = 0.55 for Omni Cell i = 0.55 for 2-sector cell i = 0.65 for 3-sector cell i = 0.75 for 4-sector cell i = 0.85 for 6-sector cell

Soft Capacity Calculation Procedure : ‹ 1. Compute the Number of Channels N in equally loaded case based on Uplink Load Factor : W    R N =η     (1 + i )v Eb    N    o   ‹ 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

© Cirta Consulting LLC 1999-2004

83

Soft Capacity „ „

Assumptions PARAM ETER B it R a te V o ic e A c tiv it y F a c to r E b /N o i N o is e R is e G oS

„

Example VA LU E S p e e c h @ 1 2 .2 k b p s 67 % S ppech : 4 dB 0 .5 5 a s s u m in g a n O m n i C e ll 3 d B (5 0 % L o a d F a c to r ) 2 %

Computation of Number of Channels : ‹ ‹ ‹ ‹ ‹ ‹ ‹ ‹ ‹

W/R = 3.84*1000/12.2 = 314.75 Eb/No = 4 dB = 2.51 v = 0.67 (67 % Voice Activity Factor) N = 60.4 channels Npool = 60.4*(1+0.55) = 93.6 Channels Offered Capacity = ErlangB (94 channels, 2% GoS) = 82.17 Erlang Soft Capacity = 82.17/(1+0.55) = 53 Erlang Hard Capacity = ErlangB (61 channels, 2% GoS) = 50.6 Erlang Soft Capacity (%) = 53.6/50.6 -1 = 4.74 approximately 5%

© Cirta Consulting LLC 1999-2004

84

CAPACITY in UMTS : Uplink Capacity - Sensitivity is affected by interference (Loading)

IUL = −10 × log(1 − ηUL ) ‹

Maximum Capacity is related to the amount of UL interference a system can tolerate I − UL 10 UL max

η

= 1 − 10

=

M

M

‹

Mmax : Maximum number of simultaneous users supported by a single cell-carrier @ 100% Loading assuming that all users are using the same service (e.g. speech or Data). Also called the Pole Capacity.

‹

For a Multi-service system :

ηUL = 1 − 10 ‹

− IUL 10

=

M3 M1 M2 + + + ... M 1,max M 2,max M 3,max

Mn : The Number of simultaneous users for the n th service

© Cirta Consulting LLC 1999-2004

85

CAPACITY in UMTS : Uplink Capacity „

Maximum UL Capacity figures

M max = 1 +

C = 10 I „

1 C  (1 + F ) ×    I  numerical

 Eb   No

 W  −   dB  R  dB 10

F Values are : ‹ 0.67 for Omni, 0.93 for 3-sector, ) 0.4 for Micro-cells (without the presence of Fast Fading) )

© Cirta Consulting LLC 1999-2004

86

CAPACITY in UMTS : Uplink Capacity F Values on the UL S ite C o n fig u ra tio n O m ni 3 -S ecto r M icro cell

P ed estrian A 3 k m /h w ith o u t F a st F a d in g

P ed estrian B 3 k m /h w ith F a st F a d in g

V eh icu la r A 120 k m /h w ith F ast F a d in g

0 .8 4 / 0 .6 7 1 .1 6 / 0 .9 3 0 .5 / 0 .4

0 .6 7 0 .9 3 0 .4

0 .6 7 0 .9 3 0 .4

Typical Uplink Mmax Values for a 3- Sector Site Service

S peech 12.2 kbps C ircuit 128 kbps C ircuit 384 kbps P acket 128 kbps P acket 384 kbps

Pedestrian A M m ax w ithout Fast F ad ing

P edestrian B M m ax w ith Fast Fading

V ehicular A M m ax W ith F ast F ad ing

94.6 / 105.9 8.5 / 9.3 4.2 / 4.6 11.1 / 12.3 4.3 / 4.8

84.4 7.9 3.7 9.9 4.0

56.3 6.8 3.2 8.1 3.4

© Cirta Consulting LLC 1999-2004

87

CAPACITY in UMTS : Uplink Capacity „

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 dB F = 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?

© Cirta Consulting LLC 1999-2004

88

CAPACITY in UMTS : Downlink Capacity „ „ „

In the DL, each user is subjected to different interference levels Therefore, no single interference level is valid in the DL For dimensioning purposes, the current solution is to use simulation. Mostly, researchers do use Monte Carlo Random Number Generation Algorithms

Urban Environment - Pedestrian A 3 km/h Assumptions : 18 dBi Antennas, PL = 134.7 + 35.2 log(R), Antenna Height 30 m, No Body Loss Considered

Relative Loading (%)

90 80 70 60

0 dB 5 dB 10 dB

50 40

15 dB 20 dB 25 dB

30 20 10 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Cell Range (km)

© Cirta Consulting LLC 1999-2004

89

CAPACITY in UMTS : Downlink Capacity

Recommended Mmax Values for 3-sector site Configuration Service Speech 12.2 kbps Circuit 128 kbps Circuit 384 kbps Packet 128 kbps Packet 384 kbps

Dense Urban & Suburban

Rural

86.4 7.3 2.7 8.4 2.8

64.1 4.1 1.6 5.2 1.7

© Cirta Consulting LLC 1999-2004

90

CAPACITY in UMTS : Downlink Capacity Open Area - Vehicular A 120 km/h Assumptions : 18 dBi Antennas, PL = 105 + 33.8 log(R), Antenna Height 50 m, 3 dB Body Loss

Relative Loading (%)

100 0 dB 5dB 10dB 15dB 20dB 25dB

80 60 40 20 0 0

5

10

15

20

25

30

35

40

45

50

Cell Range (km)

© Cirta Consulting LLC 1999-2004

91

CAPACITY in UMTS : Downlink Capacity „ „

Estimate the Number of Speech Users supported in an Urban Environment : ‹ ‹ ‹ ‹

„

Example

Cell Range : 1 km 18 dB Building Penetration Loss 4 dB Feeder Loss 18 dBi Antenna Gain

Answer ‹

1. Take the Mmax value for Speech 12.2 kbps )

‹

Find at which relative load the 25 dB curve crosses the 1 km range )

‹

It would be 86.4 users It would be 40%

Calculate the supported relative Load : )

It would be 86.4 X 0.4 = 34 simultaneous users

© Cirta Consulting LLC 1999-2004

92

CAPACITY in UMTS : Mixed Services (1/2) „

„

„

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 Services ‹ Assume 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 kbps ‹ M2max = 16.4 users for Packet Data @ 32 kbps

© Cirta Consulting LLC 1999-2004

93

CAPACITY in UMTS : Mixed Services (1/2) „

„

„

„

„

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 others Note 2 : The above Loads are the Maximum Allowed Loads for Each Service, when offered as a Single Service C) 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*M2 ‹ This Leads to M1 = 29 and M2 = 3 ; we verify that Loading = 63 % < 65%

© Cirta Consulting LLC 1999-2004

94

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 6-sector cell would have a higher capacity than a 3-sector cell.

© Cirta Consulting LLC 1999-2004

95