WCDMA Fundamentals by Dr. Hatem MOKHTARI

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This is an overview on WCDMA with some dimensioning issues

Transcript of WCDMA Fundamentals by Dr. Hatem MOKHTARI

1 Cirta Consulting LLC 1999-2004WCDMA FundamentalsDr. Hatem MOKHTARICirta Consulting LLC2 Cirta Consulting LLC 1999-2004Spread Spectrum ModulationSn ( )SWRadio PropagationChannel 1( ) SnTransmitterReceiverNarrow-Band SignalNarrow-band Signaln(t)i(t)SPREAD SPECTRUM SYSTEM BLOCK DIAGRAMSn: Narrow-band Modulated Binary Sequence (information : Speech or Data)( ) : Spreading function using high chip-rate modulation-1( ) : Despreading function using the same sequence as ( )n(t) : Gaussian White-Noise ; i(t) : Interference3 Cirta Consulting LLC 1999-2004Digital 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 4-state Modulation scheme :Used in DL because of High Data Rate demandSame Properties as the PSK : Constant Amplitude and Data in the Phase component4 Cirta Consulting LLC 1999-2004BPSK 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 : for a 0 Transmitted symbol : Where the Amplitude, , Ebis the transmitted Energy per bit The duration of this sinusoidal waveform is Tb( ) ( ) t fTEt f A t s cbbc 2 cos22 cos ) (1 = =( ) ( ) t fTEt f A t s cbbc 2 cos22 cos ) (2 = + =bbTEA2=bT t 05 Cirta Consulting LLC 1999-2004To ensure that each transmitted binary digit contains an integer number of cycles, the carrier frequency fc has to fulfill the condition : where ncis a fixed integer number. For the above example nc= 210 0 1Tcarrier= 1/fcTb= nc*TcarrierbccTnf =BPSK Modulation Example6 Cirta Consulting LLC 1999-2004Modulation 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 -1Assume data modulated carrier of power P and frequency 0and phase modulated d(t)PhaseModulatorBinary Data( ) t P0cos 2 ( ) ) ( cos 20 t t P d +) (t c( ) ) ( cos ) ( 20 t t t c P d +BPSK DS SS Transmitter7 Cirta Consulting LLC 1999-2004Modulation Example : BPSKThe Wideband Signal is transmitted througha channel having a delay tdThe received signal is mixed with interference and Gaussian NoiseDespreading is done by remodulating the wideband signal with ad-hoc delayed spreadingcode as shown bellowBandpassFilter| | ) ( ) ( cos ) ( 20 t n t t t c P d d d + + + ) ('dt c BPSK DS ReceiverData PhaseDemodulatorEstimated Data8 Cirta Consulting LLC 1999-2004Modulation Example : BPSKThe re-modulation or correlation of the received signal with the delayed spreading codeis a critical function in all SS SystemsThe signal component of the Output of the despreading Mixer is given by :| | + + = ) ( cos ) ' ( ) ( 2 ) (0'd d d d n t t t c t c P t Sdis the receivers best estimate of the Transmission delaySince c(t) equals +1 or -1 the product will be +1 if the delaysd= dthat is, if the spreading code and the despreading code are SYNCHRONIZED.) ' ( ) ( d d t c t c 9 Cirta Consulting LLC 1999-2004BPSK Spectral Density Before Spreading and for Tbas a bit duration the two-sided 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 :( ) | | ( ) | | { }b b b n T f f c T f f c PT f S0202sin sin21) ( + + =( ) | | ( ) | | { }c c c T f f c T f f c PT f Sn0202 'sin sin21) ( + + =xxx c) sin() ( sin =10 Cirta Consulting LLC 1999-2004BPSK Power Spectral Density-0.4-0.200.20.40.60.81-900-800-700-600-500-400-300-200-100 0100200300400500600700800900x (DEGREES)SIN(x)/x00.10.20.30.40.50.60.70.80.91-900-800-700-600-500-400-300-200-100 0100200300400500600700800900x (DEGREES)(sinx/x)^2xxx c) sin() ( sin =22) sin() ( sin |.|

\|=xxx c11 Cirta Consulting LLC 1999-2004QPSK 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* Esis the Transmitted Signal Energy per Symbol* Tsis the Symbol Duration. Note that each symbol can represent 2 binary digits unlike BPSK in which each symbol was just a sungle binary digit* fcis the carrier frequency equal to nc/Tcas in BPSK description( ) 4 , 3 , 2 , 141 2 2 cos2) ( =

+ = i where i t fTEt s cssi 12 Cirta Consulting LLC 1999-2004 BASIS Functions In 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 s1or s2+ Random NoiseQPSK Modulation( ) t fTt cs 2 cos2) (1 = ( ) t fTt cs 2 sin2) (2 == sTdt t t r x01 1) ( ) ( = sTdt t t r x02 2) ( ) ( 13 Cirta Consulting LLC 1999-2004 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 DIGITQPSK Modulation : Constellation14 Cirta Consulting LLC 1999-2004QPSK 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( )

=41 2 cos1 i E x s( )

=41 2 sin2 i E x sTest : Verify the above formulas for i=115 Cirta Consulting LLC 1999-2004QPSK Modulation : Answerdt t f t f t fTEdt t t r x c cT Tcsss s

= = )4sin( ) 2 sin( )4cos( ) 2 cos( ) 2 cos(2) ( ) (0 01 1 ( )dt t f t f t fTEdt t t r x s sT Tc c css = =0 021 1) 2 cos( ) 2 sin( ) 2 ( cos222) ( ) ( ( ) ) 2 cos( 121cos2 + = ) 2 sin(21cos sin =21 sEx =22 sEx =16 Cirta Consulting LLC 1999-2004QPSK Coherent Receiver17 Cirta Consulting LLC 1999-2004CDMA Multiple Access : Principal of Spread Spectrum (SS) Each User encodes its signal Code Signal Bandwidth (W) > Information Bandwidth The Receiver knows the code sequence TransmissionSpread Spectrumf ffPfReceptionDespreading18 Cirta Consulting LLC 1999-2004CDMA Multiple Access Advantages : Multiple Access Features1. All Users Signals overlap in TIME and FREQUENCY2. Correlating the Received Signal despreads ONLY the WANTED SIGNALpf fS1pS1xC1 pf fS2pS2xC2 fpS2 X C2 X C1 fpS1 = S1 X C1 X C1RECEIVER of USER 119 Cirta Consulting LLC 1999-2004CDMA Multiple Access Advantages : Interference RejectionpffS1pS1xC1 pfIfpfpIxC1 IS1Correlation Narrowband Interference Spread the power20 Cirta Consulting LLC 1999-2004CDMA Principles : MultiplexingRadio PropagationChannelD/AAAB1B2m1(t)m2(t)c1(t)c2(t)c1(t)c2(t)c1(t) and c2(t) are Orthogonal Codes : 0 ) ( ) (02 1 =Tdt t c t cm1(t)D/Am2(t)TransmitterReceiver21 Cirta Consulting LLC 1999-2004Walsh 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

=NNN NNH H H HH2

=1 00 02H

=

=

=32102 22 240 11 11 00 01 00 01 00 0WWWWH H H HH22 Cirta Consulting LLC 1999-2004Walsh 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 W0all the codes satisfy orthogonality and dot product conditions to be used in WCDMA.| || || || | 1 1 1 11 1 1 11 1 1 11 1 1 13210 = = = =WWWW23 Cirta Consulting LLC 1999-2004Walsh 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 :| || || | 1 1 1 ) (1 1 1 ) (1 1 1 ) (321+ + = + + = + + =t m t m t m-1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 11 -1 -1 1 -1 1m1(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 -1 S1(t)=24 Cirta Consulting LLC 1999-2004Walsh 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]) ( ) ( ) ( ) ( ) ( ) ( ) (3 3 2 2 1 1 t m t w t m t w t m t w t C + + =| | 1 1 3 1 1 3 1 1 3 1 1 1 ) ( = t C25 Cirta Consulting LLC 1999-2004C(t)W1(t)C(t) W1(t)bTdt t W t C01) ( ) (1 or -1| | 1 1 3 1 1 3 1 1 3 1 1 1 ) ( ) (1 = t W t CWalsh Codes : Example (4/6)) ('1 t m) ('1 t mis computed over the information period Tbusing a summation] 4 4 4 [ ) ('1 = t m4 -4 4The 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 t