Lecture 4. EITN Chapter 12, 13 Modulation and diversity. Antenna noise is usually given as a noise temperature!

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1 Lecure 4 EITN Chaper 12, 13 Modulaion and diversiy Receiver noise: repeiion Anenna noise is usually given as a noise emperaure! Noise facors or noise figures of differen sysem componens are deermined by heir implemenaion. When adding noise from several sources, remember o conver from he db-scale noise figures ha are usually given, before saring your calculaions. A passive aenuaor in (room emperaure), like a ransmission line, has a noise figure/facor equal o is aenuaion. 5 April

2 Receiver noise A final example Le s consider wo (incomplee) receiver chains wih equal gain from poin A o B: T a 1 A L f G 1 G 2 B F 1 F 2 Ta 2 A L f G 1 G 2 B F 1 F 2 5 April Receiver noise A final example T a 1 2 T a A F 1 L f G 1 G 2 F 2 B A L f G 1 G 2 Equivalen noise sources a poin A for he wo cases would have he power specral densiies: 1 2 F 1 (( 1) ( 1 ) / ( 1 ) / ) N = kt + k F - + L - G + F - L G T 0 a 1 f 1 2 f 1 0 (( 1) ( 1) ( 1 ) / ) N = kt + k L - + F - L + F - L G T 0 a f 1 f 2 f 1 0 Two of he noise conribuions are equal and wo are larger in (2), which makes (1) a beer arrangemen. This is why we wan a low-noise amplifier (LNA) close o he anenna. F 2 B 5 April

3 Receiver noise The link budge! TX POWER [db] $ f, TX' a, TX $p Noise reference level ' a, RX F [db] is he noise figure of he equivalen noise source a he reference poin and B [dbhz] he sysem bandwidh. = k(t 0 )= -204 db[w/hz] $ f, RX / 0 The receiver noise calculaions show up here. Transmier Receiver In his version he reference poin is here 5 April OPTIMAL RECEIVER AND BIT ERROR PROBABILITY 5 April

4 Wha do we mean by opimal? Every receiver is opimal according o some crierion! We would like o use opimal in he sense ha we achieve a minimal probabiliy of error. In all calculaions, we will assume ha he noise is whie and Gaussian unless oherwise saed. Transmied and received signal Transmied signals s 1 () Channel Received (noisy) signals r() 1: s 0 () s() n() r() r() 0: 5 April

5 A firs inuiive approach Look a he received signal and compare i o he possible received noise free signals. Selec he one wih he bes fi. Assume ha he following signal is received: Comparing i o he wo possible noise free received signals: r(), s 1 () r() 1: This seems o be he bes fi. We assume ha 0 was he r(), s 2 () ransmied bi. 0: 5 April Le s make i more measurable To be able o beer measure he fi we look a he energy of he residual (difference) beween received and he possible noise free signals: r(), s 1 () s 1 () - r() 1:! " = $ & " ' ) ' * +' r(), s 0 () s 0 () - r() 0:!, = $ &, ' ) ' * +' This residual energy is much smaller. We assume ha 0 was ransmied. 5 April

6 The AWGN channel The addiive whie Gaussian noise (AWGN) channel s( ) n ( ) r( ) = as( ) + n( ) s( ) a n ( ) r( ) a - ransmied signal - channel aenuaion - whie Gaussian noise - received signal In our digial ransmission sysem, he ransmied signal s() would be one of, le s say M, differen alernaives s 0 (), s 1 (),..., s M-1 (). 5 April The AWGN channel, con. I can be shown ha finding he minimal residual energy (as we did before) is he opimal way of deciding which of s 0 (), s 1 (),..., s M-1 () was ransmied over he AWGN channel (if hey are equally probable). For a received r(), he residual energy e i for each possible ransmied alernaive s i () is calculaed as! " = $ & ' α* " ' +,' = $(& ' α* " ' )(& ' α* " ' ),' = $ & ' +,' 2Re α $& ' * " ',' + α + $ * " ' +,' Same for all i Same for all i, if he ransmied The residual energy is minimized by maximizing his par of he expression. signals are of equal energy. 5 April

7 The AWGN channel, con. The cenral par of he comparison of differen signal alernaives is a correlaion, ha can be implemened as a correlaor: r( ) or a mached filer r( ) * si ( ) * i ( - ) s T s! " # * a The real par of he oupu from eiher of hese is sampled a = T s where T s is he symbol ime (duraion). * a 5 April Anipodal signals In anipodal signaling, he alernaives (for 0 and 1 ) are 0 1 ( ) = j ( ) ( ) = -j ( ) s s This means ha we only need ONE correlaion in he receiver for simpliciy: r( ) j * ( )! " # * a If he real par a T=T s is >0 decide 0 <0 decide 1 5 April

8 Orhogonal signals In binary orhogonal signaling, wih equal energy alernaives s 0 () and s 1 () (for 0 and 1 ) we require he propery: The approach here is o use wo correlaors: * s r( 0 ) ( ) $ % &, $ ( & =!$ * & $ ( &,& = 0 * s1 ( )! " #! " # * a * a Compare real par a =T s and decide in favor of he larger. (Only one correlaor is needed, if we correlae wih (s 0 () - s 1 ()) *.) 5 April Inerpreaion in signal space The correlaions performed on he previous slides can be seen as inner producs beween he received signal and a se of basis funcions for a signal space. The resuling values are coordinaes of he received signal in he signal space. Anipodal signals Decision boundaries Orhogonal signals s 1 1 ( ) 1 0 j ( ) 0 s0 ( ) 5 April

9 The noise conribuion Assume a 2-dimensional signal space, here viewed as he complex plane E s Im s i Noise-free posiions Noise pdf. E s Re Fundamenal quesion: Wha is he probabiliy ha we end up on he wrong side of he decision boundary? s j This normalizaion of axes implies ha he noise cenered around each alernaive is complex Gaussian ( s 2 ) + j ( s 2 ) N 0, N 0, wih variance σ 2 = N 0 /2 in each direcion. 5 April Pair-wise symbol error probabiliy Wha is he probabiliy of deciding s i if s j was ransmied? Im s i d ji We need he disance beween he wo symbols. In his orhogonal case: E s s j 2 2 d = E + E = 2E ji s s s Re The probabiliy of he noise pushing us across he boundary a disance d ji / 2 is E s Pr(" # " % ) = ( ( * #% 2, - 2 ) = ( (. / ), - The book uses erfc() insead of Q(). = 1 2 erfc (. / 2, - ) 5 April

10 The union bound Calculaion of symbol error probabiliy is simple for wo signals! When we have many signal alernaives, i may be impossible o calculae an exac symbol error rae. s 1 s 2 When s 0 is he ransmied signal, an error occurs when he received signal is ouside his polygon. s 6 s 7 s 5 s 0 s 4 s 3 The UNION BOUND is he sum of all pair-wise error probabiliies, and consiues an upper bound on he symbol error probabiliy. The higher he SNR, he beer he approximaion! 5 April Symbol- and bi-error raes The calculaions so far have discussed he probabiliies of selecing he incorrec signal alernaive (symbol), i.e. he symbol-error rae. When each symbol carries K bis, we need 2 K symbols. Gray coding is used o assigning bis so ha he neares neighbors only differ in one of he K bis. This minimizes he bi-error rae Gray-coded 8PSK April

11 Bi-error raes (BER) EXAMPLES: 2PAM 4QAM 8PSK 16QAM Bis/symbol Symbol energy E b 2E b 3E b 4E b BER! 2# $ % &! 2# $ % & ~ 2 3! 0.87 # $ % & ~ 3 2! # $, max 2.25% & Gray coding is used when calculaing hese BER. 5 April Bi-error raes (BER), con Bi-error rae (BER) PAM/4QAM 8PSK 16QAM Eb / N [db] 0 5 April

12 Where do we ge E b and N 0? Where do hose magic numbers E b and N 0 come from? The noise power specral densiy N 0 is calculaed according o, - =./ - 0 -, - db = db where F 0 is he noise facor of he equivalen receiver noise source. The bi energy E b can be calculaed from he received power C (a he same reference poin as N 0 ). Given a cerain daa-rae d b [bis per second], we have he relaion! " = % & "! " db = % db & " db THESE ARE THE EQUATIONS THAT RELATE DETECTOR PERFORMANCE ANALYSIS TO LINK BUDGET CALCULATIONS! 5 April Wha abou fading channels? We have (or can calculae) BER expressions for non-fading AWGN channels. If he channel is Rayleigh-fading, hen E b /N 0 will have an exponenial disribuion (N 0 is assumed o be consan) 9:;(γ. ) = 1?@ = A γ. γ. γ. -- E b /N 0 -- average E b /N 0 The BER for he Rayleigh fading channel is obained by averaging:!"# Rayleigh (γ.) = 1!"# AWGN (γ.) 9:;(γ. ):γ April

13 Wha abou fading channels? THIS IS A SERIOUS PROBLEM! 10 0 Bi error rae (4QAM) db 10 x Rayleigh fading No fading E b /N 0 [db] 5 April DIVERSITY ARRANGEMENTS 5 April

14 Diversiy arrangemens Le s have a look a fading again Received power [log scale] Illusraion of inerference paern from above A Movemen B A Posiion B Transmier Reflecor Having TWO separaed anennas in his case may increase he probabiliy of receiving a srong signal on a leas one of hem. 5 April Diversiy arrangemens The diversiy principle The principle of diversiy is o ransmi he same informaion on M saisically independen channels. By doing his, we increase he chance ha he informaion will be received properly. The example given on he previous slide is one such arrangemen: anenna diversiy. 5 April

15 Diversiy arrangemens General improvemen rend 10 0 Bi error rae (4PSK) db 10 x Rayleigh fading No diversiy db 10-3 Rayleigh fading M:h order diversiy M x 10-5 No fading E b /N 0 [db] 5 April Diversiy arrangemens Some echniques Spaial (anenna) diversiy TX We will focus on his one oday!... Signal combiner Frequency diversiy TX Temporal diversiy D D D Signal combiner Coding Inerleaving De-inerleaving De-coding (We also have angular and polarizaion diversiy) 5 April

16 Spaial (anenna) diversiy Fading correlaion on anennas Wih several anennas, we wan he fading on hem o be as independen as possible. Isoropic uncorrelaed scaering. E.g.: An anenna spacing of abou 0.4 wavelengh gives zero correlaion. 5 April Spaial (anenna) diversiy Selecion diversiy RSSI = received signal srengh indicaor 5 April

17 Spaial (anenna) diversiy Selecion diversiy, con. By measuring BER insead of RSSI, we have a beer guaranee ha we obain a low BER. 5 April Spaial (anenna) diversiy Maximum raio combining This is he opimal way (SNR sense) of combining anennas. 5 April

18 Spaial (anenna) diversiy Simpler han MRC, bu almos he same performance. 5 April Spaial (anenna) diversiy Performance comparison Cumulaive disribuion of SNR Comparison of SNR disribuion for differen number of anennas M and wo differen diversiy echniques. MRC RSSI selecion These curves can be used o calculae fading margins. [Fig ] 5 April

19 Spaial (anenna) diversiy Performance comparison, con. Comparison of 2ASK/2PSK BER for differen number of anennas M and wo differen diversiy echniques. MRC RSSI selecion [Fig ] 5 April