Lecture Slides. Lecture by Prof. Dr.-Ing. I. Willms Prof. Dr.-Ing. T. Kaiser. I. Willms. Transmission and Modulation of Signals.

Transcription

1 Leture Slides Leture by Prof. Dr.-Ing. Prof. Dr.-Ing. T. Kaiser Slide

2 Leture: :00-:30 hold by Prof. Dr.-Ing. Ingolf Willms (B 3,Tel , nts.uni-duisburg-essen.de) Exerise: 3:00-3:45 hold by Dipl.-Ing. Yao (B 39, Tel , uni-due.de) Some relevant fats Slide

3 Leture Sript/Video/Exerises Copy Shop Steeger, Bismarkstrasse or Website Examination: 90 minutes, exerises Examination olletion: available for download from the website Some relevant fats Slide 3

4 Transmitter Disrete Soure Soure enoder Channel enoder Modulator Channel Reeiver Disrete sink Soure deoder Channel deoder Demodulator Model of a ommuniation system Slide 4

5 What is Modulation? Modulation is a proess to adapt a given signal to a given hannel. Most often it means shifting in frequeny. Why it is used? For many baseband signals, the wavelengths are too large for reasonable antenna dimension (e.g. speeh signals with 5 km wavelength). Besides, transmission of signals at lower frequenies is in general more diffiult. Perfet math to allowed bands (for wireless omm. e.g.) is ahieved. Definition of Modulation Slide 5

6 xt () = mtst () () The arrier signal might be: mt () = aos( ω t+ φ) The information is hidden in the arrier amplitude a M (Linear Modulation). The information is hidden in the arrier frequeny FM (Non-linear Modulation). The information is hidden in the arrier phase PM (Non-linear Modulation). Ideal modulator Slide 6

7 Carrier Soure Signal Modulation Method brev. Sinusoidal Carrier mt () = aos( ω t+ φ) nalog Double Sideband M wc Double Sideband M sc Single Side Band M Vestigal Sideband M Phase Modulation Frequeny Modulation DSBMwC DSBMsC SSBM VSBM PM FM mplitude Shift Keying SK Pulsed Carrier mt () = digital unoded Frequeny Shift Keying Phase Shift Keying Pulse mplitude Modulation Pulse Duration Modulation Pulse Frequeny Modulation FSK PSK PM PDM PFM m( t + T ) digital oded Pulse Phase Modulation Pulse Code Modulation Delta Modulation PPM PCM M Delta-Sigma Modulation M Modulation Methods Slide 7

8 M-Modulation Old tehnique, first used last third of 9th entury Helps to understand more sophistiated tehniques ppliation in low quality M radio and analog television (LW, MW, SW, USW and VHF+ UHF bands) Usable range differs very muh (SW waves travel around the globe, USW (UKW radio), UHF only usable for regional transmission) M radio is effeted by different wave propagation effets Slide 8

9 Soure signal M signal st () = os( ω t), ω << ω s s x ( t) = ( + m os ω t)os( ω t+ φ) M M s mplitude modulated sinusoidal signal with modulation degree m M < ( + m M ) (+ m os ω t) M s ( m M ) osillation with ω osillation with (+ m os ω t) ω s M s mplitude Modulation (M) Slide 9

10 Soure signal M signal st () = os( ω t), ω << ω s s mplitude modulated sinusoidal signal with degree of modulation > envelope ( + m M ) x ( t) = ( + m os ω t)os( ω t+ φ) M M s ( m M ) Phase reversal envelope mplitude Modulation (M) Slide 0

11 Note: Demodulation an be performed by pure envelope demodulator This demodulation fails (ambiguity) with a too large degree of modulation (without detetion of phase reversals) Consideration in frequeny domain of: xm ( t) = ( + mm os ωst)os( ωt+ φ) = os( ω t+ φ) + m os( ω t)os( ω t+ φ) M s Slide

12 Fourier Transform of M Using osαos β = (os( α β) + os( α + β)) follows: mm x M ( t) = os( ω t + φ ) + (os(( ω s ω ) t φ )) mm + (os(( ω s + ω ) t + φ )) os( ω t φ) 0 jφ jθ + e e 0 π ( δω ( ω) δω ( ω)) jφ jφ X ( ω ) = π( e δ( ω + ω ) + e δ( ω ω )) M π m M jφ jφ + ( e δ ( ω + ω s ω) + e δ ( ω ω s + ω)) π mm jφ jφ + ( e δ ( ω + ωs + ω) + e δ( ω ωs ω)) Slide

13 Fourier-transform of a modulated sinusoidal signal X ( ) M ω mmπ ( ) ( ) π ( π ) mmπ ( ) mmπ ( ) mmπ ( ) ω ω ω ω + ω ω s s s ω ω ω + s ω Power of M with sinusoidal input T / lim M ( ) T T T / PM = x t dt m m m = + + = M M M Slide 3

14 Note: Sidebands have equal and full information about soure signal x M (t) is quasiperiodi Carrier ontains most part of power due to typial low value of degree of modulation Thus transmission with suppresed arrier is useful Moreover: arbitrary soure signals are onsidered Slide 4

15 Example for arbitrary soure signal with representative spetrum: ω ωsf ω+ ωsf Sr ( ω) = ret( )( β + αω) + ret( )( β αω) ω0 ω0 a where α = a ω ω β = + 0 S ( ω) r 0 ( ωsf ) ω 0 ω0 ωsf ω sf ω 0 ωsf + 0 sf Example of Spetrum ω ω ω sf ω0 ω sf + Slide 5

16 Representative signal - s () t = F S ( ω ) r { } r ω sf + ω0 / ω ω 0 sf ω0 / ω0 = si(( ωsf + ) t) si(( ωsf ) t) π π ωsf ( a) ω si t si t π 0 ( ωsf ) ( ) s () t r ω ( ) 0 a + π Representative Signal Slide 6

17 Properties of representative signal/spetrum : Band limitation with ut-off frequeny ω / o = ωsf + ω0. ω sf Theshift frequeny enables either a low pass signal or a band pass signal. This is espeially useful in order to explain Single Sideband mplitude Modulation (SSBM). Byvaryingtheparameter or lower frequenies ( a > ) either higher frequenies an be emphasized. ( a < ) The representative spetrum as well as the signal are still easy enough. Representative Signal Slide 7

18 The general transmitted M-signal is given by x () t = ( + s())os( t ω t+ φ). M For envelope demodulation we require ( + s( t)) 0, Thus it follows min st ( ) or min st ( ), Comparison with mm, gives the degree of modulation: mm = min s( t). Bandlimited Signal Slide 8

19 Double-Side-Band MwC for arbitrary spetrum The Fourier-transform X ( ) M ω { } F { } X ( ω ) = F x () t = ( + s())os( t ω t + φ) M M { os( ω t φ) } F { s( t) os( ω t φ) } = F of x t is obtained as M () = F t + + F s t F t + π { os( ω φ) } { ( )}* { os( ω φ) } jφ jφ = π ( e δ( ω + ω ) + e δ( ω ω )) S ( )*( e jφ j ( ) e φ + ω δ ω + ω + δ( ω ω)) jφ jφ = π ( e δ ( ω + ω ) + e δ ( ω ω )) ( S ( ) e jφ S j ( ) e φ + ω + ω + ω ω ). Slide 9

20 upper sideband X M r, ( ω) ω0 ωsf ω ω ω 0 lower sideband ω ω sf + + ω ω0 = π 4Hz ω = π 4Hz sf ω = πhz a = 0.75 φ = 0 x M r, () t ( + s r (0)) Representative spetrum envelopes Slide 0

21 Power of M is given by (see exerises) T / lim M ( ) T T T / PM = x t dt T / ( lim s ( t) s ( t) dt) T T T / = + + with a arrier power of and a mean power of, T / lim stdt ( ), T T T / and a power needed for transmission of T / lim s ( t) dt. T T T / Power of M signal Slide

22 pratial example (MW, 0 hannels) X ( ) M ω 30Hz 4.5kHz 50kHz 50kHz 600kHz Band spreading fator bandwidth of the modulated signal sum of bandwidth of all soure signals xt () s () n t Band spreading fator Slide

23 Demodulation of DSBMwC signals an be ahieved in two prinipal different ways: - Using a time-variant system n example is the multipliation of y () M t with a sinusoidal funtion os( ω + φ ). t 0 - Using a non-linear system n example for a non-linear system is given by y (). M t y () M t or Main priniples of these two versions to extrat information of the envelope Demodulation of DSBMwC Slide 3

24 Con. Type : Demodulation with a multiplier z () t = y ()os( t ω t+ φ) DM M with DM for Demodulation of M and y ( t) = x ( t) = x ()os( t ω t+ φ) M = ( + s( t)) os( ω t+ φ) os( ω t+ φ) = ( + s( t)) ( + os(( ωt+ φ))) = ( + st ( ))( + os(( ωt+ φ))), M M Z DM ( ω ) = ( πδ ( ω ) + S( ω )) π jφ jφ *( πδ ( ω) π δ ( ω ω) e δ ( ω ω) e ) = S ( ω) + S ( ω ω) e + S ( ω+ ω ) e 4 jφ jφ + πδω+ πδω ω + πδω+ ω jφ jφ 4 ( ) ( ) e ( ) e. Demodulation with time-variant system Slide 4

25 This results from two times multipliation of soure signals inluding DC signal by arrier signal: Z ( ω) = S( ω) + S( ω ω ) e + S( ω+ ω ) e 4 jφ jφ DM + πδω+ πδω ω + πδω+ ω jφ jφ 4 ( ) ( ) e ( ) e. Z ( ) DM ω π ( ) H ( ) LP ω π π ( ) ω ω ω + ω s ωs ω ω s s ω Demodulation-Figure Slide 5

26 Demodulated output signal is: GDM ( ω) = ZDM ( ω) HLP ( ω) = ( πδ( ω) + S( ω)) gdm () t = zdm () t hlp () t = ( + s()) t Why is the division of by two? δ ( ω) annot be suppressed by an additional highpass filter as long as st () has non-zero mean, why? hannel with a possible delay and/or a sale fator, so y () t = α x ( t t ) = α( + s( t t ))os( ω ( t t ) + φ) M M For DM the reeiver needs to know. Is it possible? φ ω t 0 Conlusions Slide 6

27 Now the influene of an unknown phase is onsidered. Slide 7

28 Demodulation is onsidered using an ideal modulator with modulating funtion os( ω + φ ) instead of os( ω + φ ) t z () t = x ()os( t ω t + φ ) DM M FOURIER-transform of it gives: Z DM ( ω) = ( πδ ( ω) + S( ω)) ( πδ ( ω)os( φ φ )) t = ( + s( t)) os( ω t+ φ) os( ω t+ φ ) = ( + st ( )) os( ) + os( + + ) [ φ φ ω t φ φ ] π j( φ+ φ ) j( φ+ φ ) + π δ( ω ω) e + δ( ω + ω) e ( ) ( ) S ( )os( ) S ( ) e j φ + φ S j ( ) e φ+ φ = ω φ φ + ω ω + ω+ ω 4 j( φ φ ) j( φ φ ) + 4 πδω ( )os( φ φ ) + πδω ( ω) e + + πδω ( + ω) e +. From this higher frequeny bands are filtered out! Unknown Phase Slide 8

29 This gives: [ S ( ω)os( φ φ ) + 4os( φ φ )] 4 = [ S ( ω) + πδ( ω)]os( φ φ ) Is it possible to exatly reonstrut from this signal (or from this spektrum) s(t) without knowing the arrier phase? Yes, by triky applying os ( x) + sin ( x) = For this we need the signal [ S ( ω ) + πδ ( ω )]sin( φ φ ). Slide 9

30 This signal is applied aording to: ω ( t) = ( + s( t))os( ω t+ φ )sin( ω t+ φ ) DM = ( + s( t)) [ sin ( φ φ ) + o s( ω t + φ + φ )]. u DM () t = ω DM () t hlp () t ( + s( t)) sin ( φ φ ). g () t = ( g ()) t + ( u ()) t QDM DM DM = ( + s( t)) sin ( φ φ ) + ( + s( t)) os( φ φ ) 4 4 = + + = + s( t) sin( φ φ ) os( φ φ ) s( t) The next figure shows the bloks of a orresponding demodulator. Unknown Phase Slide 30

31 z () t = y ()os( t ω t+ φ ) ω () t = y ()sin( t ω t+ φ ) DM M DM M g () t = z () t h () t u () t = ω () t h () t DM DM LP DM DM LP g t g t u t QDM () = DM () + DM () z g () t () DM t DM y () M t H ( ) LP ω os( ω + φ ) t sin( ω + φ ) t ( g ( t )) DM ( u ( t)) DM g () QDM t ω () t u DM () t DM QDM Slide 3

32 dvantages: Demodulation in presene of unknown phases Disadvantages: Suitable square root devie needed Moreover: Sign of + s(t) is lost Conlusions on QDM Slide 3

33 QM transmission Due to orthogonal arrier signals when applying a sine and osine it is possible to transmit signals s (t) and s (t) at one time! The following alulation shows the demodulation tehnique Slide 33

34 x () t = ( + s ())os( t ω t + φ) + ( + s ())sin( t ω t + φ) QM z () t = ( + s ())os( t ω t + φ)os( ω t + φ ) QM + ( + s ( t)) sin( ω t + φ) os( ω t + φ ) sin α os β = (sin( α β ) + os( α + β )) ( s ( t))(os( φ φ ) os( ω t φ φ )) = ( s ( t))(sin( φ φ ) sin( ω t φ φ )), g () t = z () t h () t QM QM LP ( + s( t)) os( φ φ ) + ( + s( t)) sin( φ φ ). gqm t s t () = ( + ()). φ = φ g QM t s t () = ( + / ()). φ = φ π Band spreading fator of QM: β QM () t. QM-transmission of signals Slide 34

35 Conlusions: QM may suffer from ross talk (in ase of phase distortions) QM usable for data transmission (less suitable for speeh, partly for audio stereo) Frequeny shift distortions If the reeiver exhibits a frequeny shift of os(( ω + ω ) t + φ) it follows ω z () t = y ()os(( t ω + ω ) t+ φ) DM M, desribed by = ( + s( t)) os( ωt+ φ) os(( ω + ω ) t+ φ) = ( + st ( ))(os( ω t) + os(( ωt + φ) + ω t)). gdm () t = zdm () t hlp () t = ( + s ())os( t ω t) Slide 35

36 Con. Type (Nonlinear Demodulation) - Very simple DRC-demodulator iruit is used - Simpliity is reasonable as heap reeivers for many onsumers were needed (Transmitter an be ostly) Details are disussed based on an exitation with a ret-signal and based on a DSBMwC and a DSBMsCsignal: Slide 36

37 u () in t uout () t t T / uin () t = u0ret( ), u0 > 0. T u 0 uout () t u () in t DRC-demodulator output in ase of retangular exitation and ideal diode DRC-demodulator Slide 37

38 u () t = y () t ε () t = ( + s())os( t ω t+ φ) ε(), t in M The hoie of τ = RCis essential for the performane. y () M t g () DRC t ω = H z ω s = 0.H z = φ = π m M = The DRC-demodulator output in ase of τ = 0 s. DRC-demodulation Slide 38

39 DRC-demodulation y () M t g () DRC t ω = Hz ω s = 0.Hz = π φ = m M = The DRC-demodulator output in ase of τ = 50s > / ω. s Slide 39

40 DRC-demodulation y () M t g () DRC t ω ω s = = Hz 0.Hz φ = = π m M = The DRC-demodulator output in ase of τ = 5s < / ω. s Slide 40

41 DRC-demodulation y () M t g () DRC t ω ω s = = Hz 0.Hz φ = = π m M = The DRC-demodulator output in ase of m M =.5. Slide 4

42 DRC-enhanement bsolute value demodulator (VD) uout () t it () ω ω s = H z = 0.H z = π φ = m =.5. M Slide 4

43 Output signal of DRC-enhanement VD z () t = y () t VD M = ( + s( t)) os( ω t + φ ) = ( + s( t)) os( ω t + φ ). y t () M () z t gvd () t VD ( ) n + j n( ωt+ φ ) os( ωt + φ) = e. π n = 4n Hene it results for positive + s(t) : ( ) z t s t e n + j n( ω t+ φ ) VD ( ) = ( + ( )). π n = 4n Slide 43

44 n+ ( ) ZVD ( ω) = F ( + s( t)) e π 4n n= { j n( ω )} t+ φ n+ ( ) { j n( ω )} { ( ) ( ( ) } t+ φ j n ωt+ φ = F e + F s t e ) π 4n n= ( ) Z e n S n n + j nφ VD ( ω) = ( πδ ( ω ω ) ( )). + ω ω π n = 4n (4 /3) H ( ) LP ω S( ω) Z VD r, ( ω) g VD () t = ( + s()) t π S( ω ω ) (4 /3) ω ω s ω s ω DRC-enhanement VD Slide 44

45 Usage of a square law demodulator (SLD) y t z () t M () SLD g () SLD t z t y t SLD () = M () = ( + ( )) os ( + φ ) s t ωt = ( s( t) s ( t))( os(ω t φ )). ssumption of s ( t) << s( t) or s(t) << gives: zsld ( t) ( + s( t))( + os(ω )) ( ) t + φ s t << = ( + os(ω t + φ) + s( t) + s( t) os(ωt + φ)) DRC-enhanement SLD Slide 45

46 Spektrum of SLD demodulated signal Z e e jφ jφ SLD ( ω) ( πδ ( ω) + π δ ( ω ω ) ( ) st () + π δ ω + ω << H ( ) LP ω S( ω) jφ jφ () S ω e S( ω ω) e S( ω ω)). Z SLD r, ( ω) ( π ) S( ω ω ) ( π /) ( π /) ω ω s It follows: gsld ( t) ( + s( t)) s( t) << Disadvantage of SLD demodulation: Low allowed modulation degree leads to low power effiieny of the transmission. ω s Slide 46 ω

47 SSB-Modulation and Demodulation Reduing bandwidth of DSBM transmission methods leads to SSB tehniques Problem: Low frequeny omponents (about 0 Hz e.g.) in soure signals SSB modulation needs suitable, (nearly ideal) LP or HP filtering devie with high filter seletivity Slide 47

48 Generation of an SSBM-signal upper sidebands X M r, ( ω) H ( ) LSB ω H ( ) USB ω ω ωs ω ω ω + ω s lower sidebands Problem: Suitable fast transition of frequeny response of the filter (from passband to stopband) requiring almost perfet retangular frequeny response. Possible solution: Use of Hilbert transform iruits and analyti signal SSB-Generation Slide 48

49 X SSB r, ( ω) ω + ω ωs ω ω Sr ( ω) ω s ω s Sr ω s ( ω) sign( ω) ω s ω s ( S r( ω ) + S r( ω ) sign ( ω )) δ ( ω ω ) 4 + ( S r( ω ) S r( ω ) sign ( ω )) δ ( ω + ω ) 4 ωs ω ω ω ωs + ω jφ jφ XSSB ( ω) = ( S( ω) + S( ω) sign( ω)) δ( ω ω ) e + ( S( ω) S( ω) sign( ω)) δ( ω+ ω ) e. 4 SSBM-Spetrum Slide 49

50 In the following we will make use of the analyti signal and the equivalent LP signal The relation of the analyti signal and its Fourier transform to the normal signal and its Fourier transform is as follows: 0 X ( ω) = εω ( ) X ( ω) = ( + sign( ω)) X ( ω) = ( j jsign( ω)) X ( ω) 0 xt () = xt () + jxt ˆ() with X( ˆ ω) = jsign( ω) X( ω) Slide 50

51 The analyti signal low pass signal { } 0 0 x () t = F X ( ω ) x () t = F X ( ω) ELP 0 0 x () t = x () t e ω φ ELP j( t + ) { } ELP and the equivalent are related by: 0 j 0 X ( ω) = X ( ω ω ) e φ, where φ0 ELP is an arbitrary phase. The analyti signal is also given by { 0 } xt () = Re x() t = { j ( t } 0 x ) ELP t e ω + φ Re ( ) () () ˆ(), 0 x t = x t + jx t hene = { jω t + x t + φ 0 } ELP ELP Re x ( t) e () with ELP x () t = x () t e j x () t ELP ELP = x ()os( t ω t+ x () t + φ ). ELP ELP 0 nalyti signal Slide 5

52 Generation of SSB filtering an be realized as follows: XSSB ( ω) = S( ω ω ) ε( ω ω ) + S( ω+ ω ) ε( ω ω ) 4 4 = S( ω ω)( + sign( ω ω)) + S( ω+ ω)( sign( ω+ ω)) 4 4 = S( ω ω)( j jsign( ω ω)) + S( ω+ ω)( + j jsign( ω+ ω)) 4 4 = S( ω)( j jsign( ω)) δ( ω ω ) + S( ω)( + j jsign( ω)) δ( ω+ω ) 4 4 This formula then an be rewritten using onvolution operations, Hilbert transforms and by extending it with the zero phase φ of the arrier. Slide 5

53 By use of HHILBERT ( ω) = jsign( ω), it follows j X SSB ( ω) = ( S( ω) + js( ω) H HILBERT ( ω)) δ ( ω ω ) e φ 4 j + ( S( ω) js( ω) HHILBERT ( ω)) δ( ω+ ω ) e φ jφ jφ X SSB ( ω) = ( S( ω) π( δ( ω ω ) e + δ( ω + ω ) e ) 4π jφ jφ + S( ω) HHILBERT ( ω) ( jπ( δ( ω ω ) e δ( ω + ω ) e ) This gives: x () ( ()os( ) ˆ SSB t = s t ωt+ φ s()sin( t ωt+ φ)) with sˆ () t = H s() t. { } lternative generation of SSBM Slide 53

54 x () ( ()os( ) ˆ SSB t = s t ω t + φ s()sin( t ω t + φ)), os( ω + φ ) t st () H {...} x () SSB t sin( ω + φ ) t H {...} lternative iruit for generating an SSBM-signal Generation of SSBM Slide 54

55 x () ELP t 0 0 ˆ SSB ω = F SSB = F SSB + SSB We will try to figure out in ase of SSBM. { } { } X ( ) x () t x () t jx () t = X ( ω)( + sign( ω)) SSB = X ( ω) εω ( ) with SSB jφ XSSB ( ω) = S( ω ω ) ε( ω ω ) e + S( ω+ ω ) ε( ω ω ) e 4 4 j X ( ω) = S( ω ω ) ε( ω ω ) e φ. 0 SSB jφ From this we obtain the equivalent lowpass spetrum: X X e S e 0 jφ0 j( φ φ0) ELP, SSB ( ω) SSB ( ω ω ) = + = ( ω) ε( ω) S ( 0 ) ( )( sign j ( )) e φ φ = ω + ω. j( 0 ) Hene we get: x ˆ ELP, SSB ( t) ( s( t) js( t)) e φ = + φ s φ is arbitrary we an set it to zero without problems. 0 nalyti signal for SSB Slide 55

56 x () t = x ()os( t ω t+ x () t + φ) SSB ELP, SSB ELP, SSB sˆ( t) = s ( t) + sˆ ( t) os( ωt+ artan( ) + φ) st ( ) s 0 0 = ()os( t ω t + s () t + φ) ω 0 = π Hz ω = π0.6hz sf ω = π.4hz a = φ = x () SSB t ( π x ) SSB t ω + ω sf mplitude and angle modulation of SSB signal (not always oiniding zeros) nalyti signal for SSB Slide 56

57 For the power of a SSB signal holds (due to exerise 0): P P M SSB = T / T / SSB 4 T T T / T / PSSB = lim x ( t) dt = lim s ( t) dt. T T T / (+ lim s () t s () t dt) T T + T / T / 4 lim s ( t) dt T T T / T / ( + lim st ( ) s( tdt ) ) T T + T / T / = = lim s ( t) dt T T T / 6 in ase of sinusoidal st ( ). Power of SSB Slide 57

58 Next question is how SSB signals an be demodulated In ase of SSB reeption for sinusoidal signal s( t) holds: x ( t) = os( ω t+ ω t) whih shows a onstant envelope SSB S Thus envelope demodulation will not work! lthough this introdues also disadvantages onerning power effiieny now an additive arrier is applied aording to: x () t = os( ω t+ ω t) + os( ω t+ φ) SSB S Slide 58

59 x () t = x () t + os( ω t+ φ) gives with SSB SSB WC x ˆ SSB ( t) = ( s( t)os( ωt+ φ) s( t)sin( ωt+ φ)) : x (( ( ))os( ) ˆ SSBWC = + s t ωt+ φ s( t)sin( ωt+ φ)) sˆ( t) = ( + st ( )) + st ˆ( ) os( ωt+ artan( ) + φ) + st ( ) () ˆ() ˆ() = + s( t) + st + st os( ωt+ artan( st ) + φ) st ( ) It follows for the envelope: s () t sˆ () t g SSB WC () t = + s() t Hene if st ( ) << holds it follows: g () t + s(). t SSB WC Demodulation of SSBM Slide 59

60 In addition, by use of we finally obtain x + x = +, x s() t g SSB WC () t ( + ). In order to avoid the additive arrier now synhronous demodulation with unknown arrier frequeny (or little frequeny error) and phase is onsidered: z () t = y ()os(( t ω t+ ω ) t+ φ ) with y () t = x () t SSB SSB SSB SSB z () (()os( ) ˆ SSB t = s t ωt+ φ s()sin( t ωt+ φ))os(( ωt+ ω ) t+ φ ) = ( st ( )(os( ω t + φ φ ) + os(( ω + ω ) t + φ + φ )) 4 st ˆ( )( sin( ω t+ φ φ) + sin(( ω + ω ) t+ φ + φ)). Demodulation of SSBM Slide 60

61 g () t = z () t h () t SSB SSB LP = ( st ( )os( ω ) ˆ t + φ φ + st ( )sin( ω t + φ φ )). 4 This shown an additional phase shift and an additional modulation of both s(t) and its Hilbert transform ompared with the situation without frequeny error. The interpretation using Fourier transforms (in frequeny domain) is as follows: j( φ φ) j( φ φ) GSSB ( ω) = ( S( ω ω ) e S( ω ω ) e ) ˆ j( φ φ) ˆ j( φ φ) + ( S( ω ω ) e S( ω + ω ) e ). j With Sˆ ( ω) = S( ω)( jsign( ω)) it follows: Unknown phase & arrier Slide 6

62 GSSB ( ω) = S( ω ω ) e + S( ω + ω ) e 8 j( φ φ) j( φ φ) S sign e + S + sign + e j( φ φ) j( φ φ) ( ω ω ) ( ω ω ) ( ω ω ) ( ω ω ) ) = S e sign + S + e + sign 8 j( φ φ) j( φ φ) [ ( ω ω ) ( ( ω ω ) ( ω ω ) ( ( ω ω ))] S ( ) ( ) ( ) e j φ φ j ( ) S ( ) e φ φ = ω ω ε ω ω ω ω + + ε( ω + ω) 8 S ( ) ( ) ( ) e j φ φ j ( ) S ( ) e φ φ = ω ω ε ω ω ω ω + + ε( ω + ω). 4 ω sl ω su We define and as the lower and the upper edge frequenies of respetively. Let us sketh G ( ). SSB ω st () Unknown phase & arrier Slide 6

63 S( ω) ε ( ω ω) S( ω ) ω ω su ωsl ω sl ω su S( ω + ) ε ( ω ω) ω ω + sl ω ωsu + ω S( ω ω ) ε( ω ω) e j φ φ + ( ) S( ω ω ) ε( ω ω) j( φ φ) + + e GSSB ( ω) ω < ω sl ωsu ω ωsu ω Unknown phase & arrier Slide 63

64 If the frequeny error an be negleted, it holds: G S e e = 4 j( φ φ) j( φ φ) SSB ( ω) ω 0 = ( ω) ε( ω) + ε( ω). The human ear an be modelled as a bank of filters, whih are insensitive to phase pertubations ± ( φ φ). The human ear is not thus sensitive for frequeny shifts of less than 0Hz. But suh a low frequeny error an only be ahieved with a PLL and a highly stable quartz referene lok SSBM reeives muh attention for transmission of speeh (due to little low frequeny ontent). For transmission of data QM would be better. Unknown phase Slide 64

65 VSBM aspets and appliations Important appliation of VSBM is analog transmission of video signals with 5,5 MHz bandwidth llows redued bandwidth and modest demands on the transmitter VSB filter Band spread is well below DSBM Thus nearly a double number of TV hannels an be transmitted Transmission of additional (low power) pilot arrier of nearly arbitrary frequeny allows synhronisation of reeiver with transmitter Condition of the transmitter VSB filter (applied to the DM signal) for perfet demodulation (as shown in the following slide) is: HVSB ( ω ωc ) + HVSB ( ω+ ωc ) = onstant within ωs ω + ωs This an be aomplished by a VSB BP filter with symmetry around ωc ωs + ωvsb / For the band spreading fator holds: βvsbm = ω Vestigial side-band M S Slide 65

66 Using a MsC signal and a suitable VSB band pass filter it results: x () t = s()os( t ω t+ φ) MsC VSB () = MsC () VSB () with VSB () = { VSB ( ω)} giving X ( ω) = X ( ω) H ( ω) VSBMsC MsC C x t x t h t h t F H VSB ( S j j ( C) e φ S ( C) e φ = ω ω + ω+ ω ) H VSB( ω) synhronous demodulation is applied here as desribed in the following relations: z () t = x ()os( t ω t+ φ) VSBMsC VSBMsC C ZVSBMsC ( ω) = ( XVSBMsC ( ω ωc ) + XVSBMsC ( ω+ ωc )) ( S j j ( C) e φ S ( ) e φ = ω ω + ω ) H VSB( ω ωc) 4 j j ( S ( ) e φ S ( C) e φ + ω + ω+ ω ) H VSB( ω+ ωc) giving finally 4 j j = S ( )[ e φ H VSB ( C ) e φ ω ω ω + H VSB ( ω+ ωc )] 4 Slide 66

67 Sr ( ω) ω ω ω s ωs ω X M r, ( ω) ω ω ω s ω VSB ω H ( ω ) VSB ω ω + ω s ω ω ω s ω ω ω + ω s X VSB r ( ω), ω ωs ω ω ω ωs + ω G VSB r, ( ω) ω ω ω s ωs ω The VSBMsC method Slide 67

68 Method TX Power TX osts DSBMwC env.dem. DSBMwC syn.dem. QM DSBMsC SSBMsC SSBMwC VSBMsC VSBMwC β RX osts high low low high moderate low > very low moderate/high high Purpose moderate low high Data-, stereo-sig. Low/moderate high high speeh, audio-sig. low moderate high very low moderate/high high > high moderate low TV-signals n overview on M methods Slide 68

69 Chapter. ngle modulation Basis of FM and PM Spetrum analysis of narrow/wide band FM Demodulation Comparison of FM to M Slide 69

70 . ngle Modulation For FM & PM only variations of arrier phase / frequeny are needed: x () t = os( f (()) s t t+ φ) FM FM x () t = os( ω t+ f (())) s t PM C PM Typial signals: x () t = os( ω t+ s( τ) dτ + φ) FM x () t = os( ω t+ s() t + φ) PM C C t ngle Modulation Slide 70

71 Now a omparison is made for the signal x ( t) = os( ω (t)t+ ϕ ) and the signal x ( t) = os( ϕ( t)) at a speifi time instant t. 0 For this speifi time instant t it is required: d x ( t ) = x ( t ) and x dt This leads to 0 d () t = x () t dt t= t 0 t= t 0 0 os( ω t + ϕ ) = os( ϕ( t )) or sin( ω t + ϕ ) = sin( ϕ( t )) d ω0 sin( ω0(t)t+ ϕ0) = ϕ( t) t= t sin( ϕ( t 0 0)) dt d ω0 sin( ϕ( t0)) = ϕ( t) t= t sin( ϕ( t 0 0)) dt and finally gives the definition of the instantaneous frequeny dϕ() t ω i () t = dt Slide 7

72 Usually, ϕ() t is assumed to be a stritly monotonially inreasing funtion of time. Therefore, > 0. dϕ() t dt ϕ() t Now the spetrum of a simple FM signal is analysed. nalysis annot be made for arbitrary s(t) due to nonlinear properties. ngle Modulation Slide 7

73 Example of an FM-signal with a sinusoidal soure signal s(t) : t x ( t) = os( ω t+ ω os( ω τ) dτ + φ) with s( t) = ω os( ω t) ε( t) FM C FM S FM S 0 ωfm = os( ωct+ sin( ωst) + φ) and si(t)= s( τ) dτ ω S m I = e = z t e j( ωct+ mi sin( ωst) + φ) j( ωct+ φ) Re{ } Re{ ( ) } and the omplex Fourier oeffiients (as its a periodi signal) t + T 0 S! jnωs t zn = z( t) e dt with ωst= α T S t π π jnα jmi sin( α ) jnα ( α / ωs ) α α with 0 set to - S / π π = z e d e e d t T π = π π = [os( mi sin( α) nα) + jsin( mi sin( α) nα)] dα = Jn( mi) π π due to the zero value of sin( m sin( α) nα) dα π π I t I sin( S ) S with ( ) jm t ω jn ω zt = e = ze t n n= ngle Modulation Slide 73

74 We obtain the first kind n-th order of the Bessel-funtion for whih only a Taylor-series an be derived: J ( ) 0 m I ( ) m J m n J ( ) m I J ( ) m I k I n+ k n( I) = ( ), 0. k = 0 k! n k! J ( ) 3 m I ( + ) J ( ) 4 m I J ( ) 5 m I m I Bessel funtions Slide 74

75 new writing of x FM (t) now gives: x t z t e j( ωct+ φ) FM () = Re{() }, = z e e = Re n jnωst j( ωct+ φ), n= Re{ n( I) jnωst j( ωct+ φ) }, n= J m e e = J ( m )os(( ω + nω ) t+ φ). n= n I s n xfm( t) = J0( mi)os( ωc + φ) + with J n( mi) = ( ) Jn( mi) n Jn( m ) I os(( ωc + nωs) t+ φ) + ( ) os(( ωc nωs) t+ φ). n= jφ X ( ω) = πj ( m )( δ( ω ω ) e + δ( ω+ ω ) e ) + jφ FM 0 I C C jφ π J0( m ) I δ( ω ( ωc + nωs)) e + δ( ω+ ( ωc + nωs)) e n= + δω ω ω + δω+ ω ω j( φ+ nπ) j( φ nπ) ( ( C n S)) e ( ( C n S)) e. jφ FM signal Slide 75

76 Here we see: x(t) only ontains osine expressions mplitude is given by the Bessel funtion values Frequeny values are given by: ωc + nω S The phase is not hanged This orresponds to a line spetrum with lines at the arrier frequeny and at n times the soure frequeny. Thus the FM spetrum is in priniple extended to an infinetely wide band around the arrier frequeny. This wide-spread frequeny band gives the desired threshold effet in ase of modest noise power as reeiver only ares about frequeny hanges and not for hanges in the amplitude. The FM spetrum is in detail determined in the next slides. Due to properties of the Bessel funtion essentially only the lines orresponding to the following equation need to be taken into aount. N m I x () t = J ( m )os(( ω + nω ) t + φ) FM n I s n= N m I Slide 76

77 more preise onsideration requires a summation within the limit B N ( m + α) ω with α. Sinus ω I s S ± N =± m + α I The ase α = refers to the well known Carson bandwidth: = ( + ω ). B FM, Carson ω FM s It is now assumed that s(t) onsists out of not only one but of several osines as follows: L st () = ω os( ω t+ φ ), l s, l l l= L l I() = I, lsin( ωs, l + φl), I, l = l= ωs, l s t m t m L z() t = z () t, z () t = e l= z t J m e jn ( ω t+ φ ) l s, l l () = ( ). l nl I, l n = l l l ω j( m sin( ω t) + φ ) I, l s, l l FM spetrum Slide 77

78 For narrowband FM holds with: s t and m max ( ) I I x () t = os( ω t+ s () t + φ) FM s () t I I = os( ω t+ φ)os( s ( t)) sin( ω t+ φ)sin( s ( t)) I I os( ω t+ φ) sin( ω t+ φ) s ( t) I This orresponds to an M signal with arrier! FM spetrum Slide 78

79 For the spetrum holds: X ( ω) = π( δ( ω ω ) e + δ( ω+ ω ) e jφ FM s () t C C I jφ + jφ π ( δω ( + ωc) e δω ( ωc) e ) SI( ω) jπ jφ jφ = πδω ( ( ω) e + δω ( + ω) e ) C jφ + jφ ( δ ( ω+ ωc) e δ( ω ωc) e S( ω) j jω jφ jφ = πδω ( ( ω) e + δω ( + ω) e ) C jφ S( ω+ ωc) e S( ω+ ωc) e + ( ω+ ωc ω+ω C C C jφ jφ ) This shows that a narrowband FM signal essentially has the same band spreading fator as a DSBMwC signal. Slide 79

80 Demodulation of FM dsi () t FM I st () =, dt y () t = x () t with s ( t) = m sin( ω t) x () t = os( ω t+ s () t + φ). FM FM I I s Onset: Derivation of reeived signal and subsequent envelope demodulation dyfm () t dt d( ωt+ si( t)) = sin( ωt+ si( t) + φ) dt = ( ω + s( t))sin( ω t + s ( t) + φ). I Derivation in required frequeny range an be aomplished by ω ωc system with e.g. H( ω) = jω ret( ) B FM FM demodulation Slide 80

81 M vs FM FM reeiver is very robust against additive noise (e.g. rapid amplitude flutuations) SNR of about 0dB is the treshhold Suh noise is often found in pratise Carson bandwidth is easily salable Doubling bandwidth gives doubled SNR Constant envelope ensures power effiient transmitter Slide 8

82 In broadast radios FM is the most popular analog modulation tehnique. Do you agree? Envelope detetion is possible in M as well as FM. Comment. Envelope detetion shows better performane in FM than M. Do you agree, why? M bandwidth is not adjustable as FM bandwidth. Give two reasons. M versus FM Slide 8

83 Chapter Fundamentals of Disrete Signal Transmission. short review on information theory. short review on detetion theory Chapter Disrete Signal Transmission Slide 83

84 . Information Theory Consider a disrete soure, whih emits messages x(k) We assume the alphabet X is formed by L different symbols x l One an regard xk ( ), k= () as a random proess in form of X( k ) where eah random variable X ( k) takes x χ. values of the alphabet with the probability l We assume a stationary soure (a shift of messages doesn t hange probabilities) or the soure properties never hange! p l Information theory Slide 84

85 Please note: The alphabet is represented by a aligraphi letter X It ontains the L symbols x l sample (a real message sequene) by x(k) The whole proess (set of message sequenes) is represented by X(k) Slide 85

86 How to define information? (Measure of unertainty) I( x ) > I( x ) if pl < pm l m If the events of the two symbols x l and x m are independent with { ( ) =, ( ) = } = { ( ) = } { ( ) = } P X k x X k x P X k x P X k x l m l m for all k and eah l and m we require x, x independent I( x, x ) l m = = I( x ) + I( x ) l m Therefore,we define information as def I( xl ) = log ( ) = ld( ). p p l l l m Information theory Slide 86

87 What is entropy? Defined as average information ontent H ( χ) L l= I( x ) l p l Hene, L L H( χ ) = p ld( ) = p ld( p ). E.g. for symbols with l l l l= pl l= p = p p = p and it follows H ( χ ) = pld( p) ( p) ld( p) = H ( p). def Entropy Slide 87

88 Bounds on entropy [bits/symbol] of an alphabet H( χ) ldl Example: symbols have equal probabalities giving: L= H( χ) = ld = bit/ symbol The equal sign is valid only if all symbol show the same probability /L Entropy Slide 88

89 memoryless ommuniation hannel with additive noise: Often additive white Gaussian noise is assumed (WGN hannel) + Berger s hannel diagram Slide 89

90 Note: The entropy an be speified for an alphabet of a soure or a sink (thus indiretly for a soure and sink) In the following: nalysis of a ommuniation hannel for different power levels of noise For noise with zero mean the variane desribes the power level In general Gaussian distributed noise is assumed This is not always true (impulsive noise might be present) Berger s hannel diagram Slide 90

91 H( χ y) is denoted as equivoation Soure H ( χ) T ( χ, y) ( Lost info, speifies average info needed to speify the input symbol in ase of known output symbol) H( y) Sink H( y χ) is alled irrelevane ( undesired info, e.g. noise) is alled transinformation Berger s diagram Slide 9

92 Berger s diagram The diagram shows learly the following: T( χ, y) = H( χ) H( χ y), T( χ, y) = H( y) H( y χ), Thus we obtain: T( χ, y) = T( y, χ). For noiseless hannel: Equivoation is zero, no info is lost! For useless hannel: Equivoation is the soure entropy and all info is lost (no ommuniation at all)! Slide 9

93 def H ( χ, y) p ( x, y) ld( p ( x, y)) dxdy. = X, Y X, Y (Joint entropy) The following an be shown: H( X, Y) = H( Y) + H( X Y) = H( X) + H( Y X) Berger s diagram Slide 93

94 The hannel apaity is a funtion of p X (x) and is dependant on hannel properties. It is defined as C = max T( χ, y), p X ( x) whih leads to C = (max H( y)) H( y χ). p X ( x) C gives upper bound of transferabel info! Example: hannel with additive noise and y = x + n: px, N( xn, ) = px( x) pn( n), p ( y) = p ( x) p ( y x) dx, Y X N Thus we obtain: X, Y (due to independant noise) (as sum of x and n gives y) p ( y) = p ( x, y) dx, p, ( x, y) = p ( x) p ( y x). Y XY X N Channel apaity Slide 94

95 Channel apaity Now the joint entropy is rewritten based on the last relation: H ( χ, y) = p ( x, y) ld( p ( x, y)) dxdy XY, XY, = p ( x) ld( p ( x)) dx p ( x)( H( N)) dx X X X = H( X) + H( N) (without proof) Comparing the last result with H( X, Y) = H( Y) + H( X Y) = H( X) + H( Y X) shows: HY ( X) = HN ( ) as already indiated in Berger's diagram (Conlusion: Noise is the undesired info) Slide 95

96 nother example: dditive noise is GUSSIN distributed with pn n = e πσ ( n µ ) N N σ ( ), it follows (see the orresponding exerise) C GUSS N σ = ld( + ). σ X N If the signal is bandlimited with ut-off frequeny If C σ X = f ld( + ) [bits/symbol] S GUSS o [bits/s] σ N Eah symbol needs /f o time due to. Nyquist riterion σ X 3.7 0, σ = 3.7 bit bit CS GUSS= 300 ld( + 0 ) = 3830 N s s (for 37 db S/N ratio and 300 Hz bandwidth of a memoryless telephone line) Channel apaity Slide 96

97 . Detetion theory Given are M deterministi soure signals x m (t) with a finite duration T and a finite energy E x.these waveforms and their duration must be known to the reeiver. The reeived signal is assumed to be yt () = x () t + nt (). m where the noise is n(t) whih is modeled as a random proess. Detetion problem:given is y(t), the sequene of soure signals x m (t) has to be determined. Detetion theory Slide 97

98 Let us onsider M = and in ase H 0 of a logial zero x 0 (t) = 0 and in ase H of a one x (t) = x(t) desribed by: H : ( ) ( ) 0 yt = nt H : y( t) = x( t) + n( t). The detetion is based on minimization of probabililty of a wrong deision P e 0. To onsider that y(t) is deomposed as y() t = Yn fn(), t n = 0 where f ( t) is the set of orthonormal funtions defined n in 0 t T with f ( t) = x( t) / E. xt ( ) and the set f( t) should be seleted suh that it holds: xt () = X f() t = X f() t = E f() t due to X = E n= 0 n x n n 0 0 x 0 0 x t t Example: x( t) = ret( 0.5) with Ex = T and f0( t) = ret( 0.5) T T T Suffiient statisti Slide 98

99 Moreover, it holds: nt () = Nnfn(), t n= 0 T Nn = n() t fn() tdt. t= 0 It an be shown that N l and N m are unorrelated and (as Gaussian random variables) also statistially independent. The set {N n } fully desribes the noise. Moreover it an be show that this enables the reformulation of the detetion problem to: H : 0 Yn = N H : n Yn = Xn + Nn T with Yn = y() t fn() tdt t= 0 Suffiient statisti Slide 99

100 Therefore, the whole deision should be solely based on T Y = y() t f () t dt = y() t x() t dt. 0 0 t= 0 Ex t= 0 T Due to earlier assumptions (Gaussian noise), the detetion problem further simplifies to: H : Y = N H : Y = X + N = E + N x 0 Thus Y 0 gives (or is alled) a suffiient statisti as it is the only data useful for deision-making. It inludes the whole relevant information. Suffiient statisti Slide 00

101 Error probability By dividing the range of real numbers R into two seperated subsets R R 0 we an state generally with: and R R = R Example: R : Y < 0 and R : Y > assume H :if Y R { } { } P = P H p ( y H ) dy + P H p ( y H ) dy e R Y H R 0 Y H R R { } { } = + P H p ( y H ) dy P H p ( y H ) dy Y H R 0 Y H { } { } { } = P H P H p ( y H ) P H p ( y H ) dy. R Y0 H 0 Y0 H0 0 assume due to p ( y H ) dy p ( y H ) dy = 0 + Y H Y0 H R R 0 H 0 : if Y R Slide 0

102 { } > { } Hene P H p ( y H ) P H p ( y H ) is hosen Y H 0 Y H as it maximizes the integral in the last equation 0 By introduing the likelihood ratio Λ ( y) = it follows: ssume ase H : ifλ( Y ) 0 0 { 0} { } P H P H { 0} { } P H ssume ase H: ifλ ( Y0) >. P H p p Y H Y H 0 0 ( y H ) ( y H ) 0 Problem: Both pdfs and the probabilities need to be known For GUSSIN noise holds: y p ( y H 0 0 0) = e σ Y H πσn N p ( y H 0 0) = e Y H πσ N ( y E ) σ x N Likelihood ratio Slide 0

103 Consequently, we an rewrite Λ( Y0 ) as = T Ex 0 N ( y0 ) y E x E σ x e σ N 0 Λ ( Y0 ) = = e y σ e 0 N y() t x() t dt x () t dt N 0 N 0 e σ σ T. T with Y = ytxtdt ( ) ( ) 0 E Sine the likelihood is often met in exponentional form, it is ustomary to define the loglikelihood ratio as λ T T ( Y0) = ln Λ ( Y0) = y() t x() t dt x () t dt. σ N σ 0 N 0 = ( y( txtdt ) ( ) Ex ) σ N T 0 x Likelihood ratio Slide 03

104 Then, the deision rules beomes In ase { } = P{ H } P H aept aept aept 0 { 0} { } H0 : ifλ( Y0) ln P H P H { 0} { } P H H: ifλ ( Y0) > ln. P H H : ifλ( Y ) the deision rules ould be written as aept H : ifλ ( Y ) > 0. 0 (ML-rule) Likelihood ratio Slide 04

105 Now, we onsider M signals x m (t) For simpliity we assume P{ H } { } { } 0 = P H =... = P HM. T Y = yt () f () tdt= ytx () () tdt. 0, m 0, m m 0 Ex 0 m T t the reeiver, we have first to alulate Y0, m = y() t xm() t dt. E x m T 0 Detetion of M Signals Slide 05

106 x () t E x yt () x () t T T dt... dt E x Choose the Largest xm true () t x () M t T 0... dt E M x Optimum reeiver for memoryless modulation in WGN hannel Detetion of M Signals Slide 06

107 Let us re-onsider the fundamental detetor equation. Y 0 T = E x T 0 0 y() txtdt (). This proessing an be realised using the onvolution integral: Y0 = y( τ ) x( τ) dτ E x( τ) = h( T τ) x x T = E 0 y( τ ) h( T τ) dτ Y0 = g() t = y( τ) h( T τ) dτ t= T t= T Ex 0 Thus a mathed filter with impulse response an be applied. t Mathed filter onept ht () = xt ( t). Slide 07

108 Mathed filter onept E x y() t x ( T t) x ( T t) E x Choose the Largest xm true () t x ( ) M T t E M x Optimum reeiver for memoryless modulation in WGN hannel (Mathed Filter Implementation) Slide 08

109 Some reasons for analog-to-digital onversion: Preision and Dynami Component Tolerany Temperature behaviour, ging Integration Frequeny regions Multiplexing Introdution Slide 09