2. Bandpass Signal Theory. Throughout this course, we will adopt the baseband representation of a bandpass signal.

Transcription

1 . Bandpass Signal Theory Throughout this ourse, we will adopt the baseband representation o a bandpass signal. Reerenes :. J.G. Proakis, Digital Communiations, MGraw Hill, 3 rd Edition, 995, Setion 4.. The Complex Envelop.. Introdution By deinition, a bandpass signal have null at and near DC. Any bandpass signal an be written in the orm s( t) = x( t)os( π t) y( t)sin( π t) = At ( )os( π t+ φ( t)) Inormation is ontained in the envelop (amplitude and phase), not the arrier. Distinguish between them with a onise notation : where j t { zte π } st () = Re () zt () = xt () + jyt () = Ate () jφ t ( ) -

2 is the omplex envelop, or the baseband equivalent o s(t). It is like a timevarying phasor The omplex representation o a bandpass signal is not neessarily unique. Consider a bandpass signal with the spetrum shown below. S( ) W + W W + W + W jπ t jπ t s() t = S( ) e d = Re S( ) e d W W jπ t jπβt = Re e S( β + ) e dβ W jπ t { e z t } = Re ( ) where Z( ) = S+ ( + ) and +ve requeny only. W jπ t zt () Z( ) e d =, and subsript + means W Z ( ) W W Notes. hoie o enter requeny is not unique -

3 . transorm is not neessarily symmetri about, 3. At ( ) and φ ( t) are amplitude and phase with respet to the unmodulated arrier j t e π. 4. transorm Z() twie as large as either +ve or ve omponent o S().. Generation o Bandpass Signals by Complex Envelop The band pass signal xt ( ) an be generated using the quadrature modulator shown in the diagram below xt ( ) os( π t) sin( π t) Loal Osillator s( t) yt ( ) zt ( ) Re[ ] s( t) j t e π Note that. x(t) and y(t) are generated by a modem. has to be balaned -3

4 a) 90 degree separation or rosstalk b) same gain eah branh or distortion (e.g. onstant envelop beomes non onstant envelop; SSB generation with y(t)=h[x(t)] has imperet sideband anellation)...3 Reovery o Complex Envelop Let the reeived band pass signal be r( t) = u( t)os( π t) v( t)sin( π t) = Bt ( ) os( π t+ θ( t)) j t { wte π } = Re ( ) where w() t u() t jv() t B() t e jθ t ( ) = + = is the signal we want to reover. j t Consider the multipliation o r(t) by os( t ) Re{ e π } π =. From properties o omplex numbers, we an easily show that or omplex * and Re{ α}re{ β} = Re αβ + αβ. So α β, { } Re{ wte () }Re{ e } = Re{ wt () + wte () } jπ t jπ t j4π t = ut + j4π t ( ) Re{ wte ( ) } The irst term is a low pass signal and the seond term is a double requeny term. We an thus reover ( ) u t by using a low pass ilter (LPF). -4

5 Similarly, we an reover v( t ) by multiplying r(t) by j t sin( π t) = Re je π and then low pass ilter the produt signal. { } In onlusion, the ollowing struture is known as a produt demodulator. Note that omplex envelop phase is with respet to reeiver s loal osillator. r( t) os( π t) Loal Osillator LPF u( t) sin( π t) v( t) LPF r( t) LPF w( t) e j π t Exerise : Let the transmitted omplex envelop be zt ( ) = xt ( ) + jyt ( ) and the reeived omplex envelop be w( t) = z( t) e jω. Determine the output o the quadrature demodulator shown above. -5

6 ..4 Bandpass iltering Adopt the notation that s! ( t) (or S! ( ) in the requeny domain) is a real bandpass signal and s( t ) (or S( ) in the requeny domain) its omplex envelop Let s! ( t) be the input to a bandpass ilter with an impulse response h! ( t) The output o the ilter is denoted by r! ( t).. s! ( t) h! ( t) r! ( t) Clearly R! ( ) = H! ( ) S! ( ) and hene R! ( ) = H! ( ) S! ( ) S! ( ) H! ( ) R! ( ) W + W W + W So -6

7 R( ) = H( ) S( ) where S( ) = S! + ( + ), H( ) = H! + ( + ), and R( ) = R! + ( + ). In other word, onvolution o real bandpass signals is equivalent to onvolution o the orresponding omplex envelopes. H H! +. Notes : no ator o in ( ) = ( + ) Exerise : Use the time domain approah to prove the above property. Example : In digital ommuniations, the most ommon ilter used is a mathed ilter. Let s! ( t) = x( t)os( π t) y( t)sin( π t) be a bandpass pulse. Its mathed ilter has an impulse response equal to h! ( t) = s! ( t). Determine the output o the mathed ilter and sketh its implementation struture. Solution - By deinition, h! ( t) = x( t)os( π t) + y( t)sin( π t). This means the equivalent low pass response is h( t) = x( t) jy( t). The omplex envelop o the input bandpass signal is s( t) = x( t) + jy( t) - The output o the bandpass mathed ilter is jπ t {[ ] } rt!() = Re ht () st () e jπ t { rte } = Re ( ) where the omplex envelop o the output signal is -7

8 [ ] [ ] [ xt () x( t) yt () y( t) ] j[ yt () x( t) xt () y( t) ] rt () = xt () + jyt () x( t) jy( t) = + + = at ( ) + jbt ( ) - The above equations suggest the ollowing implementation struture s( t) Quadrature Demodulator xt ( ) yt ( ) Real Filter Bank a( t) b( t) Quadrature Modulator r( t) where the ilter bank omputes a(t) and b(t) rom x(t) and y(t) aording to the last equation. - in pratie, there is no need to do the remodulation beause we are only interested in the mathed ilter output at time t = kt, where T is the bit or symbol duration, and k an integer. In this ase, the quadrature modulator in the above diagram an be replaed by the operation (whih impliitly involves sampling) {[ ] j kt akt + jbkt e π } Re ( ) ( ) Futher simpliation an be obtained i T is an integer. In this ase, the output o the bandpass ilter is simply a( t) = x( t) x( t) + y( t) y( t). In other word, the ilter bank only has the ilters x(-t) and y(-t). There is no need to dupliate these ilters...5 Fourier Symmetry Relations -8

9 Now that we re ommitted to omplex time untions, better review Fourier symmetries. They will beome handy later on. Given s( t) S( ), we have. jπ t jπ t S( ) = s( t) e dt; s( t) = S( ) e d. time/reqeny reversal : s( t) S( ) 3. * * s t S S s t ( ) *( ); ( ) *( ) 4. onjugate symmetry : s(t) real! S S * ( ) ( ) 5. onjugate odd symmetry : s(t) imag! S S * ( ) ( ) 6. s(t) real and even! S() real and even 7. I s(t) = x(t) + jy(t), x(t) and y(t) both real, then * Re{ s( t)} X( ) = S( ) + S ( ) * Im{ s( t)} Y( ) = j S( ) S ( ) 8. omplex requeny shit : s te S o jπ ot ( ) ( ) 9. onvolution : s t r t s r t d S R * * * ( ) ( ) = ( α) ( α) α ( ) ( ) 0. ross orrelation : s t r t s r t d S R * * * ( ) ( ) = ( α) ( α ) α ( ) ( ). Parsaval : * * s( α) r ( α) dα = S( ) R ( ) d -9

10 From Properties 5 & 6, we an dedue that the spetrum o a omplex envelop is not symmetrial about = 0. A non-symmetrial spetrum arries more inormation, whih simply relets the at that the omplex envelop arries two piees o inormation...6 Bandpass Random Proesses Suppose we orm a bandpass proess by up onverting a omplex lowpass proess zt ( ) = xt ( ) + jyt ( ), i.e jπ t { } zt!( ) = Re zte ( ) = xt ( )os( π t) yt ( )sin( π t) How does the seond order statistis o zt!( ) depends on those o z(t)? The autoorrelation untion o zt!( ) is deined as [ τ ] R! (, tt τ) = E ztzt!()!( ) z jπ t j π ( t τ) { } { τ } * { τ } = E Re z( t) e Re z( t ) e = E Re ztz ( ) ( t ) e + ztzt ( ) ( τ) e jπ τ j π ( t τ) Note that the irst term is wide-sense stationary (WSS) i z(t) is WSS. The seond term depends on t. Under what onditions does the seond term vanish? Assuming that x(t) and y(t) are WSS, the seond term an be expanded as : -0

11 { ( ) ( τ) ( ) ( τ) + [ ( ) ( τ) + ( ) ( τ) ]} E xtxt yt yt j xt yt ytxt e { x ( τ) y ( τ) xy ( τ) xy ( τ) } = R R + j R + R e j π ( t τ) j π ( t τ) So it should be lear that the seond term vanishes i. Rx ( τ ) = Ry ( τ) and Rxy ( τ) = Rxy ( τ), or. instead o E [ ], we use a time average in the deinition o the autoorrelation untion Rz! ( t, t τ), or j t 3. i arrier has a random phase θ, i.e instead o e π, we have j( π t+ θ) e From now on, we ignore the seond term and the autoorrelation an now be rewritten as where * jπ τ { } jπ τ { Rz τ e } R (, t t τ) = R ( τ) = E Re z() t z ( t τ) e z! z! R = = Re ( ) * z ( τ) E z( t) z ( t τ) { Rx( τ ) Ry( τ) j Rxy( τ) Rxy( τ) } = + + In other word, Rz ( τ ) is the omplex envelop o Rz ( τ )!. Note that R!!, i.e. even symmetry. * z( τ ) = Rz( τ ) -

12 The Fourier transorm o Rz ( τ ) is denoted by Sz ( ) and it is the power spetral density (PSD) o z(t). Sz ( ) is real but not neessarily even. It is so i and only i (i) Rz ( τ ) is real, i.e. Rxy ( τ ) = Rxy ( τ ) This is satisied i x(t) and y(t) are unorrelated (i.e. Rxy ( τ ) = 0 ) or idential. Conlude : i x(t) and y(t) are unorrelated, Sz ( ) is real and even. Consequently Sz! ( ), the spetrum o the bandpass signal is symmetrial about. j We have Rz( ) Re { Rz( ) e π τ! τ = τ }. This means Pz = Pz = ( Px + Py) modulation uts the power in hal.!, i.e Up to this point, we were onsidering the modulation aspet o bandpass proesses, i.e, given that the low pass proesses x(t) and y(t) are WSS, we determine the properties o the bandpass proess zt!( ). Now look at the demodulation aspet o bandpass proesses. Given that zt!( ) is WSS, what are the properties o x(t) and y(t)? The autoorrelation untion o zt!( ) is now written as : [ τ ] R! (, tt τ) = E ztzt!()!( ) z { * j } π τ j π ( t τ τ τ ) = Re E ztz ( ) ( t ) e + Eztzt { ( ) ( )} e -

13 Sine zt!( ) is WSS, R! ( t, t τ) is a untion only o τ. This means z and They imply = * E z( t) z ( t τ ) R z ( τ ) [ τ ] E z( t) z( t ) = 0 R ( τ ) = R ( τ ) and R ( τ ) = R ( τ ) x y xy xy { } { } R ( τ ) = R ( τ) + R ( τ) + j R ( τ) R ( τ) z x y xy xy = R ( τ) jr ( τ) x xy The last equation shows even more learly that, unless Sz! ( ) is symmetrial about, x(t) and y(t) are orrelated. Example : I zt!( ) is a bandlimited white noise with PSD equal to N / o, i.e. Sz(! ) N o / W/ + W/ W / + W/ then S ( ) z N o W / W/ -3

14 So x(t) and y(t) are unorrelated (independent i Gaussian), eah with PSD N o S ( ) x Sy( ) N o W / W/ N o W / W/. Sampling Theorem In the reeived modem, the reeived signal is irst iltered and then sampled. Reovery o the transmitted data an be ahieved by properly proessing the samples, using or example a DSP. So lets revisit the sampling theorem. Let x(t) be a real signal with a low pass spetrum X() shown below. The bandwidth o the signal is B X( ) Xs ( ) iii iii B B s s B B s s Impulse-sampling o x(t) yields -4

15 x ( t) = x( t) δ ( t k/ ) s k where s is the sampling requeny. Minimum s = B real samples per seond. s What about omplex z(t)? Z ( ) Zs ( ) B B iii s s B B s s iii Still sample at s B, but take omplex samples ( reals). This relets the extra inormation in non-symmetri spetrum Z(). Now look at sampling requirements or a bandpass signal zt!( ) with bandwidth B. Do we need s ( + B/)? The answer is NO and the proos are as ollow. Proo : Sine the arrier arries no inormation, bring the bandpass signal down to baseband using a quadrature demodulator : zt!( ) Quadrature Demodulator zt ( ) rate (B/) omplex samples/s or B real samples/s -5

16 Proo : I we take real samples o zt!( ) at rate B, we have (or onveniene, we set = B+ B/): Z! ( ) B +ve images iii iii iii -ve images iii Combined iii iii So inormation is preserved with B reals/se. Can generalize this result to = Bn+ B/, where n an arbitrary integer. Exerise : Consider the ase o B reals/se. Show that inormation is lost with this sampling requeny. In onlusion, need B real numbers/se to represent a real bandlimited signal. This in independent o whether the signal is lowpass or bandpass. -6

17 .3 Random Gaussian Vetors Reerene : Proakis Reall multivariate Gaussian probability density untion (pd) or reals : x = (,,#, ) t - a real (olumn) vetor, N omponents, x x x N E [ x] = m, E[( x m)( x m) t ] = Φ Φ, Φ are, real, symmetri, semi-positive deinite. ( ) t Then p ( x ) = exp / ( x m ) Φ x ( x m ) ( π ) N / Φ For omplex Gaussian : ( z, z,, z N ) t z = - a omplex (olumn) vetor, N omponents, E [ z] = m, transpose. E z m z m = Φ, [( )( ) ] ( ) represents Hermitian Now Φ is Hermitian, Φ = Φ, so it still has real eigenvalues. Sine ovariane matrix they are all non-negative, i.e. semi-positive deinite. The pd is ( ( ) ( )) p ( z z ) = exp ( π ) z m Φ z m N Φ -7

18 Alternatively, write the omplex Gaussian vetor z as a real vetor o length N, w = ( x, y, x, y,#, x, y ) t m = E[ w ], = E ( )( ) t Φ w m w m, and N t ( ( ) ( )) p ( w w ) = exp N / ( π ) Φ w m Φ w m N Let ω = ( ω, ω, #, ω ) t N. The harateristi untion o the omplex orm is : ( ) ( ) j M ( j ) E ω z e z ω = = exp jm ω exp j ω Φω Note that M z ω k = jm ωm jω = jm ω =0 k and z ( ) et. or other moment generation. Exerise : Obtain expliit expressions or the joint pd and harateristi untion o two zero mean omplex Gaussian random variables x and x. Let their ovariane matrix be φ φ Φ = * φ φ -8

19 Exerise : For the omplex Gaussian random variables x and x in the last exerise, determine the onditional pd p x x ( x x ) = p x, x p x ( x, x ) ( x ) Properly interprete the results. Exerise (Generalization o the result in the last exerise) : Consider two t t zero mean omplex Gaussian vetors x = ( x, x, #, x n ) and y = ( y, y, #, y m ) with a orrelation matrix : xx xy E x ( ) Φ Φ Φ = x y = y Φyx Φyy Determine the onditional pd p yx ( yx) = pxy, ( x, y) p ( x) x Use the ollowing identities ( denotes the determinant o a matrix) : Φ = Φ Φ Φ Φ Φ xx yy yx xx xy Φ Φ xx xxφ Φ 0 I 0 xy ΦyxΦxx I I = 0 I 0 ( Φyy ΦyxΦxxΦxy ) Quadrati orm o Gaussian RVs : -9

20 where = Gaussian RVs. D = z Fz F F, and = ( z, z,, z N ) t z is a olletion o zero mean omplex Exerise : Obtain an expression or D or the ase o N = and 0 F = 0 Charateristi untion o quadrati orm : sd sd Φ D( s) = E e = e pz( z) dz z Exerise : Determine the harateristi untion o the quadrati orm in the last exerise, assuming the orrelation matrix or z = ( z, z) t is φ φ Φ = * φ φ Deomposition o the orrelation matrix Φ : The orrelation matrix Φ o the omplex Gaussian RVs = ( z, z,, z N ) positive semi-deinite and an be written as : Φ = U Λ U z z z z is t -0

21 where U z is a unitary matrix whose olumns are orthogonal with U z = U z, and Λ z = diag ( λz,, λz,,, λz, N) is a diagonal matrix ontaining the eigenvalues o the orrelation matrix Φ. Note all eigenvalues are non-negative. Basially, what the above ormula is saying is that orrelated Gaussian RVs an be obtained by mixing independent Gaussian RVs in dierent ways. Let = ( n, n,, n N ) t n be N independent omplex Gaussian RVs with zero mean and unit variane. The transormation / z = UzΛ z n generates a set o orrelated Gaussian RVs with a orrelation matrix ( )( ) E zz = E = U Λ n U Λ n Φ / / z z z z Deomposition o the quadrati orm : - Useul in bit-error probability (BEP) analysis - The quadrati orm D = z Fz is equivalent (in a statistial sense) to N D= λ k= gk. n k where λ gk, are the eigenvalues o the matrix G = ΦF -

22 and = ( n, n,, n N ) t n are a set o independent and idential distributed (iid) omplex Gaussian RVs with zero mean and unit variane. The proo is given in Appendix A Exerise : Show that the harateristi untion o the quadrati orm D = z Fz an now be written as Hints : sd sd sd Φ D( s) = E e = e pz( z) dz= e pn( n) dn z n I sφf. G = ΦF an be written as G = AgΛ gb g where B g = A g and Λ g is a diagonal matrix ontaining the eigenvalues o G. For any square matries A, B, and C, A B C = A B C Major onlusion o this setion : - know your matrix theory, it an be your riend! -

23 Appendix A : Deomposition o a Gaussian quadrati orm The original expression or the quadrati orm is D = z Fz. From the deomposition o the orrelation matrix, we an replae z by / z = U Λ q. This means the quadrati orm an now be written as z z D = zfz ( UΛ q) FU Λ q = = / / z z z z ( z z z z ) / / q Λ UFUΛ q qbq where B = Λ UFUΛ / / z z z z Sine B is Hermitian, it an be written as B = U Λ U b b b where U b is a unitary matrix whose olumns are orthogonal, and Λ b = diag ( λb,, λb,,, λb, N) is a diagonal matrix ontaining the eigenvalues o the matrix B. This means the quadrati orm an now be written as ( b) b( b ) D = qu Λ Uq Let n = U q = ( n, n,, n ) t b N -3

24 This means the quadrati orm an now be rewritten as D Sine the ovariane matrix o n is N = n Λbn = λb. k nk k= Φn = E nn = E U bqq Ub = UIU = I b b this means the n k are iid Gaussian RVs with zero mean and unit variane. So what are the eigenvalues o the matrix B? By deinition, BVk = λbk, V k, where V k is an eigenvetor o B. This means or or or / / ( ) Λ UFUΛ V = λ V z z z z k b, k k / / / / ( )( ) = λ, ( ) U Λ Λ UFUΛ V U Λ V z z z z z z k b k z z k / / / / ( U zλz Λz Uz) F( UzΛ z Vk) = λb, k ( UzΛ z Vk) ( ΦF) Wk bk, Wk = λ In other word, the eigenvalues o B are the eigenvalues o G = ΦF -4