5.aSymmetric sideband signals (Vestigial sideband a The modulation on the vsb signal can be recovered a receiver that uses product detection or, if a large carrier is present, by the use of envelope detection. For e recovery of undistorted modulation, the transfer function for vSB filter must satify the constraint: H,(f-fC)+H,(+fc)=C<B .. Proof: SysB()=(Ac 2)[M(f-f)H,(f)+M(f+f)H,(fI the product detector output is Voutt)=[AosysB(t)coso t]*h(t) VouO=A{SsB(f)*[(1/2)6(f-f)+(1/2)6(f+fC)H( ron()=(44/4)M()H1(-f)+M()H1(f+f)/<B =(A4/4)MOH1(-f)+H1(f+f)|<B Thus only when the constraint satisfy, Vou (f)=KM(f hold true
26 5.5Asymmetric sideband signals (Vestigial sideband) • The modulation on the VSB signal can be recovered by a receiver that uses product detection or , if a large carrier is present, by the use of envelope detection. For recovery of undistorted modulation, the transfer function for VSB filter must satify the constraint: Hv (f − f c ) + Hv (f + f c ) =C, f B • Proof: S ( f ) (A / 2)[M( f f )H ( f ) M( f f )H ( f )] VSB = c − c v + + c v • Thus only when the constraint satisfy, Vout(f)=KM(f) hold true. A A M f H f f H f f f B V f A A M f H f f M f H f f f B V f A S f f f f f H f V t A s t c v c v c c v c v c out VSB c c out VSB out o = − + + = − + + = − + + = ( / 4) ( )[ ( ) ( )], ( ) ( / 4)[ ( ) ( ) ( ) ( )], ( ) { ( ) *[(1/ 2) ( ) (1/ 2) ( )]} ( ) ( ) [ ( )cos t ]* h(t) 0 0 0 c • the product detector output is
Summary AM 8/0=A. 1+m( 5/(0=AJ1 m/t/lcoso,t (O)=-[6(f-f)+M(f-fc)+δ(f+fc)+M(+fC) →DSB-SCg()=Am() s(t=A m(troso.t s(=1/2/G(f-fo)+G*+f SSB-AM g(0=A AM(f-f)f>f 0 f>-f 0f<+4M(f+f)f<-f ⅤSBg()=A4Dm(0)+m(O)S2(O)=s8-1(O)*h() SIsR(t)=A m(t)coso+m(t)sin@t]h SsB(O)=(4/2川M(-f∈)H,(O+M(+f)H,( 27
27 Summary • AM DSB-SC SSB-AM VSB g(t) A [1 m(t)] = c + g(t) A m(t) = c g(t) A [m(t) jm ˆ(t)] = c g(t) A [m(t) jm ˆ(t)] = c s (t) s (t)*h (t) VSB = SSB−AM v s(t) A m(t) t c c = [1+ ]cos s(t) A m(t) t c c = cos s (t) A [m(t)cos t m ˆ(t)sin t] SSB−AM = c c c s (t) A [m(t)cos t m ˆ(t)sin t]*h (t) VSB = c c c v S(f) 1/ 2[G(f- f ) G*(f f )] = c + + c [ ( ) ( ) ( ) ( )] 2 ( ) c c c c c f f M f f f f M f f A S f = − + − + + + + − − + + − = c c c c c c c c f f f f f f A M f f A M f f f f S f ( ) 0 0 ( ) ( ) S ( f ) (A / 2)[M( f f )H ( f ) M( f f )H ( f )] VSB = c − c v + + c v
Summary Example 5-2 if linear modulated signal is as following: e(1)s,(t=cos(Oot)cos(ot) e(2)S2(t)(1+0.5sin(@ot)) cos(ot) Where @ =6@o, please giving their spectrum Solution:()the spectrum is S()={z[8(o-00)+6(+0)*x(O-0a)+o(+2) 2丌 [(o-700)+o(+700)+6(-500)+6(+500 2 700-500 sa 7 28
28 Summary • Example 5-2 if linear modulated signal is as following: (1) s1 (t)=cos(ω0 t)cos (ωc t) (2) s2 (t)(1+0.5sin (ω0 t)) cos (ωc t) Where ωc=6ω0 , please giving their spectrum. • Solution : (1) the spectrum is: [ ( 7 ) ( 7 ) ( 5 ) ( 5 )] 2 { [ ( ) ( )]* [ ( ) ( )]} 2 1 ( ) 0 0 0 0 1 0 0 = − + + + − + + S = − + + − c + + c -7ω0 -5ω0 5ω0 7ω0
Summary (2)the spectrum is S2(O)= 6(O-0)+(+0) 0.5 {[6(-00)-8(0+00)*x[(O-02)+6(O+00)} 2丌J [(-600)+o(+600 +-{6(0-500)-6(a+500)-6(0-700)+6(0+70) 7 sa 70 29
29 Summary • (2) the spectrum is { ( 5 ) ( 5 ) ( 7 ) ( 7 )] 4 [ ( 6 ) ( 6 )] { [ ( ) ( )]* [ ( ) ( )]} 2 0.5 ( ) [ ( ) ( ) 0 0 0 0 0 0 0 0 2 + − − + − − + + = − + + + − − + − + + = − + + j j S c c c c -7ω0 -5ω0 5ω0 7ω0 j ω ω S2 (f)
5.6 Phase Modulation and Frequency Modulation (Representation of PM and FM signal) .. PM and fM are special cases of angle-modulated signaling. In this kind signaling the complex envelope is: g(t)=ace 0(t) .. Here the real envelope r(t=g(tFAc, is a constant, and the phase e(t is a linear function of the modulating signal m(t). However, g(t)is a nonlinear function of the modulation. The resulting angle-modulated signal is s(t)=A coso t+e(tI For PM, the phase is directly proportional to the modulating ignal e(t)=D,m(t) Where the proportionality constant Dn is the phase e sensitivity of the phase modulator, its units is radians per volt. For fm, the phase is proportional to the integral of m(t), so that:
30 5.6 Phase Modulation and Frequency Modulation (Representation of PM and FM signal) • PM and FM are special cases of angle-modulated signaling. In this kind signaling the complex envelope is: ( ) ( ) j t c g t A e = • Here the real envelope R(t)=|g(t)|=Ac , is a constant, and the phase θ(t) is a linear function of the modulating signal m(t). However, g(t) is a nonlinear function of the modulation. The resulting angle-modulated signal is: s(t) A cos[ t (t)] = c c + • For PM,the phase is directly proportional to the modulating signal (t) D m(t) = p • Where the proportionality constant Dp is the phase sensitivity of the phase modulator, its units is radians per volt. For FM, the phase is proportional to the integral of m(t), so that: