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第四章 模拟调制系统

第四章 模拟调制系统. 1. 引言 调制的定义 调制的分类 线性调制原理 非线性调制 ---- 角度调制 调制系统的比较（抗噪声性能分析和比较） FDM 原理 总结 重点：调制系统的抗噪声性能. 1. 调制的定义. Definition :A baseband waveform has a spectral magnitude that is nonzero for frequencies in the vicinity of the origin and negligible elsewhere.

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第四章 模拟调制系统

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  1. 第四章 模拟调制系统 • 1. 引言 • 调制的定义 • 调制的分类 • 线性调制原理 • 非线性调制----角度调制 • 调制系统的比较（抗噪声性能分析和比较） • FDM原理 • 总结 • 重点：调制系统的抗噪声性能

  2. 1.调制的定义 • Definition:A baseband waveform has a spectral magnitude that is nonzero for frequencies in the vicinity of the origin and negligible elsewhere. • Definition:A bandpass waveform has a spectral magnitude that is nonzero for frequencies in some band concentrated about a frequency f= ±fc ,where fc>>0.The spectral magnitude is negligible elsewhere. fc is called the carrier frequency.fc may be arbitrarily assigned. • Definition:Modulation is the process of imparting the source information onto a bandpass signal with a carrier frequency fc by the introduction of amplitude and/or phase perturbation.This bandpass signal is called the modulated signal s(t),and the baseband source signal is called the modulating signal m(t).

  3. Diagram of a typical modulation system • modulation m(t) s(t) Modulator Baseband signal Bandpass signal Carrier cosωct Local oscillator Modulating signal Modulated signal

  4. s(t) • Bandpass communication system m(t) channel Modulator Bandpass signal Baseband signal noise cosωct Carrier Local oscillator Modulated signal Modulating signal m’(t) Demodulator Corrupted baseband signal Corrupted bandpass signal

  5. 2.线性调制系统 • 调制系统的分类：幅度调制（线性调制），非线性调制（角度调制）和数字调制（PCM） • 线性调制：AM,DSB-SC,SSB,VSB

  6. Complex envelope representation • All banpass waveforms can be represented by their complex envelope forms. • Theorem:Any physical banpass waveform can be represented by: v(t)=Re{g(t)ejωct} Re{.}:real part of {.}.g(t) is called the complex envelope of v(t),and fc is the associated carrier frequency.Two other equivalent representations are: v(t)=R(t)cos[ωct+θ(t)] and v(t)=x(t)cos ωct-y(t)sin ωct where g(t)=x(t)+jy(t)=R(t) ejθ(t)

  7. Representation of modulated signals • The modulated signals a special type of bandpass waveform • So we have s(t)=Re{g(t)ejωct} the complex envelope is function of the modulating signal m(t): g(t)=g[m(t)] g[.]: mapping function All type of modulations can be represented by a special mapping function g[.].

  8. Complex envelope functions for various types of modulation • Type of modulation mapping functions g(m) AM Ac[1+m(t)] linear(?) DSB-SC Acm(t) linear SSB Ac[m(t)±jm’(t)] linear PM AcejDpm(t) non-linear FM non-linear

  9. Spectrum of bandpass signals • Bandpass signal’s spectrum complex envelope’s spectrum • Theorem:If a bandpass waveform is represented by: v(t)=Re{g(t)ejωct} then the spectrum of the bandpass waveform is V(f)=1/2[G(f-fc)+G*(-f-fc)] and the PSD of the waveform is Pv(f)=1/4[Pg(f-fc)+Pg(-f-fc)] where G(f)=F[g(t)], Pg(f) is the PSD of g(t). Proof: v(t)=Re{g(t)ejωct}=1/2{g(t)ejωct+g*(t)e-jωct} V(f)=1/2F{g(t)ejωct}+1/2F{g*(t)e-jωct}

  10. We have F{g*(t)}=G*(-f) Then V(f)=1/2{G(f-fc)+G*[-(f+fc)]} • The PSD for v(t) is obtained by first evaluating the autocorrelation for v(t). Rv(τ)=<v(t)V(t+τ)>=< Re{g(t)ejωct} Re{g(t+ τ)ejωc(t+ τ)} Using the identity: Re(c2)Re(c1)=1/2Re(c*2c1)+ 1/2Re(c2c*1) So we have: Rv(τ)=1/2Re<{g*(t)g(t+ τ)ejωcτ>} + 1/2Re<{g(t)g(t+ τ) ejωct ej2ωcτ>} negligible? But Rg(τ)= <{g*(t)g(t+ τ)> Rv(τ)=1/2Re<{g*(t)g(t+ τ)ejωcτ>}=1/2 Re{Rg(τ) ejωcτ} Pv(f)=F{Rv(τ)}=1/4[Pg(f-fc)+ Pg*(-f-fc)] But Pg*(f)= Pg(f),so Pv(f) is real.

  11. Linear and non-linear modulation systems • All bandpass modulated waveform can be represented by: v(t)=Re{g(t)ejωct} • The desired type of modulated waveform,s(t), is defined by the mapping function g[.]. • Linear modulation----Amplitude Modulation (AM) • Mapping function: gAM[.]=Ac[1+.] • Modulated waveform: s(t)=Re{Ac[1+m(t)] ejωct}= Ac[1+m(t)]cosωct • Spectrum:S(f)=1/2Ac[δ(f+fc)+M(f+fc)+δ(f-f0)]+M(f-fc)] • Normalized average power of s(t): <s2(t)>=1/2Ac2+1/2Ac2<m2(t)>

  12. AM system diagram----modulation: • demodulation: m(t) Ac[1+m(t)]cosωct Local oscillator Accosωct Envelope detector BPF Noisy s(t) m’(t) BPF LPF Noisy s(t) m’(t) cosωct

  13. Spectrum of AM waveform: M(f) 1 -B B f S(f) 1/2δ(f-f0) 1/2 1/2M(f-fc) f fc Where Ac=1

  14. Some definitions: • AM modulation: Ac[1+m(t)]cosω;where │m(t)│≤1 • The percentage of positive modulation on an AM signal is: %positive modulation=(Amax-Ac)/Ac*100=max{m(t)}*100 • The percentage of negative modulation is: %negative modulation=(Ac-Amin)/Ac*100=-min{m(t)}*100 • The overall modulation percentage is: %modulation= (Amax-Amin)/2Ac={max[m(t)]-min[m(t)]}/2*100 • Where Amax and Amin is Ac[1+m(t)]’s maximum and minimum values, is the level of the AM envelope when m(t)=0. • The modulation efficiency is the percentage of the total power of the modulated signal that convoys information.

  15. In AM signaling,we have: E=<m2(t)>/[1+ <m2(t)>]*100% m(t) t Ac[1+m(t)] Amax s(t) s(t) Ac Amin t

  16. If the condition │m(t)│≤1 is not satisfied and the percentage of negative modulation is over 100%,the envelope detector can not be used. • Ex. Power of an AM signal (description of the question) AM broadcast transmitter:a 5000-W transmitter is connected to a 50ohms load;then the constant Ac is given by 1/2Ac2/50=5000.So the peak voltage across the load will be Ac =707V during the times of no modulation.If the transmitter is then 100%modulated by a 100-Hz test tone, the total (carrier plus sideband) average power will be : 1.5[1/2(Ac2/50)]=7500W There we have <m2(t)>=1/2 for a sinusoidal modulation waveshape of unity (100%) amplitude. The modulation efficiency would be 33% since <m2(t)>=1/2 .

  17. Linear modulation----Double-Sideband suppressed carrier modulation (DSB) • Mapping function:gDSB[.]=Ac. • Modulated waveform: s(t)=Re{Acm(t)] ejωct}=Acm(t)cosωct • Spectrum:S(f)=1/2Ac[M(f+fc)+M(f-fc)] • Normalized average power of s(t): <s2(t)>=1/2Ac2<m2(t)> • Diagram of DSB system: channel BPF LPF m(t) s(t) m’(t) s’(t)+n(t) Accosωct cosωct

  18. Linear modulation----Single-Sideband modulation (SSB) • Definition:An upper sideband (USSB) signal has a zero-valued spectrum for │f│<fc , where fc is the carrier frequency. A lower sideband (LSSB) signal has a zero-valued spectrum for │f│>fc , where fc is the carrier frequency. • Mapping function: • Modulated waveform: s(t)=Re{Acg(t)] ejωct}=Ac[m(t)cosωct m^(t) sinωct] where the upper (-) sign is used for USSB and the lower (+) is for LSSB. m^(t) denotes the Hilbert transform of m(t). m^(t)=m(t)*h(t) ±

  19. h(t)=1/πt and H(f)=-f for f>0 and H(f)=f for f<0 M(f) │G(f)│ 2Ac 1 f B B f USSB │S(f)│ Ac f -fc-B -fc fc fc+B

  20. USSB’s Spectrum: S(f)=Ac{M(f-fc),for f> fc and 0 for f< fc} + Ac{0 for f>-fc and M(f+fc),for f<- fc } • Normalized average power of SSB: <s2(t)>=1/2<│g(t)│2>= 1/2Ac2<m2(t)+ [m^(t)]2 > = Ac2<m2(t)> • Diagram of SSB system: m(t) m(t) s(t) L.O. SSB signal -90o -90o phase shift across band of m(t) Phasing method m^(t)

  21. Diagram of SSB system(con.): • Demodulation: s(t) Sideband filter m(t) SSB signal Accosωct Filter method m’(t) s(t) channel BPF LPF SSB signal s’(t)+n(t) cosωct

  22. Vestigial sideband modulation • DSB spectrum resource • SSB too expensive to implement • A compromise between two systems:VSB • The vestigial sideband modulation is obtained by partial suppression of one sideband of a DSB signal. • If the bandwidth of the modulating signal m(t) is B,the VSB signal has a bandwidth between B and 2B. sVSB(t) VSB filter (Bandpass filter) m(t) DSB modulator s(t) Baseband signal DSB signal VSB signal Hv(f)

  23. Spectrum of VSB signal: S(f) 1/2 1/2M(f-fc) f fc Where Ac=1,DSB signal VSB Filter (USSB) SVSB(f) fΔ Hv(f-fc)+Hv(f+fc)

  24. The spectrum of VSB signal: SVSB(f)=S(f)HV(f) • The VSB filter must satisfy the constraint: HV(f-fc)+HV(f+ fc)=C, │f│≤B where B is the bandwidth of modulating signal. • So: SVSB(f)=Ac/2[M(f-fc)HV(f)+M(f+fc) HV(f)]] • Demodulation: sVSB(t) m’(t) Low pass filter cosωct

  25. 3.Non-linear modulation----angular modulation • Non-linear modulation systems----phase modulation and frequency modulation • Representation of PM and FM signals • Complex envelope for angular modulation: g(t)=Ace jθ(t) , • the modulated signal is: s(t)= Accos[ωct+θ(t)] for PM system, the modulated signal’s phase is directly proportional to the modulating signal mp(t): θ(t)=Dpmp(t) where Dp is the phase sensibility of phase modulator and constant.

  26. For FM, the phase of modulated signal is proportional to the integral of the modulating signal mf(t): θ(t)=Df -∞∫tmf(t)dt Df is the frequency deviation constant. • With the angular modulated waveform’s representation, we can not distinguish which is PM or FM.So if we have a PM signal modulated by mp(t),it is possible to represent it by a frequency modulation modulated by a different waveform mf(t).mf(t) is given by: mf(t)= Dp/Df[dmp(t)/dt] • similarly,if we have an FM signal modulated by mf(t),the corresponding phase modulation on this signal is mp(t)=Df/Dp -∞∫tmf(t)dt

  27. Generation of FM from a phase modulator and vice versa mp(t) s(t) mf(t) Integrator gain=Df /Dp Phase modulator FM signal out FM from a phase modulator mf(t) s(t) mp(t) Differentiator gain=Dp/Df Frequency modulator PM signal out PM from a frequency modulator

  28. Some definitions: • Definition:If a bandpass signal is represented by s(t)=R(t)cosψ(t) where ψ(t)=ωct+θ(t),then the instantaneous frequency of s(t) is fi(t)=1/2πωi(t)=1/2π[dψ(t)/dt] or fi(t)=fc+1/2π[dθ(t)/dt] For the case of FM,we have fi(t)=fc+1/2π[dθ(t)/dt]= fc+1/2πDfmf(t) the peak frequency deviation: ΔF=max{1/2π[dθ(t)/dt]}

  29. For FM signaling: ΔF=max{1/2π[dθ(t)/dt]}= 1/2πDfVp where Vp=max[mf(t)] • the peak phase deviation: Δθ=max[θ(t)] • so for PM signaling: Δθ=max[θ(t)]=DpVp where Vp=max[mp(t)]. • Definition:The phase modulation index is given by: βp= Δθ and the frequency modulation index is: βf= ΔF/B B is the bandwidth of modulating signal.

  30. Spectra of angular modulated signals • s(t)= Re{Acg(t) ejωct}= Accos[ωct+θ(t)] • the spectrum is S(f)=1/2[G(f-fc)+G*(-f-fc)] where G(f)=F[g(t)]=F[Ace jθ(t) ] In general,it is impossible to have an analytic form of angular modulation signal’s spectrum.We will use some special modulating signal (sinusoid) to estimate the spectrum.

  31. Ex. Spectrum of a FM or PM signal with sinusoidal modulation • PM case: The PM modulating waveform: mp(t)=Amsinωmt Then θ(t)=βsinωmt where β=DpAm=βp is the phase modulation index. For FM case : mf(t)=Amcosωmt and β=DfAm/ωm=βf So the complex envelope:g(t)= Ace jθ(t) = Ace jβsinωmt g(t) is a periodic function,so it can be represented by Fourier series. We have: g(t)= Σcnejnωmt where cn is Fourier coefficients.

  32. Cn=AcJn(β) • The Bessel function Jn(β) can not be evaluated in analytic form,but it is well-known numerically. • We have G(f): G(f)= Σcnδ(f-nfm)= Σ AcJn(β)δ(f-nfm) The G(f)’s distribution depends greatly on β. Conclusion:the bandwidth of the angle modulated signal will depends on β and fm.In general,that will be infinite. In fact, it can be shown that 98% of the total power is contained in the bandwidth: Carson’s Rule BT=2(β+1)B where β is phase or frequency modulation index. So we can estimate the bandwidth of an angle modulated signal by Carson’s Rule with sufficient precision.

  33. Narrowband angle modulation • when θ(t) is restricted to a small value, │θ(t)│<0.2rad, the complex envelope g(t)= Ace jθ(t) may be approximated by a Taylor’s series where only the first two terms are used. g(t)=Ac[1+j θ(t)] So we have the narrowband angle modulated signal: s(t)=Accosωct-Acθ(t) sinωct we see that the narrowband angle modulation can be considered an AM-type. Discrete carrier term Sideband power

  34. Diagram of the narrowband Angle modulation system: • Spectrum of NBFM S(f)=Ac/2{[δ(f-fc)+δ(f+fc)]+j[Θ(f-fc)-Θ(f+fc)]} Θ(f)=F[θ(t)]=DpM(f) for PM and (Df/j2πf)M(f) for FM m(t) - Σ Integrator gain=Df s(t) NBFM + Local oscillator -90o

  35. Wideband frequency modulation • Theorem:for WBFM signaling,where s(t)= Accos[ωct+Df-∞∫tm(t)dt] βf =(Df/2πB)max[m(t)]>>1 and B is the bandwidth of m(t).The normalized PSD of the WBFM signal is approximated by: Where fm(*) is the PSD of the modulating signal m(t).

  36. Summary: • it is a non-linear function of the modulation,and consequently the bandwidth of the modulated signal increases as the modulation index increases. • The real envelope of an angle-modulation signal is constant. • The bandwidth can be approximated by Carson’s rule.It depends on the modulation index and the bandwidth of the modulating signal.

  37. 4.调制系统的比较（抗噪声性能分析和比较） 信噪比：通信系统抗噪声性能的体现 信噪比：信号与噪声平均功率之比 S/N 分析方法：在相同的信号传输功率和相同的Gauss白噪声功率谱密度的条件下，调制系统的解调器输出信噪比。 分析模型： 其中：ni(t)是带通（窄带）Gauss白噪声 解调器 BPF LPF mo(t)+no(t) sm(t)+n(t) sm(t)+ni(t)

  38. 我们有ni(t)： ni(t)= nc(t)cosωct - ns(t)sinωct =V(t) cos[ωct+θ(t)] ni(t)， nc(t)，ns(t)有相同的平均功率，即σi= σc =σs或〈ni2(t)〉=〈nc2(t)〉=〈ns2(t)〉 解调器前的带通滤波器的带宽为B（与调制方式及m(t)有关），故解调器的输入噪声功率为： Ni=〈ni2(t)〉=noB (no:为噪声的单边带功率谱密度） 解调器的输出信噪比为： So/No=<mo2(t)>/ <no2(t)> 系统的调制制度增益G: G=输出信噪比/输入信噪比=[So/No]/[Si/Ni] 不同的调制方式，可以获得不同的G。G越大，调制方式就抗噪声而言就越佳。

  39. 系统的调制制度增益G与采用的解调方式有密切的关系系统的调制制度增益G与采用的解调方式有密切的关系 • AM系统的性能： • 同步检测 si(t)=Ac[1+m(t)] cosωct Si =1/2+1/2<m2(t)>, Ni=<ni2(t)>=noB so(t)=1/2m(t),So=1/4< m2(t)> no=1/2nc(t) ,N0=1/4Ni G=2 <m2(t)>/[1+ <m2(t)>] 若m(t)为方波，G=2/3 mo(t) BPF LPF si(t)+ni(t) cosωct

  40. 包络检测（电路图） • R1C1构成低通滤波器，当B≤1/ R1C1 ≤fc, R1C1电路只对vin的峰值的变化有响应。 • B≤fc是为了包络清楚，此时C1在载波的两个峰值间只有微小的放电，因此v近似等于vin的包络（除高频锯齿外）， R2C2起隔离v中的直流成分。 C1 Ac[1+m(t)] vout vin R1 C1 v R1 s(t) t

  41. 包络检测（续） v s(t) m(t)

  42. 信噪比计算： Si =Ac2/2+Ac2/2<m2(t)>, Ni=<ni2(t)>=noB • 检波器输入端的信号和噪声合成为： si(t) +ni(t) =Ac[1+m(t)] cosωct+ nc(t)cosωct - ns(t)sinωct =E(t)cos[ωct +ψ(t)] 其中：E(t)={[Ac+ Ac m(t)+ nc(t)]2 + ns2(t)}1/2 ψ(t)=arctg{ns(t)/[Ac+ Ac m(t)+ nc(t)]} E(t)的信号和噪声存在非线性关系。 分析：大信噪比情况，即Ac+ Ac m(t)>> ni(t) 则：E(t)≈ Ac+ Ac m(t)+ nc(t) 故： So= Ac2 < m2(t)>， N0=noB G=2 Ac2 < m2(t)>/[Ac2+Ac2<m2(t)>] (-1≤m(t)≤1) 当max[m(t)]=1（100%调制）且为正弦波，我们有G=2/3 包络检测能达到的最大信噪比增益。

  43. 小信噪比情况，即Ac+ Ac m(t)<< ni(t) • 则包络E(t)为： E(t)≈ R(t)+[Ac+ Ac m(t)]cosθ(t) • 包络中的信号部分完全被噪声所淹没。门限效应。 • 结论：包络法在大信噪比情况下，性能与同步检测法相似，在小信噪比时系统不能解调出信号。 噪声项

  44. DSB-SC系统的抗噪声性能 si(t)=Acm(t)cosωct，Si= Ac2/2<m2(t)> 经同步检测后，输出信号和噪声为： mo(t)=1/2Acm(t)， no=1/2nc(t) 因此： So= Ac2/4<m2(t)>， N0=1/4noB= 1/4Ni G=2 • SSB系统的抗噪声性能 • 单边带解调器与双边带相同，因此有： N0=1/4noB= Ni si(t)=Ac[m(t)cosωct m^(t) sinωct]，Si=Ac2/4<m2(t)> So= Ac2/16<m2(t)> G=1 双边带（G=2）是否优于单边带？No. why? ±

  45. 角度调制系统的抗噪声性能 • FM的抗噪声性能 • 解调法：鉴频法 sFM(t)=Acos[ωct+ φ(t)]，φ(t)=Df -∞∫tmf(t)dt • 设sFM(t)的带宽为B（不是m(t)的带宽），则鉴频法的输入信噪比为： Si/Ni=A2/2noB • G=? sFM(t)+ nc(t)cosωct - ns(t)sinωct = Acos[ωct+ φ(t)]+V(t)cos[ωct+θ(t)]=V’(t)cosψ(t) 带通限幅器 鉴频器 Low-pass filter m(t) sFM(t) 信号项 噪声项ni(t)

  46. 经限幅带通滤波器后，有：Vocosψ(t) • ψ(t)=？（信号和噪声的合成） • 令： Acos[ωct+ φ(t)]=a1cosΦ1 • V(t)cos[ωct+θ(t)]= a1cosΦ2 • a1cosΦ1+a1cosΦ2= acosΦ • 利用矢量表示法得： a a a2 a1 Φ2-Φ1 Φ1-Φ2 a1 a2 Φ Φ Φ1 Φ2 Φ2 Φ1 任意参考相位 任意参考相位 图b 图a

  47. 由图a得：tg(Φ- Φ1)=asin(Φ2- Φ1)/[a1+a2cos (Φ2- Φ1)] Φ= Φ1+arctg{a2sin(Φ2- Φ1)/[a1+a2cos (Φ2-Φ1)]} • 由图b得： Φ= Φ2+arctg{a1sin(Φ1- Φ2)/[a2+a1cos (Φ1-Φ2)]} • 根据设定的关系，有： ψ(t)=ωct+φ(t)+arctg{V(t)sin(θ(t)-φ(t))/[A+V(t)cos (θ(t)-φ(t))]} 或：ψ(t)=ωct+θ(t)+arctg{Asin(φ(t)-θ(t))/[V(t)+cos (φ(t)-θ(t))]} 鉴频器的输出正比于输入信号的瞬时频率偏移，以上表达式无法直接给出有用信号m(t)。特例分析。 两种情况：大Si/Ni和小Si/Ni情况。 分别讨论。

  48. 大Si/Ni：即A>>V(t)，因此有： V(t)sin(θ(t)-φ(t))/[A+V(t)cos (θ(t)-φ(t))]≈0 则图a成为： ψ(t)≈ωct+φ(t)+V(t)/Asin(θ(t)-φ(t)) 输出电压：vo(t)=1/2π[dψ(t)/dt]-fc = 1/2π[dφ(t)/dt]+1/(2πA)dni(t)/dt 输出的有用信号为： mo(t)= 1/2π[dφ(t)/dt]= Df/2πmf(t) So= Df2/4π2< mf2(t)> 输出噪声：no(t)= 1/(2πA)dns(t)/dt 求ns(t)的功率？ 信号 噪声 Ans(t)

  49. ni(t)= V(t)cos[ωct+θ(t)]:带通噪声 • ns(t)为低通型噪声，带宽由低通滤波器的截止频率确定[0，B’/2] • ns(t) =V(t)sin(θ(t)-φ(t)): φ(t)信号，因此ns(t) Gauss型 • 有：〈 ns2(t) 〉=〈 ni2(t) 〉=noB’ • 因此d ns(t)/dt的PSD Pi(f)为ns(t)的PSD乘以理想微分器的功率传递函数│H(f)│2=4π2f2 • n’s(t)的PSD为Po(f): Po(f)=4π2f2 Pi(f) Pi(f)=<ns2(t) >/B’=no f≤B’ (单边带PSD) P0(f)= 4π2f2 no f≤B’ • 结论： n’s(t)的PSD与频率f有关，即与f2 成正比。 d ns(t)/dt ns(t) 理想微分器

  50. 解调器的输出噪声功率为： No= <no2(t) >= <ns’2(t) >/ 4π2A2 =1/ 4π2A2 0∫fmP0(f) df So/ No= 4A2 Df2 <m2(t) >/[8π2nofm3] 讨论：m(t) is a tone signal. sFM(t)=Acos[ωct+βf sinωmt], βf=Df/ ωm βf=Δf/fm So/ No=3/2 βf2(A2/2)/(nofm) Si= A2/2, nofm为(0, fm)的白噪声记作Nm So/ No=3/2 βf2(Si /Nm) fm≠B, Ni≠Nm B=2(Δf+fm) 因此： So/ No=3 βf2(βf+1) (Si /Ni) G= 3 βf2(βf+1) 例： βf=5 则G=450，此时B=2 (βf+1) fm=12 fm P0(f) f B’

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