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Figure 15 message spectrum and dsb spectrum for

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Figure 15: Message spectrum and DSB spectrum for Problem 3.10. Both signals are real-valuedin time domain, and hence conjugate symmetric in frequency domain. We only show positivefrequencies for the DSB spectrum.10Downloaded by Agrim Chaudhry ([email protected])lOMoARcPSD|11644210
Figure 16: Spectrum of Q component of VSB signal in Problem 3.10.Figure 17: Using differentiation to find the time domain expression for the Q component of theVSB signal in Problem 3.10.11Downloaded by Agrim Chaudhry ([email protected])lOMoARcPSD|11644210
The results are sketched in Figure 14.(d) The message signal is given bym(t) = 4sinc4t2 cos 2πtM(f) =I[2,2](f)δ(f1)δ(f+ 1)The DSB spectrum (ignoring constant factors) is given byUDSB(f) =M(ffc) +M(f+fc).The message spectrum and the DSB spectrum are sketched in Figure 15.(e) The complex envelope of the VSB signal is given byy=mh=m(hc+jhs) =mhc+jmhs,so that the I component is given byyc=mhcand the Q component byys=mhs. Going tothe frequency domain, the I and Q components are given byYc(f) =M(f)Hc(f) andYs(f) =M(f)Hs(f). From Figure 14, we see thatHc(f) is flat over [2,2], the message band, so thatYc(f) =M(f). This is already sketched in Figure 15. For the Q component, we multiplyHs(f)in Figure 14 andM(f) in Figure 15 to obtain the spectrum shown in Figure 16.(f) From Figure 16, we see that the Q component can be written asYs(f) =jA(f) +j(δ(f1)δ(f+ 1))whereA(f) is as shown in Figure 17. The inverse Fourier transform of the impulsive term is asinusoid:j(δ(f1)δ(f+ 1))j(ej2πtej2πt)=j×2jsin 2πt=2 sin 2πtTo find the inverse Fourier transform of the non-impulsive term (i.e., to findja(t)) we employdifferentiation.B(f) =dA(f)/dfb(t) = (j2πt)a(t)From the graph ofB(f) in Figure 17, we can immediately write down its inverse Fourier trans-form:b(t) =ej4πt+ej4πt2sinc2t= 2 cos 4πt2sinc2tThus,ja(t) =jb(t)j2πt=b(t)2πt=sinc2tcos 4πtπtThe time domain expression for the Q component is therefore given byys(t) =ja(t)2 sin 2πt==sinc2tcos 4πtπt2 sin 2πtProblem 3.11(a) The messagem(t) is a sawtooth waveform.It, and the AM signalx(t)generated from it, are plotted in Figure 18.(b) For the modelx(t) =Accos 2πfct+m(t) cos 2πfct, the power of the carrier term isA2c/2 andthe power of message term ism2/2, so the power efficiency is given byηAM=m2A2c+m2We haveAc= 4 andm2=12integraldisplay11t2dt=13Plugging in, we haveηAM=1/342+ 1/3=149or a power efficiency of about 2%.12Downloaded by Agrim Chaudhry ([email protected])lOMoARcPSD|11644210
-5-4-3-2-1012345-1-0.8-0.6-0.4-0.200.20.40.60.81tMessage m(t)(a) Message signal-5-4-3-2-1012345-5-4-3-2-1012345tAM signal x(t)(b) AM signalFigure 18: The sawtooth message signal and the corresponding AM signal in Problem 3.11. Thewaveform is sampled slowly enough that we can see the carrier waveform.fIm(M(f))-111/2-1/21/2-11M(f)Figure 19: Hilbert transform of a sinc function.

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Term
Spring
Professor
Rappaport
Tags
Fourier Series, Hertz, Agrim Chaudhry

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