6.003: Signals and SystemsModulationMay 6, 2010Course VI Underground GuidePlease give us and future students feedback on 6.003 by participatingin the Course VI Underground Guide Survey:https://sixweb.mit.edu/student/evaluate/6.003-s2010Communications SystemsSignals are not always well matched to the media through which wewish to transmit them.signal applicationsaudio telephone, radio, phonograph, CD, cell phone, MP3video television, cinema, HDTV, DVDinternet coax, twisted pair, cable TV, DSL, optical fiber, E/MAmplitude ModulationAmplitude modulation can be used to match audio frequencies toradio frequencies. It allows parallel transmission of multiple channels.x1(t)x2(t)x3(t)z1(t)z2(t) z(t)y(t)z3(t)cosw1tcosw2tcoswctcosw3tLPFSuperheterodyne ReceiverEdwin Howard Armstrong invented the superheterodyne receiver,which made broadcast AM practical.Edwin Howard Armstrong also invented andpatented the “regenerative” (positive feedback)circuit for amplifying radio signals (while he wasa junior at Columbia University). He also in-vented wide-band FM.Amplitude, Phase, and Frequency ModulationThere are many ways to embed a “message” in a carrier. Here arethree.Amplitude Modulation (AM): y1(t) = x(t) cos(ωct)Phase Modulation (PM): y2(t) = cos(ωct + kx(t))Frequency Modulation (FM): y3(t) = cosωct + kRt−∞x(τ )dτFrequency ModulationIn FM, the signal modulates the instantaneous carrier frequency.y3(t) = cosωct + kZt−∞x(τ )dτ| {z }φ(t)ωi(t) = ωc+ddtφ(t) = ωc+ kx(t)Frequency ModulationCompare AM to FM for x(t) = cos(ωmt).AM: y1(t) = (cos(ωmt) + 1.1) cos(ωct)tFM: y3(t) = cos(ωct + m sin(ωmt))tAdvantages of FM:• constant power• no need to transmit carrier (unless DC important)• bandwidth?Frequency ModulationEarly investigators thought that narrowband FM could have arbitrar-ily narrow bandwidth, allowing more channels than AM. Wrong!y3(t) = cosωct + kZt−∞x(τ )dτ= cos(ωct) × coskZt−∞x(τ )dτ− sin(ωct) × sinkZt−∞x(τ )dτIf k → 0 thencoskZt−∞x(τ )dτ→ 1sinkZt−∞x(τ )dτ→ kZt−∞x(τ )dτy3(t) ≈ cos(ωct) − sin(ωct) ×kZt−∞x(τ )dτBandwidth of narrowband FM is the same as that of AM!(integration does not change bandwidth)Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .m0−mm sin(ωmt)t10−1cos(m sin(ωmt))tincreasing mPhase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 0Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 1Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 2Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 5Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 10Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 20Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 30Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 40Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore cos(m sin(ωmt)) is periodic in T .10−1cos(m sin(ωmt))t|ak|k0 10 20 304050 60m = 50Phase/Frequency ModulationFourier transform of first part.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt)))| {z }ya(t)− sin(ωct) sin(m sin(ωmt)))|Ya(jω)|ωωcωc50ωmm = 50Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore sin(m sin(ωmt)) is periodic in T .m0−mm sin(ωmt)t10−1sin(m sin(ωmt))tincreasing mincreasing mPhase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m sin(ωmt))= cos(ωct) cos(m sin(ωmt))) − sin(ωct) sin(m sin(ωmt)))x(t) is periodic in T =2πωm, therefore sin(m sin(ωmt)) is periodic in T .10−1sin(m sin(ωmt))t|bk|k0 10 20 304050 60m = 0Phase/Frequency ModulationFind the Fourier transform of a PM signal.x(t) = sin(ωmt)y(t) = cos(ωct + mx(t)) = cos(ωct + m
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