EE 102 spring 2001-2002 Handout #25Lecture 12Modulation and Sampling• The Fourier transform of the product of two signals• Modulation of a signal with a sinusoid• Sampling with an impulse train• The sampling theorem12–1Convolution and the Fourier transformsuppose f(t), g(t) have Fourier transforms F (ω), G(ω)the convolution y = f ∗ g of f and g is given byy(t)=∞−∞f(τ)g(t − τ) dτ(we integrate from −∞ to ∞ because f(t) and g(t) are not necessarilyzero for negative t)from the table of Fourier transform properties:Y (ω)=F (ω)G(ω)i.e.,convolution in the time domain corresponds to multiplication in thefrequency domainModulation and Sampling 12–2Multiplication and the Fourier transformthe Fourier transform of the producty(t)=f(t)g(t)is given byY (ω)=12π∞−∞F (λ)G(ω − λ)dλY =12π(F ∗ G)i.e.,multiplication in the time domain corresponds to convolution in thefrequency domainModulation and Sampling 12–3example:f(t)=e−|t|,F(ω)=21+ω2g(t)=cos20t, G(ω)=πδ(ω − 20) + πδ(ω + 20)the Fourier transform of y(t)=e−|t|cos 20t is given byY (ω)=12π∞−∞F (λ)G(ω − λ) dλ=12∞−∞F (λ)δ(ω − λ − 20) dλ +12∞−∞F (λ)δ(ω − λ + 20) dλ=12F (ω − 20) +12F (ω + 20)=11+(ω − 20)2+11+(ω + 20)2Modulation and Sampling 12–4−3 −2 −1 0 1 2 300.20.40.60.81tf(t)−30 −20 −10 0 10 20 3000.511.52ωF (ω)−3 −2 −1 0 1 2 3−1−0.500.51ty(t)−30 −20 −10 0 10 20 3000.20.40.60.81ωY (ω)Modulation and Sampling 12–5Sinusoidal amplitude modulation (AM)cos ω0tu(t) y(t)=u(t)cosω0tFourier transform of yY (ω)=12πU(ω) ∗ (πδ(ω − ω0)+πδ(ω + ω0))=12U(ω − ω0)+12U(ω + ω0)• cos ω0t is the carrier signal• y(t) is the modulated signal• the Fourier transform of the modulated signal is the Fourier transformof the input signal, shifted by ±ω0Modulation and Sampling 12–6Sinusoidal amplitude modulationωU(ω)0ω0ω(1/2)U(ω−ω0)0πδ(ω−ω0)πδ(ω+ω0)(1/2)U(ω+ω0)*=Baseband SignalCarrierModulatedSignalModulation and Sampling 12–7example: u(t)=2+cost, ω0=200 2 4 6 8 1011.522.53tu(t)U(ω)=4πδ(ω)+πδ(ω−1)+πδ(ω+1)−10 1π4ππω0 2 4 6 8 10−3−2−10123ty(t)Y (ω)=12U(ω − 20) +12U(ω + 20)−21 −19 01921π/22ππ/2π/22ππ/2ωModulation and Sampling 12–8demodulationcos ω0tz(t)lowpassfiltery(t)=u(t)cosω0tu(t)Fourier transform of y and z:Y (ω)=12U(ω − ω0)+12U(ω + ω0)Z(ω)=12πY (ω) ∗ (πδ(ω − ω0)+πδ(ω + ω0))=12Y (ω − ω0)+12Y (ω + ω0)=14U(ω − 2ω0)+12U(ω)+14U(ω +2ω0)if U is bandlimited, we can eliminate the 1st and 3rd term by lowpassfilteringModulation and Sampling 12–9Sinusoidal amplitude demodulationω0πδ(ω−ω0)πδ(ω+ω0)*=ω(1/2)U(ω)0ω(1/2)U(ω−ω0)0(1/2)U(ω+ω0)(1/4)U(ω−2ω0)(1/4)U(ω+2ω0)LowpassFilterModulatedSignalDemodulationCosineDemodulated SignalModulation and Sampling 12–10ApplicationSuppose for example that u(t) is an audio signal (frequency range10Hz − 20kHz)We rather nottransmitu directly using electromagnetic waves:• the wavelength is several 100 km, so we’d need very large antennas• we’d be able to transmit only one signal at a time• the Navy communicates with submerged submarines in this bandModulating the signal with a carrier signal with frequency 500 kHz to 5GHz:• allows us to transmit and receive the signal• allows us to transmit many signals simultaneously (frequency divisionmultiplexing)Modulation and Sampling 12–11Sampling with an impulse trainMultiply a signal x(t) with a unit impulse train with period Tx(t) y(t)p(t)=∞k=−∞δ(t − kT )Sampled signal: y(t)=x(t)∞k=−∞δ(t − kT)=∞k=−∞x(kT)δ(t − kT)tx(t)ty(t)T 2T(a train of impulses with magnitude . . . , x(−T ), x(0), x(T ), x(2T ),...)Modulation and Sampling 12–12The Fourier transform of an impulse traintrain of unit impulses with period T : p(t)=∞k=−∞δ(t − kT)t0 T2T3T 4T5T6T−T1Fourier transform (from table): P (ω)=2πT∞k=−∞δ(ω −2πkT)ω02πT4πT6πT8πT10πT20πT−2πT2π/TModulation and Sampling 12–13Consequences of Samplingtttω << 2π/Tω = 2π/Tω = 4π/T3Τ 4ΤΤ 2Τ0Τ 2Τ03Τ 4ΤΤ 2Τ03Τ 4Τ• Frequencies well below the sampling rate (ω<<2π/T)are“sampled”in the sense we expect.• Frequencies at multiples of the sampling rate (ω =2πn/T)looklikethey are constant. We can’t tell them from DC. These frequencies“alias” as DC.Modulation and Sampling 12–14Frequency domain interpretation of samplingThe Fourier transform of the sampled signal isY =12π(X ∗ P ),i.e.,theconvolution of X with P (ω)=2πT ∞k=−∞δ(ω −2πkT)Y (ω)=12π∞−∞X(λ)P (ω − λ) dλ=1T∞−∞X(λ)∞k=−∞δ(ω − λ −2πkT)dλ=1T∞k=−∞∞−∞X(λ)δ(ω − λ −2πkT) dλ=1T∞k=−∞X(ω −2πkT)Modulation and Sampling 12–15example: sample x(t)=e−|t|at different rates−3 −2 −1 0 1 2 300.20.40.60.81tx(t)−30 −20 −10 0 10 20 3000.511.52ωX(ω)x sampled with T =1(2π/T =6.3)−3 −2 −1 0 1 2 300.20.40.60.81ty(t)−30 −20 −10 0 10 20 3000.511.52ωY (ω)Modulation and Sampling 12–16x sampled with T =0.5 (2π/T =12.6)−3 −2 −1 0 1 2 300.20.40.60.81ty(t)−30 −20 −10 0 10 20 3000.511.522.533.54ωY (ω)x sampled with T =0.2 (2π/T =31.4)−3 −2 −1 0 1 2 300.20.40.60.81ty(t)−50 0 500246810ωY (ω)Modulation and Sampling 12–17The sampling theoremcan we recover the original signal x from the sampled signal y ?example: a band-limited signal x (with bandwidth W )ωX(ω)1W−WFourier transform of y(t)= kx(kT)δ(t − kT):ωY (ω)1/T2π/T 4π/TW−WModulation and Sampling 12–18suppose wefiltery through an ideal lowpass filter with cutoff frequency ωc,i.e.,wemultiply Y (ω) withH(ω)=1 −ωc≤ ω ≤ ωc0 |ω|≥ωcif W ≤ ωc≤ 2π/T − W ,thenthe result is H(ω)Y (ω)=X(ω)/T , i.e.,werecover X exactlyωY (ω)H(ω)1/T2π/T4π/TW−WY (ω)H(ω)1/TW−WModulation and Sampling 12–19same signal, sampled with T = π/WωY (ω)2π/T4π/Twe can still recover X(ω) perfectly by lowpass filtering with ωc= Wsample with T>π/WωY (ω)2π/T 4π/TX(ω) cannot be recovered from Y (ω) by lowpass filteringModulation and Sampling 12–20the sampling theoremsuppose x is a band-limited signal with bandwidth W , i.e.,X(ω)=0for |ω| >Wand we sample at a rate 1/Ty(t)=∞k=−∞x(kT)δ(t − kT)then we can recover x from y if T ≤ π/W• the sampling