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MIT 6 003 - Lecture Notes

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6.003: Signals and SystemsSamplingNovember 24, 2009SamplingConversion of a continuous-time signal to discrete time.tx(t)0 246 8 10nx[n]0 246 8 10We have used sampling a number of times before.Today: new insights from Fourier representations.SamplingSampling allows the use of modern digital electronics to process,record, transmit, store, and retrieve CT signals.• audio: MP3, CD, cell phone• pictures: digital camera, printer• video: DVD• everything on the webSamplingSampling is pervasive.Example: digital cameras record sampled images.xyI(x, y)mnI[m, n]SamplingPhotographs in newsprint are “half-tone” images. Each point isblack or white and the average conveys brightness.SamplingZoom in to see the binary pattern.SamplingEven high-quality photographic paper records discrete images. WhenAgBr crystals (0.04 − 1.5µm) are exposed to light, some of the Agis reduced to metal. During “development” the exposed grains arecompletely reduced to metal and unexposed grains are removed.SamplingEvery image that we see is sampled by the retina, which contains ≈100 million rods and 6 million cones (average spacing ≈ 3µm) whichact as discrete sensors.http://webvision.med.utah.edu/imageswv/sagschem.jpegCheck YourselfYour retina is sampling this slide, which is composed of 1024×768pixels.Is the spatial sampling done by your rods and cones ade-quate to resolve individual pixels in this slide?Check YourselfThe spacing of rods and cones limits the angular resolution of yourretina to approximatelyθeye=rod/cone spacingdiameter of eye≈3 × 10−6m3 cm≈ 10−4radiansThe angle between pixels viewed from the center of the classroomis approximatelyθpixels=screen size / 1024distance to screen≈3 m/102410 m≈ 3 × 10−4radiansLight from a single pixel falls upon multiple rods and cones.SamplingHow does sampling affect the information contained in a signal?SamplingWe would like to sample in a way that preserves information, whichmay not seem possible.tx(t)Information between samples is lost. Therefore, the same samplescan represent multiple signals.tcos7π3n? cosπ3n?Sampling and ReconstructionTo determine the effect of sampling, compare the original signal x(t)to the signal xp(t) that is reconstructed from the samples x[n].Uniform sampling (sampling interval T ).tnx[n] = x(nT )Impulse reconstruction.tnxp(t) =Xnx[n]δ(t − nT )SamplingImpulse reconstuction produces a signal xp(t) that is equal to theoriginal signal x(t) multiplied by an impulse train.xp(t) =∞Xn=−∞x[n]δ(t − nT )=∞Xn=−∞x(nT )δ(t − nT )=∞Xn=−∞x(t)δ(t − nT )= x(t)∞Xn=−∞δ(t − nT )| {z }≡ p(t)xp(t) is motivated by impulse reconstruction (top line)– can be understood entirely within CT framework (bottom line)SamplingMultiplication by an impulse train in time is equivalent to convolutionby an impulse train in frequency.→ generates multiple copies of original frequency content.ωX(jω)−W W1ωP (jω)−ωsωs2πTωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωsωs=2πT1TCheck YourselfWhat is the relation between the DTFT of x[n] = x(nT )and the CTFT of xp(t) =Px[n]δ(t − nT ) for X(jω) below.ωX(jω)−W W11. Xp(jω) = X(ejΩ)|Ω=ω2. Xp(jω) = X(ejΩ)|Ω=ωT3. Xp(jω) = X(ejΩ)|Ω=ωT4. Xp(jω) = X(ejΩ)|Ω=ω5. none of the aboveCheck YourselfDTFTX(ejΩ) =∞Xn=−∞x[n]e−jΩnCTFT of xp(t)Xp(jω) =Z∞−∞∞Xn=−∞x[n]δ(t − nT )e−jωtdt=∞Xn=−∞x[n]Z∞−∞δ(t − nT )e−jωtdt=∞Xn=−∞x[n]e−jωnT= X(ejΩ)Ω=ωTCheck YourselfXp(jω) = X(ejΩ)Ω=ωTωX(jω)−W W1ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωsωs=2πT1TΩX(ejΩ) = Xp(jω)ω=ΩT−2π 2π1TCheck YourselfWhat is the relation between the DTFT of x[n] = x(nT )and the CTFT of xp(t) =Px[n]δ(t − nT ) for X(jω) below.ωP (jω)−ωsωs2πT1. Xp(jω) = X(ejΩ)|Ω=ω2. Xp(jω) = X(ejΩ)|Ω=ωT3. Xp(jω) = X(ejΩ)|Ω=ωT4. Xp(jω) = X(ejΩ)|Ω=ω5. none of the aboveSamplingThe high frequency copies can be removed with a low-pass filter(also multiply by T to undo the amplitude scaling).ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)1T−ωs2ωs2TImpulse reconstruction followed by ideal low-pass filtering is calledbandlimited reconstruction.The Sampling TheoremIf signal is bandlimited → sample without loosing information.If x(t) is bandlimited so thatX(jω) = 0 for |ω| > ωmthen x(t) is uniquely determined by its samples x(nT ) ifωs=2πT>2ωm.The minimum sampling frequency, 2ωm, is called the “Nyquist rate.”SummaryThree important ideas.Samplingx(t) → x[n] = x(nT )Bandlimited Reconstruction−ωs2ωs2ωTx[n] xr(t)ImpulseReconstructionxp(t) =Px[n]δ(t − nT )LPFSampling Theorem: If X(jω) = 0 ∀ |ω| >ωs2then xr(t) = x(t).Check YourselfWe can hear sounds with frequency components between 20 Hzand 20 kHz.What is the maximum sampling interval T that can be usedto sample a signal without loss of audible information?1. 100 µs 2. 50 µs3. 25 µs 4. 100π µs5. 50π µs 6. 25π µsCheck Yourself2πfm= ωm<ωs2=2π2TT <12fm=12 × 20 kHz= 25 µsCheck YourselfWe can hear sounds with frequency components between 20 Hzand 20 kHz.What is the maximum sampling interval T that can be usedto sample a signal without loss of audible information?1. 100 µs 2. 50 µs3. 25 µs 4. 100π µs5. 50π µs 6. 25π µsCT Model of Sampling and ReconstructionSampling followed by bandlimited reconstruction is equivalent tomultiplying by an impulse train and then low-pass filtering.−ωs2ωs2ωT×x(t)p(t)xr(t)xp(t)LPFtp(t) = ”sampling function”0TAliasingWhat happens if X contains frequencies |ω| >πT?ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωs2ωs21TωP (jω)−ωsωs2πTωX(jω)AliasingWhat happens if X contains frequencies |ω| >πT?ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωs2ωs21TωP (jω)−ωsωs2πTωX(jω)AliasingWhat happens if X contains frequencies |ω| >πT?ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωs2ωs21TωP (jω)−ωsωs2πTωX(jω)AliasingWhat happens if X contains frequencies |ω| >πT?ωXp(jω) =12π(X(j · ) ∗ P (j · ))(ω)−ωs2ωs21TωP (jω)−ωsωs2πTωX(jω)AliasingThe effect of aliasing is to wrap frequencies.ω−ωs2ωs21TωωX(jω)Input frequencyOutput frequencyωs2ωs2AliasingThe effect of aliasing is to wrap frequencies.ω−ωs2ωs21TωωX(jω)Input frequencyOutput frequencyωs2ωs2AliasingThe effect of aliasing is to wrap frequencies.ω−ωs2ωs21TωωX(jω)Input frequencyOutput


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MIT 6 003 - Lecture Notes

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