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MIT 6 003 - Sampling and Quantization

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6.003: Signals and SystemsSampling and QuantizationApril 29, 2010What to do with a billion transistorsGene Frantz, Texas InstrumentsSeminar, today, 32-155, 4pmWe are getting closer to a time when we will be able to cost ef-fectively integrate billions of transistors on an integrated circuit. Infact, we are seeing the beginning of this era with the broad adoptionof multi-processing system-on-chips, which has both advantages anddisadvantages that should be considered. This talk will discuss theoptions we have, the issues we must face and the future we can lookforward to.Last Time: 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 webLast Time: SamplingTheorySampling:x(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).PracticeAliasing → anti-aliasing filterTodayDigital recording, transmission, storage, and retrieval requires dis-crete representations of both time (e.g., sampling) and amplitude.• audio: MP3, CD, cell phone• pictures: digital camera, printer• video: DVD• everything on the webQuantization: discrete representations for amplitudesQuantizationWe measure discrete amplitudes in bits.-1-10011Input voltageOutput voltage2 bits 3 bits 4 bits0001100 0.5 1-101Time (second)0 0.5 1Time (second)0 0.5 1Time (second)-101Input voltage-101Input voltageBit rate = (# bits/sample)×(# samples/sec)Check YourselfWe hear sounds that range in amplitude from 1,000,000 to 1.How many bits are needed to represent this range?1. 5 bits2. 10 bits3. 20 bits4. 30 bits5. 40 bitsCheck YourselfHow many bits are needed to represent 1,000,000:1?bits range1 22 43 84 165 326 647 1288 2569 51210 1, 02411 2, 04812 4, 09613 8, 19214 16, 38415 32, 76816 65, 53617 131, 07218 262, 14419 524, 28820 1, 048, 576Check YourselfWe hear sounds that range in amplitude from 1,000,000 to 1.How many bits are needed to represent this range? 31. 5 bits2. 10 bits3. 20 bits4. 30 bits5. 40 bitsQuantization DemonstrationQuantizing Music• 16 bits/sample• 8 bits/sample• 6 bits/sample• 4 bits/sample• 3 bits/sample• 2 bit/sampleJ.S. Bach, Sonata No. 1 in G minor Mvmt. IV. PrestoNathan Milstein, violinQuantizationWe measure discrete amplitudes in bits.-1-10011Input voltageOutput voltage2 bits 3 bits 4 bits0001100 0.5 1-101Time (second)0 0.5 1Time (second)0 0.5 1Time (second)-101Input voltage-101Input voltageExample: audio CD2 channels × 16bitssample× 44, 100samplessec× 60secmin× 74 min ≈ 6.3 G bits≈ 0.78 G bytesQuantizing ImagesConverting an image from a continuous representation to a discreterepresentation involves the same sort of issues.This image has 280 × 280 pixels, with brightness quantized to 8 bits.Quantizing Images8 bit image 7 bit imageQuantizing Images8 bit image 6 bit imageQuantizing Images8 bit image 5 bit imageQuantizing Images8 bit image 4 bit imageQuantizing Images8 bit image 3 bit imageQuantizing Images8 bit image 2 bit imageQuantizing Images8 bit image 1 bit imageCheck YourselfWhat is the most objectionable artifact of coarse quantization?8 bit image 4 bit imageDitheringOne very annoying artifact is banding caused by clustering of pixelsthat quantize to the same level.Banding can be reduced by dithering.Dithering: adding a small amount (±12quantum) of random noise tothe image before quantizing.Since the noise is different for each pixel in the band, the noisecauses some of the pixels to quantize to a higher value and some toa lower. But the average value of the brightness is preserved.Quantizing Images with Dither7 bit image 7 bits with ditherQuantizing Images with Dither6 bit image 6 bits with ditherQuantizing Images with Dither5 bit image 5 bits with ditherQuantizing Images with Dither4 bit image 4 bits with ditherQuantizing Images with Dither3 bit image 3 bits with ditherQuantizing Images with Dither2 bit image 2 bits with ditherQuantizing Images with Dither1 bit image 1 bit with ditherCheck YourselfWhat is the most objectionable artifact of dithering?3 bit image 3 bit dithered imageRobert’s TechniqueOne annoying feature of dithering is that it adds noise.The noise can be reduced using Robert’s technique.Robert’s technique: add a small amount (±12quantum) of randomnoise before quantizing, then subtract that same amount of randomnoise.Quantizing Images with Robert’s Method7 bits with dither 7 bits with Robert’s methodQuantizing Images with Robert’s Method6 bits with dither 6 bits with Robert’s methodQuantizing Images with Robert’s Method5 bits with dither 5 bits with Robert’s methodQuantizing Images with Robert’s Method4 bits with dither 4 bits with Robert’s methodQuantizing Images with Robert’s Method3 bits with dither 3 bits with Robert’s methodQuantizing Images with Robert’s Method2 bits with dither 2 bits with Robert’s methodQuantizing Images with Robert’s Method1 bits with dither 1 bit with Robert’s methodQuantizing Images: 3 bits8 bits 3 bitsdither Robert’sQuantizing Images: 2 bits8 bits 2 bitsdither Robert’sQuantizing Images: 1 bit8 bits 1 bitdither Robert’sProgressive RefinementTrading precision for speed.Start by sending a crude representation, then progressively updatewith increasing higher fidelity versions.Discrete-Time Sampling (Resampling)DT sampling is much like CT sampling.×x[n]xp[n]p[n] =Pkδ[n − kN]nxp[n]0nx[n]0np[n]0Discrete-Time SamplingAs in CT, sampling introduces additional copies of X(ejΩ).×x[n]xp[n]p[n] =Pkδ[n − kN]02π34π32π−2π3−4π3−2π13ΩXp(ejΩ)0 2π−2π1ΩX(ejΩ)02π34π32π−2π3−4π3−2π2π3ΩP (ejΩ)Discrete-Time SamplingSampling a finite sequence gives rise to a shorter sequence.nxb[n]0nxp[n]0nx[n]0Xb(ejΩ) =Xnxb[n]e−jΩn=Xnxp[3n]e−jΩn=Xkxp[k]e−jΩk/3= Xp(ejΩ/3)Discrete-Time SamplingBut the shorter sequence has a wider frequency representation.02π34π32π−2π3−4π3−2π13ΩXb(ejΩ) = Xp(ejΩ/3)00 2π−2π13ΩXp(ejΩ)2π−2π1ΩX(ejΩ)0Discrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingDiscrete-Time SamplingInsert zeros between samples to upsample the images.nxb[n]0nxp[n]0nx[n]0Xb(ejΩ) =Xnxb[n]e−jΩn=Xnxp[3n]e−jΩn=Xkxp[k]e−jΩk/3=


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