6.003: Signals and Systems Lecture 22 April 29, 201016.003: 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)6.003: Signals and Systems Lecture 22 April 29, 20102Check 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 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.Check YourselfWhat is the most objectionable artifact of coarse quantization?8 bit image 4 bit imageDitheringDithering: 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.6.003: Signals and Systems Lecture 22 April 29, 20103Check YourselfWhat is the most objectionable artifact of dithering?3 bit image 3 bit dithered imageRobert’s TechniqueRobert’s technique: add a small amount (±12quantum) of randomnoise before quantizing, then subtract that same amount of randomnoise.Quantizing Images with Robert’s Method3 bits with dither 3 bits with Robert’s methodQuantizing Images: 3 bits8 bits 3 bitsdither 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]06.003: Signals and Systems Lecture 22 April 29, 20104Discrete-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 Sampling: Progressive RefinementJPEGExample: JPEG (“Joint Photographic Experts Group”) encodes im-ages by a sequence of transformations:• color encoding• DCT (discrete cosine transform): a kind of Fourier series• quantization to achieve perceptual compression (lossy)• Huffman encoding: lossless information theoretic codingWe will focus on the DCT and quantization of its components.• the image is broken into 8 × 8 pixel blocks• each block is represented by its 8 × 8 DCT coefficients• each DCT coefficient is quantized, using higher resolutions forcoefficients with greater perceptual importance6.003: Signals and Systems Lecture 22 April 29, 20105JPEGDiscrete cosine transform (DCT) is similar to a Fourier series, buthigh-frequency artifacts are typically smaller.Example: imagine coding the following 8 × 8 block.For a two-dimensional transform, take the transforms of all of therows, assemble those results into an image and then take the trans-forms of all of the columns of that image.JPEGPeriodically extend a row and represent it with a Fourier series.x[n] = x[n + 8]n0 8There are 8 distinct Fourier series coefficients.ak=18Xn=<8>x[n]e−jkΩ0n; Ω0=2π8JPEGDCT is based on a different periodic representation, shown below.y[n] = y[n + 16]n0 16Check YourselfWhich signal has greater high frequency content?x[n] = x[n + 8]n0 8y[n] = y[n + 16]n0 16JPEGPeriodic extension of an 8 × 8 pixel block can lead to a discontinuousfunction even when the “block” was taken from a smooth image.original rown08 pixel ”block”n0x[n] = x[n + 8]n0 8JPEGPeriodic extension of the type done for JPEG generates a continuousfunction from a smoothly varying image.original rown08 pixel ”block”n0y[n] = y[n + 16]n0 166.003: Signals and Systems Lecture 22 April 29, 20106JPEGAlthough periodic in N = 16, y[n] can be represented by just 8 distinctDCT coefficients.y[n] = y[n + 16]n0 16bk=7Xn=0y[n] cosπkNn +12This results because y[n] is symmetric about n = −12, and this sym-metry introduces
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