Sampling and QuantizationLecture #5January 21, 2002Jan 21, 2003Sampling2Sampling and Quantization• Spatial Resolution (Sampling)– Determines the smallest perceivable image detail.– What is the best sampling rate?• Gray-level resolution (Quantization)– Smallest discernible change in the gray level value.– Is there an optimal quantizer?Jan 21, 2003Sampling3Image sampling and quantizationIn 1-Df(x,y)(Continuous image)Samplerfs(m,n)Quantizeru(m,n)To ComputerJan 21, 2003Sampling41-D1−Dx(t)Time domainX(u)FrequencyTs(t)xs(t) = x(t) s(t) = Σ x(kt) δ (t-kT) s(t)1/T1/TXs(f)Jan 21, 2003Sampling52-D: Comb functionycomb(x,y;∆x,∆y)x∆x∆yComb( , ; , ) ( , )xy x y x m xy n ynm∆∆ ∆ ∆≅−−=−∞∞=−∞∞∑∑δJan 21, 2003Sampling6Sampled Imagef xy f xy xy x yfmxny x mxy nyxy x y uvxyuvxysnm(,) (,) (, ; , )(,)( , )(, ; , ) (,),; , )==−−←→==−∞∞=−∞∞ℑ∑∑ comb comb COMB comb(∆∆∆∆ ∆ ∆∆∆∆∆ ∆ ∆δ111Jan 21, 2003Sampling7Sampled SpectrumF uv Fuv uvxyFuv ukxvlyxyFukxvlysklkl(, ) (, ) (, )(, ) ,,,,=∗=∗−−FHGIKJ=−−FHGIKJ∑∑∑∑=−∞∞=−∞∞COMB11∆∆ ∆ ∆∆∆ ∆ ∆δJan 21, 2003Sampling8Sampled Spectrum: ExampleJan 21, 2003Sampling9Bandlimited ImagesA function f(x,y) is said to be band limited if the Fourier transformF(u,v) = 0 for | u | > u0, | v | > v0u0, v0Band width of the image in the x- and y- directionsv0u0region of supportJan 21, 2003Sampling10Foldover FrequenciesSampling frequencies:Let usand vsbe the sampling frequenciesThen us > 2u0 ;vs> 2v0 or ∆ x <1/2u0 ; ∆ y <1/2v0 Frequencies above half the sampling frequencies are called fold over frequencies.Jan 21, 2003Sampling11Sampling TheoremA band limited image f(x,y) with F(u,v) as its Fourier transform; and F(u,v) = 0 |u| > u0 |v| > v0 ; and sampled uniformly on a rectangular grid with spacing ∆∆∆∆x and ∆∆∆∆y, can be recovered without error from the sample values f(m ∆∆∆∆x, n ∆∆∆∆y) provided the sampling rate is greater than the nyquist rate.i.e 1/ ∆∆∆∆x = us > 2 u0, 1/ ∆∆∆∆y = vs > 2 v0The reconstructed image is given by the interpolation formula:fxy fm xn yxu mxu myv nyv nmnssss(,) ( , )()()()(),=−−−−∑∑=−∞∞∆∆sin sinππππJan 21, 2003Sampling12Reconstruction0R1R2δRJan 21, 2003Sampling13Reconstruction via LPFF(u,v) can be recovered by a LPF withHuvxy uv RRRRRFuvHuvFuvFuvfxy Fuvs(, )(, )~(,) (,) (,) (,)(,) [ (,)]=∈RST===ℑ−∆∆0121 Other wise is any region whose boundary is containedwithin the annular ring between the rectangles and in the figure. Reconstructed signal is∂Jan 21, 2003Sampling14AliasingNote: If us and vsare below the Nyquist rate, the periodicreplications will overlap, resulting in a distorted spectrum.This overlapping of successive periods of the spectrumcauses the foldover frequencies in the original image toappear as frequencies below us/2, vs/2 in the sampled image. This is called aliasing.Jan 21, 2003Sampling15Jan 21, 2003Sampling16Examplethere will be aliasing.fxy x yFuv u v u vuvxy uv u vss(,) ( ( ))(,) ( , ) ( , ),.,.=+=--+++fi= === fi== =<223434 3 43402102520000 cos Let ,< 2 pddDDJan 21, 2003Sampling17Example:(contd.)FLet Otherwise coss(, ) ( , )[( , ) ( , )](, )..,..(, ) (, ) (, )(,)(,)~(,) (2 (2 )),,uv Fu ku v lvukvlukvlHuvuuFuv Huv F uvuv uvfxy x yssklkls=−−=−−−−++−+−=−≤≤ −≤≤RST∴==+++−−∴= +∑∑∑∑=−∞∞=−∞∞2525 35 45 35 4525 25 25 25021 212125δδδδπJan 21, 2003Sampling18ExamplesOriginal and the reconstructed image from samples.Jan 21, 2003Sampling19Another exampleSampling filtersampled imageJan 21, 2003Sampling20AliasingProblems (real
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