Image SamplingLecture Slide #10Week 5Sampling 2Sampling and Quantization• Spatial Resolution (Sampling)– Determines the smallest perceivable imagedetail.– What is the best sampling rate?• Gray-level resolution (Quantization)– Smallest discernible change in the gray levelvalue.– Is there an optimal quantizer?Sampling 3Image sampling and quantizationIn 2-Df(x,y)(Continuous image)Samplerfs(m,n)Quantizeru(m,n)To ComputerSampling 41-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)Sampling 52-D: Comb functionycomb(x,y;Δx,Δy)xΔxΔyComb ( , ; , ) ( , )x y x y x m x y n ynm! ! ! !" # #= #$$= #$$%%&Sampling 6Sampled Imagef x y f x y x y x yf m x n y x m x y n yx y x y u vx yu vx ysnm( , ) ( , ) ( , ; , )( , ) ( , )( , ; , ) ( , ), ; , )== ! !" #$ == !%%= !%%&'' comb comb COMB comb(( (( ( ( (( (( ( ( ()1 1 1Sampling 7Sampled SpectrumFs(u, v) = F(u, v) ! COMB (u,v)=1"x"yF#(u, v)k,l = $%%#!&u $k"x, v $l"y'()*+,=1"x"yF#k,l = $%%#u $k"x, v $l"y'()*+,Sampling 8Sampled Spectrum: ExampleSampling 9Bandlimited ImagesA function f(x,y) is said to be band limited if the FouriertransformF(u,v) = 0 for | u | > u0, | v | > v0u0, v0Band width of the image in the x- and y- directionsv0u0region of supportSampling 10Foldover FrequenciesSampling frequencies:Let us and vs be the sampling frequenciesThen us > 2u0 ; vs > 2v0 or Δ x <1/2u0 ; Δ y <1/2v0 Frequencies above half the sampling frequenciesare called fold over frequencies.Sampling 11Sampling TheoremA band limited image f(x,y) with F(u,v) as its Fourier transform;and F(u,v) = 0 |u| > u0 |v| > v0 ; and sampled uniformly on arectangular grid with spacing Δx and Δy, can be recoveredwithout error from the sample values f(m Δx, n Δy) provided thesampling rate is greater than the nyquist rate.i.e 1/ Δx = us > 2 u0, 1/ Δy = vs > 2 v0The reconstructed image is given by the interpolation formula:f x y f m x n yx u mx u my v ny v nm nssss( , ) ( , )( )( )( )( ),=!!!!""= !##$ $sin sin%%%%Sampling 12Reconstruction0R1R2δRSampling 13Reconstruction via LPFF(u,v) can be recovered by a LPF with H (u, v) =!x !y (u, v) "R0 Other wise#$%R is any region whose boundary &R is containedwithin the annular ring between the rectangles R1 andR2 in the figure. Reconstructed signal is!F(u , v) = H (u, v) Fs(u, v) = F(u , v)f (x, y) = '(1[F(u,v)]Sampling 14AliasingNote: If us and vs are 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 sampledimage. This is called aliasing.Sampling 15Sampling 16Example there will be aliasing.f (x, y) = 2 cos (2!(3x + 4y))F(u, v) ="(u # 3,v # 4) +"(u + 3, v + 4)$ u0= 3, v0= 4Let %x = %y = 0.2, $ us= vs=10.2= 5 < 2u0,<2v0Sampling 17Example:(contd.) Fs(u, v) = 25 F(u ! kus, v ! lvs"k,l = !##")= 25 [$(u ! 3 ! 5k, v ! 4 ! 5l) +$(u + 3 ! 5k, v + 4 ! 5l)]"k,l = !##"Let H(u,v ) =125! 2.5 % u % 2.5, ! 2.5 % u % 2.50 Otherwise &'() F(u, v) = H (u, v) Fs(u, v)=$(u + 2,v + 1) +$(u ! 2,v ! 1))!f (x, y) = 2 cos(2*(2x + y))Sampling 18ExamplesOriginal and the reconstructed image from samples.Sampling 19Another exampleSampling filtersampled imageSampling 20Aliasing Problems (real
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