1-D Rate Conversion. Decimation

Transcription

1 C. A. Bouman: Digital Image Processing - January 8, Decimation 1-D Rate Conversion Reduce the sampling rate of a discrete-time signal. Low sampling rate reduces storage and computation requirements. Interpolation Increase the sampling rate of a discrete-time signal. Higher sampling rate preserves fidelity.

2 C. A. Bouman: Digital Image Processing - January 8, D Periodic Subsampling Time domain subsampling ofx(n) with period D y(n) = x(dn) x(n) D y(n Frequency domain representation Y(e jω ) = 1 D D 1 k= X (e j(ω 2πk)/D ) Problem: Frequencies aboveπ/d will alias. Example when D = 2 aliased signal aliased signal π π/2 π/2 π ω Solution: Remove frequencies aboveπ/d.

3 C. A. Bouman: Digital Image Processing - January 8, Decimation System jω x(n) H(e ) D y(n) Apply the filter H(e jω ) to remove high frequencies For ω < π For allω H(e jω ) = Impulse response H(e jω ) = rect k= rect = prect 2π/D (ω) ( ) Dω 2π h(n) = 1 D sinc(n/d) Frequency domain representation Y(e jω ) = 1 D D 1 k= H ( D ω k2π ) 2π ) ) (e j(ω 2πk)/D X (e j(ω 2πk)/D

4 C. A. Bouman: Digital Image Processing - January 8, Graphical View of Decimation ford = 2 Spectral content of signal X ( e jω) 1 π π/2 π/2 π 3π/2 2π 5π/2 ω Spectral content of filtered signal H ( e jω) X ( e jω) 1 π π/2 π/2 π 3π/2 2π 5π/2 ω Spectral content of decimated signal Y(e jω ) = 1 D 1 ) H (e j(ω 2πk)/D X D k= (e j(ω 2πk)/D ) 1/D 2π π π 2π 3π 4π 5π ω

5 C. A. Bouman: Digital Image Processing - January 8, Decimation for Images Extension to decimation of images is direct Apply 2-D Filter f(i,j) = h(i,j) x(i,j) Subsample result y(i,j) = f(di,dj) Ideal choice of filter is h(m,n) = 1 D 2sinc(m/D)sinc(n/D) Problems: Filter has infinite extent. Filter is not strictly positive.

6 C. A. Bouman: Digital Image Processing - January 8, Alternative Filters for Image Decimation Direct subsampling h(m,n) = δ(m,n) Advantages/Disadvantages: Low computation Excessive aliasing Block averaging h(m,n) = δ(m,n)+δ(m+1,n) +δ(m,n+1)+δ(m+1,n+1) Advantages/Disadvantages: Low computation Some aliasing Sinc function h(m,n) = 1 D 2sinc(m/D,n/D) Advantages/Disadvantages: Optimal, if signal is band limited... High computation

7 C. A. Bouman: Digital Image Processing - January 8, Decimation Filters 2 D Filter Impulse Response used for Block Averaging Block averaging filter 2 D Sinc Filter used for Ideal Subsampling Sinc filter

8 C. A. Bouman: Digital Image Processing - January 8, Original Image Full resolution

9 C. A. Bouman: Digital Image Processing - January 8, Image Decimation by 4 using Subsampling Severe aliasing

10 C. A. Bouman: Digital Image Processing - January 8, Image Decimation by 8 using Subsampling More severe aliasing

11 C. A. Bouman: Digital Image Processing - January 8, Image Decimation by 4 using Block Averaging Sharp, but with some aliasing

12 C. A. Bouman: Digital Image Processing - January 8, Image Decimation by 4 using Sinc Filter Theoretically optimal, but not necessarily the best visual quality

13 C. A. Bouman: Digital Image Processing - January 8, D Up-Sampling Up-sampling by L { x(n/l) if n = KL for somek y(n) = otherwise or the alternative form y(n) = x(m)δ(n ml) Example forl = 3 m= Discrete time signal x(n) x(n) up sampled by

14 C. A. Bouman: Digital Image Processing - January 8, Up-Sampling in the Frequency Domain Up-sampling by L y(n) = x(m)δ(n ml) m= In the frequency domain Y(e jω ) = X(e jωl ) Example forl = 3 15 DTFT of x(n) DTFT of y(n)

15 C. A. Bouman: Digital Image Processing - January 8, Interpolation in the Frequency Domain x(n) L jω H(e ) y(n) Interpolating filter has the form Example forl = 3 Y(e jω ) = H(e jω )X(e jωl ) 15 DTFT of y(n) Interpolated Signal in Frequency

16 C. A. Bouman: Digital Image Processing - January 8, Interpolating Filter x(n) L jω H(e ) y(n) In the frequency domain H(e jω ) = Lprect 2π/L (ω) In the time domain ( n h(n) = sinc L) 1 Interpolating filter in Time Interpolating filter in Frequency

17 C. A. Bouman: Digital Image Processing - January 8, Interpolation in the Time Domain x(n) L jω H(e ) y(n) In the frequency domain Example forl = 3 Y(e jω ) = X(e jωl ) 1 Discrete time signal x(n) Interpolated Signal in Time

18 C. A. Bouman: Digital Image Processing - January 8, D Interpolation x(n) (L,L) H(e jυ,e jµ ) y(n) Up-sampling y(m,n) = x(k,l)δ(m kl,n ll) k= l= In the frequency domain H(e jµ,e jν ) = prect 2π/L (µ)prect 2π/L (ν) h(m,n) = sinc(m/l,n/l) Problems: sinc function impulse response Infinite support; infinite computation Negative sidelobes; ringing artifacts at edges

19 C. A. Bouman: Digital Image Processing - January 8, Pixel Replication x(n) (L,L) H(e jυ,e jµ ) y(n) Impulse response of filter { 1 for m L 1 and n L 1 h(m,n) = otherwise Replicates each pixell 2 times

20 C. A. Bouman: Digital Image Processing - January 8, Bilinear Interpolation x(n) (L,L) H(e jυ,e jµ ) y(n) Impulse response of filter h(m, n) = Λ(m/L)Λ(n/L) Results in linear interpolation of intermediate pixels