C. A. Bouman: Digital Image Processing - January 20, 2021 1 1-D Rate Conversion • Decimation – 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. C. A. Bouman: Digital Image Processing - January 20, 2021 2 1-D Periodic Subsampling • Time domain subsampling of x(n) with period D y(n)= x(Dn) x(n)D y(n) • Frequency domain representation 1 D−1 Y (ejω)= X ej(ω−2πk)/D D Xk=0 • Problem: Frequencies above π/D will alias. • Example when D =2 aliased aliased signal signal ω −π −π/2 π/2 π • Solution: Remove frequencies above π/D. C. A. Bouman: Digital Image Processing - January 20, 2021 3 Decimation System ω x(n) H(e j ) D y(n) • Apply the filter H(ejω) to remove high frequencies – For |ω| <π Dω H(ejω)= rect 2π – For all ω ∞ ω − k2π H(ejω) = rect D 2π kX=−∞ = prect2π/D(ω) – Impulse response 1 h(n)= sinc(n/D) D • Frequency domain representation 1 D−1 Y (ejω)= H ej(ω−2πk)/D X ej(ω−2πk)/D D Xk=0 C. A. Bouman: Digital Image Processing - January 20, 2021 4 Graphical View of Decimation for D =2 • Spectral content of signal X ejω 1 ω −π −π/2π/2π 3π/2 2π 5π/2 • Spectral content of filtered signal H ejω X ejω 1 ω −π −π/2π/2π 3π/2 2π 5π/2 • Spectral content of decimated signal 1 D−1 Y (ejω)= H ej(ω−2πk)/D X ej(ω−2πk)/D D Xk=0 1/D ω −2π −ππ 2π 3π 4π 5π C. A. Bouman: Digital Image Processing - January 20, 2021 5 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 1 h(m,n)= sinc(m/D)sinc(n/D) D2 • Problems: – Filter has infinite extent. – Filter is not strictly positive. C. A. Bouman: Digital Image Processing - January 20, 2021 6 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 1 h(m,n)= sinc(m/D, n/D) D2 – Advantages/Disadvantages: ∗ Optimal, if signal is band limited... ∗ High computation C. A. Bouman: Digital Image Processing - January 20, 2021 7 Decimation Filters 2−D Filter Impulse Response used for Block Averaging 1 0.8 0.6 0.4 0.2 0 40 30 35 30 20 25 20 15 10 10 5 0 0 Block averaging filter 2−D Sinc Filter used for Ideal Subsampling 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 −0.01 40 30 35 30 20 25 20 15 10 10 5 0 0 Sinc filter C. A. Bouman: Digital Image Processing - January 20, 2021 8 Original Image • Full resolution C. A. Bouman: Digital Image Processing - January 20, 2021 9 Image Decimation by 4 using Subsampling • Severe aliasing C. A. Bouman: Digital Image Processing - January 20, 2021 10 Image Decimation by 8 using Subsampling • More severe aliasing C. A. Bouman: Digital Image Processing - January 20, 2021 11 Image Decimation by 4 using Block Averaging • Sharp, but with some aliasing C. A. Bouman: Digital Image Processing - January 20, 2021 12 Image Decimation by 4 using Sinc Filter • Theoretically optimal, but not necessarily the best visual quality C. A. Bouman: Digital Image Processing - January 20, 2021 13 1-D Up-Sampling • Up-sampling by L x(n/L) if n = KL for some K y(n)= 0 otherwise or the alternative form ∞ y(n)= x(m)δ(n − mL) mX=−∞ • Example for L =3 Discrete time signal x(n) 1 0 −1 0 5 10 15 20 x(n) up−sampled by 3 1 0 −1 0 10 20 30 40 50 60 C. A. Bouman: Digital Image Processing - January 20, 2021 14 Up-Sampling in the Frequency Domain • Up-sampling by L ∞ y(n)= x(m)δ(n − mL) mX=−∞ • In the frequency domain Y (ejω)= X(ejωL) • Example for L =3 DTFT of x(n) 15 10 5 0 −4 −2 0 2 4 DTFT of y(n) 15 10 5 0 −4 −2 0 2 4 C. A. Bouman: Digital Image Processing - January 20, 2021 15 Interpolation in the Frequency Domain ω x(n) L H(e j ) y(n) • Interpolating filter has the form Y (ejω)= H(ejω) X(ejωL) • Example for L =3 DTFT of y(n) 15 10 5 0 −4 −2 0 2 4 Interpolated Signal in Frequency 40 20 0 −4 −2 0 2 4 C. A. Bouman: Digital Image Processing - January 20, 2021 16 Interpolating Filter ω x(n) L H(e j ) y(n) • In the frequency domain jω H(e )= Lprect2π/L(ω) • In the time domain n h(n)= sinc L Interpolating filter in Time 1 0.5 0 −0.5 0 10 20 30 40 50 60 Interpolating filter in Frequency 3 2 1 0 −4 −2 0 2 4 C. A. Bouman: Digital Image Processing - January 20, 2021 17 Interpolation in the Time Domain ω x(n) L H(e j ) y(n) • In the frequency domain Y (ejω)= X(ejωL) • Example for L =3 Discrete time signal x(n) 1 0 −1 0 5 10 15 20 Interpolated Signal in Time 1 0 −1 0 10 20 30 40 50 60 C. A. Bouman: Digital Image Processing - January 20, 2021 18 2-D Interpolation x(n) (L,L) H(e j υ ,e j µ ) y(n) • Up-sampling ∞ ∞ y(m,n)= x(k,l)δ(m − kL, n − lL) kX=−∞ l=X−∞ • In the frequency domain jµ jν H(e ,e )= prect2π/L(µ)prect2π/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 C. A. Bouman: Digital Image Processing - January 20, 2021 19 Pixel Replication x(n) (L,L) H(e j υ ,e j µ ) y(n) • Impulse response of filter 1 for 0 ≤ m ≤ L − 1 and 0 ≤ n ≤ L − 1 h(m,n)= 0 otherwise • Replicates each pixel L2 times C. A. Bouman: Digital Image Processing - January 20, 2021 20 Bilinear Interpolation x(n) (L,L) H(e j υ ,e j µ ) y(n) • Impulse response of filter h(m,n)=Λ(m/L)Λ(n/L) • Results in linear interpolation of intermediate pixels.