SC434L_DVCC-Tutorial 6 Subband Video Coding

Transcription

1 SC434L_DVCC-Tutorial 6 Subband Video Coding Dr H.R. Wu Associate Professor Audiovisual Information Processing and Digital Communications Monash University hrw@csse.monash.edu.au and Computer Engineering Nanyang Technological University ashrwu@ntu.edu.sg

2 T6. Downsampling and Upsampling[Strang & Nguyen, 996] A filter bank: x( ) x( ) x(0) x() x() H ( ) / The signal x[n] is split into two parts, yl[n] and yh[n], which can be compressed separately and efficiently. The signal can be reconstructed using a synthesis bank, which is called a perfect reconstruction (PR) filter bank if it reconstruct the input signal exactly. There exists redundancy in yl[n] and yh[n], and the downsampling is required, which is not an invertible process. H L H H ( ) ( ) / = + / yl ( ) yl ( ) yl(0) yl () yl() yh ( ) yh ( ) yh (0) yh () yh ()

3 [Downsampling:] DVCC_Tutorial-6 T6. Downsampling and Upsampling 3 To avoid doubling storage space to keep two full length output vectors, yl and yh, we downsample them by a factor of, (represented by ( ) ( down two )) removing every other component, which is called decimation, i.e., yl( ) yl( 4) yh( ) yh( 4) yl( ) yl( ) yh( ) yh( ) = yl(0) = yl(0) = = yh(0) = yh(0) = y yl() yl() yh() y () H yl() yl(4) yh() yh (4) ( ) yl ( ) ydl, or similarly ( ) yh ( ) dh i.e., ydl[n]=yl[n] and ydh[n]=yh[n], where n = 0, ±, ±,. (T6.-)

4 T6. Downsampling and Upsampling [Downsampling in matrix form:] Downsampling matrix is obtained by removing odd-numbered rows from the identity matrix I, i.e., 4 x( ) x( ) x( ) 0 0 x(0) = x(0) 0 0 x() x() x() (T6. -)

5 T6. Downsampling and Upsampling [Shannon Down-Sampling Theorem:] The halfband digital signal x[n] with its spectrum defined in Eq. (T6. -3) for 0 < ; X ( ) = 0 for <. (T6.-3) can be reconstructed from its downsampled signal, x, using Eq. (T6. -4): sin(( n k) / ) xn [ ] = + x[0]sinc( n) + x[]sinc( n ) + = x[ k] (T6.-4) where k = ( n k) / sin( n / ) sin c( n) =. n / ( ) The signal yl[n] or yh[n] can be reconstructed using downsampled ydl[n] or ydh[n], if yl[n] is bandlimited to the lower half-band or yh[n] the upper halfband: Y ( ) 0 ( ) 0 0 (T6.-5) L = for < or YH = for < 5

6 T6. Downsampling and Upsampling Downsampling in the frequency domain [Theorem T6.-:] Assume that the discrete-time Fourier transform (DTFT) of x is X ( ). The the discrete-time Fourier transform of x = x d is: ( ) X ( ) = X + X +. d (T6.-6) [Proof:] Construct a new vector u=(,x(-),0,x(0),0,x(),0, ) = ( (/)x(-) +(/)x(- ), (/)x(-) -(/)x(-), (/)x(0) +(/)x(0), (/)x() -(/)x(), (/)x() +(/)x(), ). It is easy to see u= x. The DTFT of u is Since xd[n]=u[n], the DTFT of xd is k n jk jn jn Xd( ) = xd[ n] e = u[ n] e = u[ k] e = U( ) n= n= k= = X X + + ( ) ( ) jn jn jn( + ) U( ) = DTFT( u) = une [ ] = xne [ ] + xne [ ] n= n= n= X( ) X( + ) = [ X( ) + X( + ) ] (T6.-7) (T6.-8) 6

7 T6. Downsampling and Upsampling Downsampling in the frequency domain-aliasing effect If X ( ) has period of, X ( ) has period of. The second term of d 4 X ( ), ie.., X /+ causes aliasing d ( ) X ( ) 7 0 X ( /) ( /+ ) X 0

8 T6. Downsampling and Upsampling [Upsampling:] Upsmapling performs zero-padding or places zeros into the odd-numbered components into given signal vector x, and the upsampled vector x = ( ) x u. The operation is called interpolation by a factor of, and defined as, x u x[ ] x [ ] u x[ ] 0 xu[ ] = ( ) x= ( ) x[0] = x[0] = xu[0] x[] 0 xu[] x[] xu[] (T6.-9) The transpose of downsampling is upsampling and vice versa: T ( ) = ( ). (T6.-0) 8

9 T6. Downsampling and Upsampling [Upsampling in matrix form:] Upsampling matrix is obtained by insert zeros into odd-numbered rows of the identity matrix I, i.e., x( ) 0 x( ) 0 0 x(0) = x(0) 0 x() 0 0 x() (T6.-)

10 T6. Downsampling and Upsampling The product of matrix by matrix is the identity matrix I, i.e., ( ) ( ) ( )( ) =I (T6.-) = (T6.-3)

11 T6. Downsampling and Upsampling However, the product of ( ) matrix by ( ) matrix is not the identity matrix I, i.e., (T6.-4) 0 = 0 ( )( ) ( )( ) I x[ ] 0 x[ ] x[0] x[0] = x= 0 x[] x[] 0 ( ) x and ( )( ) (T6.-5) (T6.-6)

12 T6. Downsampling and Upsampling Upsampling in the frequency domain [Theorem T6.-:] Assume that the discrete-time Fourier transform (DTFT) of x is X ( ). The the discrete-time Fourier transform of x = u ( ) x is: X ( ( ) (T6.-7) u ) = X. [Proof:] Since x u xu[ k] = x[ k] = ( ) x xu[k+ ] = 0, (T6.-8) the DTFT of xu is jn j k j(k + ) u( ) = u[ ] = u[ ] + u[ + ] n= k= k= xk [ ] 0 X x n e x k e x k e = xke = X k = ( ) jk [ ] (T6.-9)

13 T6. Downsampling and Upsampling Upsampling in the frequency domain-imaging effect If X ( ) has period of, X ( u ) has period of. The original spectrum is compressed into /. An image of the compressed spectrum appears next to it, causing imaging effect. X ( ) 3 0 Image Xu ( ) = X ( ) / 0 /

14 T6. Downsampling and Upsampling Upsampling immediately after after downsampling in the frequency domain [Theorem T6.-3:] Assume that the discrete-time Fourier transform (DTFT) of x is X ( ). The the discrete-time Fourier transform of x = r x is: [Proof:] Since DVCC_Tutorial-6 ( )( ) ( ) ( ) Xr ( ) = X + X +. ( ), where ( ) x = x x = x r d d And from Therorems T5.3- and, Xd ( ) = X X, + + (T6.-0) (T6.-) (T6.-) 4 Xr( ) = Xd ( ) = X ( ) + X ( + ). (T6.-3)

15 T6. Downsampling and Upsampling Upsampling immediately after downsampling in the frequency domain- Aliasing and Imaging effects X ( ) 5 0 X d ( /) X 0.5 ( ) X ( / ) + Aliasing 0 Imaging X r ( ) = X ( ) d / /

16 T6. Downsampling and Upsampling Upsampling and downsampling in the z-domain j Setting z = e in the DTFT, we have the z-transform of x[n]: n X( z) = x[ n] z 6 (T6.-4) The DTFT is z-transform evaluated on the unit circle,, in the z-plane. j( k) j/ / j j( ) z = e +, for k Z 0, ±, ±,... ; e = z ; e = z ; and e + = z. [Theorem T6.-4:] Assume that the z-transform of x is X( z). The the z-transform of x = xand x = xare: d ( ) u ( ) And, z-transform of x = r x is DVCC_Tutorial-6 n { } z = / / X d( z) = X ( z ) + X ( z ) and X u( z) = X z ( )( ) X ( z) = X ( z) + X ( z) r (T6.-5) (T6. -6)

17 T6. Downsampling and Upsampling Upsampling and downsampling by a factor of M In the time domain, xd = ( M ) xand xu = ( M ) x, lead to n x[ ] for M divides n; xd[ n] = x[ Mn] and xu[ n] = M 0 otherwise. In the frequency domain, ( M ) + + Xd ( ) = X + X + + X M M M M Xu ( ) = X ( M) 7 (T6.-7) (T6.-8) (T6.-9) ( )( ) For x x, r = ( M ) Xr ( ) = X ( ) + X X + M M M (T6.-30)

18 T6. Downsampling and Upsampling Upsampling and downsampling by a factor of M / M In the z-domain, M z ; [Theorem T6.-5:] / M M z ; + / M j/ M ze Assume that the z-transform of x is X( z). The the z-transform of xd = ( ) xand xu = ( ) x are: M / M jk/ M M Xd( z) = X ( z e ) and Xu( z) = X z M (T6.-3) And, z-transform of x = x is r DVCC_Tutorial-6 k = 0 ( )( ) j M j M M / ( )/ ( ) = ( ) + ( ) + + ( ) Xr z X z X ze X ze M 8 (T6.-3) NOTE: The decimator (or downsampling) causes aliases and the expander (or upsampling) causes images. To minimize the effects, we usually use decimation filters to suppress aliasing and interpolation filters to suppress imaging effects. x Decimation Filter xlf M xlfd M xlfr Interpolation Filter x R