Discrete Cosine Transform (DCT)

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Where C C 2 6 π 2π = 2cos ; C4 = 2cos = 8 8 3π π = 2cos = 2sin 8 8 The matrix L can be factored as: L = B1 x M x A1 x A2 x A3 with: 2 26
27 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − − − = 1 1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 B ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 2 6 4 6 2 4 C C C C C C M
Page 1 and 2: Discrete Cosine Transform (DCT) 1
Page 3 and 4: Two related transform are the discr
Page 5 and 6: Formulation: DCT is a linear and i
Page 7 and 8: DCT-II: f j = ∑ − n 1 k = 0 x k
Page 9 and 10: DCT-IV: f j = ∑ − n 1 k = 0 x k
Page 11 and 12: Inverse Transforms: IDCT-I is DCT-
Page 13 and 14: References: 1. Rao and Yip, Discre
Page 15 and 16: Then the 1-D DCT is applied to the
Page 17 and 18: 4 C u πu cos S 16 8 ( u) = 7 ∑ x
Page 19 and 20: We have the following DCT v.s. DFT
Page 21 and 22: Let X M = (f(0), f(1), f(2), f(3),
Page 23 and 24: F M = X M x (1/8)T M Because cosα
Page 25: 25 X M = F M x (8T M -1 ) = FM x L
Page 29 and 30: 29 So the IDFT can be performed by
Page 31 and 32: 31 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪
Page 33: If you count the operations you wil

Discrete Cosine Transform (DCT)

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