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Consider the following orthogonal transform in @ = = AX, VI-2 A= √1-2² Show that the average coefficient variance is equal to ok, 1 -- T i.e. [E(03)+ E(0*)]=o Note: & is the coefficient vector, e = [0,0]T A orthogonal transform X the input vector note o is a vector 0 - [6.07T A is a 2 by 2 matrix. a

— Orthogonal Transformation A transform A is orthogonal if A+ = A. Orthogonal transforms are energy preserving, i.e. the sum of squares of the transformed sequence is the same as the sum of the squares of the original sequence. For Example O = AX 0(1) (2) O= : O(N) Ca = 8' = (4x)' AX = x 4'AX = YA-4x = 2x = x N-1 N-1 i=0 i=0 NOTE: DCT can obtained from the FFT: 2C(0) Žf(m)cos((2m +1)+7 N-1 DCT(u) = N 2N m=0 Let fe Sf(m) m = 0,1,..., N-1 m = N N + 1.... 2N-1 fp denotes padded 0 2N-1 sequence. Then, 2C(u) (2m +1)ur DCT(u) = Σf, m)cos N 2N (2N-1 -|(2m+1) DCT(u) = Re: Σ. (m)e N m=0 2C(u) 2N m=0 - JUN 2N-1 2C(u) DCT(u) = N i2mu 2N 2010 Ree 2N Ef (me m=0 2N-1 Σf(m)e 2N -j2 mu is a 2N element DFT m=0