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01_CM0340Tut_Frequency_Space

# 01_CM0340Tut_Frequency_Space - CM0340 Tutorial 1 Frequency...

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1 JJ II J I Back Close CM0340 Tutorial 1: Frequency Space Refer to Lecture notes for main point of reference (also recapped during this tutorial) Some additional explanation here: DCT revisited and practical computation explained (MATLAB) Demo MATLAB JPEG implementation discussed.

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2 JJ II J I Back Close DCT Revisited DCT (Discrete Cosine Transform) is actually a cut-down version of the Fourier Transform or the Fast Fourier Transform (FFT): For N data items 1D DCT is defined by: F ( u ) = 2 N 1 2 . Λ( u ) N - 1 X i =0 cos h π.u 2 .N (2 i + 1) i f ( i ) where Λ( ξ ) = 1 2 for ξ = 0 1 otherwise
3 JJ II J I Back Close DCT Example Let’s consider a DC signal that is a constant 100, i.e f ( i ) = 100 for i = 0 . . . 7 (see DCT1Deg.m ): So the domain is [0 , 7] for both i and u We therefore have N = 8 samples and will need to work 8 values for u = 0 . . . 7 . We can now see how we work out F ( u ) : As u varies we work can work for each u a component or a basis . F ( u ) . Within each F ( u ) , we cam work out the value for each F i ( u ) to define a basis function Basis function can be pre-computed and simply looked up in DCT computation.

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4 JJ II J I Back Close f ( i ) and F ( U ) Plots 1 2 3 4 5 6 7 8 0 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 0 50 100 150 200 250 300 f ( i ) = 100 for i = 0 . . . 7 F ( u ) : F (0) 283 , F (1 . . . 7) = 0
5 JJ II J I Back Close DCT Example: F (0) So for u = 0 : Note: Λ(0) = 1 2 and cos(0) = 1 So F (0) is computed as: F (0) = 1 2 2 (1 . 100 + 1 . 100 + 1 . 100 + 1 . 100 + 1 . 100 +1 . 100 + 1 . 100 + 1 . 100) 283 Here the values F i (0) = 1 2 2 ( i = 0 . . . 7 ). These are a bases of F i (0)

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6 JJ II J I Back Close F (0) Basis Function Plot 1 2 3 4 5 6 7 8 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 F (0) Basis Function
7 JJ II J I Back Close DCT Example: F (1 . . . 7) So for u

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01_CM0340Tut_Frequency_Space - CM0340 Tutorial 1 Frequency...

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