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Quantitative Example of Energy Compaction and Decorrelation

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# 17 ECE 253a Digital Image Processing Pamela Cosman 11/17/10<br />

<strong>Quantitative</strong> <strong>Example</strong> <strong>of</strong> <strong>Energy</strong> <strong>Compaction</strong> <strong>and</strong> <strong>Decorrelation</strong><br />

Let u be a 2 × 1 vector with mean zero:<br />

<strong>and</strong> covariance<br />

Ru =<br />

=<br />

=<br />

u =<br />

<br />

u(0)<br />

u(1)<br />

<br />

<br />

E[(u(0) − µ0)(u(0) − µ0)] E[(u(0) − µ0)(u(1) − µ1)]<br />

E[(u(0) − µ0)(u(1) − µ1)] E[(u(1) − µ1)(u(1) − µ1)]<br />

<br />

E[u(0) 2 ] E[u(0)u(1)]<br />

E[u(1)u(0)] E[u(1) 2 <br />

]<br />

<br />

1 ρ<br />

ρ 1<br />

for 0 < ρ < 1. From the expression for Ru, the variances<br />

<br />

σ 2 u(0) = σ 2 u(1) = 1<br />

that is, the total average energy <strong>of</strong> 2 is distributed equally between u(0) <strong>and</strong> u(1).<br />

The parameter ρ gives an indication <strong>of</strong> the correlation between u(0) <strong>and</strong> u(1). The correlation is<br />

by definition<br />

corr[u(0), u(1)] = E[u(0)u(1)]<br />

= ρ<br />

σu(0)σu(1)<br />

So in this case, the <strong>of</strong>f-diagonal elements <strong>of</strong> the covariance matrix are in fact exactly equal to the<br />

correlation, but that is only because the σ’s are both equal to one.<br />

Now we transform u as follows:<br />

The covariance matrix <strong>of</strong> v is then<br />

Rv =<br />

v = 1<br />

√<br />

3 1<br />

2 −1 √ 3<br />

<br />

<br />

u<br />

1 + √ 3ρ/2 ρ/2<br />

ρ/2 1 − √ 3ρ/2<br />

The entries <strong>of</strong> this matrix Rv are computed as follows:<br />

1


√<br />

3 1<br />

v(0) = u(0) +<br />

2 2 u(1)<br />

E[v(0) 2 ] = 3<br />

√<br />

1 3<br />

+ + 2<br />

4 4 4 ρ<br />

etc. Now if we compare the energy <strong>of</strong> the two coordinates:<br />

σ 2 v(0) = 1 + √ 3ρ/2<br />

σ 2 v(1) = 1 − √ 3ρ/2<br />

we see that the total average energy is still 2, but now the energy in v(0) is greater than in v(1). If<br />

ρ = 0.95, then 91.1% <strong>of</strong> the total average energy has been packed in the first sample.<br />

σ 2 v(0) = 1.82 σ2 v(1)<br />

The correlation between v(0) <strong>and</strong> v(1) is given by<br />

ρv(0,1) = E[v(0)v(1)]<br />

=<br />

σv(0)σv(1)<br />

ρ<br />

= 0.18<br />

2 <br />

(1 − 3<br />

4 ρ2 )<br />

which is less in absolute value than | ρ | for | ρ |< 1 For ρ = 0.95, we find ρv(0,1) = 0.83. Hence the<br />

correlation between the transform coefficients has been reduced.<br />

Suppose instead we use the transform matrix<br />

then we find<br />

A = 1 √ 2<br />

σ 2 v(0)<br />

σ 2 v(1)<br />

<br />

1 1<br />

1 −1<br />

= 1 + ρ<br />

= 1 − ρ<br />

ρv(0,1) = 0<br />

For ρ = 0.95, now 97.5% <strong>of</strong> the energy is packed into v(0), <strong>and</strong> the two components v(0) <strong>and</strong> v(1)<br />

are uncorrelated.<br />

2


DCT basis images for 8x8 blocks:<br />

Luminance Quantization Table:<br />

16 11 10 16 24 40 51 61<br />

12 12 14 19 26 58 60 55<br />

14 13 16 24 40 57 69 56<br />

14 17 22 29 51 87 80 62<br />

18 22 37 56 68 109 103 77<br />

24 35 55 64 81 104 113 92<br />

49 64 78 87 103 121 120 101<br />

72 92 95 98 112 100 103 99<br />

Chrominance Quantization Table:<br />

17 18 24 47 99 99 99 99<br />

18 21 26 66 99 99 99 99<br />

24 26 56 99 99 99 99 99<br />

47 66 99 99 99 99 99 99<br />

99 99 99 99 99 99 99 99<br />

99 99 99 99 99 99 99 99<br />

99 99 99 99 99 99 99 99<br />

99 99 99 99 99 99 99 99<br />

3

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