It is found to be appropriate and convenient to model the
distribution of many types of transformed image coefficients by
Laplacian distributions. It is appropriate because much real
data is approximately modeled by the Laplacian probability
density function (PDF), and it is convenient because the
mathematical form of the Laplacian PDF is simple enough to allow
some useful analytical results to be derived.
A Laplacian PDF is a back-to-back pair of exponential decays and
is given by:
px=12x0ⅇ-|x|x0
p
x
1
2
x0
x
x0
(1)
where
x0
x0
is the equivalent of a
time
constant which defines the
width
of the PDF from the centre to the
1ⅇ
1
points. The initial scaling factor ensures that the
area under
px
p
x
is unity, so that it is a valid PDF.
Figure 1 shows the shape of
px
p
x
.
The mean of this PDF is zero and the variance is given by:
vx0=∫-∞∞x2pxdx=2∫x22x0ⅇ-xx0dx=2x02
v
x0
x
x
2
p
x
2
x
x
2
2
x0
x
x0
2
x0
2
(2)
(using integration by parts twice).
Hence the standard deviation is:
σx0=vx0=2x0
σ
x0
v
x0
2
x0
(3)
Given the variance (power) of a subimage of transformed pels, we
may calculate
x0
x0
and hence determine the PDF of the subimage, assuming
a Laplacian shape. We now show that, if we quantise the subimage
using a uniform quantiser with step size
QQ, we can calculate the entropy of
the quantised samples and thus estimate the bit rate needed to
encode the subimage in bits/pel. This is a powerful analytical
tool as it shows how the compressed bit rate relates directly to
the energy of a subimage. The vertical dashed lines in
Figure 1 show the decision thresholds
for a typical quantiser for the case when
Q=2x0
Q
2
x0
.
First we analyse the probability of a pel being quantised to
each step of the quantiser. This is given by the area under
px
p
x
between each adjacent pair of quantiser thresholds.
-
Probability of being at step 0,
p0=Pr-12Q<x<12Q=2Pr0<x<12Q
p0
1
2
Q
x
1
2
Q
2
0
x
1
2
Q
-
Probability of being at step kk,
pk=Prk−12Q<x<k+12Q
pk
k
1
2
Q
x
k
1
2
Q
First, for
x2≥x1≥0
x2
x1
0
, we calculate:
Prx1<x<x2=∫x1x2pxdx=-12ⅇ-xx0|x1x2=12ⅇ-x1x0−ⅇ-x2x0
x1
x
x2
x
x1
x2
p
x
x1
x2
1
2
x
x0
1
2
x1
x0
x2
x0
Therefore,
p0=1−ⅇ-Q2x0
p0
1
Q
2
x0
(4)
and, for
k≥1
k
1
,
pk=12ⅇ-k−12Qx0−ⅇ-k+12Qx0=sinhQ2x0ⅇ-kQx0
pk
1
2
k
1
2
Q
x0
k
1
2
Q
x0
Q
2
x0
k
Q
x0
(5)
By symmetry, if
kk is nonzero,
p
-
k
=pk=sinhQ2x0ⅇ-|k|Qx0
p
-
k
pk
Q
2
x0
k
Q
x0
Now we can calculate the entropy of the subimage:
H=-∑k=-∞∞pklog2pk=-p0log2p0−2∑k=1∞pklog2pk
H
k
pk
2
pk
p0
2
p0
2
k
1
pk
2
pk
(6)
To make the evaluation of the summation easier when we
substitute for
pk
pk
, we let
pk=αrk
pk
α
r
k
where
α=sinhQ2x0
α
Q
2
x0
and
r=ⅇ-Qx0
r
Q
x0
. Therefore,
∑k=1∞pklog2pk=∑k=1∞αrklog2αrk=∑k=1∞αrklog2α+klog2r=αlog2α∑k=1∞rk+αlog2r∑k=1∞krk
k
1
pk
2
pk
k
1
α
r
k
2
α
r
k
k
1
α
r
k
2
α
k
2
r
α
2
α
k
1
r
k
α
2
r
k
1
k
r
k
(7)
Now
∑k=1∞rk=r1−r
k
1
r
k
r
1
r
and, differentiating by
rr:
∑k=1∞krk−1=11−r2
k
1
k
r
k
1
1
1
r
2
. Therefore,
∑k=1∞pklog2pk=αlog2αr1−r+αlog2rr1−r2=αr1−rlog2α+log2r1−r
k
1
pk
2
pk
α
2
α
r
1
r
α
2
r
r
1
r
2
α
r
1
r
2
α
2
r
1
r
(8)
and
p0log2p0=1−rlog21−r
p0
2
p0
1
r
2
1
r
(9)
Hence the entropy is given by:
H=-1−rlog21−r−2αr1−rlog2α+log2r1−r
H
1
r
2
1
r
2
α
r
1
r
2
α
2
r
1
r
(10)
Because both
αα and
rr are functions of
Qx0
Q
x0
, then
HH is a function of
just
Qx0
Q
x0
too. We expect that, for constant
QQ, as the energy of the subimage
increases, the entropy will also increase approximately
logarithmically, so we plot
HH
against
x0Q
x0
Q
in dB in
Figure 2. This
shows that our expectations are born out.
We can show this in theory by considering the case when
x0Q≫1
≫
x0
Q
1
, when we find that:
α≈Q2x0
α
Q
2
x0
r≈1−Qx0≈1−2α
r
1
Q
x0
1
2
α
r≈1−α
r
1
α
Using the approximation
log21−ε≈-εln2
2
1
ε
ε
2
for small εε, it is
then fairly straightforward to show that
H≈-log2α+1ln2≈log22ⅇx0Q
H
2
α
1
2
2
2
x0
Q
We denote this approximation as
Ha
Ha
in Figure 2, which shows
how close to HH the approximation
is, for
x0>Q
x0
Q
(i.e. for
x0Q>0
x0
Q
0
dB).
We can compare the entropies calculated using Equation 10 with those that were calculated
from the bandpass subimage histograms, as given in these figures
describing Haar transform energies and entropies; level 1
energies, level 2 energies, level 3 energies, and level 4
energies. (The Lo-Lo subimages have PDFs which are more
uniform and do not fit the Laplacian model well.) The values of
x0
x0
are calculated from:
x0=std. dev.2=subimage energy2 (no of pels in subimage)
x0
std. dev.
2
subimage energy
2 (no of pels in subimage)
The following table shows this comparison:
Table 1
|
Transform level
|
Subimage type
|
Energy (×
106
10
6
)
|
No of pels
|
x0
x0
|
Laplacian entropy
|
Measured entropy
|
|
1
|
Hi-Lo
|
4.56
|
16384
|
11.80
|
2.16
|
1.71
|
|
1
|
Lo-Hi
|
1.89
|
16384
|
7.59
|
1.58
|
1.15
|
|
1
|
Hi-Hi
|
0.82
|
16384
|
5.09
|
1.08
|
0.80
|
|
2
|
Hi-Lo
|
7.64
|
4096
|
30.54
|
3.48
|
3.00
|
|
2
|
Lo-Hi
|
2.95
|
4096
|
18.98
|
2.81
|
2.22
|
|
2
|
Hi-Hi
|
1.42
|
4096
|
13.17
|
2.31
|
1.75
|
|
3
|
Hi-Lo
|
13.17
|
1024
|
80.19
|
4.86
|
4.52
|
|
3
|
Lo-Hi
|
3.90
|
1024
|
43.64
|
3.99
|
3.55
|
|
3
|
Hi-Hi
|
2.49
|
1024
|
34.87
|
3.67
|
3.05
|
|
4
|
Hi-Lo
|
15.49
|
256
|
173.9
|
5.98
|
5.65
|
|
4
|
Lo-Hi
|
6.46
|
256
|
112.3
|
5.35
|
4.75
|
|
4
|
Hi-Hi
|
3.29
|
256
|
80.2
|
4.86
|
4.38
|
We see that the entropies calculated from the energy via the
Laplacian PDF method (second column from the right) are
approximately 0.5 bit/pel greater than the entropies measured
from the Lenna subimage histograms. This is due to the heavier
tails of the actual PDFs compared with the Laplacian
exponentially decreasing tails. More accurate entropies can be
obtained if
x0
x0
is obtained from the mean absolute values of the pels
in each subimage. For a Laplacian PDF we can show that
Mean absolute value=∫-∞∞|x|pxdx=2∫0∞x2x0ⅇ-xx0dx=x0
Mean absolute value
x
x
p
x
2
x
0
x
2
x0
x
x0
x0
(11)
This gives values of
x0
x0
that are about 20% lower than those calculated from
the energies and the calculated entropies are then within
approximately 0.2 bit/pel of the measured entropies.