This module introduces the use of Laplacian PDFs in image compression.
It is found to be appropriate and convenient to model the
distribution of many types of transformed image coefficients byLaplacian distributions. It is appropriate because much real
data is approximately modeled by the Laplacian probabilitydensity function (PDF), and it is convenient because the
mathematical form of the Laplacian PDF is simple enough to allowsome useful analytical results to be derived.
A Laplacian PDF is a back-to-back pair of exponential decays and
is given by:
where
${x}_{0}$ is the equivalent of a
time
constant which defines the
width of the PDF from the centre to the
$\frac{1}{e}$ points. The initial scaling factor ensures that the
area under
$p(x)$ is unity, so that it is a valid PDF.
shows the shape of
$p(x)$ .
The mean of this PDF is zero and the variance is given by:
Given the variance (power) of a subimage of transformed pels, we
may calculate
${x}_{0}$ and hence determine the PDF of the subimage, assuming
a Laplacian shape. We now show that, if we quantise the subimageusing a uniform quantiser with step size
$Q$ , we can calculate the entropy of
the quantised samples and thus estimate the bit rate needed toencode the subimage in bits/pel. This is a powerful analytical
tool as it shows how the compressed bit rate relates directly tothe energy of a subimage. The vertical dashed lines in
show the decision thresholds
for a typical quantiser for the case when
$Q=2{x}_{0}$ .
First we analyse the probability of a pel being quantised to
each step of the quantiser. This is given by the area under
$p(x)$ between each adjacent pair of quantiser thresholds.
Probability of being at step 0,
${p}_{0}=(-(\frac{1}{2}Q)< x< \frac{1}{2}Q)=2(0< x< \frac{1}{2}Q)$
Probability of being at step
$k$ ,
${p}_{k}=((k-\frac{1}{2})Q< x< (k+\frac{1}{2})Q)$
First, for
${x}_{2}\ge {x}_{1}\ge 0$ , we calculate:
$$({x}_{1}< x< {x}_{2})=\int_{{x}_{1}}^{{x}_{2}} p(x)\,d x=(, , -\left(\frac{1}{2}\right)e^{-\left(\frac{x}{{x}_{0}}\right)})=\frac{1}{2}(e^{-\left(\frac{{x}_{1}}{{x}_{0}}\right)}-e^{-\left(\frac{{x}_{2}}{{x}_{0}}\right)})$$ Therefore,
To make the evaluation of the summation easier when we
substitute for
${p}_{k}$ , we let
$${p}_{k}=\alpha r^{k}$$ where
$\alpha =\sinh \left(\frac{Q}{2{x}_{0}}\right)$ and
$r=e^{-\left(\frac{Q}{{x}_{0}}\right)}$ . Therefore,
Because both
$\alpha $ and
$r$ are functions of
$\frac{Q}{{x}_{0}}$ , then
$H$ is a function of
just
$\frac{Q}{{x}_{0}}$ too. We expect that, for constant
$Q$ , as the energy of the subimage
increases, the entropy will also increase approximatelylogarithmically, so we plot
$H$ against
$\frac{{x}_{0}}{Q}$ in dB in
. This
shows that our expectations are born out.
We can show this in theory by considering the case when
$\gg (\frac{{x}_{0}}{Q}, 1)$ , when we find that:
$$\alpha \approx \frac{Q}{2{x}_{0}}$$$$r\approx 1-\frac{Q}{{x}_{0}}\approx 1-2\alpha $$$$\sqrt{r}\approx 1-\alpha $$ Using the approximation
$\log_{2}(1-\epsilon )\approx -\left(\frac{\epsilon}{\ln 2}\right)$ for small
$\epsilon $ , it is
then fairly straightforward to show that
$$H\approx -\log_{2}\alpha +\frac{1}{\ln 2}\approx \log_{2}\left(\frac{2e{x}_{0}}{Q}\right)$$ We denote this approximation as
${H}_{a}$ in
, which shows
how close to
$H$ the approximation
is, for
${x}_{0}> Q$ (i.e. for
$\frac{{x}_{0}}{Q}> 0$ dB).
We can compare the entropies calculated using
with those that were calculated
from the bandpass subimage histograms, as given in these figuresdescribing 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
${x}_{0}$ are calculated from:
$${x}_{0}=\frac{\mathrm{std.dev.}}{\sqrt{2}}=\sqrt{\frac{\mathrm{subimageenergy}}{\mathrm{2(noofpelsinsubimage)}}}$$ The following table shows this comparison:
Transform level
Subimage type
Energy (×
$10^{6}$ )
No of pels
${x}_{0}$
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) areapproximately 0.5 bit/pel greater than the entropies measured
from the Lenna subimage histograms. This is due to the heaviertails of the actual PDFs compared with the Laplacian
exponentially decreasing tails. More accurate entropies can beobtained if
${x}_{0}$ is obtained from the mean absolute values of the pels
in each subimage. For a Laplacian PDF we can show that
This gives values of
${x}_{0}$ that are about 20% lower than those calculated from
the energies and the calculated entropies are then withinapproximately 0.2 bit/pel of the measured entropies.
Questions & Answers
can someone help me with some logarithmic and exponential equations.
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Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
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Damian
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Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.