2.4 Use of laplacian pdfs in image compression

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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:

$p(x)=\frac{1}{2{x}_{0}}e^{-\left(\frac{\left|x\right|}{{x}_{0}}\right)}$
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:

$v({x}_{0})=\int_{()} \,d x$ x 2 p x 2 x x 2 2 x 0 x x 0 2 x 0 2
(using integration by parts twice).

Hence the standard deviation is:

$\sigma ({x}_{0})=\sqrt{v({x}_{0})}=\sqrt{2}{x}_{0}$
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,
${p}_{0}=1-e^{-\left(\frac{Q}{2{x}_{0}}\right)}$
and, for $k\ge 1$ ,
${p}_{k}=\frac{1}{2}(e^{-\left(\frac{(k-\frac{1}{2})Q}{{x}_{0}}\right)}-e^{-\left(\frac{(k+\frac{1}{2})Q}{{x}_{0}}\right)})=\sinh \left(\frac{Q}{2{x}_{0}}\right)e^{-\left(\frac{kQ}{{x}_{0}}\right)}$
By symmetry, if $k$ is nonzero, ${p}_{-k}={p}_{k}=\sinh \left(\frac{Q}{2{x}_{0}}\right)e^{-\left(\frac{\left|k\right|Q}{{x}_{0}}\right)}$

Now we can calculate the entropy of the subimage:

$H=-\sum_{k=()}$ p k 2 logbase --> p k p 0 2 logbase --> p 0 2 k 1 p k 2 logbase --> p k
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,
$\sum_{k=1}$ p k 2 logbase --> p k k 1 α r k 2 logbase --> α r k k 1 α r k 2 logbase --> α k 2 logbase --> r α 2 logbase --> α k 1 r k α 2 logbase --> r k 1 k r k
Now $\sum_{k=1}$ r k r 1 r and, differentiating by $r$ : $\sum_{k=1}$ k r k 1 1 1 r 2 . Therefore,
$\sum_{k=1}$ p k 2 logbase --> p k α 2 logbase --> α r 1 r α 2 logbase --> r r 1 r 2 α r 1 r 2 logbase --> α 2 logbase --> r 1 r
and
${p}_{0}\log_{2}{p}_{0}=(1-\sqrt{r})\log_{2}(1-\sqrt{r})$
Hence the entropy is given by:
$H=-((1-\sqrt{r})\log_{2}(1-\sqrt{r}))-\frac{2\alpha r}{1-r}(\log_{2}\alpha +\frac{\log_{2}r}{1-r})$
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: 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

x p x 2 x 0 x 2 x 0 x x 0 x 0
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.

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