Everyone knows that JPEG2000 can obtain much better compression result than JPEG. The fundamental transform behind JPEG2000 is wavelet while the basic transform in JPEG is DCT. So my question is why wavelet is more suitable for image compression compared to DCT. Thanks.

  • $\begingroup$ DCT is a wavelet transform. $\endgroup$ – geometrikal Jul 21 '15 at 7:14

Both JPEG and JPEG 2000 use the change of basis compression type.

Namely, we transform the data into a different representation assuming in this representation the number of parameters needed to describe to data is lower.
Or to the least, most of the information is gathered within few parameters.

Now, if you look at the energy level of the DCT coefficients of real world images you'd see most of the energy is limited to the very few coefficients.
This is what JPG does, keeps only the few dominant coefficients and throws the rest.

How efficient is DCT?
It depends on how less coefficients are needed to describe the image in an acceptable quality.

It turns out that for Wavelets the situation is better.
Namely on real world images less coefficients are needed to describe the image with the same perceived quality.

This property is called the ability to decorrelate the data / being energy densed, etc...

At the end, it is all up to how much less parameters are needed to describe the same data.

Using SVD / KLT (They are equivalent in this sense) would be even more efficient in the sense of how much less coefficients are needed for the same quality (In the $ {\ell}_{2} $ sense).
The problem is those bases are adaptive to the data and hence if you use them you have to include them to be able to decompress the data.
What's so nice, you can ask which, in general, approximate the SVD better?
It turns out that Wavelets are better doing so.

  • $\begingroup$ I agree with everything you said, but I think the OP is more asking why wavelets need fewer coefficients to describe an image than the DCT does. $\endgroup$ – Jim Clay Jul 21 '15 at 1:07
  • $\begingroup$ @JimClay, I also answered this. If we choose $ {\ell}_{2} $ to be our measurement, the best which can be done in this approach is the SVD. If you compare which pre defined basis is closer to the SVD in average for real world images you'll find the Wavelets. $\endgroup$ – Royi Jul 21 '15 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.