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Asked by I Made
on 13 Mar 2013

I tried to test out the dwt and idwt, before i used this code i've build my own formula to work on dwt and idwt. But i figured that my own formula for dwt and idwt the reconstructed value are not the same as the input. so i tried with the built in function.

A=wavread('tes.wav'); leftchanel=A(1:size(A),1); [cA,cD] = dwt(leftchanel,'haar'); X = idwt(cA,cD,'haar'); check=leftchanel-X;

from this code i got check not all value are 0, that mean leftchanel is not the same as X ( the input for dwt and the output of idwt are mostly same but some are different ). The main question from me is "Aren't the after we decompose(DWT) and reconstruct(IDWT) the input and the output should be same (exactly same)?"

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Answer by Wayne King
on 13 Mar 2013

Edited by Wayne King
on 13 Mar 2013

When you say they are not zero how large is the difference? You have to keep in mind that there may be small numerical differences. For example:

x = randn(1024,1); [A,D] = dwt(x,'haar'); xrec = idwt(A,D,'haar'); max(abs(x-xrec))

For the above particular random signal, the largest absolute value of the difference is 10^{-16}. That is as equal as you can get.

I Made
on 13 Mar 2013

e.g i got : -3.46944695195361e-18, -1.73472347597681e-18, 6.93889390390723e-18

I don't know if that a large or small difference, is it small or large, will it make a big difference on the wav file i rewrite it?

I Made
on 14 Mar 2013

Hey i miss look (forget to look at the leftchanel and X), before i use

check=leftchanel-X;

The leftchanel and X is has the same value, but after i do above code i become has a very small difference that you say in check variable. how it can be? was it if

A[1 2 3 4 5] B[1 2 3 4 5]

then C=A-B; which should be 0? can have small difference because of datatype or anything?

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