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input of DWT and output of IDWT not same?

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I Made
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)?"

Answers (1)

Wayne King
Wayne King on 13 Mar 2013
Edited: 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.
  3 Comments
Wayne King
Wayne King on 14 Mar 2013
e-18 means 10^(-18). That is an very small difference.
I Made
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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