Discrete wavelet transform-wavedec

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Dear all, Could you please answer my question? First, I use [C,L]=wavedec(signal(1024x1),m(1:10),'haar') (Eq.1) to find cD1 (512x1),cD2(256x1)... after that, i want to check how does wavedec function work? And i try with haar=[1/sqrt(2) -1/sqrt(2)](for example m=1) and apply cD1'=conv(signal,haar) (2). Eq.(2) gave me cD1'= minus(cD1) in Eq.1 (I just talk about value) Why is there difference between wavelet coefficients in Eq1 VS.Eq.2 ?
Thank you so much for your consideration to my question

Accepted Answer

Wayne King
Wayne King on 10 May 2016
Hi Hai,
The Haar high pass analysis filter used in DWT() which is called by WAVEDEC() is
[-1/sqrt(2) 1/sqrt(2)]
To demonstrate the equivalence between convolution as implemented by conv() and the wavelet transform, you should set the DWTMODE to 'per'
dwtmode('per');
Lo = [1/sqrt(2) 1/sqrt(2)];
Hi = [-1/sqrt(2) 1/sqrt(2)];
% generate a test signal
x = randn(16,1);
[A,D] = dwt(x,'haar');
% Now compare A to
dyaddown(conv(x,Lo))
% Compare D to
dyaddown(conv(x,Hi))
  1 Comment
Hai Tran
Hai Tran on 11 May 2016
Thank you so much Dr.Wayne King

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More Answers (1)

Asfaw Alem
Asfaw Alem on 28 Jul 2022
how can plot PAPR for haar wavelet

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