Discrete wavelet transform-wavedec

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

Answer 1

[Wayne King]

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))

[Hai Tran]

Thank you so much Dr.Wayne King

Answer 2

[Asfaw Alem]

how can plot PAPR for haar wavelet