moment of wavelet


by omer21
Tags: moment, wavelets
omer21
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#1
Jan10-13, 03:47 PM
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I don't understand how to get the equation [itex]\sum_k (-1)^kk^pc_k=0[/itex] from [itex]∫t^pψ(t)dt=0[/itex] from here on page 80. Can somebody explain it?
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Simon Bridge
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Jan11-13, 01:29 AM
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It's an approximation method - follows from the definitions in eqs. 5.12-13 on p79.
Can you see how 5.14 and 5.15 are obtained?
omer21
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#3
Jan11-13, 12:28 PM
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i can see 5.14 but i can not obtain 5.15 and 5.16.
Actually i get [itex]\sum_k (-1)^kc_k.0=0[/itex] instead of [itex]\sum_k (-1)^k k^p c_k=0[/itex] from the integral [itex]∫t^p [\sum_k (-1)^kc_k \phi(2t+k-N+1)]dt=0[/itex].

Simon Bridge
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Jan12-13, 07:35 PM
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moment of wavelet


Then you should go back over the derivations and make sure you understand what they are saying ... starting from the latest that you appear to be able to follow:

5.14 was the case for p=0.
i.e. ##\phi(t)## is capable of expressing only the zeroth monomial.
therefore, the zeroth-order moment of the wavelet function must be zero.
Which is to say...

$$\int_{-\infty}^\infty \psi(t)dt=0$$ ... which, to my mind, means that ##\psi## must be odd... which places limits on the expansion coefficients as the book says.

5.15 was the case for p=1 ... so ##\phi(t)## is capable of expressing up to the first monomial.
therefore, the zeroth and first-order moment of the wavelet function must be zero.
the zeroth order moment is above. The first order is given by...

$$\int t\psi(t)dt =0$$ ... so what steps did you follow from here?
omer21
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#5
Jan13-13, 05:15 AM
P: 25
I substituted

[itex]\psi(t)=[∑_k(−1)^kc_k\phi(2t+k−N+1)][/itex]

into

[itex]∫t\psi(t)dt[/itex].

Then i got

[itex]∫t[∑_k(−1)^kc_k\phi(2t+k−N+1)]dt=∑_k(−1)^kc_k∫t\phi(2t+k−N+1)]dt=0[/itex].

Actually i am not sure how to find the above integral but i did some change of variables and used integration by parts however i could not reach [itex]∑_k(−1)^kkc_k[/itex].


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