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Re: Describing a filter in a wavelet basis

Started by Rune Allnor May 23, 2007
Corrected a bad typo below...

Rune


On 23 May, 21:13, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 23 May, 09:23, Palle Uldenborg <palle.uldenb...@gmail.com> wrote: > > > Suppose that I have used the same orthogonal (Dauberchies) wavelet > > basis to describe a filter I(t) and a set of signales x1(t), x2(t), > > x3(t). Now I want to calculate the convolutions (I*x1)(t), (I*x2)(t), > > (I*x3)(t).. and express them in the same wavelet basis as the signals > > and the filter. > > > Is there a smart way to do this within the wavelet formalism? > > Before discussing wavelets, let's see how time-domain > convolution and the DTFT are related (view with fixed-width > font): > > [All sums from -infinite to infinite. sum means "sum over k"] > k > > y[n] = x[n] (x) h[n] = sum x[k] h[n - k] [1] > k > > Y(w) = sum y[n] exp(jwn) > n > > = sum sum x[k] h[n-k] exp(jwn) [2] > n k > > set p = n-k, find n = p+k and substitute: > > Y(w) = sum sum x[k] h[p] exp(jw(k+p)) [3] > p k > > = sum sum x[k] h[p] exp(jwn)exp(jwp) [4] > p k > > = sum h[p] exp(jwp) sum x[k] exp(jwp) [5] > p k > > = X(w)H(w) [6] > > Equation [6] is the key why the DTFT is so useful when > computing convolutions, since the numercal work of > computing [6] often is far less than computing [1] directly. > > But do take a closer look into how we arrived at [6], and you > will find one utterly essential step: > > Transforming from equation [3] to equation [4]. > > Here, one uses the property of the exponential function that > > F(a+b) = F(a)F(b). [7] > > I don't know of any other function than the exponential > function that has this property. Not that this is the
NotE that this is the...
> reason why one can set up equation [5] without any (p,k) > cross terms, which in turn is essential for the whole result. > > > Is it better to do inverse wavelet transform, FFT, and then wavelet > > transform the result? > > If the reason is to save FLOPs, then yes. Unless your > wavelet function satisfies [7] above. > > Rune