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I'm trying to reproduce some results from a paper and I need an explicit formula for a specific quadratic spline to do so.

The problem is, I've only got a plot of it. The quadratic spline is from figure 2 (a) of Stephane Mallats paper "Singularity Detection and Processing with Wavelets" in IEEE Transactions on Information Theory, Vol. 38, No. 2. March 1992. Link to journal. Link to PDF.

The spline has a positive lobe lobe from $[-1,0]$ and a negative lobe from $[0,1].$ It appears to go to zero from $[-2,-1]$ and from $[1,2].$ It also acheives a peak values of about $0.66$ or $-0.66$ in either lobe as measured in imageJ. This spline is bound to be a wavelet so it's necessarily $0$ on the average and is compactly supported from $[-2,2]$ (possibly a smaller interval if it goes to zero where I think it does.)

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That doesn't look very quadratic to me. Does "quadratic spline" have some technical meaning in your field that is different from the usual "curve parameterized on the general form $t\mapsto(at^2+bt+c, dt^2+et+f)$"? –  Henning Makholm Jul 5 '12 at 2:32
    
This is a spline, and is piecewise quadratic. It should be thought of as a piecewise quadratic polynomial smoothly joined. There is an upward parabola prior to t =0 and a downward parabola following t= 0. –  ncRubert Jul 5 '12 at 4:55
    
Have you seen the Mallat-Zhong paper by any chance? –  J. M. Jul 5 '12 at 5:34
    
A function of the form $-ax \exp(-bx^2)$ with $a,b >0 $ will yield a very similar curve. –  Pedro Tamaroff Jul 5 '12 at 19:23

2 Answers 2

It looks something like this. $$ \cases{0 & $t < -1$\cr 1.481481482+ 2.962962963\,t+ 1.481481482\,{t}^{2}&$-1 \le t<- 0.55$\cr 1.285416667+ 2.250000000\,t+ 0.833333333\,{t}^{2}&$-0.55 \le t<- 0.35$\cr 0.3666666622- 3.000000024\,t- 6.666666700\,{t}^{2}&$-0.35 \le t<- 0.25$\cr - 5.933333324\,t- 12.53333330\,{t}^{2}&$-0.25 \le t<0$\cr - 5.933333324\,t+ 12.53333329\,{t}^{2}&$0 \le t< 0.25$\cr - 0.3666666564- 3.000000069\,t+ 6.666666780\,{t}^{2}&$0.25 \le t< 0.35$\cr - 1.285416675+ 2.250000040\,t- 0.8333333750\,{t}^{2}&$0.35 \le t< 0.55$\cr - 1.481481475+ 2.962962942\,t- 1.481481467\,{t}^{2}&$0.55 \le t < 1$\cr 0 & $t \ge 1$} $$

enter image description here

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So, the spline comes from the reference in the comment above? Should I be able to work out the formula in the time domain by examining the appendix of that paper and doing some Fourier transforms? –  ncRubert Jul 5 '12 at 16:09
    
In Maple 16: > CurveFitting[Spline]([[-1,0],[-0.55,0.3],[-0.35,0.6],[-0.25,0.7],[0,0],[0.25,-0.‌​7],[0.35,-0.6],[0.55,-0.3],[1,0]],t,degree=2,knots=data); –  Robert Israel Jul 5 '12 at 16:19
up vote 0 down vote accepted

From appendix A of the Mallat paper linked to in the comment by J.M., the smoothing spline figure is the Fourier transform of the following function:

$\hat{\psi}(\omega) = i \omega (\mathrm{sinc}(\omega/4) )^4$

where

$\mathrm{sinc}(\omega) = \frac{\sin(\omega)}{\omega}$ and $\omega = 2 \pi f$

The FT of a sinc function is simply a rectangle of width 1

$\mathcal{F}^{-1}\{sinc(\omega)\} = Rect(\omega)$

Multiplication becomes convolution:

$\mathcal{F}^{-1}\{sinc(\omega)^4\} = Rect(t) \star Rect(t) \star Rect(t) \star Rect(t)$

Working out these four convolutions gives a curve defined piecewise from $t \epsilon [-2,2]$:

$\mathcal{F}^{-1}\{sinc(\omega)^4\} =$

$t^3/6 + t^2 + 2t + 4/3,\text{ } -2 \le t \le -1$

$-t^3/2 - t^2 + 2/3,\text{ } -1 <t \le 0$

$t^3/2 - t^2 + 2/3,\text{ } 0<t \le 1$

$-t^3/6 + t^2 - 2t + 4/3,\text{ } 1 <t \le 2$

The FT of a scaled sinc is a scaled Rect having width 1/4 and height 4:

$\mathcal{F}^{-1}\{sinc(\omega/4)\} = 4 Rect(4t)$

The extra factor of $i\omega$ is equivalent to a derivative in the time domain according to:

$\frac{d}{dt}(f(t) ) = i \omega \mathcal{F}^{-1}\{\hat{f}(\omega)\}$

Scaling the convolved sincs gives a curve defined piecewise from $t \epsilon [-1/2,1/2]$, which we then take a derivative of to get the final result:

$\mathcal{F}^{-1}\{i \omega sinc(\omega)^4\} =$

$4*((4t)^2/2 + 2*(4t) + 2), \text{ } -1/2 \le t \le -1/4$

$4*(-3/2*(4t)^2 - 2(4t) ) ,\text{ } -1/4 <t \le 0$

$4*(3/2*(4t)^2 - 2*(4t)) ,\text{ } 0<t \le 1/4$

$4*(-(4t)^2/2 + 2*(4t) - 2),\text{ } 1/4 <t \le 1/2$

The plot looks like this:

enter image description here

The problem with this spline is that it is half the width of the plot shown in figure 2 of Mallat's paper. I believe the difference between the two is just a scaling factor though.

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