Analytic expession for the width of a wavelet

1. Mar 18, 2010

nkinar

Hello--

I've been working with a time-domain function to calculate the Ricker (Mexican Hat) wavelet for $$-0.2 \leq t \leq 0.2$$. This function is given as $$s(t)$$:

$$s(t) = \left( 1 - \frac{1}{2} \omega_0^2 t^2 \right) \mbox{exp} \left ( -\frac{1}{4} \omega_0^2 t^2 \right )$$

In the equation above, $$t$$ is time (s), and $$\omega_0$$ is the angular frequency (1/s).

What I would like to do is find an analytical expression to determine the width of this wavelet, which is the time between the two side-lobes of the wavelet. The wavelet consists of a maximum value at $$t = 0$$, and the side-lobes are situated on either side. I think that it is a very pretty wavelet.

I am wondering if it would be possible to do this type of calculation in the frequency domain.

A paper that I am reading informs me that the width $$w$$ of the Ricker wavelet is

$$w = \sqrt(6) / \pi / f0$$

where $$f0$$ is the peak frequency (Hz) of the wavelet. However, I am uncertain as to how this expression is derived.

2. Mar 18, 2010

hamster143

By differentiating s with respect to t, you find that locations of secondary peaks are

$$t = \pm \sqrt{6} / \omega_0$$

and width is obviously $w = 2t$.

Assuming that $\omega_0$ is the peak angular frequency of the wavelet (that is the conventional notation; this can probably be proved if you compute the Fourier transform of s), $f_0 = \omega_0 / 2\pi$, whence you get the result you need.

3. Mar 18, 2010

nkinar

Hello hamster143--

Yes, $$\omega_0$$ is the peak angular frequency of the wavelet.

Let me add to the discussion here on this thread. Following your procedure, to obtain the locations of the secondary peaks, I take the derivative of $$s(t)$$:

$$s'(t) = \frac{1}{4} \mbox{exp} \left( -\frac{1}{4} \omega_0^2 t^2 \right) \omega_0^2 t \left( \omega_0^2 t^2 - 6 \right)$$

Then by finding the roots of $$s'(t) = 0$$, it follows that:

$$t = \pm \frac{\sqrt{6}}{\omega_0}$$

Since

$$w = 2t$$

$$\omega_0 = 2 \pi f_0$$

So

$$w = \frac{ \sqrt{6}}{\pi f_0}$$

Now I have one more lingering question.

As $$t \rightarrow \infty$$ and $$t \rightarrow -\infty$$ beyond the secondary peaks, it is apparent that $$s(t) \rightarrow 0$$. How might I analytically find the turning point of the curve when

$$s(t) \approx 0$$

Does this point have an actual name in the terminology of wavelets?

4. Mar 18, 2010

hamster143

I'm not sure I understand this question.

5. Mar 18, 2010

nkinar

Thank you very much for your response!

Hmm, well...I don't think that I'm phrasing this question very well.

A plot of $$s(t)$$ will taper off beyond the secondary peaks. That is, the curve of $$s(t)$$ will approach zero. What I would like to do is to analytically find the location where the curve becomes close to zero.

For $$s(t)$$ as given above, this occurs near $$t = \pm 0.02$$, on the interval $$-0.2 \leq t \leq 0.2$$, with $$\omega_0 = 2 \pi (50)$$.