Analytic expession for the width of a wavelet

[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):

<br /> 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 )<br />

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

<br /> w = \sqrt(6) / \pi / f0<br />

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

 

[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.

 

[nkinar]

Hello hamster143--

Many thanks for your reply! Your response is very helpful!

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):

<br /> s&#039;(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)<br />

Then by finding the roots of s&#039;(t) = 0, it follows that:

<br /> t = \pm \frac{\sqrt{6}}{\omega_0}<br />

Since

<br /> w = 2t<br />

<br /> \omega_0 = 2 \pi f_0<br />

So

<br /> w = \frac{ \sqrt{6}}{\pi f_0}<br />

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?

 

[hamster143]

I'm not sure I understand this question.

 

[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).