Analytic expession for the width of a wavelet

In summary, the width of the Ricker wavelet is w = 2t and it is obtained by differentiating s with respect to time. The secondary peaks are located at t = \pm \sqrt{6} / \omega_0 and w = 2t.
  • #1
76
0
Hello--

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

[tex]
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 )
[/tex]

In the equation above, [tex]t[/tex] is time (s), and [tex]\omega_0[/tex] 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 [tex]t = 0[/tex], 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 [tex]w[/tex] of the Ricker wavelet is

[tex]
w = \sqrt(6) / \pi / f0
[/tex]

where [tex]f0[/tex] is the peak frequency (Hz) of the wavelet. However, I am uncertain as to how this expression is derived.
 
Physics news on Phys.org
  • #2
By differentiating s with respect to t, you find that locations of secondary peaks are

[tex]t = \pm \sqrt{6} / \omega_0[/tex]

and width is obviously [itex]w = 2t[/itex].

Assuming that [itex]\omega_0[/itex] 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), [itex] f_0 = \omega_0 / 2\pi[/itex], whence you get the result you need.
 
  • #3
Hello hamster143--

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

Yes, [tex]\omega_0 [/tex] 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 [tex]s(t)[/tex]:

[tex]
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)
[/tex]

Then by finding the roots of [tex]s'(t) = 0[/tex], it follows that:

[tex]
t = \pm \frac{\sqrt{6}}{\omega_0}
[/tex]

Since

[tex]
w = 2t
[/tex]

[tex]
\omega_0 = 2 \pi f_0
[/tex]

So

[tex]
w = \frac{ \sqrt{6}}{\pi f_0}
[/tex]

Now I have one more lingering question.

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

[tex]s(t) \approx 0 [/tex]

Does this point have an actual name in the terminology of wavelets?
 
  • #4
I'm not sure I understand this question.
 
  • #5
Thank you very much for your response!

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

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

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

1. What is an analytic expression for the width of a wavelet?

An analytic expression for the width of a wavelet is a mathematical equation that describes the size or scale of a wavelet. It takes into account parameters such as the center frequency, bandwidth, and shape of the wavelet to determine its width.

2. Why is an analytic expression for the width of a wavelet important?

An analytic expression for the width of a wavelet is important because it allows us to accurately characterize and compare different wavelets. It also helps us understand how the properties of a wavelet affect its behavior in signal processing applications.

3. How is the width of a wavelet related to its frequency spectrum?

The width of a wavelet is inversely proportional to its frequency spectrum. This means that a wider wavelet will have a narrower frequency spectrum and vice versa. The center frequency and bandwidth of a wavelet also play a role in determining its width.

4. Can the width of a wavelet be adjusted?

Yes, the width of a wavelet can be adjusted by changing its parameters such as the center frequency, bandwidth, and shape. Different applications may require different wavelet widths, so it is important to have the ability to adjust it.

5. How does the width of a wavelet affect its time-frequency localization?

The width of a wavelet directly affects its time-frequency localization. A narrower wavelet will have better time localization, meaning it can accurately detect and represent rapid changes in a signal. On the other hand, a wider wavelet will have better frequency localization, meaning it can accurately detect and represent different frequencies in a signal.

Suggested for: Analytic expession for the width of a wavelet

Back
Top