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Analytic expession for the width of a wavelet

  1. Mar 18, 2010 #1

    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]:

    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, [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

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

    where [tex]f0[/tex] is the peak frequency (Hz) of the wavelet. However, I am uncertain as to how this expression is derived.
  2. jcsd
  3. Mar 18, 2010 #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.
  4. Mar 18, 2010 #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]:

    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 [tex]s'(t) = 0[/tex], it follows that:

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


    w = 2t

    \omega_0 = 2 \pi f_0


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

    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?
  5. Mar 18, 2010 #4
    I'm not sure I understand this question.
  6. Mar 18, 2010 #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].
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