Scaling a wavelet in continuous wavelet transform

The mexican hat wavelet, obtained from the second derivative of a Gaussian, has a functional form of

$$\frac{2}{\sqrt{3 \sqrt{\pi} \sigma \ }} \left(1-\frac{t^2}{\sigma^2} \right) e^{-t^2/(2\sigma^2)}$$

where "$$\sigma$$ is the standard deviation of a Gaussian and $$t$$ is the independent variable.

If we graphically wish to graphically see its dilations, say at scale $$1 \le a \le 5$$, the typical form of a wavelet function $$\psi$$ has a scale parameter $$a$$ and a translation parameter $$b$$. The functional form of the Mexican hat given in MATLAB does not explicitly have them both.

The standard deviation $$\sigma$$ is the only variable here that controls with width, but this is not strictly equal to the scale $$a$$. How should we generate scaled versions of the Mexican hat?

Thanks.

• i think you substitute: $$t \leftarrow \frac{t-b}{a}$$ to get this in terms of the wavelet params $a$ and $b$. but $a$ and $\sigma$ always team up, so i would set $\sigma=1$ and just normalize the mother wavelet and let $a$ be the sole scaling factor. – robert bristow-johnson Jan 15 at 5:22
• I am familiar with the leftward arrow notation. Do you mean to say that we replace in the wavelet equation with (t-b)/a? It seems $a$ and $\sigma$ are inversely related. Larger the $a$ value, wider is the wavelet, in the wavelet literature. If this is the case, then this substitution would be valid. As per wavelet criterion, $\sigma$ has to be 0.68 in the Mexican hat. – M. Farooq Jan 15 at 6:50
• yes, i am saying replace $t$ with $\frac{t-b}{a}$. both $\sigma$ and $a$ divide into $t$, so i would think that they team up. if one increases by a factor of 5 and the other decreases by the inverse ratio, the result will be the same wavelet. "$b$" and "$a$" are simply other symbols to express the same translation and scaling that "$\mu$" and "$\sigma$" do. – robert bristow-johnson Jan 15 at 7:37
• You can do Latex for equations by yourself. Start by right clicking on an equation and selecting "View math as tex". You can also see RB-J's helpful addition to your post in edit mode. – Cedron Dawg Jan 15 at 15:13

Okay, I am not quite understanding your question. Let's start with the definition of the Gaussian, aka the Bell Curve, in its general form.

$$f(t) = \frac{1}{ \sigma \sqrt{2\pi}} e^{ -\frac{(t-\mu)^2}{2\sigma^2} }$$

$$\mu$$ is the mean, and represents where the peak occurs.

$$\sigma$$ is the standard deviation, and identifies where the inflection points are.

Taking the first derivative:

\begin{aligned} f'(t) & = \frac{1}{ \sigma \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \right] \\ & = - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( t-\mu \right) \right] \\ \end{aligned}

Taking the second derivative.

\begin{aligned} f''(t) &= - \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( -\frac{(t-\mu)}{\sigma^2} \right) \left( t-\mu \right) + e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \left( 1 \right) \right] \\ f''(t) &= \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ \frac{(t-\mu)^2}{\sigma^2} - 1 \right] e^{ -\frac{(t-\mu)^2}{2\sigma^2} } \\ f''(t) &= \frac{1}{ \sigma^3 \sqrt{2\pi}}\left[ \left( \frac{t-\mu}{\sigma} \right)^2 - 1 \right] e^{ - \frac{1}{2} \left( \frac{t-\mu}{\sigma} \right)^2 } \\ \end{aligned}

In your example $$\mu=0$$ and $$\sigma=1$$. In your graph, the zero is where the peak is, and the one is the +/-1 on your horizontal scale. Notice that the second derivative is zero at these points (inflection in the original function.)

You can translate the graph by setting $$\mu$$. You can stretch the graph horizontally by setting $$\sigma$$.

I am not familiar with wavelets, so I don't know what this $$a$$ and $$b$$ are that you are talking about. You can definitely to another coordinate conversion if you want:

$$t = a \tau + b$$

Gosh, but that is awfully similar to

$$\frac{1}{\sigma} t - \frac{\mu}{\sigma}$$

$$\mu = b$$
$$\sigma = 0.63628 a$$
• Thanks but understanding the translation part is easy that we shift the Mexican hat to a new time position which was originally $\mu$. Would you mind having a look at this section on Wikipedia en.wikipedia.org/wiki/… – M. Farooq Jan 15 at 4:44
• i think it's, with $\mu=0$ and $\sigma=1$ then usually the only variable of a mother wavelet would be $$t = \frac{\tau - b}{a}$$. or maybe $t$ and $\tau$ are swapped. – robert bristow-johnson Jan 15 at 5:10
• i dunno, Ced, the wikipedia definition for Mexican hat is $$\psi(t) = \frac{2}{\sqrt{3\sigma}\pi^{1/4}} \left(1 - \left(\frac{t}{\sigma}\right)^2 \right) e^{-\frac{t^2}{2\sigma^2}}$$ which is consistent with the OP. – robert bristow-johnson Jan 15 at 5:19
• @M.Farooq When you plug $$\frac{t-b}{a}$$ in for $t$ within $\frac{t}{\sigma}$ you simply get: $$\frac{t-b}{a\sigma}$$ So adjusting $a$ and $\sigma$ have reciprical effect from each other. – Cedron Dawg Jan 15 at 15:07