### Minimum Scale

To determine the minimum scale, first obtain the Fourier transform, $${\omega}_{x}$$ of
the base wavelet. Then, find the peak frequency of that transformed
data. For Morse wavelets, dilate the wavelet so that the Fourier transform
of the wavelet at $$\pi $$ radians
is equal to 10% of the peak value. The smallest scale occurs at the
largest frequency:

$${s}_{0}=\frac{{\omega}_{x}^{\text{'}}}{\pi}$$

As a result, the smallest
scale is the minimum of (2, $${s}_{0}$$).
For Morse wavelets, the smallest scale is usually $${s}_{0}$$.
For the Morlet wavelet, the smallest scale is usually 2.

### Maximum Scale

Both the minimum and maximum scales of the CWT are determined
automatically based on the energy spread of the wavelet in frequency
and time. To determine the maximum scale, CWT uses the following algorithm.

The standard deviation of the Morse wavelet in time, $${\sigma}_{t}$$,
is approximately $$\sqrt{\frac{P}{2}}$$,
where $$P$$ is
the time-bandwidth product. The standard deviation in frequency, $${\sigma}_{f}$$,
is approximately $$\frac{1}{2}\sqrt{\frac{2}{P}}$$.
If you scale the wavelet by some $$s>1$$,
he time duration changes to $$2s{\sigma}_{t}=N$$,
which is the wavelet stretched to equal the full length (*N* samples)
of the input. You cannot translate this wavelet or stretch it further
without causing it to wrap, so the largest scale is $$floor\left(\frac{N}{2{\sigma}_{t}}\right)$$.

Wavelet transform scales are powers of 2 and are denoted by $${s}_{0}{\left({2}^{\frac{1}{NV}}\right)}^{j}$$. *NV* is
the number of voices per octave, and *j* is from
0 to the largest scale. For a specific small scale, $${s}_{0}$$:

$${s}_{0}{\left({2}^{\frac{1}{NV}}\right)}^{j}\le \frac{N}{2{\sigma}_{t}}$$

Converting to log2:

$$j{\mathrm{log}}_{2}\left({2}^{\frac{1}{NV}}\right)\le {\mathrm{log}}_{2}\left(\frac{N}{2{\sigma}_{t}{s}_{0}}\right)$$

$$j\le NV{\mathrm{log}}_{2}\frac{N}{2{\sigma}_{t}{s}_{0}}$$

Therefore, the maximum
scale is

$${2}^{floor\left(NV{\mathrm{log}}_{2}\frac{N}{2{\sigma}_{t}{s}_{0}}\right)}$$

### L1 Norm for CWT

Wavelet transforms commonly use L2 normalization of the wavelet.
For the L2 norm, dilating a signal by 1/*s*, where *s* is
greater than 0, is defined as follows:

$${\Vert x\left(\frac{t}{s}\right)\Vert}_{2}^{2}=s{\Vert x\left(t\right)\Vert}_{2}^{2}$$

The energy is now *s* times
the original energy. To preserve the energy of the original signal,
you must multiply the CWT by $$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{s}$}\right.$$.
When included in the Fourier transform, multiplying by this factor
produces different weights being applied to different scales, so that
the peaks at higher frequencies are reduced more than the peaks at
lower frequencies.

In many applications, L1 normalization is better. The L1 norm
definition does not include squaring the value, so the preserving
factor is 1/*s* instead of $$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{s}$}\right.$$.
Instead of high-frequency amplitudes being reduced as in the L2 norm,
for L1 normalization, all frequency amplitudes are normalized to the
same value. Therefore, using the L1 norm shows a more accurate representation
of the signal.