Intuition behind the Continuous Wavelet Transform?
I think that the best way to explain the CWT is to start by explaining the Fourier Transform, then move on to explaining the Short-Time Fourier Transform, and then finally explain the CWT as a variation of the STFT.

The Fourier Transform exploits the fact that any decently behaved function can be represented as a sum of sinusoids (i.e. a Fourier series) and that the sinusoid basis posses the property of orthogonality:

$$ \int_{-\infty}^{\infty}  \sin(nx)\,\sin(mx)\,dx = \int_{-\infty}^{\infty}  \cos(nx)\,\cos(mx)\,dx = \begin{cases}
1,  &amp; \text{if $n=m$} \\
0, &amp; \text{if $n\neq m$}
\end{cases}$$

So, since:

$$ e^{iax} = \cos(ax)+i\,\sin(ax)$$

the Fourier Transform is simply doing this integration for all frequencies and keeping track of which outputs are zero (i.e. that frequency is not in the signal) and which are non-zeros (i.e. that frequency is in the data and its output is scaled by how much of it is in there):

$$ S(f) = \int_{-\infty}^{\infty} s(t)\,e^{i2\pi ft}dt $$

In this case you are doing this integration over the entire signal so you can't really tell if the frequency content is changing from the beginning of the signal to the end. One way around this is to compute the Short-Time Fourier Transform: i.e. window the signal, calculate the Fourier transform of the windowed signal, store it, then shift the window down a bit and repeat for all shifts:

$$ S(\tau,f) = \int_{-\infty}^{\infty} w(t- \tau)\,s(t)\,e^{i2\pi ft}dt  $$

where

$$ w(\tau,t)=\begin{cases}
1,  &amp; \text{if $\tau\approx t$} \\
0, &amp; \text{if $\tau \not\approx t$}
\end{cases} $$

The key thing here is that you are calculating the typical Fourier transform but of a new signal that only exists in a localized part of the t-axis. To emphasize this, you can see the new signal whose Fourier transform we are calculating by associating:

$$ S(\tau,f) = \int_{-\infty}^{\infty} [w(t-\tau)\,s(t)]\,e^{i2\pi ft}dt  $$

And here's a graphical example of this showing the new signal for different $\tau$ values and on the right are a few sinusoids to represent what we are using to decompose the signals (i.e. our basis, or kernel).



But, we can also change the association as such without changing the outcome:

$$ S(\tau,f) = \int_{-\infty}^{\infty} s(t)\,[w(t-\tau)\,e^{i2\pi ft}]dt  $$

So this means that instead of windowing our signal, we are windowing our basis functions. But here's the kicker, if we are windowing our basis functions, we don't have to use a constant-size window since we know that a basis function of high frequency will need a shorter window than a basis function of low frequency. This is the whole point of the CWT. It is a decomposition of a signal by "wavelets" (i.e. windowed sinusoids in this case) where the windowing is adaptive to the sinusoid frequency. If the we choose a Gaussian window (as I have chosen in these examples), then our wavelets are called Morlet wavelets (or Gabor wavelets, in some literature).



Finally, you can generalize this for any choice of wavelet that you want. In that generalization, you can describe your wavelet basis functions as being "stretched" and "squeezed" versions of some arbitrary "mother" wavelet, $ \psi $. And so the previous equation can now be written as the final form of the CWT:

$$ S(a,b) = \frac{1}{\sqrt b}\int_{-\infty}^{\infty} s(t)\,\psi (\frac{t-a}{b})dt $$

where $a$ is now what used to be our $\tau$ (i.e. a time shift), and $b$ is called the "scale" which is just a parameter to stretch and squeeze the wavelet (similar to what our parameter $f$ was except now the interpretation is more difficult). And the only reason you have the $\frac{1}{\sqrt b}$ up front is to normalize the wavelets so that they have the same "energy" and you end up comparing apples to apples in your time-frequency representation.

I hope this helps!

-Antonio