Before we dive into why we need wavelets, we must first understand the central mathematical tool that all ofthese transforms (including Fourier) are built on: the inner product.

In 2D, the inner product (or dot product) of $\vec{v}$ and $\vec{w}$ is:

$$\vec{v} \cdot \vec{w} = |\vec{v}| |\vec{w}| \cos(\theta)$$

The reader probably recalls from high-school trigonometry that:

  • If $\vec{v}$ and $\vec{w}$ point in the same direction, $\theta=0$ and $\cos(\theta)=1$. The inner product is at its positive maximum.

  • If they are perpendicular (orthogonal), $\theta=90^\circ$ and $\cos(\theta)=0$. The inner product is zero.

  • If they point in opposite directions, $\theta=180^\circ$ and $\cos(\theta)=-1$. The inner product is at its negative maximum.

Now, a function $f(t)$ can be understood as a vector of infinite dimensionality. It should be clear then that the inner product converges to a convolution as the number of dimensions increase, and most textbooks include a proof of this fact.

$$\langle f, g \rangle = \int_{-\infty}^{\infty} f(t) g(t) dt$$

This effectively measures how "similar" the two functions are, in the following sense:

  • If $f(t)$ and $g(t)$ are "in sync" (where $f$ is positive, $g$ is positive; where $f$ is negative, $g$ is negative), their product $f(t)g(t)$ will be positive everywhere. The integral of all this positive area will be a large positive number.

  • If $f(t)$ and $g(t)$ are "out of sync" (they are orthogonal), the product $f(t)g(t)$ will have exactly as much positive area as it has negative area. The integral of these cancelling areas will be zero.

  • If $f(t)$ and $g(t)$ are "perfectly opposite" (where $f$ is positive, $g$ is negative), their product $f(t)g(t)$ will be negative everywhere. The integral of all this negative area will be a large negative number.

Now, consider the definition of the continuous wavelet transform and the Fourier transform:

$$\text{CWT: } W_f(u, s) = \int_{-\infty}^{\infty} f(t) \psi^* \left( \frac{t-u}{s} \right) dt$$

$$\text{Fourier: } \hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$$

It is not important for now if you understand what the functions involved represent: it suffices for the moment to appreciate that these transforms define families of inner products. They do not ask "How similar is our signal $f(t)$ to the function $\psi(t)$?", but rather:

"How similar is our signal $f(t)$ to the wavelet $\psi(t)$...

  • ...when the wavelet is at position $u_1$ and scale $s_1$?"

  • ...when the wavelet is at position $u_2$ and scale $s_1$?"

  • ...when the wavelet is at position $u_3$ and scale $s_2$?"

  • ...for all possible $u$ and $s$?"

The resulting "scaleogram" (or "spectrogram" for Fourier) is a giant map where every single pixel's brightness is the result of one inner product, telling you the "similarity score" at that specific time ($u$) and scale ($s$). In summary, an inner product is a similarity score; a transform is a map of similarity scores.

The Fourier Transform is built upon the fact that any complex signal $f(t)$ can be perfectly reconstructed as a sum of simple sine and cosine waves (or complex exponentials).

The transform $\hat{f}(\omega)$ answers the question: "How much of frequency $
omega$ is in my signal?" $$\hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$$

This is, as we just discussed, an inner product. We are asking: "How similar is our signal $f(t)$ to a 'pure' infinite wave $e^{-i\omega t}$?"

A fundamental problem is that the basis functions (sines and cosines) are infinite in time. They have perfect "frequency localization" (we know exactly what frequency they are) but zero "time localization" (they exist everywhere, so we don't know when they are).

The first logical step to fix this is the Short-Time Fourier Transform (STFT), or Gabor Transform. The idea is simple:

  1. Choose a "window" function $g(t)$ (like a Gaussian) that is localized in time.

  2. Slide this window along your signal $f(t)$.

  3. Perform a Fourier Transform only on the "windowed" part of the signal.

$$\text{STFT}f(u, \omega) = \int{-\infty}^{\infty} f(t) g(t-u) e^{-i\omega t} dt$$

This works and produces a valid spectrogram, i.e. a representation of frequency components over time. However, it introduces the Gabor Uncertainty Principle. The size of our window $g(t)$ is fixed, and:

  • A narrow window gives good time localization (we know when it happened) but poor frequency localization (the smearing from the window blurs all the nearby frequencies).

  • A wide window gives good frequency localization (we can resolve close frequencies) but poor time localization (we only know it happened "sometime in this wide window").

This fixed window size is the critical flaw. We can't analyze high-frequency (short-lived) and low-frequency (long-lived) events with the same "ruler." We need a more flexible tool.

Wavelet analysis solves the fixed-window problem by using a "window" (the wavelet) that can stretch and shrink.

We start with a single "Mother Wavelet" function, $\psi(t)$. This function must be localized in both time and frequency and must have a zero mean (it must be a "wave").

$$\int_{-\infty}^{\infty} \psi(t) dt = 0$$

A classic example is the Morlet wavelet, which is just a sine wave inside a Gaussian window. From this one "mother," we generate all our basis functions (the "children") by two operations:

  1. Translation ($u$): Sliding the wavelet in time, just like in STFT.

  2. Scaling ($s$): Stretching or shrinking the wavelet.

This gives us the wavelet family $\psi_{u,s}(t)$: $$\psi_{u,s}(t) = \frac{1}{\sqrt{s}} \psi \left( \frac{t-u}{s} \right)$$

  • $u$: The position (time).

  • $s$: The scale (inverse of frequency).

  • $\frac{1}{\sqrt{s}}$: A normalization factor to ensure all wavelets have the same "magnitude."

Going back to our analogy with a "ruler" which we slide across our signal:

  • For a small $s$, the wavelet is compressed (shrunk). This gives a very narrow, high-frequency "ruler" for pinpointing the exact time of a high-frequency spike.
  • For a large $s$, the wavelet is stretched. This gives a very wide, low-frequency "ruler" for analyzing the frequency of a long, slow oscillation.

The CWT is the full, mathematically-rich version of the wavelet transform. It is defined as the inner product of our signal $f(t)$ with the wavelet family $\psi_{u,s}(t)$ for all possible scales $s$ and translations $u$.

$$W_f(u, s) = \langle f, \psi_{u,s} \rangle = \int_{-\infty}^{\infty} f(t) \frac{1}{\sqrt{s}} \psi^* \left( \frac{t-u}{s} \right) dt$$

The result, $W_f(u,s)$, is a 2D map of "similarity scores," termed a scleogram. It shows the energy (the squared coefficient) of the signal at every time $u$ and scale $s$.

The reader should have noticed immediately from its definition that the CWT involves a huge computation. For every time step, we calculate an inner product with hundreds of scales!

The Discrete Wavelet Transform (DWT) provides a fast, efficient, and non-redundant way to compute wavelet coefficients. It is the workhorse behind JPEG2000 image compression, denoising, and other data analysis tools based on wavelets. The DWT is a very complex algorithm which is worth studying on its own, but falls beyond the scope of this entry. I encourage to read about it on their own.