A tribute to erudite pursuits.
Motivation: Fourier Transforms have good frequency resolution but no time localisation.
What does this mean? Let there be a signal
\[f(t) = \begin{cases} \cos(2 \pi t) & -5 < t < 0 \\ \cos(5 \pi t) & 0 \leq t < 0.4 \\ \cos(10 \pi t) & 0.4 \leq t < 5 \\ 0 & \text{otherwise} \end{cases}.\]We would not be able to distinguish this signal from
\[g(t) = \begin{cases} \cos(2 \pi t) & -4 < t < 1 \\ \cos(10 \pi t) & 1 \leq t < 0.8 \\ \cos(5 \pi t) & 0.8 \leq t < 1.2 \\ \cos(10 \pi t) & 1.2 \leq t < 5\\ 0 & \text{otherwise} \end{cases}\]with only Fourier Transforms. Another good example is that $\mathcal{F}[\delta]$ is white noise, i.e. we lose all time-localisation. In a sense, this is Heisenberg’s Uncertainty Principle at play.