How to demonstrate that the Gaussian window minimizes the uncertainty principle in the time-frequency analysis?

Question

I need to show that the Gaussian window in the time-frequency analysis minimizes the uncertainty principle. I know that for the demonstration, it is necessary to define the resolution in time and frequency. I have read the resolution can be expressed as: $$ \Delta t^2 = \frac{\int t^2 {|g(t)|}^2 dt }{\int {|g(t)|}^2 dt} $$ and $$ \Delta {\omega}^2 = \frac{\int {\omega}^2 {|g(\omega)|}^2 d\omega }{\int {|g(\omega)|}^2 d\omega} $$ But I do not know why it is so. I need to know how to get to that expression.

Answer 1

The time spread $\Delta_t$ at zero is defined by $$\Delta_t^2 = \frac{||tf(t)||^2}{||f(t)||^2}$$ and the frequency spread $\Delta_\gamma$ is defined by $$\Delta_\gamma^2 = \frac{||\gamma\widehat{f}(\gamma)||^2}{||f(t)||^2} =\frac{||\gamma\widehat{f}(\gamma)||^2}{||\widehat{f}(\gamma)||^2}$$ where the second equality is due to Parseval's and/or Plancherel's identity and $$\widehat{f}(\gamma):= \int_{\mathbb{R}} f(t)e^{-2\pi i \gamma t}dt.$$ With this definition of the FT, we have the classical uncertainty principle: $$\Delta^2_t\Delta^2_\gamma \geq \frac{1}{16\pi}.$$

Let $f(t)=e^{-\pi t^2}$. Then we have $$\widehat{f}(\gamma)=\int_\mathbb{R} f(t)e^{-2\pi i \gamma} dt = e^{-\pi \gamma^2},$$ and we can say that $f(t)$ is it's own Fourier transform. Namely, it is an eigenfunction of the Fourier transform with eigenvalue $1$. Using the classic integration over $\mathbb{R}^2$ trick, we can evaluate $||f(t)||^2$: \begin{align} ||f(t)||^4 &= \left(\int_{\mathbb{R}} e^{-2\pi t^2}dt\right)^2\\ & = \int_{\mathbb{R}^2} e^{-2\pi (x^2+y^2)} dA\\ & = \int_0^{2\pi} \int_0^\infty re^{-2\pi r^2}dr d\theta\\ & = 2\pi \frac{1}{4\pi} = \frac{1}{2}, \end{align} and we have that $||f(t)||^2=\frac{1}{\sqrt{2}}.$ Applying integration by parts to $||tf(t)||^2$ yields \begin{align} ||tf(t)||^2 & = \int_{\mathbf{R}} t^2 f(t)^2 dt\\ &= \int_{\mathbf{R}} t^2 e^{-2\pi t^2} dt\\ &= \int_{\mathbf{R}} t \cdot t e^{-2\pi t^2}dt\\ (IBP) & = \frac{1}{4\pi}\int_{\mathbf{R}} e^{-2\pi t^2}dt\\ &= \frac{1}{4\pi}\frac{1}{\sqrt{2}}. \end{align} Obviously, we then have $||\gamma f(\gamma)||^2= \frac{1}{4\pi}\frac{1}{\sqrt{2}}$ as well. Putting it all together we get $$\frac{||tf(t)||^2}{||f(t)||^2}\frac{||\gamma\widehat{f}(\gamma)||^2}{||\widehat{f}(\gamma)||^2}=\frac{1}{16\pi}$$ which is the lower limit in the uncertainty principle.