Sech laser pulse time-bandwidth product calculation

Question

Find the time bandwidth product for a $\operatorname{sech}$ shaped pulse.

Attempt:

I know that the time bandwidth product $\Delta t \Delta \omega$ for a $\operatorname{sech}$ shaped pulse must be $0.315.$

Here is a calculation for the time bandwidth product of a Gaussian shaped pulse. How exactly did they obtain the expressions for $\Delta t$ and $\Delta \omega$? Because when I try to solve for $t$ and $\omega$ by equating each of the Fourier transform pairs to $1/x,$ I get different answers.

So, for the $\operatorname{sech}$ function we have the Fourier transform pair:

$$\operatorname{sech} (\alpha x) \leftrightarrow \frac{1}{\alpha} \sqrt{\frac{\pi}{2}} \operatorname{sech} \left( \frac{\pi}{2 \alpha} \omega\right)$$

How can I apply the method used in the link above to show that the time bandwidth product of a $\operatorname{sech}$ pulse has to be $0.315$?

Answer 1

Your problem is more math related than physics. To find proper full-width at half maximum $\Delta t$ and $\Delta \omega$ you have to equate the functions to $x$, with some proportionality coefficient so that when $x=1$ you're at the maximum of the function.

This way after solving for $t$ or $\omega$ you'll find the half-width half-maximum by plugging in $x=0.5$, and then the full-width half maximum (in amplitude) by taking $2t$ or $2\omega$.

For instance, the gaussian $e^{-\alpha t^2}$ reaches $1$ at $x=0$, so you can solve $e^{-\alpha t_0^2}=x$ and then take $\Delta t = 2t_0$.

This $\Delta t$ will be expressed as a function of $x$, but it is an amplitude. What you want is the Full-Width Half-Maximum for intensity, so you take $y=x^2=0.5$ as if you solved the equations for the squares of the amplitude ( $(e^{-\alpha t^2})^2=x^2=y$ ).

From there you should be good by folling their steps. Watch out for $\Delta \omega$ where you will have to solve : $$\frac{1}{\sqrt{2\alpha}}e^{-\omega_0^2/4\alpha}=\frac{1}{\sqrt{2\alpha}} x$$

Finally, i did the same for the $sech$ pulse and found $0.315$. If you find the good result for the Gaussian, you should have no trouble with this one using hyperbolic functions and their inverses.

Edit : For the sech i used the inverse hyperbolic functions $arcsech$. For instance $sech(\alpha t_0) = x$ (only $x$ since the maximum of $sech$ is $1$ gives : $$ \Delta t = 2t_0 = \frac{2}{\alpha} arcsech(x) $$

The time bandwidth at the end is : $\Delta t \Delta\nu = \frac{8}{2\pi^2} arcsech^2(\sqrt{y}) = 0.3148$