What is time badwidth product for pulses with GDD?

Question

For a gaussian pulse the time bandwidth product is dwdt = 0.4.

What is the time-bandwidth product for chirped pulse with group delay dispersion (GDD) added only ?

Answer 1

If you convert GDD to group velocity dispersion ($\beta_{2}$, ordinarily in units of fs$^{2}/$mm) then you can define both the dispersion length (the distance over which a Gaussian pulse broadens in the time domain by a factor of $\sqrt{2}$) and a chirp parameter (which quantifies how much second order dispersion has been applied to a Gaussian pulse). These can then be used to answer your question.

Start with the equation for a Gaussian pulse propagating in a medium with second order dispersion only:

(1) $U(z,T) = \frac{T_{0}}{(T_{0}^{2} - i\beta_{2}z)^{1/2}}\exp\{-T^{2} / [2(T_{0}^{2} - i\beta_{2}z)]\}$.

The pulse duration increases with propagation distance according to:

(2) $T(z) = T_{0}[1+|\beta_{2}|^{2}z^{2}/T_{0}^{4}]^{1/2}$.

The dispersion length can then be defined as follows:

(3) $L_{D} = \frac{T_{0}^{2}}{|\beta_{2}|}$

It is often convenient to define a chirp parameter to make things more compact:

(4) $C = z/L_{D}$,

giving:

(5) $T(z) = T_{0}[1+C^{2}]^{1/2}$.

Plugging these definitions into the time bandwidth product formula that you give for a Gaussian pulse:

(6) $\Delta f T_{0} = \Delta f\frac{T}{[1+c^{2}]^{1/2}} = 0.44$

(7) $\Delta f T = 0.44\sqrt{1+C^{2}}$.

Although the chirp parameter is dimensionless, using equations (1) and (5) you can put it in a form which allows you to think of it as having units of Hz/s, which makes it very useful and intuitive when thinking about (second order) dispersion.

Hope that helps!

Source: 'Nonlinear Fiber Optics', G. P. Agrawal.