# Time-Bandwidth Product (Ideal Mode-Locking)

 P: 265 So, I'm trying to prove $$\Delta\nu\Delta\tau\approx0.44$$ where; $$\Delta\nu$$ is the FWHM in freq domain for a gaussian pulse and $$\Delta\tau$$ is FWHM in time domain for a gaussian pulse. I do the problem by taking a standard gaussian exponential and finding the FWHM in both the time and frequency domain. BUT, my answer is always EXACTLY a factor of 2 off. Ie. I get $$\approx0.88$$. No matter what form of a gaussian I use it is a factor of 2 off. 0.44 is the right answer (its quoted in many books). Anyone familiar with this derivation?
 P: 265 Because this is a gaussian pulse the FWHM is defined the same way in both time and freq domain (the same goes for $$sech^2(t)$$). Ie. FWHM is just the 'Full Width' at 'Half the Maximum'. It is based on the intensity distribution. So if we had: $$I(t)=Aexp(\frac{-t^2}{2c^2})$$ then we just make $$I(t) = \frac{A}{2}$$ solve for 't', and say $$t_+ - t_-$$. Thus, we have the FWHM. It is a good point as to why we have FWHM, I never thought about it...
 P: 265 Yes, if I calculate the 'half-width' too, I get 0.44. But $$\Delta\nu\Delta\tau$$ is defined as the FWHM ie. the 'full-width'. In which case the full width would be 0.88. Unless the time-bandwidth product is actually defined as the HWHM (half-width) then I have no problems. Would you know if it is?
 P: 1,235 Sorry, but I get lost with all the abreviations FWHM, HWHM, ... Sorry too if I used the wrong words ... Here is precisely what I did: I used the Fourier transform table table from wiki. There you find the following entry: $$e^{-\alpha t^2}$$ == FT ==> $$\frac{1}{\sqrt{2 \alpha}}\cdot e^{-\frac{\omega^2}{4 \alpha}}$$ Therefore, the full width at a certain height "x" of the "signal amplitude" is: $$Dt = 2 \sqrt{\frac{1}{\alpha} ln(\frac{1}{x})}$$ and the full width at a certain height "x" of the "fourier amplitude" is: $$D\omega = 2 \sqrt{\frac{1}{\beta} ln(\frac{1}{x})}$$ where $$\beta = \frac {1}{4\alpha}$$ Now, the intensity is the square of the amplitude, therefore cutting the intensity at height y is the same as cutting the amplitude at height x, if y=x². Therefore on this basis the formula above become: $$Dt = 2 \sqrt{\frac{1}{\alpha} ln(\frac{1}{\sqrt{y}})}$$ $$D\omega = 2 \sqrt{\frac{1}{\beta} ln(\frac{1}{\sqrt{y}})}$$ where $$\beta = \frac {1}{4\alpha}$$ And therefore successively: $$Dt.D\omega = 2 \sqrt{\frac{1}{\alpha} ln(\frac{1}{\sqrt{y}})} * 2 \sqrt{\frac{1}{\beta} ln(\frac{1}{\sqrt{y}})}$$ $$Dt.D\omega = 8 ln(\frac{1}{\sqrt{y}})$$ $$Dt.Df= \frac{8 ln(\frac{1}{\sqrt{y}})}{2\pi}$$ and for y=0.5, you get Dt.Df = 0.44
 P: 265 Oh, I completely see where I went wrong. I was treating the intensity as having a gaussian function like the electric field function you had. I also had put in numbers at the start of my equations eg: $$I(t)=Aexp(\frac{-t^2}{2c^2})$$ $$\frac{A}{2}=Aexp(\frac{-t^2}{2c^2})$$ $$ln(\frac{1}{2})=\frac{-t^2}{2c^2}$$ and so on... but thanks, I see what I've done wrong and corrected it.