Time-Bandwidth Product (Ideal Mode-Locking)

by n0_3sc
Tags: ideal, modelocking, product, timebandwidth
 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: 1,235 How is the FWHM defined in freq domain and in the time domain? Is it based on the power spectrum and power (in time) of the signal? Or is it based on the amplitude? What definition did you take in your calculation? Personally I would preffer using standard deviations ...
 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: 1,235

Time-Bandwidth Product (Ideal Mode-Locking)

n0_3sc,

I did the calculation myself and found Dt.Df = 8*ln(2^0.5)/2Pi = 0.44.

You probably made a mistake by confusing half-width on the amplitude spectrum (DfA) and half-width on the intensity spectrum (DfI), and the same for the time-signal.

If you take the half-width on the amplitude spectrum you get: Dt.Df = 8*ln(2)/2Pi = 0.44 .
If you take the half-width on the intensity spectrum you get: Dt.Df = 8*ln(2^0.5)/2Pi = 0.44 .
The square root corresponds simply to a conversion from heigth in ampltude to heigth in intensity.
 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.

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