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Shortest possible pulse duration

  • Thread starter ATY
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    laser
  • #1
ATY
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Hey guys,
I really need your help.
I got the following task:
"Sketch the output spectrum of a mode-locked Ti-Sapphire laser (assuming eg. gaussian shape) and calculate the shortest possible pulse duration by Fourier transformation."
So the spectrum is gauss shaped but I have no clue how to get the shortest possible pulse duration.
I thought about starting with something like
c363b850c5.jpg
and get something that I can use with the time bandwidth product
5be9dd0593.jpg
.
But the first equation does not contain the mode-locking thing and I have no idea how to calculate the stuff.

I am an absolute beginner without any knowledge about lasers (and bad english).
I hope that somebody understands my question and is able to help me.
Have a nice day
ATY
 

Answers and Replies

  • #2
blue_leaf77
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Yes, you can use the time bandwidth product analysis by fixing the spectrum width ##\Delta \omega##. This way you can express the shortest duration ##\Delta t_{min}## in terms of ##\Delta \omega##. Just that, which parameters are you given?
 
  • #3
ATY
34
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The only parameter we got are the wavelength of the laser. From 690nm to 1080nm. this would give me the spectrum width, but how do I get the shortest duration if I have to use the fourier transformation ?
 
  • #4
blue_leaf77
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From 690nm to 1080nm
So, is this the full width at half maximum (FWHM) in terms of wavelength?
 
  • #5
ATY
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Not sure. The text says:
"With properly chosen optics, the TI:Sa laser is in general capable of delivering pulsesin the wavelength reange from 690nm to 1080nm and pulse durations down to 6fs"
The task itself does not give any information about he laser and I just took this from the text a few sites before the task.
 
  • #6
blue_leaf77
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In general, there are a number of ways to characterize the width of a localized function, as for the case of the spectrum of a laser it's usually defined as the FWHM of the power spectrum. Anyway, whatever measure of the widths used, the widths in time and frequency domain always satisfy the relation ##\Delta \omega \Delta t \geq K## where ##K## is a constant that depends on the definition of the widths and the particular shapes of the functions in time and frequency domains. For a Gaussian functions in frequency (and also in time) along with FWHM to describe the widths, ##K=0.441##. From this, it should be straightforward to determine ##\Delta t_{min}##.
 
  • #7
ATY
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you are right, but the task requires to get the shortest possible amount by fourier transformation
calculate the shortest possible pulse duration by Fourier transformation
 
  • #8
blue_leaf77
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but the task requires to get the shortest possible amount by fourier transformation
It asks for ##\Delta t_{min}##, which is the shortest pulse duration.
 

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