Tuesday, November 20, 2007
Excitation Profiles
The range of frequency over which NMR resonances are excited depends on the duration of the monochromatic radio frequency pulse applied to the sample. Very long pulses will excite a very narrow range of frequencies whereas very short pulses will excite very wide frequency ranges. For example, presaturation pulses used for solvent suppression, are typically several seconds long while non-selective pulses are typically micro seconds in duration. It is essential to use very short pulses to provide even excitation over large spectral widths. The shape of the excitation profile is related to the Fourier transform of the pulse.
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7 comments:
Dear Glenn
thanks a lot for these posters which are of great helpful for new beginners like me. may i ask the reason why a relative longer pulse can only excite a narrow width? thank you very much for your time.
Liming,
In simple terms, the excitation profile of a pulse is the Fourier transform of the pulse. Short monochromatic pulses transform to broad excitation profiles whereas long monochromatic pulses transform to narrow excitation profiles. In the limit, an infinitely long pulses will excite only a single frequency.
Glenn
GLEN,
YOU ARE A HELP TO MILLIONS OF PEOPLE AND YOU SHOULD THINK THIS EVERYDAY BEFORE GETTING UP OF THE BED.
THANK YOU!
Vindana
Dear Glenn, I'm a student from Argentina and I would like to do a question.
What is the problem that occurs in RMN spectrum if the excitation pulse is too long?
Thank you very much!
Unknown,
Long 90 degree excitation pulses will excite narrower frequency ranges and limit the spectral width one can use. If the excitation pulse is longer than the duration of the 90 degree pulse, you will obtain signals that are reduced in intensity due to B1 inhomogeneity and a phase dependent on the flip angle. See this post:
https://u-of-o-nmr-facility.blogspot.com/2009/11/b-1-homogeneity.html
-Glenn
Dear Glenn,
Although I know that it is possible to simulate the excitation profile of a rectangular pulse from the "stdisp" command in TopSpin, I am looking to calculate this profile to obtain the graph's shape.
To do this, I saw on another blog that the formula to use in the case of a rectangular pulse is the following:
f(v) = (10^6/pi*v*p)*sin(pi*v*p/10^6)
Indeed, by using this formula I manage to obtain the graph on excel for a P1= 10 µs and yB1/2pi = 25 KHz.
My question is to understand this equation. How did they come up with this formula? (https://nmrfacilities.chem.kuleuven.be/applet/excitation_profile.php?spectrometer_frequency=500&pulse_length=10&graph_range=500#graph) -
I know that a rectangular pulse depends on the sin (x) function. But how to predict other excitation profiles?
best regards
Dear Unknown,
In simple terms, the excitation profile of a pulse is the Fourier transform of the pulse. There are however complications with phase of the excitation profile. One can measure the excitation profile experimentally. See the two links below.
http://u-of-o-nmr-facility.blogspot.com/2009/12/defining-excitation-profile.html
http://u-of-o-nmr-facility.blogspot.com/2011/01/excitation-profiles-for-shaped-pulses.html
-Glenn
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