I´a m chemical engineer and I can say that the content of this video resumes almost 3 university courses or more depending the schoool, amazing how much talent is here!
One of the best presentations of Fourier transform applied to spectroscopy. Not only that, but apodization, resolution, etc. Thank you for making this so well!!
Your video really helps me a lot! Fourier transform is a bunch of integration and I don't understand it just by reading journal articles. You made it easy and simple! Thank you!
Thanks for your comment. I should emphasize that treating it as a series of integrated cross products (multiplication) of sample signals and test frequencies is a simple way to explore the idea but the actual math is much more powerful. I don't have a clue how to do real math so I stick with calculus. I am glad that my interpretation was helpful.
I dont know exactly what is going on, but some of the early math is wrong (or stated incorrectly). for example the integral of cos(5x)*cos(5x) from -.5 to .5 does not equal 1, in fact cos(whatever)^2 is equal to 1 - sin(whatever)^2 so it is always less than one. It is still a really good explanation wise, but something is wrong with the early math (around 2:40).
That is amazing lecture. Thank you so much ! I have a question. What can the intensity of FTIR peaks tell us about ? What is the meaning of the intesity in FTIR ?
The ratio of observed intensity to original intensity of IR light is what is reported in a typical IR spectrum output. If a peak is at 20% intensity that means that 80% of the IR light was observed at that wavelength. By converting the ration to absorbance we can obtain a number that is directly proportional to concentration. Abs = -log[I(out)/I(in)]. If we know the molar absorptivity for a given substance at a given wavelength we could use the abs value to calculate concentration. This would only work where we have a dilute solution (usually in chloroform). Other IR methods using thin films or KBr pellets to mount the sample would not give results where concentration has much meaning.
One of the best presentations of FT, but i still din't understand why need to use interferometer, can we just use different frequencies of light are shone on the sample simultaneously then do FT? why need interferometer.
That is where real mathematics comes in. I used a continuous function made up of a set of combined cosine waves and tested it with a single cosine wave of a chosen frequency. The integration of the product will give an answer of 1 if the frequency is present and zero otherwise. However you will see in the presentation that there are serious limitations to this approach and it is not a "real world" application. In the FTIR, the spectrum received is a series of data points taken at set distance intervals (a multiple of the wavelength of the laser you see passing through the sample chamber). So you have a list of power received at each distance and need to turn that into a list wavelengths in the signal and their power. This is where the "Fast Fourier transform" (FFT) algorithm is used. It is designed for manipulating data points, not continuous functions. I sadly can tell you little about it, but it is famous in math and computer science and there are many videos dedicated just to the FFT available out there. My method demonstrates the idea of the Fourier transform but is useless otherwise.
I made the slides in Adobe Illustrator and then copied them over to Apple Keynote to set up the slides. All the transitions are "magic move" transitions where I copy the slide, change elements and the transition takes care of any movement, scaling or fading/appearing.
One of the best presentations of FT, but i still din't understand why need to use interferometer, can we just use different frequencies of light are shone on the sample simultaneously then do FT? why need interferometer.
That could be done. However then you need either a diode array detector or the machinery to move a diffraction grating so that the light crosses a detector as the grating moves. In that case you would not need FT as you are analyzing the energy at each discrete wavelength. UV-vis spectrophotometers use this method. IR wavelengths make the use of the interferometer convenient and less expensive than continuous wave systems (like UV-vis). Diffraction gratings for IR are difficult to make but not impossible. I myself am not familiar with all the limitations and advantages of both approaches but clearly for IR the industry has chosen interferometry for its effectiveness, cost and convenience in this wavelength range.
I´a m chemical engineer and I can say that the content of this video resumes almost 3 university courses or more depending the schoool, amazing how much talent is here!
One of the best presentations of Fourier transform applied to spectroscopy. Not only that, but apodization, resolution, etc. Thank you for making this so well!!
Do you know why FTIR need interferometer?
Thank you! It really helped me prepare for my presentation. The quality of the presentation is outstanding!
Your video really helps me a lot! Fourier transform is a bunch of integration and I don't understand it just by reading journal articles. You made it easy and simple! Thank you!
Thanks for your comment. I should emphasize that treating it as a series of integrated cross products (multiplication) of sample signals and test frequencies is a simple way to explore the idea but the actual math is much more powerful. I don't have a clue how to do real math so I stick with calculus. I am glad that my interpretation was helpful.
Really enjoyed this presentation. Many thanks for your work, I greatly appreciate it!
I wish I could give you a high five and buy you a beer for these incredible lectures!
The best presentation I've ever seen!
taking the red pill is never easy but I loved the lecture! Thanks so much! :)
Lovely, should have been on youtube back when I was studying!
Fantastic lecture on UV/vis and IR spectroscopy! Would love to donate to your channel.
Best explanation I have ever seen
After taking the lecture from the Prof. Linkletter, I really want to go to his university and take lectures from him.
Thank you very much!! Very clear ans concise explanation of the content🔝
finally I understand fourier transformation!! 😍
I dont know exactly what is going on, but some of the early math is wrong (or stated incorrectly). for example the integral of cos(5x)*cos(5x) from -.5 to .5 does not equal 1, in fact cos(whatever)^2 is equal to 1 - sin(whatever)^2 so it is always less than one. It is still a really good explanation wise, but something is wrong with the early math (around 2:40).
that is integartion with span=1 ( the area)
This is lovely. Thank you so much
Awesome video. Thank you so much!!
That is amazing lecture. Thank you so much ! I have a question. What can the intensity of FTIR peaks tell us about ? What is the meaning of the intesity in FTIR ?
The ratio of observed intensity to original intensity of IR light is what is reported in a typical IR spectrum output. If a peak is at 20% intensity that means that 80% of the IR light was observed at that wavelength. By converting the ration to absorbance we can obtain a number that is directly proportional to concentration. Abs = -log[I(out)/I(in)]. If we know the molar absorptivity for a given substance at a given wavelength we could use the abs value to calculate concentration. This would only work where we have a dilute solution (usually in chloroform). Other IR methods using thin films or KBr pellets to mount the sample would not give results where concentration has much meaning.
@@keynotechem Thank you very much for your answer !
One of the best presentations of FT, but i still din't understand why need to use interferometer, can we just use different frequencies of light are shone on the sample simultaneously then do FT? why need interferometer.
For FTIR instrument, the IR source come with single or multi wave length?
but, how can we get a function of the FTIR interferogram that we can put into the Fourier transformation?
That is where real mathematics comes in. I used a continuous function made up of a set of combined cosine waves and tested it with a single cosine wave of a chosen frequency. The integration of the product will give an answer of 1 if the frequency is present and zero otherwise. However you will see in the presentation that there are serious limitations to this approach and it is not a "real world" application. In the FTIR, the spectrum received is a series of data points taken at set distance intervals (a multiple of the wavelength of the laser you see passing through the sample chamber). So you have a list of power received at each distance and need to turn that into a list wavelengths in the signal and their power. This is where the "Fast Fourier transform" (FFT) algorithm is used. It is designed for manipulating data points, not continuous functions. I sadly can tell you little about it, but it is famous in math and computer science and there are many videos dedicated just to the FFT available out there. My method demonstrates the idea of the Fourier transform but is useless otherwise.
This is AMAZING
hey! Super nice video. Could I ask you how do you make these kind of presentations, like what program?
I made the slides in Adobe Illustrator and then copied them over to Apple Keynote to set up the slides. All the transitions are "magic move" transitions where I copy the slide, change elements and the transition takes care of any movement, scaling or fading/appearing.
@@keynotechem thank you, it looks very nice
Great vdo!
Thank you
Excellent!
Great!
very interesting
Good
you are Jesus
One of the best presentations of FT, but i still din't understand why need to use interferometer, can we just use different frequencies of light are shone on the sample simultaneously then do FT? why need interferometer.
That could be done. However then you need either a diode array detector or the machinery to move a diffraction grating so that the light crosses a detector as the grating moves. In that case you would not need FT as you are analyzing the energy at each discrete wavelength. UV-vis spectrophotometers use this method.
IR wavelengths make the use of the interferometer convenient and less expensive than continuous wave systems (like UV-vis). Diffraction gratings for IR are difficult to make but not impossible. I myself am not familiar with all the limitations and advantages of both approaches but clearly for IR the industry has chosen interferometry for its effectiveness, cost and convenience in this wavelength range.