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Part 3: The Contrast Transfer Function - G. Jensen 

caltech
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17 сен 2024

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Комментарии : 24   
@alcyonecrucis
@alcyonecrucis 9 лет назад
Totally revolutionized the way I think about contrast transfer function from three boring letters "CTF" in my head to a beautiful mathematical concept
@grace96342
@grace96342 Год назад
The clearest demonstration ever. Thank you you saved my exam scores
@capitus1295
@capitus1295 7 лет назад
It had taken me a very long while and a lot of research to get a grasp on what was behind this CTF, but this video just amazed me by how simply it managed to explain it. And the orchestra metaphor at the end is just perfect to compare!
@HawaiianBlack
@HawaiianBlack 6 лет назад
Same here, I've spent months asking people and reading papers trying to wrap my head around the CTF, but this explained it perfectly
@AlpayBurakSeven
@AlpayBurakSeven 6 лет назад
Unfortunately, it is not an accurate description. According to this explanation, we should observe oscillating contrast transfer even without defocusing, which is not right.
@clarkmichael7459
@clarkmichael7459 6 лет назад
No wonder!
@lalitasharma6687
@lalitasharma6687 4 месяца назад
Best explanation
@Marc11121
@Marc11121 8 лет назад
That was incredibly helpful! :)
@RicardoRighetto
@RicardoRighetto 7 лет назад
Great explanation of why the CTF exists.
@peasant12345
@peasant12345 4 года назад
How the photon-atom scattering example relates to elastic scattering which preserves the speed of electron? I don't see the pic clearly.
@ptarmigan1356
@ptarmigan1356 5 лет назад
The part about scattered waves being phase shifted by 90 degrees reminds me of what I am just learning about anomalous scattering in x-ray diffraction. This occurs when the x-rays have the same energy as a particular electronic transition. So I am wondering, for this scattering of electrons with associated phase shift, does their energy have to match something? And does any scattering without phase shift occur?
@lfestrozi
@lfestrozi 6 лет назад
I think this explanation is wrong in the sense that it depicts the CTF as independent of defocus. In the end he shows the correct formula where defocus appears as the main cause of CTF oscillations. Moreover, if the drawings were correct, the CTF would oscillate according to sin(l*defocus*k+...) but the formula is sin(l*defocus*k^2+...)
@AlpayBurakSeven
@AlpayBurakSeven 6 лет назад
I agree with you Leandro Farias Estrozi. This representation cannot be accurate since it does not depend on defocus. Everyone here should watch the way Marin van Heel explains the same phenomena.
@ambitionsatangel3453
@ambitionsatangel3453 5 лет назад
I agree with you.
@dvivek07
@dvivek07 5 лет назад
If you vary the defocus, the path length would vary and hence the phase shift also would. That would change the CTF curve. I don't see why this explanation does not take defocus into account.
@lfestrozi
@lfestrozi Год назад
A great explanation by Fred Sigworth can be found here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-m2Hm1ziZFZg.html
@somnathganapa5789
@somnathganapa5789 7 лет назад
Shouldn't we worry that some rays are lost in the production of the final image?
@brendawilliams8062
@brendawilliams8062 2 года назад
Thankyou
@Anonymous-nf1zt
@Anonymous-nf1zt 4 года назад
26:06 Awesome
@somnathganapa5789
@somnathganapa5789 7 лет назад
Why would all of the scattered rays converge at one pixel instead of many? Isn't the interaction of a single electron with the sample depicted across many pixels?
@RicardoRighetto
@RicardoRighetto 6 лет назад
This effect is represented in reciprocal (Fourier) space. When you go to real space, i.e. your actual image, the interaction is indeed "depicted" across many pixels.
@jongseokchoi4038
@jongseokchoi4038 6 месяцев назад
24:40
@robertstroud7793
@robertstroud7793 5 лет назад
Regrettably this is completely incorrect and misleading. The zero crossings in the ripples in the CTF depend on the amount of defocus, not on the path difference of electrons at different scattering angles to the single pixel used in this depiction. The single pixel of the detector is implied to be in the direction of the direct beam (straight through beam) which would be at k=0. There is no treatment here of how the defocus distance plays into the equation, -until it magically appears in the final formula, - after the statement that 'you can find that in books'.
@alexanderstrubel4988
@alexanderstrubel4988 4 года назад
It is explained in the next video.
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