| Author | Message | |||
     Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 506 Registered: 9-2001 |
The compromises that have to be made between spectral resolution, S/N and measurement time, have been known for a long time, and have been called the spectroscopic "trading rules" by the FTIR coomunity. Peter Griffiths has a moderately extensive discussion of these "trading rules" in his 1975 book "Chemical Infrared Fourier Infrared Spectroscopy". Zenqui does well in reminding us of that. There is a point that Zenqui is possibly missing, though, when he expects his measurements to correspond exactly to the instrument settings. The time that appears in the equations, and that are operative for the "trading rules", refers to the actual active measuremnt time during which data is measured and collected. If you measure the total time needed to collect some number of spectra, say by using an external stopwatch, however, that will also include "dead" time, time needed to operate the instrument in order to do the data collection, but during which no actual data collection occurs. There are both electronic, software and mechcanical contributions to the "dead" time, including, for example, the time needed for the interferometer mirror to accelerate to speed, have the speed stabilize, decelerate at the end of a scan and retrace the path to the starting point. Most of those activities take a fixed time to perform, which is independent of the length of the scan. Therefore they occupy a larger fraction of the total scan time when the scans are short than when the scans are long. \o/ /_\ | |||
| ZHENQI SHI (shizhenq) New member Username: shizhenq Post Number: 4 Registered: 3-2010 |
Hi, I think we may miss another important factor which Eric early referred to a bit . That is the measurement time and/or co-adds on the instrument. According the original FT-IR theory, the s/n is proportional to the product between resolution and square root of measurement time. Assuming the measurement time is only a function of co-adds here, when changing an instrument setting of 32cm-1 with 16 co-adds to resolution of 4cm-1, the co-adds need to be at least 1024 in order to match the s/n, which will make the measurement time extremely long and impractical. This is what we found when we tried to optimize the instrument setting for an NIRS-based content uniformity method for API content in pharmaceutical products. Another interesting thing we observed in the same study is when we compared the noise level between two instrument settings of 32cm-1 with 16 co-adds and 4cm-1 and 256 co-adds. What we found was that in spite of the fact that the resolution setting was improved 8 times (i.e., from 32cm-1 to 4cm-1), the noise level of the former setting was found to be better than the later one. The reason is that there was not enough measurement time compensation in order to fully demonstrate the advantage of improved resolution setting. Hope it helps. Pete | |||
| Gabi Levin (gabiruth) Senior Member Username: gabiruth Post Number: 78 Registered: 5-2009 |
Hi guys, Let me see if I can simplify. Resolution when all is considered is the ability to differentiate between two neighboring peaks, The closer the peaks are to each other and the better our ability to differentiate the better is the resolution. Of course, this is a sum total of numerous factors, and these are different for different basic technologies and well discussed in the posts. However, to the subject of this string - there is a need to clarify Doubling the resolution to me means that we go from 16 cm-1 to 8cm-1, rather than from 8 to 16. However, since the question was put in the context of going from 8 to 16, it means that from the simple aspect of slit width the width become larger. Thus, the flux of light increases. If we go back to my previous post, we can see that the shot noise, while increasing in absolute size by the square root of the flux, the ratio of S to the square root of S is getting larger. As said before if S is 100 and the N is square root of 100, i.e., 10 the ratio S/N is 10. If S is 10000 the N is 100 and the ratio is 100. With all other factors well explained in other posts being constant for a given instrument, the nominal resolution - i.e., the amount of energy allowed to get to the sample is the main factor affecting the S/N value. I hope this simplifies what I tried to say in my previous post. Gabi Levin | |||
| Eric LALOUM (elaloum) New member Username: elaloum Post Number: 4 Registered: 5-2006 |
Bonjour Pierre, If the resolution goes from 16 to 8, then the mo.ving mirror travel length is divided by two ; thus for the same analysis time, the number of scans is multiplied by two, and finally noise in the resulting spectrum is divided by square-root of 2. So S is the same and N is reduced and S/N is increased. Bien a toi, Eric | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 505 Registered: 9-2001 |
Tony - yes, that's a good example of what happens whan you change settings. My only quarrel with it is the use of the word "resolution". At best, the term is poorly defined. In practice, the numbers assigned to various settings of "resolution" are not exact descriptors of any physical quantity. "Bandwidth", at least, has distinct physical meaning, even when you need to specify how it is to be measured. At best, values for "resolution" are approximations, and besides hardware issues such as exactly how many laser fringes are counted for an interferogram, "resolution" depends on software issues such as the apodization function used and on whether the final computations of the spectra provide spectral data points at "natural" fequencies (i.e., those corresponding to the actual Fourier inverse of the interferogram) or interpolated frequencies (set to make the spectral data points correspond to some integral number of wavenumbers). Purists tend to prefer the "natural" frequencies, but those are tantamount to being irrational numbers, and correspondingly difficult to use. People who use FTIR for routine analysis tend to prefer the interpolated frequencies, simply because it's easier to write (and think about, and remember) 4530.1 cm-1 than, say, 4530.09752 when identifying a data point. Trust me, Tony: I did FTIR for a long time before getting into NIR, and you're entering terrritory that make the hardware and physics issues I described previously look trivial by comparison. At least those have a fixed basis in the real world. The software issues you start to deal with in the FT world depend on some programmer's decision about how to do something, or maybe even some marketing manager's. Do you REALLY want to have to start trying to guess why one of them made the decisions they did - even if you can figure out what that decision was? Even to this day, FTIR experts atill argue over which apodization algorithm is "best" just as we argue about which data transform is "best"! Best for the novice to stay away from all that quicksand! \o/ /_\ | |||
| Tony Davies (td) Moderator Username: td Post Number: 288 Registered: 1-2001 |
Pierre, I attach a copy of Ed Starks Figure 10 from Proceedings of ICNIRS-91. As you "double" the resolution, detail is lost and and the noise decreases. These were produced from a spectrum measured at 4cm-1 and the Fourier transformation was varied to give the desired effective resolution. If I understand this correctly, more information is being averaged at the higher resolutions so the noise is decreased but information is lost. Best wishes, Tony
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| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 504 Registered: 9-2001 |
Pierre - It occurs to me that we seem to be getting away from your original question. You wanted to know about S/N. But your comment to Gabi indicated that you were concentrating on the noise, when you said "Anyway when doubling resolution (8 to 16), Noise decreases." But that's not what happens, and I was trying to give you the pieces of information you needed to work through what actually does happen. When you change the "resolution" from 8 to 16 (cm-1) what those numbers actually describe is the bandwidth, not the resolution. "Resolution" is simply a label for an instrument setting that affects the bandwidth (more-or-less inversely: "high" resolution means small bandwidth). And as I mentioned before, what actually changes is that the signal increases when you increase the bandwidth. The noise is fixed by the detector at a constant value, so the S/N increased when you increased the signal. If you want to know how S/N is affected, you need to know individually about both parts: how each one naturally behaves, and how they can change. You also seemed to be under the misconception that your question was "simple". It wasn't, though, because digging down to the fundamentals of what actually goes on is not simple, and misunderstanding that can mislead you, as seems to have occured. \o/ /_\ | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 503 Registered: 9-2001 |
Maybe I should also point out that that equation describes the fundamental behavior of light (unless you want to dig deep into Maxwell's equations). It doesn't matter how your instrument makes the changes to the quantities involved, the effect is the same. The equation is based on the First Law of thermodynamics, and simply says that if there's no absorption or scattering of the light from the beam, the energy passing through a portion of the beam (and by extension, the entire beam) is constant throughout the length of the beam, from source to detector (otherwise the First Law would not be holding, because energy would have disappeared). The product dA x d[omega] is known as the "optical invariant" for the beam. That also describes why, for example, when you magnify or demagnify an image (change A), the light approaching the image must converge less or more strongly (change [omega]). \o/ /_\ | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 502 Registered: 9-2001 |
Maybe I should also point out that that equation describes the fundamental behavior of light (unless you want to dig deep into Maxwell's equations). It doesn't matter how your instrument makes the changes to the quantities involved, the effect is the same. The equation is based on the First Law of thermodynamics, and simply says that if there's no absorption or scattering of the light from the beam, the energy passing through a portion of the beam (and by extension, the entire beam) is constant throughout the length of the beam, from source to detector (otherwise the First Law would not be holding, because energy would have disappeared). The product dA x d[omega] is known as the "optical invariant" for the beam. That also describes why, for example, when you magnify or demagnify an image (change A), the light approaching the image must converge less or more strongly (change [omega]). \o/ /_\ | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 501 Registered: 9-2001 |
Pierre - if you go back to the basic equation I sent: dE = [epsilon] dA d[omega] d[lambda] dt Increasing the resolution (as Tony says, go from 16 cm-1 to 8 cm-1) is decreasing d[lambda](the bandwidth). If you decrease d[lambda], with everything else remaining the same, then, from the equation, dE (which represents the signal) is decreased proportionately. \o/ /_\ | |||
| Tony Davies (td) Moderator Username: td Post Number: 287 Registered: 1-2001 |
Hi Pierre et al. (Hope you had a good Thanks Giving, Howard), When you say "when doubling the resolution ...". Have you gone from (say) 16 to 8 cm-1 or 16 to 32 cm-1? Best wishes, Tony | |||
| Pierre Dardenne (dardenne) Senior Member Username: dardenne Post Number: 74 Registered: 3-2002 |
Howard, I did not say Gabi was wrong. You explained the noise sources. But everything else being constant (temp.,..) within the FT instrument, why S/N increases when going from 8 to 16cm-1 ? If S>>, why ? if N<<, why ? Pierre | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 500 Registered: 9-2001 |
Pierre - Gabi is not wrong, but he omitted a point or two in the explanation: 1) Since shot noise (what I called "Photon noise") is proportional to the square root of the number of photons falling on the detector (not necessarily all those emitted by the source), the signal (the number of photons itself) increases faster than the noise does, as more light reaches the detector. Therefore the ratio S/n (where S = signal strength, n = noise level) increases as S/sqrt(S), which is an increase proportional to sqrt(S). 2) The dark current is not itself necessarily a noise source. Only fluctuations in the dark current would contribute to what we conventionally call "noise". Dark current can, however, be a contributor to other sources of spectroscopic errors; for example it can be one of the contributions to what we call "stray light". 3) Gabi gives a good description of thermal noise. What he fails to mention, however, is that any resistor in an electrical circuit creates thermal noise. This includes all the resistors in the detector circuit as well as any input resistance in the first detector amplifier. This noise is proportional to the resistance and, as Gabi said, exponentially proportional to the temperature. It occurs to me that you might be posting your question to the wrong forum. The noise observed in any instrument is dependent on the design of the instrument and how the noise contributions are balanced in thae design. This is usually within the bailiwick of electrical engineering, therefore you should perhaps be asking the question on a discussion group for electical engineers. Another group that deals heavily with these matters are astronomers, you might want to ask them. \o/ /_\ | |||
| Pierre Dardenne (dardenne) Senior Member Username: dardenne Post Number: 73 Registered: 3-2002 |
Gabi, S/N is (for me) signal devided by noise. That�s perhaps a misunderstanding of the foreign language. Anyway when doubling resolution (8 to 16), Noise decreases. Thanks for your explanations. Anyway it seems for me quite confusing and complicated. Could you make it simpler?  Pierre | |||
| Gabi Levin (gabiruth) Senior Member Username: gabiruth Post Number: 77 Registered: 5-2009 |
Hi guys, From what I have learned, noise in photodetectors is made of 3 major contributions: 1. Shot noise that is due to the random nature of the photon emission in the light source. The final equation arrived at by my sources is that the square of the "shot noise" is equal to the flux of photons, so the "shot noise" is the square root of the flux. Therefore, the smaller the flux, the higher the effect of the shot noise. if flux is = 100 the noise is 10, i.e., 10%, if the flux is 10000 - the noise is 100, but it is only 1%. 2. The second part is the dark current. The important aspect is that it is quite a constant for a given detetcor type and independent of the flux of photons. 3. The third contributor is the thermal noise. This is simply reflecting the fact that the electrons in the semi-conductor - our detector - have energies distributed according to the Boltzman distribution. Some of them have enough energy to "jump" across the barrier and create a signal. This is also a random process and depends strongly on temperature in an exponential manner. This is why we stabilize the detector's temperature, and sometimes cool it. This noise is also independent of the flux of photons arriving at the detector. Therefore, when we have two factors that are independent of flux, and one that tells us that the ratio of the signal to the noise is increasing with the increase in signal it becomes clear that as we reduce the signal by increasing the resolution by narrowing the physical size of the beam, we reduce the S/N. | |||
| Howard Mark (hlmark) Senior Member Username: hlmark Post Number: 499 Registered: 9-2001 |
Pierre - Gabi and David have pieces of the answer, but they're not fundamental. There are two pieces to the question: the signal and the noise. Let's start with the signal: The fundamental equation describing the energy through an optical system is: dE = <epsilon> dA d<omega> d<lambda> dt where: E is the energy <epsilon> is the intensity of the light at a given wavelength A is the cross-sectional area of the optical beam <omega> is solid angle the beam subtends <lambda> is the bandwidth of the beam t is time You can simplify the formula somewhat by setting dP = dE / t, where dP is now the optical power density, and is now assumed to be time-invariant. The actual behavior of a given optical system depends on its design, and the effect of that on the optical throughput. For example, assuming a grating system with two slits, closing down only the entrance slit may or may not reduce the bandwidth, whereas closing down the exit slit will. Closing down the entrance slit will, however, reduce the beam area (the A term). The determine the actual effect, you have to integrate the respective derivative terms over the pertinent range, which, as I said, depends on the instrument design. There are similar considerations for FT instruments. Peter Griffith's book deals with those in detail. As for the noise, that depends on the detector and on the noise behavior of the detector. Most of the detectors used in IR and NIR have a noise contribution that is indpendent of the signal level. That's not necessarily true for measurements in other spectral regions. If the detector is sensitive enough, as it is in the UV and visible regions, for example, then the noise is not limited by the detector noise, but by the photon noise, which increases with the square root of the intensity of the light. The effect of all this on S/N is complicated. I worked it out for these two cases, and some others, however, and published it in a series of Spectroscopy columns back in 2000-2001. It's also in our book "Chemometrics in Spectroscopy". You can make some simplifications and approximations but I can't deal with that now, the family just arrived for Thanksgiving dinner. \o/ /_\ | |||
| Tony Davies (td) Moderator Username: td Post Number: 286 Registered: 1-2001 |
Hi Pierre, Gabi and David, Before I started working in NIR I worked with prism and grating spectrometers which had variable slit controls and the pen would shake more if you closed the slits too far! Do any of you have databases containing spectra of the same sample run on dispersive and FT systems that I could borrow? Best wishes, Tony | |||
| David W. Hopkins (dhopkins) Senior Member Username: dhopkins Post Number: 221 Registered: 10-2002 |
Hi Pierre, I think - I'm not sure, I hope someone can confirm this - that the noise goes up with the square root of the light intensity reaching the detector. Assuming a triangular slit function, raising the resolution by a factor of 2 halves the base of the triangle and halves the height, so the area goes down by 4. This would then give the increase in noise of a factor of 2 that you mention. I'm sure this is discussed in basic IR texts. I don't think it is limited to FTIR, it should be observed in grating instruments too. I hope someone can confirm this. Best regards, Dave | |||
| Gabi Levin (gabiruth) Senior Member Username: gabiruth Post Number: 76 Registered: 5-2009 |
Hi Pierre, I think that you meant that the signal to noise is reduced - i.e., the signal to noise gets worse rather than improve, please confirm. I don't really know the tech reason for this - but I have seen many cases where this phenomenon is exhibited. We participated long ago in a comparative study by Roche Switzerland that included the Bomem FTNIR, Foss bench top as standard and our Luminar. The study was later published in the Ph.D. dissertation of Lars Sukowski. The SEP (which reflects the S/N) of the regressions definitely showed higher values due to lower S/N value. Gabi Levin | |||
| Pierre Dardenne (dardenne) Senior Member Username: dardenne Post Number: 72 Registered: 3-2002 |
Hi, A fondamental question; what is the reason S/N is doubled when doubling the resolution with a FT instrument ? Pierre |
Ed's toluene spectra