 Let us continue our discussion with Fourier transform NMR and its various aspects. We are talking about the various practical aspects of Fourier transform NMR and we had reached up to some aspects and one of the important aspects which we had started discussing was the so called folding of signals. You recall that the FID has points because we collect the data in a digital form and the sampling theorem dictates how fast we should collect the data, how the thing should be digitized and what should be the time in between the two consecutive points and that is called as the dwell time and this is determined by what is our spectral width, where we are kept over offset. So the sampling theorem tells us or the next frequency tells us that the sampling rate should be equal to 2 times W max, where W max is the maximum frequency in your spectrum, I mean omega max not W max, omega max is the maximum frequency in your spectrum. So the sampling theorem therefore poses a difficulty that one has to know the frequency range in the spectrum even before you collecting the data. So where to place the offset, what is the spectral range and what is the maximum frequency in that spectrum with respect to the offset is something one has to know. However, this is not easy to know. What happens as a consequence of this is that some artifacts appear in your spectrum. Let me try and explain that a little bit here and this is shown in this particular slide in a picturized way. So if you have a spectral width chosen from here to here, you can see that in my arrow there is from here the blue this line indicates what is the spectral width you have chosen and if your frequency is kept offset is kept here and if there is a line somewhere outside here then it will fold into your spectral region into this and this is in the case of a quadrature detection. And the same thing let me try and explain. So if we have a, this is your carrier frequency and if your spectral region is here, you have lots of lines here, here and here and here and here, you think this is your spectral region. So therefore to excite these uniformly what you do is you shift your carrier frequency to the center of the spectrum here, this is what we said is called the offset. Now if this is properly chosen then of course all your signals will be inside the region what you have chosen, this is the spectral region what you have chosen from here to here. If the offset is properly chosen and you have chosen the spectral width which is determined by the sampling rate, your dwell time what you have given then there is no issue. But suppose you have a signal which is somewhere here which is not covered in this region which you have chosen. In that case this fellow will actually fold into your region of interest in some manner and that is called as the folding of the signal. How does that happen? This happens as a consequence of digitization theorem which we discussed earlier and we will look at that once more. Of course the noise also will fold from here to here, it has its another consequence which we will also discuss in the next few minutes. So what is the consequence of the digitization theorem? Sampling rate is wrongly chosen, as I mentioned you earlier we place the offset somewhere and you choose your spectral width wrongly and then you collect the digitized data in this manner and do a digital Fourier transformation then it is a consequence of this digital Fourier transformation that you get lines to fold. Let us look at that little bit more carefully, if we have an FID which is digitally collected like this and you do a digital Fourier transformation, you remember this from this theorem that you get a spectrum which is in given by this kind of an equation. This is 1 by tau summation minus infinity to infinity f omega minus n by tau. So n is, this is the running index is the n here, now we get series of spectra, f w is your spectrum, f w is your spectrum and you get as many sets of such spectra as there are n values. However, always we choose n is equal to 0, if n is equal to 0 you have F of w. So this whole range from here to here is 1 F of w and from here to here is another F of w. And what is the range of this, this will be 1 by tau, the distance between here to here will be 1 by tau that is from this center to this center it will be 1 by tau because each of them is F of omega is separated by 1 by tau. If n goes from 0 to 1 you increment by omega minus 1 by tau that means the whole spectrum is shifted by 1 by tau region likewise if n is equal to 2 you shift it by 2 and so on. So therefore, there will be a whole range of peaks you will have spectra f w, n is equal to plus 1 will give you 1 spectrum, n is equal to 0 will give you 1 spectrum, n is equal to minus 1 will give you 1 spectrum, minus 2 will be here, plus 2 will be here and so forth. The whole range of frequencies you will have in your digitized data, Fourier transform data. However, of course we always choose one particular spectral region by using what are called as the filters. So when you do a filtering process it eliminates all the others and you pick up only this much region of the spectral width. So this shaded area here is the spectral width. Now if you are wrongly chosen your spectral width or you are offset then you may have a signal which is somewhere present here. So this is not covered inside your spectral region. So what will happen? So this peak will appear from each of these regions from this F w also there will be peak that comes here. The one which actually belongs to here will appear here, the one which actually belongs to this one will appear here and the one which belongs to this one here will appear here. This is an additional signal which is not covered in the spectral region. Now if you therefore you will have a peak here which is coming from this spectral region n is equal to plus 1 peak will appear in this region and this is therefore called as the folded signal. Now depending upon what sort of a detection system you choose of course there will also be folding of noise. So this first of all explains how the folding of signals happens. This is the consequence of digitization of your FID. Now there will also be folding of noise. Suppose your spectral region is from here to here and you put your offset in this region. You put your carrier offset here and your peaks are here and this is so called the single channel detection you have put the carrier at one end of the spectrum. But as a result of Fourier transformation of course you will generate both positive and negative frequencies and those negative ones may appear in this area but that does not matter to us because that does not interfere with your any of your signals. But these peaks will not fold because you have used a filter which goes like this and that will eliminate this and it will not these peaks will not matter to you. But then this of course the noise which is present here which will fold into this area. These peaks are eliminated so the noise will be added into this area. So noise will get doubled up whereas the signals are not disturbed you will have the signals here. The signal to noise ratio will be different than what happens if you put your carrier in the middle. If you put your carrier in the middle this is your region of interest and if you put the carrier in the middle the positive and negative frequencies are distinguished in the case of quadrature detection this is single channel and this is quadrature. So if the carrier is put here your positive and negative frequencies are distinguished positive and negative noise will be distinguished therefore there is no additional noise folding into a spectral region. As a consequence if you do quadrature detection your signal to noise will be enhanced by a factor of root 2. Root 2 enhance maintenance signal to noise for quadrature. This is the consequence of folding of noise different types of folding of noise in the case of single channel and quadrature. In the case of quadrature there is no folding of noise in single channel there is a folding of noise and therefore there is a loss of signal to noise by a factor of root 2. So this is the folding problems and we are now going to see other parameters which are specific to the Fourier terms of NMR. The next parameter is the acquisition time how much data you collect acquisition time is defined as the time for which data is collected in the FID. So here is the FID so this goes from starts from here and you collect the data all the way up till here. What happens if you go beyond this? Of course there will be no signal here and there will be mostly noise which is coming here. So therefore you do not want to collect noise but typically therefore you are limited by the relaxation of the signals in as to how much data you will collect. This is the time axis here to remember. Therefore this is dictated by the transverse relaxation time of your spin system. So by about 3 times the transverse relaxation time this signal would have to get almost to 0. If you collect any data beyond that you will be basically collecting noise therefore you do not want to collect any data beyond that up till that time only you will collect. So this time is called as the acquisition time. Total from here to here is called as the acquisition time. Let us be more quantitative in this sense. Now suppose you have collected N data points and if the time interval between two consecutive data points is tau then your total acquisition is N times tau and where tau is 1 by 2 times omega max, omega max is the maximum frequency in your sampling in the spectral region and therefore the acquisition time will be equal to N into tau that is N divided by 2 times omega max this is for single channel. In the single channel we have kept the carrier at one end of the spectrum therefore the maximum frequency is the entire spectral region, spectral width. In the case of quadrature you keep the carrier in the middle as I explained to you before. Therefore the spectral width is suppose it is SW then that SW is equal to 2 times W max because you have put the carrier in the middle there are positive and negative frequencies you have omega max on the positive side and omega max on the negative side therefore there is total spectral width therefore is equal to 2 times W max. The sampling rate will be 1 by 2 times W max and so therefore your acquisition time will be N divided by 4 times. Why is it 4 factor here your sampling rate is 1 by 2 times omega max yes but how many data points you collect when you collect data you actually collect real and imaginary points in quadrature they are collected simultaneously if you are collecting a total of N data points you collect half of them as real points and half of them as imaginary points. So therefore you can see here if you are collecting N data points this is for the Fourier transformation and you collect N by 2 points as real points and N by 2 as imaginary points that is you collect X magnetization and Y magnetization you have N by 2 points for each one of those. So in quadrature you collect N by 2 real points that is if you call this as Y magnetization and you collect N by 2 imaginary points and if you call this as X magnetization. So therefore your acquisition time N by 2 into 1 by 2 times omega max and therefore that becomes N divided by 4 times omega max. Now the next parameter that is related to this is the so called digital resolution. Suppose your FID is represented by this equation S0 into e to the i omega t into the e to the minus RT. This omega is actually your sample frequency we can collect here only let us say there is only one frequency what is the result of the Fourier transformation here. Now you remember the spectrum also will be digital in nature therefore there will be a certain number of points in your spectrum as well. If you are collected N data points in your FID then the Fourier transformation of this FID will result in two components this is the first component is the real component and the second component is the imaginary component. This is given by these equations here and this imaginary component has results in a line shape which is like this and there is results in a line shape which is like this. This is your absorptive spectrum and this is a dispersive spectrum and the total N points which you have collected will now be distributed between these two spectrum. So you will have N by 2 points to represent your real data and N by 2 points to represent your imaginary data. And we pick up generally only the real points because it results in an absorptive line shape. So remember both the frequency informations are entirely present in both the real points and the imaginary points just that they have different phase relationships and we will be needing these at some other point later but for presentation we generally choose absorptive line shapes which is indicated here. N data points which are collected in the FID which consists of N by 2 along the x axis, N by 2 along the y axis and that Fourier transformation will result in N by 2 points for the real part of the spectrum and N by 2 points for the imaginary part of the spectrum. So this is the consequence of the Fourier transformation as I indicated here you see I indicates here the imaginary component and this has a line shape which is like this this has a line shape which is like this. And what is R? R is 1 by T2 this is a relaxation rate okay transverse relaxation rate and that is multiplying your FID how fast your FID is decaying is dictated by this factor and that appears as your as a line width in your data later on. If this is equal to if R is equal to T2 then this is roughly will be equal to yield to the minus T and then it is called as a matched decay or one we will see later that it is also generally use such a kind of a multiplication later on to call what is called as a matched filter that will come in the next few minutes okay. Therefore your digital resolution will be defined in this manner you are in the single channel direction your total spectral width is omega max because you kept the carrier at one end so our maximum frequency is equal to your spectral width and the number of data points which are representing this spectral region is N by 2 therefore this is 2 times omega max divided by N and that is equal to 1 by acquisition time T acquisition. In the case of quadrature detection also it happens in the same manner now you have 2 times omega max is your spectral width and you are collecting N by 2 data points there and therefore this is also equal to 1 by acquisition time. So this is what is explained here in quadrature detection N by 2 real points and N by 2 imaginary points are collected simultaneously and dwell time is twice that in single channel detection since the largest frequency is half of the spectral width. So in both cases digital resolution is inversely proportional to the acquisition time and we have to optimize acquisition time to get maximum signal to noise and maximum resolution. The next point to consider is signal averaging and pulse repetition rate we say signal averaging is easy in the case of Fourier terms of NMR because you can improve you collect the data for a very short period of time and you can repeat this process so that you can increase your signal to noise ratio. This is schematically indicated here you have a particular relaxation delay and you apply a pulse acquire the data and repeat this relaxation delay here again apply the pulse acquire the data same thing continues several times. So each one of them is called as a scan. So from here to here is a scan 1, here to here is scan 2 and here to here is scan 3 and so on. And you add all these FIDs these are all individual FIDs, FIDs of scan 1, FID of scan 2, FID of scan 3 all these are added and then you Fourier transform at the end. So this is the signal averaging process. Now the question is how much relaxation delay one should give because we want that the magnetization should recover back to equilibrium or should it we will see that in few minutes and after that you reach a signal data collection. So what is the best value of the relaxation delay one should choose or the TP from here to here capital TP how much is the value of TP one should choose and what are the constraints in deciding on this parameter. Detailed calculations have been done because in the end we actually we reach a steady state here when do repetition this several times every scan has to be identical to the previous one and then we call that as a steady state. You it in every scan you get the same kind of a signal and then that is the steady state signal in the repetitive experiment. So when you do this one can do a calculation we will not go into the details of this calculation this has been done when equilibrium or a steady state is reached the MX magnetization after each pulse is given by this equation. This is MX plus this is the magnetization is proportional to M naught M naught is the equilibrium magnetization this is the maximum magnetization sin beta where beta is a flip angle what you apply we said we apply a pulse. So what should be the flip angle of the pulse should be 90 degrees 30 degrees 45 degrees what it should be. So this is the question one has to optimize and then it is proportional to 1 minus E 1 divided by 1 minus E 1 cosine beta as I said here beta is the flip angle and E 1 is related to TP and the longitudinal relaxation time T 1 or the spin lattice relaxation time T 1. Maximum amplitude is obtained for an optimum flip angle which is given by cosine beta optimum is equal to E 1 when you match this condition you will get maximum signal to noise ratio. So that is the optimization one has to do so therefore you will see depending upon what is the value of TP by T 1 because E 1 is equal to exponential minus TP by T 1 you do not necessarily have a beta of 90 degree you do not necessarily require a 90 degree flip angle. A detailed calculation will show you what is the dependence of the beta optimum on the T 1 relaxation time or the ratio of TP by T 1 and this is indicated here you plot here Mx by M naught as a function of the flip angle for various values of TP by T 1. So your TP by T 1 is 0.01 and TP by T 1 is 0.1, 0.25, 0.5, 1, 2, 3, 5 and 10 so and this is see the 10 you give a TP which is 10 times the T 1 value and this particular curve here and you notice for this the flip angle is 90 degree. So after you apply the 90 degree pulse which you have been assessing all along you give enough time for the magnetization to recover back to equilibrium which is along the Z axis and then you will get the maximum signal to noise ratio. However, if the T 1 is very long then you will have to wait for such a long time before you actually apply the pulse. So if T 1 is let us say 10 seconds then you will have to wait for 100 seconds in between 2 pulses in between 2 scans and that is actually not necessarily economical from the point of view of spectrometer time or signal to noise ratios per unit time. So and here is a plot for example if you are having a TP by T 1 this curve if you take this is 0.01. So if the T 1 is very long then if you apply a TP which is TP by T 1 is equal to 0.01 then you do not need a 90 degree flip angle you will get a maximum signal to noise ratio when the flip angle is this much only 15 degrees. This is 15 degrees this may be only 10 degrees. If you go to use the right curve TP by T 1 is equal to 0.1 then you need you get this much you get a maximum signal when you are using approximately 30 degrees. However notice that you do not get the maximum signal here this is the maximum signal per unit time this when you are this is the best one we will get. So therefore this signal averaging will depend upon what is it you want to get for the maximum signal intensity in your spectrum how much to optimize. And this same curve if you plot it in a different style you plot the beta optimum versus TP by T 1 and you can see here for very large T 1's you need to have actually optimum value of flip angle is given by this manner. So you have TP by T 1 is very small then you have floppy flip angle is here and TP by T 1 is very long then you have a 90 degree flip angle. So therefore this takes this sort of a shape in your experimental setup this have to be optimized so that you get the best signal to noise ratio per unit time in your spectrum. Now the next thing is the data processing in Fourier transform NMR. The number of data points in your FID is restricted as I mentioned to you before because the signal to noise considerations there is no point collecting data after this stage because there is only noise there therefore you often end up truncating your FID at this point you collected this many points. Now if I do a Fourier transformation of this data this is truncated FID then I get a signal like this which will have some wiggles of this type here there are small wiggles at this point and that is I mean this is not wiggles here actually this is the line shape is not very good because you are suddenly a truncated the FID here you have this kind of a problem. Now to improve this digital resolution what one does is you add artificial points you add some additional zeros here this is called a zero filling you add some additional points here this do not have a noise these are just zeros. So instead if you collected 1024 points here for example you add another 1024 zeros here therefore your total number of points becomes 2048 if you add more of course you will get more points essentially this is to improve your digital resolution not the inherent resolution. So your number of points per hertz in your spectrum will be better that is called as a digital resolution. So if you do that now if you add zeros and then a Fourier transformation then you get the same thing but you see your digital resolution is improved your signals are better represented in this spectrum however you are getting some wiggles here by the side of each signal and there is more explicitly indicated here you have truncated a FID with zero filling then you get this wiggles here these are the form of sync or the sin x by x wiggles appearing on either side of your lines and this is not desirable. So therefore what one has to do we have to get over this problem we have to get over this problem and this is done by what is called as upwardization or window multiplication and or it is also called filtering digital filtering. Okay so there are various ways one can do it. So there are various functions one uses to multiply your FID and I will show you a few of those examples here. So the first thing that one could do is multiply by an exponential function which is of this form window multiplication digital filtration or window multiplication or all the three terms mean the same thing okay different times you use different terminologies. So you multiply an exponential function exponential function is given by this manner e to the minus t by t max so you have the FID truncated FID with zero filling you multiply by this exponentially decaying function which goes like this all the way till the end of your FID and then you will get a spectrum which is looking like this. So this is the result of multiplication you notice the sudden the termination here is removed and you get a smoothly going FID here and it comes to zero at this point nearly zero and this results in a cleaner spectrum all those wiggles which were present here have vanished but you notice one more thing as a consequence of this your line resolution which was present here has disappeared or rather it has reduced. So it causes the line broadening so the lines get broadened by an amount 1 by pi times t max and the resultant line weights will be given by this delta nu is equal to t2 prime plus t max divided by pi t2 prime t max. So this is the price one has to pay for improving the appearance of the spectrum in terms of the removing of the wiggles by the side of each of the lines. Then you have other functions the next function which is commonly used is the cosine function when you do the cosine function you multiply your FID by such a kind of a function. So this will be a function which adjusted in such a way that you start from here the zero point is here and it will slowly come over and come to zero at this point. So you multiply a function essentially you apply the function with a certain frequency in other words the wave has a certain frequency or it has a certain shape so that it starts at zero here and comes to zero at this point. So that is the last point and once you multiply with such a function then your FID becomes like this see it is smoothly going down to zero at this point the wiggles have gone and also there is not much loss in your resolution in your spectrum. Therefore this is the optimally used multiplication function and by and large one uses this sort of a function to improve your signal to noise ratio or digital resolution as well. Now sometimes you want to use a function of a different type which is called as the sine bell function. The sine bell function has this sort of a formula this is the pi t by t to make this was the same as before in the case of cosine but now add here a phase. This is the phase so now you see if this phi if it is equal to pi by 2 this becomes the same as the cosine function but this putting this phase would allow you different possibilities. Now suppose you apply phi is equal to zero and then it is a total sine function that means the sine function is zero at time t is equal to zero therefore this function will go like this and come down and then comes to zero at this point. So this is your wave this is your sine wave which is applied for multiplication. So when you do this you get significant enhancement in the resolution in your spectrum but you also get distorted line shape you get many negative peaks here you get each line has some kind of a wiggle like this and you have a distorted line shape. If you use phi is equal to 30 degrees then it is the kind of a shifted cosine. So you start from here so it is the you do not start from zero but you start from here so your signal to noise is not sacrificed as much but you have improved resolution but still you have some of this week the line shape distortions you get some negative peaks which is present here. Phi is equal to 60 produces peaks like this and phi is equal to 90 is the same as the cosine function which was shown in the previous case. And one more function which I want to show you this is called the Lorentz Gauss this function as this sort of a formula is e to the power t by t2 star minus sigma square t square divided by 2. Sigma is a parameter which has to be adjusted for a given t max and an estimated t2 star. So when you multiply your FID with such a kind of a function so your FID looks smoother as and it comes to zero at this point and when you Fourier transformation you have a reasonably good signal to noise ratio and resolution in your spectrum resolution therefore combining with zero filling and multiplication by suitable functions you can get better signal to noise ratio as well as better resolution in your spectrum. So these are typical features and we may call this as cosmetics of data collection and processing and that is what is practically used and one has to optimize these parameters for better signal to noise ratio and better resolution in your spectrum. So with that we come to close to this session and we will continue with the data processing and other aspects of Fourier transform NMR in the next class. Thank you.