FI109238B - Multicomponent analysis of FT-IR spectra - Google Patents

Description

1 109238 Multicomponent Analysis of FT-IR Spectra

BACKGROUND OF THE INVENTION The present invention relates to a method for analyzing the FT-IR spectrum of an unknown gas mixture whose gases are not known, not to mention their partial pressures. Instead, a large number of pure molecular gas library spectra are known, measured at known pressures and with the same interferometer as an unknown sample.

1. Field of the Invention The FT-IR spectroscopy used in multicomponent analysis has an IR spectrum of an unknown gas mixture, the gases of which are components of the gas mixture, not to mention their partial pressures. Instead, a large number of pure molecular gas library spectra are known, measured at known pressures and with the same interferometer as an unknown sample. Using these pure spectra, the partial pressures of the pure gases in the mixture and their error limits should be calculated. Errors in the values obtained are due to measurement noise in the spectra. 25 Since the computation of partial pressures that best explain the spectrum of the mixture is a relatively simple task, it is only considered briefly and focuses on the calculation of their error limits. Consideration shall also be given to the choice of a solution which gives a minimum error margin of 30 and the application of a non-negativity constraint to partial pressures. In addition to the spectra of gases, all calculations are also applicable to the spectra of non-interfering fluids.

2 109238 2. Description of Prior Art

Let s denote the measured spectrum of the mixture to be analyzed and let {KJ | j = 1, ..., M} be the set of library spectra by which s is attempted to be explained. Assume that all spectra are linearized, i.e. that the y-axis always has a negative logarithm of transmittance (possibly with the exception of a factor). This means that an absorbance scale is used. Then, according to Beer's law, s 10 is a linear combination of spectra Kj, or S = (1) j-i

Our task now is to determine the coefficients x ^ with which 15 can explain s as well as possible. The partial pressures of the pure gases in the mixture are then obtained by multiplying the measurement pressures of the respective library spectra by their coefficients in a linear combination.

Since all spectra are known only at spaced intervals of υ1, ..., υΗ, s and KJ can be treated as vectors, and equation 1 can be represented in vector form as follows 25 / Kn \ (\ f \ f Sl \

Kn K22, K2M ^ I]. + 12. + ... + XM - · '(2)

\ Kni / \ Kff2 '\ KtfM' \ StfJ

30 This equation can be further simplified by aggregating all Kj vectors into a matrix K, which gives the matrix equation

Kx = s, (3) 35 3 109238 where f Κιι K12 · * · Kim \

K21 K22 · · K2M

K =

5 \ Κηι Ktf2 ··· KtfM J

and x is a coefficient vector (vertical vector) containing the pure spectra coefficients Xj.

Summary of the Invention

Now, all those spectra that can be accurately explained by the set of pure spectra M used can be represented as Kx. Although the set {KJ | j = 1, ..., M) would contain all the components in the mixture, but there will always be some noise in the spectra, so that s is never exactly in the vector space developed by the vectors Kj. Thus, a linear combination of pure spectra Kx that is as close as possible to vector s must be satisfied. This means that the norm of the unexplained residual vector s-Kx must be minimized. Our problem can thus be written in the form of the following optimization problem 25 II s - Kx II min! . (4)

The optimization parameters are then the components of the coefficient vector x.

The optimum value of the coefficient vector depends on which vector standard is used. This norm is now defined by the condition ιμκ ^ Σ ^ · (5)> 1 1 = 1 109238 4 This is the standard sum vector standard, except for the standard factor 1 / N. The purpose of this coefficient is to make the vector standard as independent as possible of the number of data N (and thus independent of the solution).

5 It has no effect on the solution of optimization problem 4. The solution to Problem 4 is now obtained by setting the gradient of the ice-thrust vector to zero, i.e. V II s - KX I! = 0. (6) 10

A very simple calculation shows that this condition is satisfied by the coefficient vector xopt = (KTK) '1KTs (7) 15 (provided, of course, that N> M). This is known in the literature as a pseudo-inverse solution of the matrix equation Kx = s (see e.g. [1]). Due to the choice of the norm, see equation 5 - this can also be called the solution of the smallest ne-20 excess. The matrix KTK is reversible if and only if the vectors Kj are linearly independent. If they are not, then the best solution is not unambiguous. If error limits are not required, the following 25 methods are proposed for the application of Equation 7.

Let us now define the scalar product <a | b> of two vectors a and b from (a | b) = - Τ 'a.b,.

N h 1 (8) 30

A factor of 1 / N again ensures that the scalar product is as independent as possible from the number of data. Similarly, the normal relation between the norm and the scalar product is Il v II = <v | v> 1/2. We now see that equation 7 can be written as 5 109238 xopt = A_1y, (9) where 5 Aij = <KijKj> (10) and

Yj = <Kj I s> (11) 10 Matrix A thus contains the mutual scalar product of the used library spectra and can be called a scalar product matrix. If we are not interested in the error limits of the coefficients, then any possible scalar product can be calculated in advance by the library. Thereafter, in each of the 15 analyzes, it is only necessary to extract the scalar products corresponding to the set of library spectra used in this analysis, to compute the matrix A, to compute the vector y and to solve the matrix equation Ax = y. If this is done using the Gaussian elimination method (see ·. ·, 20 e.g. [2, p. 179]), there is no need to translate A, and • * · analysis is a very fast procedure.

All noise in the spectra produces errors in the optimal coefficient vector of Equation 9. In this explanation, only the 25 noise in the spectrum of the mixture s is discussed. Also, the noise in the library spectra Kj could be overcome. takes into account roughly, but here it is wasted for two reasons: First, the research would be too. ·, long and only approximate. Second, in practical applications, it makes sense to register the library spectra 'I' at a much higher resolution (i.e. ; V a larger number of individual sequences) than the spectrum of the mixture i 'obtained by rapid measurement, so virtually all noise is in the spectra of the mixtures, and the library spectra can be considered to be non-noise.

109238 6

The library spectra KJ parent vector Reina have been used above to explain all measurements in the form E '= 1x.jKj or Kx. However, when calculating the error limits, it is preferable to use an orthogonal base that generates the same vector space as the original library spectra.

These orthogonal base vectors are constructed by the known Gram-Schmidt orthogonalization method (see a course book on linear algebra, e.g. [2, p. 138]). This means that the new base vectors {K'j} are defined by 10 recursive sets of equations K'1 = K 'τΛ (Κ'ΙΚ'1) ,,,, K -K w 15 <:. · (12) TCii ^ (^ \ K'r) r

y. K -K 2y || K, rj | 2K

* ·%> «Φ · · 20: * · t ·

Thus, the old base vectors are obtained from the new base vectors y as follows * * 25. : f K1 = K'1: ··; 2_ {k2 | k '·) ,,,

::: k- | k «i» KtK

* *: (13) ··· 30 ρ-γΙΚΠ. ,,,. ,,; έίΙΙκ- || 2Κ + Κ \ · 109238 7

Next, the transform matrix Q is defined as' 0, i> j

Qij = | l, '. . i =. (14) (ΚΊΚ '·) || K' · IJ3 J> 15 (Q is therefore an upper triangle matrix.) The transformation of the base can be expressed in a very concise form as K = K'Q. (15) 10 The conversion of the coefficient vectors between these two base vectors is controlled by x = Q_1x '· (16) 15 Thus, an inverse matrix of Qn is required. After quite laborious calculations, it is found that it is an upper-triangle matrix obtained by the following {0, i> j 1,. . · =; · <17> -EL + .e «« v. i> · »tii · 'I The elements with j> i must be calculated in the order i = Ml, *» · • 1 »* · · · ^ 1 ·, ί · The determination of the best coefficient vector is very simple in strain K '. Since this strain is orthogonal,: * ·, · CK '^ K' ^ = δ1;) and the scalar product matrix A '(see. ···. Equation 10) is diagonal, «· · A' = diag (|| K '1 || 2, || K'2 || 2, -, || K'M || 2). (18) •: 30 v. Thus, the best coefficient vector χς t gives (see equation 9) / 11 K'1 r 'o \ x'.,. = A'-y = (Κ ^ Κ'Γ'Κ'3 / = ± KlTs (19) V ol | K "v '|| -v 5 e 109238 eli *' - <K ,, ls) ept ' 3 II K'1 II2 '(20)

The vector x'opt gives the orthogonal base vectors K'J

optimikertoimet. The corresponding vector x of the true pure spectra KJ is then Q'1xJ) pt. As intuitively very clear, this equals the optimal coefficient vector xopt in the original strain. This fact 10 can also be formally proved in the following way: x'opt = (K 'TK') _1K 'Ts = [(Q' ^ VKQ - ^ - 'iQ · 1) 1 ^ = = Qi ^ K ^ QWr ^ s = Q (KTK) _1KTs = Qxopt.

15 This means that * oPt = Q_1x'opt · (21)

Let us now consider the effect of measurement noise on the optimal coefficient vector of equation 9 or 21. For this purpose, s is divided into two different parts as follows V s = s ° + se, (22) «a · * 25 where s ° is the correct spectrum of the noiseless mixture and contains noise. Since the dependence between s and xopt is linear • ·: '., So also the coefficient vector xopt can be divided into the right coefficient vector x ° pt and the error vector x * pt, which follow the equations 30' ·: · x ° pt = (K'K) · 1 ^ 0 = Q- ^ K ^ K ') _ 1K, Ts ° (23a): ··: and 35 XoV = (KTK) _1KTse = Q "1 (K' TK ') 'XK' Tse (24a) 9 109238 or respectively <t = Ay = Q-'A'-y0 (23b) and 5 x: pt = Ay = Q-'AV * (24b)

Thus, the errors of the coefficients depend linearly on the measurement results. The coefficients of the noise vector it are normally distributed and the mean is zero. Even if the noise data of a single measurement 10 is not normally distributed for one reason or another, according to the central limit theorem of probability calculation, the sum of multiple noise data is always normally distributed, and in practice several individual result sequences are always combined to obtain one spectrum. Thus, if their standard deviation is denoted os, then s! W (0, a2s) (25)

* · · I

20 (This expression indicates that s * follows the normal distribution. The mean is 0 and the variance a \. Note that NI represents the normal distribution while N is the number of data.) Since the noise data distribution is known, * * · can now be calculated distributions of the errors of the coefficients using equation 24 using 25 *. This is done next.

The known normal distribution of probability • ·: / ·· calculations without proof is first presented.

': Let z ± * N (] ii, o2i) be independent and aA € R. Then' ..I 30 Σα «ζ« · * · Ν (Υ ^ α · 'Μή Y. Α1σϊ) - (26)'. .. * '* * • This means that the linear combination of independent normal: · · *: random variables is also normal. Using this result, x * pt; in the expression 109238 10 sa, the components of the vector y'e in equation 24b are: y'i = + N (0, ± £ (*;>) '<,;) = N (0, § Il K «f), 27, 1 = 1 5

A second result of probability calculation is now needed, showing that the two normally distributed random variables z1 and z2 are independent if and only if the correlation between them is zero. The correlation is defined by 10 equations _ / _ _ \ _ cov (ri, 22) e (* 1, Z2 "where 15 COv (z1 (z2) = E [(Zj - E (z1))) (z2 - E (z2)) ] and E is a wait operator Because the expectation values E (y '*) are zero; the two components y' * and y '^ are independent towers if and only if E (y'jy' ^) is • zero.

, '! Now = ^ £ ((K'V) (K'V) j 25 - = i £ [(äf) 2j (K'J | K '*), i, l ly> t ·

• I

: for E (s®s ^) = fijjfis]) 2] and all random variables s * follow the same distribution. The independence of y '* and y' * now follows from the fact that the strain K 'is orthogonal. · From 19, 26 and 27, it follows that, c. M / n 1 1 2 \ (28) xoPt, j · N (° »jyj | K / j | ρσ ^ 109238 11 and that the components of x '* pt are independent, so that again 26 can be used to obtain f 2 M (Q71) 2 \ 5 *:, <J = (Q_1x '. »0i - N (o, jL Σ l ^ TTjjJ J · (29)

If a non-orthogonal base K is used, the left-handed equation of equation 24 gives

M

xlpt, j + N (0, σ] ^ Pj,) ((30) 1 = 1 where P is a pseudo-inverse matrix 15 P = (KTK) _1KT = j-j A_1KT. (31)

Now that the distributions of the errors in the coefficients are known, the coefficients of error for the coefficients can be given. Since these errors are normally distributed random variables, no upper bound can be given to them. Instead ·. for the correct coefficients we can calculate the error limits I, within which the coefficients are with a given probability. "Suppose now that such an error limit vjf · · · ;;; is desired, that the coefficient xopt j lies within the range [x ° pt d ~ + * 25 with probability p. This means that p (xoPt.je [-Vj, -Uj]) = p.

If the distribution function 30 of the standard normal distribution (iV (0, l)) is denoted by Φ, then we get: V ^ t, / xlpt, jz -Vj vj j \ y / YaxixCoptj) [yVarix ^.) 'YVar (x ^ t) J. ) J) = i (v / v ^; "., ·)) (V ^ o) ~ 2 * -1 '109238 12 of which

J

Vx / ^ wv 5 ie (32)

For example, the 50% error limit for the library spectrum number j10 multiplier is given by 7Var (* '', Jj-1 »0.67449o7Var (x: '> J.).

The variance of x®pt Jin is obtained from result 29 or 30 as follows: 15 2 Al (n-1) 2 Al

Var (* W) = ^ Ejf ^ = -; i: ^. (33) 20 At first it may seem that the original · _ ·, strain K would be the most sensible choice for analysis,! because the odds and their variations are then obtained from very simple results 9 and 30. Every time you enter - ;;; however, a new library spectrum is added to the lysis, however, the scalar product Rice A changes, and the inverse matrix required at least in Equation 31 needs to be recalculated, and a simple formula such as 17 is not available when translating the general matrix. In addition, the calculation of P; In order to compute, we need to recalculate the matrix product Α'1 ^. The previous value of xopt 30 is thus useless. When using the orthogonal strain K ', the equations 20, 21 and 29 give 'Actions and their variations. Let us now take a closer look at what happens if a new library spectrum is to be added to the set of library spectra used in the analysis, and if 35 strain K 'is used.

109238 13

First of all, you have to calculate the new orthogonal spectrum K, Mtl from equation group 12. While calculating the coefficients in K'M + 1 in equation group 12, we also get the elements of column M + 1 that must be added to the transform matrix Q. (New horizontal line items QMtl) rj> M + 1 are zero according to equation 14.) This new vertical row means that the inverse transform matrix Q'1 also changes. However, from Equation 17, it is seen that an arbitrary element of Q_1 depends only on previously counted Q_1 elements on the same column and Q elements on the same or previous columns. Thus, the previous elements of Q_1 do not change and only need to calculate M new elements Qm ^ m + i, · · ·, Q ^ m * i, obtained from equation 17 as follows 15 m + l Q7, M + 1 = - Σ * (34) J = i + 1 (Since Q "1 is an upper triangle matrix, all elements of the new horizontal oblique are either zero or one.) 20 If the same measurement s • as before the update of Q_1 is still being analyzed, then the previous calculations- Note that the following corrections have to be made: 1) A new component x'optM + 1 must be added to the vector x'opt.

2) According to Equation 21, xopt.M + 1 = Q ^ x'opt = x'opt.M + 1. For other components xopt ^, the term Q ^ + 1x 'opt M + 1 must be added.

, '· · 3) According to Equation 29, Var (x * ptM + 1) = σ2 / (N || k' M + 11 | 2). For other variations, Var (xopt) = Var (x | pt}) must include the term 30 (Φ /, λ / + ι) 2 JV || K'M + l || 2 '

Because the squares of norms || K, J || 2 are repeatedly needed, so it makes sense to store them in a vector after being calculated from equation group 12. In computer programs 14, 109238, it is highly recommended to use double-precision real numbers.

Brief Description of the Drawings

The invention will now be described in detail by way of example with reference to the drawings, in which Fig. 1 is a mixture spectrum s, a linear combination of library spectra Kxopt that best explains s, and a 10 residual spectrum s-Kxopt which is then not explained; Figure 2 illustrates what happens if the library spectrum used is deficient; the same spectrum of the mixture as in Figure 1 has been analyzed, but now the spectrum of 2-butanone 15 has been removed from the used library spectrum; Figure 3 shows the situation when the interferometer is properly tuned but the radiation source is circular with an ideal point ·. ·. instead of the source, the signal at each different wave number ____: 20 d0 is uniformly distributed between u0 (l-a * ax / 2) - v0; . . Figure 4 illustrates a situation where the observed line shape e ° is; always the convolution of the true line form e, the quadratic expansion function w ° shown in Fig. 3 and the Fourier transform wL of the interfero ···· 'gram quadratic truncation function; '25 Figure 5 illustrates a situation where for each component Xj of the optimal non-negative coefficient vector xopt, one of the following conditions applies: * ». ·; (a) = 0 and xj> 0 or (b)> 0 and x; = 0.

30 '•: Description of the Preferred Embodiment This section describes a method for analyzing the spectra of a simple 35 gas mixture. The coefficients are calculated using 109238 15 using equations 20 and 21 and their error limits using equations 29 and 32. The 50% error limits are used here because they give a very good picture of the magnitude of the errors. Figure 1 shows the spectrum s of the mixture, which is ana-5 lysed using a set of 12 library spectra. Table 1 shows the components of xopt with error limits. The first three components are the background spectra. These are all spectra or functions that are also involved in the background measurement or that may be caused by the error information in the interferogram 10. They can have negative as well as positive coefficients. The spectrum s of the mixture and the library spectra KJ are measured in similar arrangements, but additional artificial noise has been added to s in order to simulate the lower accuracy assumed in this specification. In addition, a linear combination of these library spectra, Kxopt, is best seen to explain s. The figure also shows the residual spectrum s-Kxopt, which then remains unexplained. As can be seen, the remaining spectrum is formed by white noise alone. This indicates that * * tii; 20 analysis was sufficient and 12 libraries were used. . spectra are sufficient to explain the measurement. In Table 1 • * '; we see the result of the analysis, i.e. the optimal coefficient vector 'V * opf

IM

. * * 25 Optimal coefficient vector

Base Function Multiplier Error Limit (50%) • aiaa • '*: # 1: Standard Function (1) 0.0001 0.0001 30 Background # 2: Water -0.0062 0.0029 V. # 3: carbon dioxide -0,0002 0.0037 No. 4: methanol -0.0013 0.0027 / · No. 5: ethanol 0.0724 0.0023 No. 6: 2-butanone 0.2193 0.0017 No. 7: Chloroform -0.0011 0.0005 *. No. 8: Acetone 1,2805 0.0028 '' No. 9: Toluene 0.0014 0.0015 N: O0: Methyl acetate 0.0006 0.0004 N: O1: Methyl formate 0.0035 0.0008 40 N : ol2: methyl propanoate 0.1351 0.0007 109238 16

The partial pressure of the library gas number j is now obtained by multiplying its measurement pressure by its coefficient xopt j. Those components that are not present in the mixture have small positive or negative coefficients of the order of magnitude 5 equal to their error limits. If some background spectra are added to the analysis, the optimum background spectral removal is also performed automatically. In fact, the first three spectra (water, carbon dioxide, and a constant function of one everywhere) in the table are the 10 background spectra. Because we are not interested in accurately multiplying them, but simply wanting to get rid of them, they do not even have to be pure, but can contain each other. The library spectra may also contain some carbon dioxide and water without any influence on the value of their coefficients. The only effect is that the coefficients in the background spectra and the error bars are not reliable. The coefficients in the background spectra can be negative as well as positive values.

A. In reality, for example, there may be • »carbon dioxide during the measurement of a * 20 tan interferogram. . more than what it was during the measurement of the interferogram of the sample.

· '/ Figure 2 shows what happens if a library is used

• t I

; ·: The set of spectra is not sufficient to explain the composite · '' 25 spectra; Table 2 shows the results of the analysis, i.e. the optimal coefficient vector x.

> ·> · · * · · · ·

1 I I

> »* * *

<I M

109238 17

Optimal Coefficient Vector Base Function Coefficient Error Limit (50%) 5 N ° 1: Standard Function (1) -0,0001 0.0001 N ° 2: Water 0.0249 0.0029 N ° 3: Carbon Dioxide -0.0197 0, 0037 No. 4: Methanol 0.0446 0.0027 10 No. 5: Ethanol 0.1184 0.0022 No. 6: 2-Butanone ------ ------ No. 7 : chloroform -0.0053 0.0005 No 8: acetone 1.5193 0.0021 No 9: toluene 0.01212 0.0015 15 N: ol0: methyl acetate -0.0098 0.0004 N: oll: methyl form 0.0512 0.0007 N: ol2: methyl propanoate 0.1111 0.0007 The spectrum of the mixture used is the same as in Figure 1, 20 but the spectrum of 2-butanone, which is an important component of the mixture, is not included in the analysis. Now, the minimized residual spectrum is not pure noise, and the coefficients on the remaining library spectra have also changed. As can be seen, the now minimized residual spectrum s-Kxopt is no longer white: ··· 25 noise, but has a distinct structure and, ·. what is very important is that it has some kind of structure at the same wave numbers that the missing spectrum has' V spectral lines. Thus, it is possible to deduce from the remaining spectrum what kind of spectrum should be added to the '30 analysis. Because part of the missing spectrum has to be explained as much as possible by the remaining spectra, ·: their coefficients are also distorted and the error limits are no longer reliable. Thus, new library spectra must always be added to the analysis until there is no structure left in the remaining spectrum. If s and * libraries have different pressure decays, it may be · useful to use a pair of libraries with one compound. However, a better method would be to reduce the resolution so that all lines are in the form of a sinc-40 function (see next section).

109238 18

In Fig. 4, the function wL is a sinc function, and the interval between its two consecutive zeroes is equal to the data interval Δυ in the spectral region, which in turn is equal to 1 / (2νΔχ). wL's FWHH is approximately the same. The width W of wQ is directly adjacent to ver-5 at u0 (the position of the spectral line) and the area of the radiation source. Its height H is inversely proportional to u0 and directly proportional to the surface brightness (luminance) of the source. Under optimal conditions, the FWHHs of the three left curves are approximately the same. The total area of e ° 10 is the product of the domains e, wQ and wL. The area of wL is always 1 and can be considered dimensionless.

After specifying the wave number range to be processed, the sampling interval in the interferogram is also determined according to the Nyqvist sampling theorem.

However, there is still a choice of the length of the recorded interferogram, the amplitude of the mirror period, which in turn is determined by the number of data to be used. Note that N denotes the number of data to one side ·, *. interferogram. The corresponding number used in the fast Fourier transform algorithm is then 2N. Now, let's examine .. how should the number of data be selected with the error limits mini '; mise. As we can see from Equation 32, the error limits are * · _ · 'directly proportional to the standard deviations of the coefficients, where * · ·> ·! standard deviation means the square root of their variance.

. '· 25 Equation 29 or 33 yields I,) = -¾ .. o5> V for' II2 30 'I · ··' The error limits are thus directly proportional to the standard deviation os of the spectrum noise and inversely proportional to the square root of the number of data N. By the definitions of the scalar product and norm of equations 8 and 5, the remaining 109238 19 square root expression is not an explicit function of N. However, it depends on the forms of the library spectrum.

Now let's look at what happens when the number of data N is reduced by a factor of 1 / k. A negative effect is immediately seen that the coefficient 1 // N increases by a factor k1 / 2. However, the standard deviation of the spectrum noise of the mixture has also changed. This change is illustrated by Parseval's theorem 10 7, 7 J nl (x) dx = J | n, (i /) j2 du, (36)

-oo -OO

where n * and ns are noise functions in the interferogram and spectrum respectively. (These two random processes are Fourier transform pairs.) Because noise nA is completely white, its "amplitude" is the same everywhere. Thus, when the length of the first integral is truncated to one-kth of its original value, the value of the integral small i i 20 decreases by the same factor k'1. Thus, the second integral is,. change by the same factor. Since the wave number range * '' being studied is not changed, the only possibility is that · · the so-called "amplitude" of the noise, i.e. its standard deviation σβ, is reduced by a factor k'1 / 2. This effect completely reverses the dependency; :; 25 of 1 // N in equation 35.

As noted above, when reducing the difference · ,; the precision factor in os // n equation 35 remains constant. Thus, the only possible source of error boundary changes is the expression so n k · 1 p.

As mentioned earlier, the definitions of the scalar product and the norm, · - ·, mean that this expression depends solely on its own form of library spectrum when M is fixed. The number of data itself is not important. All linear transforms of 109238, in which all library spectra are multiplied by some constant factor C, change this square root expression by a constant C'1. Spectra are now practically always computed from corresponding interferograms using the Fast Fourier Transform (FFT) Fou-5 rier conversion algorithm. The basic feature of this algorithm is that the data interval in the spectrum is 1 / (2ΝΔχ), where Δχ is the sampling interval in the interferogram. Thus, when the number of data is reduced by a factor of 1 / k of the intermediate data in the spectral domain 10 is increased by a factor of k As long as the intermediate data ( "separation accuracy / 1.21) remains less than the spectral lines FWHH. (Full width at half-height, full-width half-height of the probe); each line has at least one information and the shape of the spectral lines does not vary significantly. 15 This means that using a resolution much higher than the width of the spectral lines is of little use. In the interferogram area, this means that the interferogram can be safely interrupted as long as - *. ·, Not cutting off a substantial part of the signal.

20 Now we define the interferograram cleavage function,. to a vertex function with a value of 1 between x = ± NΔχ and 0 elsewhere. Because only the finite region of the • · ferogram can be registered, the real infinite length of the interferogram is always multiplied by this function.

•: 25 In the spectral range, this means that the convolution of the Fourier transform wL of the spectra and the cutoff function is calculated, i.e. '* · ♦ Λ e ° = e "* wL, (37) 30!» · Where> · wL (t) = 2NAxsinc ( 2ΝΔχπυ), (38) 109238 21 e ° is the spectrum of the mixture or library and e "is the spectrum that would be obtained by converting the entire interferogram. The FWHH of this sinc function is approximately 1.21 / (2ηΔχ), and this is what is called the resolution here. As long as N remains greater than (2 Δχ x FWHH of the spectral lines) '1, wL is narrower than the lines of e' and has no significant effect on their shape. If the resolution is further reduced after this point, then the spectral lines begin to suddenly widen, and their shape 10 becomes essentially determined by wL instead of their actual forms. This means that the convolution of equation 37 then changes the spectra in a non-linear fashion, so its effect is not just multiplying the square root expression by a constant factor. If apodization (Apodi-15 zation) is not used, then the lines begin to resemble sinc curves. (If apodization is performed, the interferogram noise data is no longer identically distributed, and error analysis is invalid.) Due to the widening · ._, the lines start to overlap, making it more difficult to separate them. This, in turn, means that the sum expression of equation 35 begins to increase.

· '·' However, the rate of this growth depends on the number of library spectra M * used in the analysis and how close the lines are to each other. For example, if the lines were-= * *,: · 25 were originally grouped into sets of overlapping lines, the growth rate would not be as high as it would be if the lines were originally located approximately. to an equal degree. The same raw results can be given anyway. For example, if no more than 50 sets of booklet · · · '30 spectra are used, then the square root expression normally has a growth factor between k1 / 3 and k1 / 2, depending on how the' 'lines are located. The square root expression also depends on the number of lines in the spectrum. This dependence follows an approximate law such that the square root value is approximately inversely proportional to the square root of the average number of 109238 lines in one spectrum. Thus, it can be considered as a constant independent of N.

As stated above, the best choice for resolution resolution would be the FWHH of the spectral lines. Thus, the gram of registered interfero-5 should range from -1 / (2 x FWHH) to 1 / (2 x FWHH). However, this is only true if the interferometer settings cannot be tuned. If all the parameters of the device can be freely set, there are two additional benefits of lowering the resolution. These are discussed in more detail in the following 10.

As is known, a non-point radiation source causes the broadening of all spectral lines. Typically, the radiation source (aperture) is circular, in which case each monochromatic spectral line is scattered as a rectangular line as shown in Figure 3. The width of the box is then directly proportional to the area of the radiation source. This means that the spectrum e 'in equation 37 is, in fact, a square function of the real spectrum e and the non-zero area of the light source.

l * I

At 20, wQ convolution. Thus, equation 37 can be rewritten »· · · as follows> ·> · • * I ·: e ° = e * w ° * wL. (39) * »»,! · 25 Since the width of w ° depends on the wavelength υ0 of the spectral line under consideration, accurate processing would require the use of a different w ′ for each line. An example illustrating equation 39 is shown in Figure 4. Since the FWHH of the line determined by convolution is approximately the sum of the FWHHs of the convolutional components, the distortions w0 and wL have a significant effect only if their widths are greater. than the natural widths of the spectral lines. The signal can thus be safely magnified by increasing the radius of the radiation source until the (maximum) le-35 of the warp distortion equals the FWHH of the undistorted spectral lines.

109238 23

Similarly, the number of operations can be reduced by decreasing the resolution until the FWHH of the sinc distortion equals the sinc distortion of the spectral lines. (This represents the optimum cut-off of the interferogram.) However, for a gas-5 sample, the natural width of the lines may be so small that this situation cannot be achieved. In any case, it is still wise to make the distortions wQ and wL equal. This situation can therefore be taken as a starting point. Let us now reduce the number of data 10 by a factor k'1. As discussed above, this reduction widens the lines by wl increasing w by a factor k, thereby increasing the square root expression of Equation 35 by up to a factor k1 / 2 if a few dozen library spectra are used. If M is in the order of a couple hundred, then this factor can be in the order of k. However, it is now possible to enlarge the area of the radiation source by a factor k without significantly increasing the line widths. Since the area under the spectral lines must then increase, with a factor k due to the increase in the signal, the only possibility »·» · * t 20 is to increase the line heights. * i · with the same factor k and that the change in spectra is approximately V linear. Multiplying spectra by a constant factor k pi- *:; Subtract the square root expression by the factor k'1. This is more than sufficient to reverse the growth of the square root expression in the nonlinear cut operation of the Γ: 25 interferogram.

However, in practice there may be difficulties with the source of the radiation '; centering the magnified image on the detector.

i »

Another added benefit of lowering the resolution is that it is now possible to register k interferograms in the same time as previously only i I | one. Since the Fourier transform is a linear operation, t adding these interferograms to each other means IM; that the corresponding spectra are also added to each other.

The errorless spectra e remain the same for each measurement, 35 which means that when added together, they become multiples of 109238 24. This implies a simple linear change in the spectra, which in turn means that the square root expression is multiplied by k'1. On the other hand, the noise of s is different each time and does not add up linearly. Results 25 and 26 show that the summed noise distribution is 2V (0, ka2). The standard deviation of noise is thus increased by a factor k1 / 2. The overall effect is that the error limits are multiplied by a factor k'1 / 2.

Finally, when all the various effects mentioned above are summed up, it becomes clear that if all the parameters of the interferometer can be freely changed, then as little resolution as possible should be used. However, the number N of the data should be at least two or three times the maximum number of spectral spectra of the booklet-15 so that the structure of the residual spectrum s-Kx can be investigated.

If the library spectra are measured with a different resolution than that of the mixture spectra, the analysis may fail and large negative IV 20 coefficients may occur. A similar situation may occur if the shapes of the lines t · "· in the library spectra and the spectrum of the mixture are different due to nonlinearities or different pressure distributions. Then some improvement can be achieved by calculating the best non-negative solution instead of the top 25. A non-negative solution means a coefficient vector x, which is the solution to problem 4, provided that each component of x is complemented by a negative one, which brings the analysis

I

more information because the a priori information is applied to the multipliers. An algorithm is now derived to find a solution to Problem 4 with the non-negativity constraint in effect.

Let d denote the residual norm 35 d (x) = s-Kx 109238 25

Since the norm is always a non-negative quantity, so the norm || cl || is exactly the same minimum as its square || d || 2, so its square can be minimized instead of the norm. Now || d || 2 is a convex function of x. This means that 5 for each x1 # x2 and λ, 0 <λ <1 H d (λχχ + (1 - λχ2) 112 <Il d (Χχ) Il2 + (1 - λ) || d (x2) || 2.

This is seen by using the triangular inequality and the fact that the geometric mean is always less than or equal to the arithmetic mean. Convexity means that the square norm has only one minimum point, which makes minimization much simpler.

In particular, if other coefficients are kept constant and only one factor Xj is varied, || d || 2 is a single variable convex function. Thus, there are two possibilities for the optimum xopt. Joko d II d (xopt) II2 _ 0. . dxj 20 1 (where all components except j are attached to .V: xopt) or if the derivative zero is not allowed in the range Xj> 0, 25 xopt.j = 0, ♦ · ·. . which means that xopt j is at the border of the allowed and denied;> # * ranges. This can be proved as follows: 1) If the zero of the partial derivative IId (xopt) || 2 / x5 is · ': 30 in the allowed range x> 0, then obviously xopt Jin must be equal to this zero.

2) If the zero of the partial derivative || <3 (xopt) || 2 / Xj is in the forbidden region Xj <0, then the derivative is positive due to the convexity of || d (x) || 2 if Xj> 0. Thus 35 if xopt.j> 0, so reducing xopt jj would reduce the value of || d || 2 109238 26 without leaving the allowed range. So the only possibility is that xopt j = 0.

Now, the following condition can be proposed for optimality: 5 In the unique solution point of the minimization problem 4, by the non-negative coefficients of xopt x, for each component of x., One of the following conditions applies: = 0 and x,> o) or ("= 0 axd>"). (4Q, 10 v 3 This is shown in Figure 5. The so-called Kuhn-Tucker criteria (see some optimization course book, e.g. [3]), which is a necessary and sufficient condition for optimality in the case of a convex target function, would give 15 exactly the same condition.

The calculation of the partial derivatives is very simple, and we get V (|| d (x) || 2) = 2 (Ax-y), (41) 20 where A is the scalar product matrix defined in Equation 10 and V: y is given by Equation 11. partial derivatives are components of this gradient, i.e., 25 ^ yiQ = (V> || d (x) || J), = 2 ^ f;% Ii-Vi) <«),, If some background spectra are included in the set of library spectra used. , it makes sense not to set a non-negativity limit on their coefficients. The background spectra consist of all the compounds that are also involved in the measurement of the background (such as water and carbon dioxide), plus the standard function and simple co-curves that some incorrect data can generate in the interferogram. When this is also taken into account 109238 27 o, the minimum point can be found, for example, by the following algorithm: 1) Select some starting point x = x0, e.g. x0 = (0,0,..., 0) T, or x ° = A_1y. Place j = l.

5 2) Calculate

Vj ~ Σ «9ί; AjiXi

Xj ~> which according to equation 42 is the zero of the partial derivative || d (x) || 2 / Xj 10. If x ^ cO, and if xd has a nonnegativity constraint, reposition Xj = 0. Place j = j + l.

3) If j <M (where M is the number of library spectra used in the analysis) then return to step 2). Otherwise, proceed to Step 4).

15 4) Calculate G = 5 ^ (11 d (x) || :) = Ajc - y.

If for each component of G Gj either | Gj | <εχ or (G.,> 0 • 20 and Xj <ε2), where ε1 and ε2 are suitable small real numbers, then stop. Otherwise, place j = 1 and .V: return to step 2).

• ·

References 25 1. C. L. Lawson & R. J. Hanson, Solving least squares, Problems (Prentice-Hall, New Jersey, 1974), p.36.

•] · 2. B. Noble & J. W. Daniel, Applied Linear Algebra (Prentice-Hall, New Jersey, 1977), 2nd ed.

*: 30 3. M. S. Bazaraa & C. M. Shetty, Nonlinear Programming - Theory and Algorithms (John Wiley & Sons, New York, 1979), p.137.

Claims (4)

A method for analyzing an unknown gas mixture multicomponent FT-IR spectrum with an analyzer, 5. which stores a group of pure molecular gases FR-IR library spectra measured at known pressures as vectors k with N components, with which the analyzer is calculated and stored the scalar products of all possible combinations of two vectors k, characterized in that the method comprises the following steps for analysis of the unknown gas mixture, M library vectors k and thus a base matrix K, an orthogonal base matrix is formed. K 'is formed by transformation of the tili matrix K selected vectors k tili orthogonal vectors k', that the scalar products of these vectors k with each other are zeros but not S zeros with them selves, and the k of these orthogonal vectors 'scalar products with themselves are calculated and stored, a base transformation matrix Q is formed and stored such that K = K'0, r-': matrix inverse matrix Q_1 is formed and stored, ·. the FT-IR spectrum of the unknown gas mixture is measured. as a vector with N components, an optimal coefficient vector x 'is calculated for the spectra in orthogonal base, for its components x'j using the equation - and, a coefficient vector x is calculated such that x = Q "1x and the partial pressures of the pure V gases present in the analyzed gas mixture are calculated by multiplying the measurement pressure of the respective library spectra by their coefficients Xj and thus the analytical result of the spectra is p. 109238 32 2. Method according to claim 1, characterized in characterized in that it further comprises comparing the measured spectrum s with its analysis results if there is a difference, a new library vector k1 + 1 is added to the base matrix K, calculation of a new orthogonal vector k'M + 1, calculation and storage of the new orthogonal vector k, M + 1 scalar product with itself, addition of a new vertical row in the transformation matrix Q and storage of this new matrix, formation and storage of a new inverse matrix Q'1 on the basis of the new matrix Q by adding a new vertical row in the inverse matrix Q'1, forming and storing a new inverse matrix Q "1 on the basis of the new matrix Q and adding a new vertical row in the inverse matrix Q" 1, calculation of the analyzed spectrum's orotogonal coefficient vector x 'new component x'Mtl using the equation ..V · 20' 0c 1 s) "0:" ** 1! And adding the addition to the vector x ', as well as calculating a new coefficient vector x such that x = Q_1x'. 3. A method according to claims 2 and 3, characterized in that it further comprises - evaluation of the noise spread σβ in the spectrum x, *, ·. Calculating the variance in the vector x coefficients Xj using the equation <· · V,,. And # calculation of the error limits in the coefficients x} using the equation ^ = JVäFCxS Φ_1 (-¾2) 5 where p is the probability with which the library spectrum j error lies between the boundaries (-v .., Vj), and Φ'1 is the inverse of a one-dimensional standardized normal distribution distribution function. Method according to claims 2 and 3, characterized in that it further comprises calculating the variance in the new coefficient XM + 1 using the equation 15 Vr (x 1 N (k w <1 | k'11 * 1) and calculation of the values such should be added to the coefficients Xj, j = 1, ..., M variances using eq. 20 vation •: * 2 N (k'Htl \ k'H ^) '/ Addition of said values to the variances in coefficients x} / j = 1, ..., M and' computation of the coefficients Xj new error bounds Vj (p) using the equation ·. Vj = JVariXj) Φ "1 30 where p is the probability that the library spectrum j's error; ·" lies between the boundaries (-v. ,, vi) and Φ-1 is the inverse value. for a one-dimensional standardized normal distribution · distribution function. 35