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Generalized Rank Annihilation Factor Analysis Anal Chem 58(1986)496. E Sanchez B R Kowalski
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Bilinear data 2 1 3 2.1.2.3.2 e (Excit.) f (Emiss.) X 1 (Fluoresc.) = One component Rank =1 Conc. 0.40.81.20.8 0.20.40.60.4 0.61.21.81.2
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21 12 31 10 03.1.2.3.2.4.3 21.1 +13.2 21.1 +13.2 11.1 +23.2 31.2 +13.3 Two components EFConc. = X 2 (Fluoresc.)
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0.2 +0.60.4 +1.20.6 +1.20.4 +.9 0.1 +1.20.2 +2.40.3 +2.40.2 +1.8 0.3 +0.60.6 +1.20.9 +1.20.6 +0.9 Two components X 2 = Rank = 2 0.40.81.20.8 0.20.40.60.4 0.61.21.81.2 one component (calibration matrix) Rank = 1 X 1 = Lorber, 1984 X 2 - 0.5 X 1 = E Rank=2Rank=1 Quantification of one component.
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0.2 +0.60.4 +1.20.6 +1.20.4 +0.9 0.1 +1.20.2 +2.40.3 +2.40.2 +1.8 0.3 +0.60.6 +1.20.9 +1.20.6 +0.9 Two components(sample) X 2 = Rank = 2 0.4+0.20.8+0.41.2+0.40.8+0.3 0.2+0.40.4+0.80.6+0.80.4+0.6 0.6+0.21.2+0.41.8+0.41.2+0.3 Two components(calibration) Rank = 2 X 3 = What about quantific. of more than one component?
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Generalized RAFA [Anal Chem 1986, 58, 496-499. B.R. Kowalski] 1. Non-iterative [Lorber, 1984]. 2. Simultaneous detn. of analytes using Just one bilinear calibration spectrum from one mixture of standards. a.Bilinear spectrum of each analyte b. Relative conc.s
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Theory E F T = X 2 E F T = X 1 sample : Calibration : E = X 2 (F T ) + -1 E = X 1 (F T ) + -1 X 1 (F T ) + -1 = X 2 (F T ) + -1 X 1 Z = U S V T Z -1 Z = V S -1 Z * (definition) Common F and E Trilinearity
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X 1 V S -1 Z * = U S V T V S -1 Z * -1 I I U T X 1 V S -1 Z * = Z * -1 R V = V (eigenvector analysis) F T = (V S -1 Z * ) + E = U Z * -1 -1 = ?
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Simult. detn. of two acids in a sample H 2 A HA A H 2 B HB B using pH-metric titration
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H 2 A HA A H 2 B HB A sample C 0A ? C 0B ? H 2 A HA A H 2 B HB A calibr. C 0A =0.02 M C 0B =0.04 M Data matrices
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sample calibration
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Only HA - and HB - are optically active.
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sample calibration
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[Zstar,λ]=eig(Usm‘ * Xcl‘ * Vsm* inv(Ssm)) [Usm,Ssm,Vsm] = svd(Xsm')
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0.6669 1.9998 λ=λ= (C oA ) cl (C oA ) sm (C oB ) cl (C oB ) sm, (C oA ) cl =0.02 M => (C oA ) sm =0.03 M, (C oB ) cl =0.04 M => (C oB ) sm =0.02 M 0.03 0.02 β =β = 15
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F = pinv( Vsm * inv( Ssm ) * Zstar) Conc. profiles
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E = Usm * Zstar * inv(β) spectral profiles
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What if: The calibration sample includes some components that are not present in unknown sample, And there be some components in unknown sample not present in the calibration sample. HPLC-DAD chromatogram for A,B, and C (as CL), for ?,?,and ? (as SM) Example: The General Condition
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Xcl C Acl = 1 mM C Bcl = 3 mM C Ccl = 2 mM
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Xsm ?, ?, and ?,..
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[Zstar,λ]=eig(Utot‘ * Xsm‘ * Vtot* inv(Stot)) [Utot,Stot,Vtot] = svd(Xtot') Xtot = Xcl + Xsm The total space, rank =4 (includes A, B, C,and D)
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0.9999 0 0 0 0 0.0003 0 0 0 0 0.5000 0 0 0 0 0.3334 λ= β / ( β + ξ ) C?sm C?sm+C?cl =0.9999 0.0003 0.5000 0.3334 C?cl=0 Only in sm C?sm=0 Only in cl C?sm= C?cl 2C?sm= C?cl C B A D CBsm= 3 mM CCsm= 1 mM
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F = pinv( Vtot * inv( Stot ) * Zstar) Conc. profiles
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E = Utot * Zstar spectral profiles
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Non-bilinear RA Analyte detn...in the presence of unaccounted spectral interference.. Rank for the pure component >1
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H2A HA A
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One compon, but Rank=…3 Xcl
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H 2 A and H 2 B
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Rank(Xsm)=5 H 2 A and H 2 B Interference
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Conc. Prof.s Spect. Prof.s
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0.94150000 00.0003000 00-0.00300 0002.00440 00002.0010 λ of H 2 B
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Direct Exponential Curve Resolution Algorithm J. Chemom. 14 (2000) 213-227. DECRADECRA
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Model base: an exponential decay 162 54 18 6 2 162 54 18 6 2 x2x2 x1x1 162/54= 54/18= 18/6= 6/2= 3 3 3 3 shift x
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C 1 = e –k t C 2 = e –k (t+S) C1C1 C2C2 = = = e –kt +k(t+S) = e –k S e –k t e –k (t+S) k = ln( ) / S x 2 : x 1 : Shift
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Shift=7 x2x2 x1x1 1 st Ord Data From 1 sample
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k = ln( ) / 7 =0.1
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cPcP sPTsPT cQcQ sQTsQT cRcR sRTsRT =++ X Expon. Decay 2 st Ord Data From 1 sample
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Trilinear structure N X1X1 X2X2 =+ E Gives k1 and k2 X 2-way (MN) X 3-way ((M-S) N 2) Stacking E F λ 1 M-S 1+S M
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Decomposition of a number of colorants to colorless products.. A A’ B B’ C C’ … 1 st order reactions
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svd(X)= 6279.5 294.0 34.4 0.7 0.6 … Three components
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Shift = 10 min
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Estimated F
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estmated E
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k = ln( λ ) / shift 3.320100 02.22550 001.4918 λ = 0.1200 00.080 000.04
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A consecutive reaction:
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No Expon. Decaying concn. A B D k1k2
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Reaction model First order, consecutive C A,i = C A,0 e –k1 ti C B,i = (e –k1 ti - e –k2 ti ) k1 C B,0 k2- k1 C D,i = C A,0 - C A,i - C B,i A B D k1k2 Columns of C matrix cAcA cBcB cDcD
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X* = c A s A T + c B s B T + c D s D T + c L s L T = (e -k1 t ) s A T + k(e -k1 t ) s B T - k(e -k2 t ) s B T - (e -k1 t )s D T + e 0 t s D T - k(e -k1 t )s D T + k(e -k2 t )s D T + e 0 t s L T = e -k1 t ( s A + k s B - s D – k s D ) T + e -k2 t ( - k s B + ks D ) T + e 0 t (s D + s L ) T Sum of exponentially decaying functions
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Unique decomp. But not result into actual spectra and concn. profiles = e -k1 t ( s P + k s Q - s R – k s R ) T + e -k2 t ( - k s Q + ks R ) T + e 0 t (s R + s L ) T e 1 e 2 e 3
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Trilinear structure 1 M-s 1+s M N X1X1 X2X2 =+ X 2-way X 3-way E Gives k1 and k2 (MN) ((M-S) N 2) Stacking E F λ Expon. Decaying
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X* fAfA eATeAT fBfB eBTeBT fDfD eDTeDT =+ ++ 1..11..1 N+1 eLTeLT =[0 0.. 0 1] fLfL =[e 0 e 0.. e 0 ]
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What if : Not applying the ones column?
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An Example for consecutive reaction
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0.9999700 05.770 0019.848 λ=λ= k = ln( ) / shift 0.000000 00.07010 000.1195
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Not proper pure spectra !
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Not proper pure conc. Prof.s !
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What about estimation of spectral and concentration profiles?
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An NMR example
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PGSE NMR Pulsed Gradient Spin Echo NMR A Mixture, with exponential decay of the contribution of each component A series of spectra A function of diffusion coefficient of component
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Low-MW Poly(dimethylsiloxane) PDMS
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MRI 14 images (echo times (TEs) from 15 to 210ms) Exponential decay of signal from each component =f(sp.-sp. relax. Time of compon.)
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Thanks
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