Generalized Rank Annihilation Factor Analysis Anal Chem 58(1986)496. E Sanchez B R Kowalski.

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Presentation transcript:

Generalized Rank Annihilation Factor Analysis Anal Chem 58(1986)496. E Sanchez B R Kowalski

Bilinear data e (Excit.) f (Emiss.) X 1 (Fluoresc.)  = One component  Rank =1 Conc

1.1 +13.2 21.1 +13.2 11.1 +23.2 31.2 +13.3 Two components EFConc.  = X 2 (Fluoresc.)

Two components X 2 = Rank = one component (calibration matrix) Rank = 1 X 1 = Lorber, 1984 X X 1 = E Rank=2Rank=1 Quantification of one component.

Two components(sample) X 2 = Rank = Two components(calibration) Rank = 2 X 3 = What about quantific. of more than one component?

Generalized RAFA [Anal Chem 1986, 58, 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

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

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  =  ?

Simult. detn. of two acids in a sample H 2 A  HA  A H 2 B  HB  B using pH-metric titration

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

sample calibration

Only HA - and HB - are optically active.

sample calibration

[Zstar,λ]=eig(Usm‘ * Xcl‘ * Vsm* inv(Ssm)) [Usm,Ssm,Vsm] = svd(Xsm')

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 β =β = 15

F = pinv( Vsm * inv( Ssm ) * Zstar) Conc. profiles

E = Usm * Zstar * inv(β) spectral profiles

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

Xcl C Acl = 1 mM C Bcl = 3 mM C Ccl = 2 mM

Xsm ?, ?, and ?,..

[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)

λ= β / ( β + ξ ) C?sm C?sm+C?cl = 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

F = pinv( Vtot * inv( Stot ) * Zstar) Conc. profiles

E = Utot * Zstar spectral profiles

Non-bilinear RA Analyte detn...in the presence of unaccounted spectral interference.. Rank for the pure component >1

H2A  HA  A

One compon, but Rank=…3 Xcl

H 2 A and H 2 B

Rank(Xsm)=5 H 2 A and H 2 B Interference

Conc. Prof.s Spect. Prof.s

λ of H 2 B

Direct Exponential Curve Resolution Algorithm J. Chemom. 14 (2000) DECRADECRA

Model base: an exponential decay x2x2 x1x1 162/54= 54/18= 18/6= 6/2= shift x

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

Shift=7 x2x2 x1x1 1 st Ord Data From 1 sample

k = ln(  ) / 7 =0.1

cPcP sPTsPT cQcQ sQTsQT cRcR sRTsRT =++ X Expon. Decay 2 st Ord Data From 1 sample

Trilinear structure N X1X1 X2X2 =+ E Gives k1 and k2 X 2-way (MN) X 3-way ((M-S) N 2) Stacking E F λ 1 M-S 1+S M

Decomposition of a number of colorants to colorless products.. A  A’ B  B’ C  C’ … 1 st order reactions

svd(X)= … Three components

Shift = 10 min

Estimated F

estmated E

k = ln( λ ) / shift λ =

A consecutive reaction:

No Expon. Decaying concn. A  B  D k1k2

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

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

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

Trilinear structure 1 M-s 1+s M N X1X1 X2X2 =+ X 2-way X 3-way E Gives k1 and k2 (MN) ((M-S) N 2) Stacking E F λ Expon. Decaying

X* fAfA eATeAT fBfB eBTeBT fDfD eDTeDT = N+1 eLTeLT =[ ] fLfL =[e 0 e 0.. e 0 ]

What if : Not applying the ones column?

An Example for consecutive reaction

λ=λ= k = ln( ) / shift

Not proper pure spectra !

Not proper pure conc. Prof.s !

What about estimation of spectral and concentration profiles?

An NMR example

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

Low-MW Poly(dimethylsiloxane) PDMS

MRI 14 images (echo times (TEs) from 15 to 210ms) Exponential decay of signal from each component =f(sp.-sp. relax. Time of compon.)

Thanks