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HPLC-DAD. HPLC-DAD data t w t 2 Cw 2 S = Suppose in a chromatogram obtained with a HPLC-DAD there is a peak which an impurity is co-eluted with analyte.

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Presentation on theme: "HPLC-DAD. HPLC-DAD data t w t 2 Cw 2 S = Suppose in a chromatogram obtained with a HPLC-DAD there is a peak which an impurity is co-eluted with analyte."— Presentation transcript:

1 HPLC-DAD

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4 HPLC-DAD data t w t 2 Cw 2 S = Suppose in a chromatogram obtained with a HPLC-DAD there is a peak which an impurity is co-eluted with analyte and you know analyte. Apply orthogonal projection concept and obtain the chromatographic profile of impurity.

5 HPLC-DAD m.file

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26 ? Apply the HPLC-DAD m.file on noised data and check the accuracy of the method

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31 From a geometrical point of view, we can interpret a matrix with n rows and p columns either as a pattern of n points in a p-dimensional space, or as a pattern of p points in n-dimensional space Matrices p n … …

32 1 2 3 4 5 6 Two vectors in a three dimensional space Three vectors in a two dimensional space

33 p n xiTxiT xjxj x ij PpPp SnSn vnvn v1v1 vivi PnPn SpSp upup u1u1 ujuj xixi xjxj Geometrical interpretation of an n x p matrix X

34 p n X The rank of matrix X is equal to the number of linearly independent vectors from which all p columns of X can be constructed as their linear combination Geometrically, the rank of pattern of p point can be seen as the minimum number of dimension that is required to represent the p point in the pattern together with origin of space rank(P p ) = rank(P n ) = rank(X) < min (n, p) Rank

35 Rank of the real chemical matrix 1 2 3  0.1 0.2 0.3 1 2 Conc. s 2 = 2s 1 s1s1 0.1 0.2 0.3 0.2 0.4 0.6 s2s2 3 1 2 l 3 = 3l 1 l 2 = 2l 1 rank of a ideal chemical matrix = number of chemical species

36 Determination of rank with MATLAB

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38 Anal.m file Constructing the data matrix for further analysis

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56 ? Apply the anal m.file and determine the rank of an absorbance data matrix which created from several three component mixtures

57 Rank Annihilation Methods A B k [A]=[A] 0 exp(-kt) [B]=[A] 0 (1 - exp(-kt)) A=  A [A] +  B [B] = A [A] [B] AA BB A A [A] = + [A] AA A [B] BB [B] =+

58 = + = + =

59 = - rank(A)=2rank(D)=1 rank(F)=1 A A [A] - = F - =

60 Kin.m file

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82 ? Simply modify the Kin.m file and show that using the spectrum of product component instead of reagent, can not decrease the rank of data

83 ? Does the rank of matrix F decrease if the applied spectrum of the reagent is not correct in its intensities?


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