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HPLC-DAD. w. 2. w. S. HPLC-DAD data. 2. C. t. t. =. 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. - PowerPoint PPT Presentation
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HPLC-DAD
HPLC-DAD data
t
w
t
2
C
w
2S
=
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.
HPLC-DAD m.file
?
Apply the HPLC-DAD m.file on noised data and check the accuracy of the method
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
…
1 2 3
4 5 6
Two vectors in a three dimensional space
Three vectors in a two dimensional space
p
n
xiT
xj
xij
Pp
Sn
vn
v1
vi
PnSp
up
u1
uj
xi xj
xijxij
Geometrical interpretation of an n x p matrix X
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(Pp) = rank(Pn) = rank(X) < min (n, p)
Rank
Rank of the real chemical matrix
1 2 3
0.1 0.2 0.3 1 2Conc.
s2= 2s1
s1 0.1 0.2 0.3
0.2 0.4 0.6s2
31 2
l3= 3l1
l2= 2l1
rank of a ideal chemical matrix = number of chemical
species
Determination of rank with MATLAB
Anal.m file
Constructing the data matrix for further analysis
?
Apply the anal m.file and determine the rank of an absorbance data matrix which created from several three component mixtures
Rank Annihilation Methods
A Bk
[A]=[A]0exp(-kt)
[B]=[A]0 (1 - exp(-kt))A=A[A] + B[B]
=A [A][B]
A
B
A A[A]= +
[A]
A
A[B]
B
[B]= +
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)
Ab
so
rba
nc
e
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)
Ab
so
rban
ce
0
0.1
0.2
0.3
400 450 500 550 600Wavelength (nm)
Ab
so
rban
ce
=
+
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250
Time (s)
0
0.050.1
0.15
0.2
0.250.3
0.35
0.4
400 450 500 550 600
Wavelength (nm)
=
0
0.2
0.4
0.6
0.8
1
1.2
0 50 100 150 200 250
Time (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 50 100 150 200 250
Time (s)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
400 450 500 550 600Wavelength (nm)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
400 450 500 550 600Wavelength (nm)
+=
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)
Ab
so
rba
nc
e
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)A
bs
orb
ance
0
0.1
0.2
0.3
400 450 500 550 600Wavelength (nm)
Ab
so
rban
ce
=
-
rank(A)=2 rank(D)=1 rank(F)=1
A A[A]- = F
0
0.1
0.2
0.3
400 450 500 550 600Wavelength (nm)
Ab
so
rban
ce
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)
Ab
so
rba
nc
e
0
0.1
0.2
0.3
400 450 500 550 600
Wavelength (nm)
Ab
so
rban
ce
-
=
Kin.m file
?
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
?
Does the rank of matrix F decrease if the applied spectrum of the reagent is not correct in its intensities?
