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Perfectly noise-free spectrum

You start with a perfectly noise-free spectrum ... [Pg.146]

Next, let s consider the spectrum of a sample that contains 2 concentration units of Component 1 and none of Component 2. In this perfectly linear, noise-free case, when we double the concentration of Component I, the absorbance at each of the wavelengths will also double. We have also plotted this second spectrum in Figure 29. It lies along the same direction from the origin as the first spectrum and at twice the distance. [Pg.82]

Next, let s consider the spectrum of a sample that contains both components together 2 concentration units of Component 1 and 3 concentration units of Component 2. Figure 31 contains a plot of this sample. The heavy X are plotted to indicate the absorbance contribution from each of the pure components in the sample. Since the contribution to the absorbance at each wavelength from each component adds linearly, the spectrum of the mixture is identical to the vector addition of the spectra of the pure components. Thus, it is apparent that, if we were to plot the spectrum of any mixture of these two components, it must be located somewhere in the plane determined by the lines which lie along the directions of the two pure component spectra. Notice that these lines that define the plane do not have to be perpendicular to each other. Indeed, they will usually not be mutually orthogonal. Figure 32 shows a plot of a number of such samples for this noise-free, perfectly linear case. [Pg.83]


See other pages where Perfectly noise-free spectrum is mentioned: [Pg.146]    [Pg.150]    [Pg.146]    [Pg.150]    [Pg.146]    [Pg.150]    [Pg.146]    [Pg.150]    [Pg.81]    [Pg.134]    [Pg.74]    [Pg.156]    [Pg.280]    [Pg.3276]    [Pg.247]    [Pg.248]    [Pg.415]   
See also in sourсe #XX -- [ Pg.146 , Pg.150 ]

See also in sourсe #XX -- [ Pg.146 , Pg.150 ]




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