# Solution 3.2 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

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## Education Article

• Published: Jan 1, 2000
• Channels: Chemometrics & Informatics

1. The graphs are as follows.

A, B and D are triangular peaks of increasing width but similar integrated intensity. C has the same features as B but is shifted in position.

2. The transforms are presented below.

 1 1 1 1 0.383 0.379 0.823 0.373 -0.707 -0.68 0.368 -0.637 -0.924 -0.846 -0.179 -0.727 0 0 -0.604 0 0.924 0.719 -0.763 0.457 0.707 0.489 -0.639 0.245 -0.383 -0.229 -0.332 -0.082 -1 -0.5 0 -0.111 -0.383 -0.154 0.224 -0.016 0.707 0.218 0.285 0.004 0.924 0.205 0.218 0.001 0 0 0.104 0 -0.924 -0.078 0.016 -0.045 -0.707 -0.027 -0.015 -0.056 0.383 0.004 -0.008 0.039

If you obtain different numbers, carefully check your trigonometric function and also that the integers n and m start from 0.

3. The graphs are as follows.

Function A

Function B

Function C

Function D

4. There are three oscillations in the transform of function A. This is because it is centred at datapoint 6, the number of oscillations equalling 6/2. There is no decay because the peak is a speak a single point wide in the raw data.

5. The cosine waves in the transform are identical except that they decay faster in the transforms for B and D. This is because a broader line in the raw data (e.g. a wider peak) corresponds to a faster decay rate in the transform.

6. The difference is that the peak in the original data for C is centred at half the distance from the origin, so oscillates at half the frequency (1.5 oscillations) in the transform, compared to B.

7. The imaginary transform is presented below.

It is simply shifted by 90o (p /2 radians) from the real transform.

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