# Solution 3.9 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

## Education Article

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

1. There are 72 points in the time series, therefore the acquisition time equals 72 ´ 0.008124 = 0.5849s. Hence

• since M = 2 ´ S ´ T,
• 72 = 2 ´ S ´ 0.5849,
• so S = 61.546 Hz.

Therefore the digital resolution = 61.546 / 36 = 1.7096 Hz / datapoint.

2. The graph is presented below. Notice how it is advisable to set the first point at t = 0.

3. The parameter

• M equals the total number of datapoints (72)
• n corresponds to a number between 0 and 35 corresponding to the spectral frequencies,
• m corresponds to the 72 raw points in time,
• f (m) corresponds to the mth reading in the FID.

The equivalent Imaginary transform is

IM(n) =

4. The two graphs are presented below. To obtain the correct answer it is essential to use an appropriate measure of angular frequency, in most environments (such as Excel) this involves multiplying n m / M by 2p.

The axis in Hz is sometimes labelled from right to left rather than left to right.

5. The phasing of the imaginary part is almost 180o out. However the phase error changes across the spectrum. The absolute value spectrum is presented below.

6. The imaginary spectrum is approximately 1800 (or p radians) out of phase. Thus sin(y ) » -1, so that y is around -900 (or 2700) or -p /2 radians. Hence the new spectrum can be obtained by

ABS = cos(-p /2) RL + sin(-p /2) IM

and is presented below.

Notice that the phased spectrum has sharper peakshapes and so better resolution than the absolute value spectrum, which is one advantage of phasing.

7. It is possible to include a small second order correction. This means that y changes across the spectrum. There are many ways of expressing this mathematically.

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