Synthetic chromatogram dataset: peak overlap and resolution

Before you start

Estimated time 25–35 minutes.

Prerequisites Basic graph reading and arithmetic. Use a spreadsheet or plotting tool that can draw XY charts. A calculator is enough for the analytic area option.

Learning objectives

• Plot component and combined signals for each scenario on the same axes.
• Calculate resolution from the retention times and the model baseline widths.
• Compare component signal areas and identify which assumptions make the area fraction constant.

What is in the files?

Time runs from 0.00 through 10.00 minutes in 0.02-minute steps. AU means arbitrary signal units. Each scenario contains the two component curves and their sum. The component signals are available because the example generated them; a real detector trace may provide only the combined signal.

• Component 1
• Component 2
• Combined signal

Practice: plot, resolve, integrate

1. Filter the trace CSV by scenario. Make an XY line plot with time_min as the horizontal axis and all three signal columns as the vertical series. Compare the three panels on a common scale.
2. Use the parameter table and the width approximation w = 4σ to compute the model resolution for each case. Which change makes the peaks overlap most in these examples?
3. Integrate each component using the trapezoid rule on the CSV grid, or compute its analytic area. Calculate the first component’s share of the total modeled area. Can this fraction tell you the amount of material?

Reproduce the data

For each component, compute signal = height × exp(−0.5 × ((time − center)/σ)²). Add the two unrounded components for the combined signal. The CSV rounds each series independently to nine decimal places.

Use Rₛ = 2(t₂ − t₁)/(w₁ + w₂). For an ideal Gaussian, the tangents at the inflection points meet the baseline 4σ apart. The Gaussian tails continue beyond that width. Compute the trapezoid area by adding 0.02 × (previous signal + current signal)/2 over the grid. The full Gaussian integral is height × σ × √(2π).

Worked answer key

Model results (rounded)

Scenario	Each baseline width (min)	Rₛ	Area 1 (AU·min)	Area 2 (AU·min)
reference	1.40	1.143	0.8773	0.6141
closer_centers	1.40	0.429	0.8773	0.6141
wider_peaks	2.60	0.615	1.6293	1.1405

For the reference trace, the centers differ by 1.60 min and each baseline width is 1.40 min, giving Rₛ ≈ 1.143. The closer-centers trace has the lowest resolution of these three. The time grid gives component areas that round to the same four decimals shown here.

Each pair shares a width, so the first component contributes 100 × 1.0/(1.0 + 0.7) ≈ 58.82% of the modeled area in all three cases. Broader peaks here have larger area because their model heights were held fixed. The area fraction describes simulated signal. Material amounts require a suitable quantitative procedure and calibration.

Instructor notes

• Ask learners to compare the combined traces before looking at the individual component curves. Discuss which model change explains each trace.
• Keep the axis limits fixed across plots. Have learners check the units of time, width and integrated signal when explaining their results.