Interpolate an expected peak from saturated data

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Jonathan Wells
Jonathan Wells 2017-7-27
编辑： Image Analyst 2017-7-27
Hi,
I have a set of data in which the intensity pattern is maxed out at many values, causing it to look like a plateau instead of a nice rounded peak. Is there any way to interpolate the value and location of the peak if it were properly rounded? I'm hoping to create a gaussian type curve fitted to the shape of the data. A sample is given below.
0.407775878906250 0.485290527343750 0.562133789062500 0.664245605468750 0.760009765625000 0.860107421875000 0.963195800781250 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.983215332031250 0.999938964843750 0.984191894531250 0.919006347656250 0.904113769531250 0.914733886718750 0.831604003906250 0.763122558593750 0.677856445312500 0.597961425781250 0.503234863281250 0.420288085937500 0.353881835937500 0.294616699218750 0.236633300781250 0.183837890625000 0.138244628906250 0.108459472656250 0.0681762695312500 0.0527954101562500 0.0538940429687500 0.0429077148437500 0.0131835937500000 0.0164794921875000 0.0164794921875000 0.0186767578125000 0.0363159179687500 0.0197753906250000 0.0153808593750000 0.0308227539062500 0.0186767578125000 0.0131835937500000 0.0286254882812500 0.0308227539062500 0.00878906250000000 0.0142822265625000 0.0208740234375000 0.0219726562500000 0.0264282226562500 0.0252685546875000 0.0208740234375000 0.0186767578125000 0.0186767578125000 0.0264282226562500 0.0164794921875000 0.0131835937500000 0.0330200195312500 0.0275268554687500 0.0264282226562500 0.0219726562500000 0.0153808593750000 0.0120849609375000 0.0264282226562500 0.0208740234375000 0.00988769531250000 0.0164794921875000 0.0186767578125000 0.0208740234375000 0.0308227539062500 0.0153808593750000 0.0219726562500000 0.0175781250000000 0.0131835937500000 0.0219726562500000 0.0175781250000000 0.00769042968750000 0.0252685546875000 0.0252685546875000 0.0175781250000000 0.0319213867187500 0.0275268554687500 0.0164794921875000 0.0109863281250000 0.0230712890625000 0.00549316406250000 0.0175781250000000 0.0241699218750000 0.0131835937500000 0.0219726562500000 0.0208740234375000 0.0109863281250000 0.0230712890625000 0.0131835937500000 0.0286254882812500 0.0164794921875000 0.0109863281250000 0.00878906250000000 0.0197753906250000 0.0208740234375000 0.0264282226562500
Thanks, Jon
[EDITED, Jan]
  2 个评论
Jan
Jan 2017-7-27
编辑：Jan 2017-7-27
This is a rather unclear example, which looks like a graveyard of numbers. It would require a little bit of work to create a diagram out of it, and the drawback is, that all readers, who want to help you, have to perform this work in parallel. It might be useful, if you post a diagram instead. I've done this for you this time.
The curve does not look like a Gauss curve. But you could crop the trailing and saturated data manually and fit the parameters of a Gauss curve to the data. Did you try this already? Do you have the Optimization Toolbox?
Image Analyst
Image Analyst 2017-7-27
编辑：Image Analyst 2017-7-27
m=[0.407775878906250 0.485290527343750 0.562133789062500 0.664245605468750 0.760009765625000 0.860107421875000 0.963195800781250 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.999938964843750 0.983215332031250 0.999938964843750 0.984191894531250 0.919006347656250 0.904113769531250 0.914733886718750 0.831604003906250 0.763122558593750 0.677856445312500 0.597961425781250 0.503234863281250 0.420288085937500 0.353881835937500 0.294616699218750 0.236633300781250 0.183837890625000 0.138244628906250 0.108459472656250 0.0681762695312500 0.0527954101562500 0.0538940429687500 0.0429077148437500 0.0131835937500000 0.0164794921875000 0.0164794921875000 0.0186767578125000 0.0363159179687500 0.0197753906250000 0.0153808593750000 0.0308227539062500 0.0186767578125000 0.0131835937500000 0.0286254882812500 0.0308227539062500 0.00878906250000000 0.0142822265625000 0.0208740234375000 0.0219726562500000 0.0264282226562500 0.0252685546875000 0.0208740234375000 0.0186767578125000 0.0186767578125000 0.0264282226562500 0.0164794921875000 0.0131835937500000 0.0330200195312500 0.0275268554687500 0.0264282226562500 0.0219726562500000 0.0153808593750000 0.0120849609375000 0.0264282226562500 0.0208740234375000 0.00988769531250000 0.0164794921875000 0.0186767578125000 0.0208740234375000 0.0308227539062500 0.0153808593750000 0.0219726562500000 0.0175781250000000 0.0131835937500000 0.0219726562500000 0.0175781250000000 0.00769042968750000 0.0252685546875000 0.0252685546875000 0.0175781250000000 0.0319213867187500 0.0275268554687500 0.0164794921875000 0.0109863281250000 0.0230712890625000 0.00549316406250000 0.0175781250000000 0.0241699218750000 0.0131835937500000 0.0219726562500000 0.0208740234375000 0.0109863281250000 0.0230712890625000 0.0131835937500000 0.0286254882812500 0.0164794921875000 0.0109863281250000 0.00878906250000000 0.0197753906250000 0.0208740234375000 0.0264282226562500]
plot(m, 'LineWidth', 2)
grid on;
What kind of curve do you want to fit it to? Quadratic? Gaussian? Sawtooth (linear)? Something else?

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