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Hi I have a 1D radial profile of a sample across a pipe (fig_1). One data point (along the orange straight line) is acquired at each "band" from the image. The resolution (x,y,z) of each data point is 100um x 100um x 1000um.

fig_1 (fig_1)

However in order to produce a quantitative image, each data point in the 1D profile must be re-scaled by multiplying with a constant C. Where C is:

fig_2 (fig_2)

Where, phi0 = 0.35, A = cross-section A of a circular pipe, v = velocity across a pipe cross-section. S = signal (data points). At the end they produce a 1D profile as such.

fig_3 (fig_3)

My question is, how do I calculate the integral over the area (highlighted in red box, fig_2)?

In case I have missed any details I have provided the snippet of the text.

Comparison of radially averaged profile from a quantitative 2D image and 1D image is:

enter image description here

Ps: I have gotten around this problem by acquiring a 2D image from which I have calculated a 1D radially averaged intensity profile. 2D profile is quantitative and works fine, however I would like to know how I can make my 1D data quantitative as well.

dr_j_doe
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  • How do you compute the 1D radially averaged data, and how is the result there different? The radially averaged result will have less noise. Does that data otherwise look like the 1D data? – Cris Luengo Jan 21 '18 at 19:05
  • Can't you just sum the integrand over all pixels of the 2D image? – Amos Egel Jan 21 '18 at 19:54
  • An integral may be numerically approximated using a Reimann sum or the [trapezoid rule](https://en.wikipedia.org/wiki/Riemann_sum#Trapezoidal_rule). Take a look at MATLAB's [`trapz`](https://www.mathworks.com/help/matlab/ref/trapz.html) function. – jodag Jan 21 '18 at 21:35
  • @CrisLuengo: I use radialavg[1] to radial average a 2D image. I have attached an figure which shows that the intensity reported by 1D profile is lower than that acquired by radially averaging a 2D image which is quantitative. I hope I have answered your question. Many thanks for your help. [1] https://uk.mathworks.com/matlabcentral/fileexchange/46468-radialavg-zip?focused=6478826&tab=function [2] Please see the new image I have added to the question. – dr_j_doe Jan 22 '18 at 22:49
  • You compute averages along perfect circles, but your data does not contain perfect circles. The white lines will end up across multiple radial bins. Your 1D data in contrast is showing the expected jaggedness of going across those white lines. I'd trust your 1D data more! – Cris Luengo Jan 23 '18 at 00:00
  • @CrisLuengo: Perhaps that could be attributed to the the number of averages used to acquire each data set. The 2D image was acquired using 64 averages which eliminates the slight variation of signal over time, while the 1D profile was acquired using 4 averages. Moreover, the 2D image has been verified by other technique and is proven to be correct. Hence, I would assume the 1D profile require correction. – dr_j_doe Jan 23 '18 at 11:12
  • If the 2D data you are using is the image you posted, then your 2D results are not as great as you seem to think. That pipe is not even round! Your radial averaging needs to take that into account, but doesn't. – Cris Luengo Jan 23 '18 at 14:10
  • Hi 2D data is not from the image posted. I have posted that image as an example. Many thanks for all your help. – dr_j_doe Jan 23 '18 at 17:03

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