Using interp1 with semilog dataset

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Alex
Alex 2022 年 8 月 31 日
編集済み: Alex 2022 年 8 月 31 日
I have 3 arrays:
  • 'Dose' contains 27 values ranging from 1x10^2 to 1x10^6
  • 'AlThick' contains 27 values ranging from 0.0001 to 20
  • 'Rays' contains 10000 values ranging from 0 to 20
The points in each data set are not equally spaced. Dose represents y values; AlThick and Rays both represent x values. I want to find the value of Dose for each non-zero value of Rays. I am doing the following and getting, as expected, 10000 values.
DoseMat = interp1(AlThick, Dose, rays(rays>0), 'linear')
However, I'm not sure if I need to be accounting for the fact that the y-axis data is plotted as a base-10 logarithm. I.e. it is a semilog plot with base-10 log values on the y-axis and linear values on the x-axis, as shown below.
Any advice is appreciated!
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Torsten
Torsten 2022 年 8 月 31 日
編集済み: Torsten 2022 年 8 月 31 日
And plotting "Dose" against "AlThick" in semilogy gives you a linear relationship ?
Could you include this plot ?
Alex
Alex 2022 年 8 月 31 日
I have updated the OP with the plot of Dose vs AlThick

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Cris LaPierre
Cris LaPierre 2022 年 8 月 31 日
編集済み: Cris LaPierre 2022 年 8 月 31 日
You do not need to take account of the fact that you are plotting your data in a semilog plot. That doesn't impact how you would interpolate the data. The 'linear' here is the interpolation method.
However, you are not using interp1 correctly. Interp1 is for (x,y) data. The syntax is
  • newY = interp1(knownX, knownY, newX, method).
You probably want to use interp2.
  • newZ = interp1(knownX, knownY, knownZ, newX, newY, method).
  4 件のコメント
Image Analyst
Image Analyst 2022 年 8 月 31 日
@Alex, attaching data in a .mat file, and screenshots of it plotted, would help.
Alex
Alex 2022 年 8 月 31 日
I've updated the OP with the plot of Dose vs AlThick

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