Interpolation with multiple intersection points

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0 个投票

I have phase data from a bode plot. I want to interpolate this data and find the frequency where phase equals -135 deg.

Example:

f=[0.1 1 10 100 1000];
phase=[-90 -140 -80 -130 -180];
interp1(phase,f,-135)
ans =
50.5000

Clearly this is not the correct answer. To start there are 3 intersection points as shown by:

I am especially interested in the last intersection point around 200Hz. Is the use of interp1 incorrect in this case? Can somebody easily solve this issue. Only solution I see is using the find function and defining the intersection point as a range.

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Answer 1

John D'Errico 2018-1-15

编辑：John D'Errico 2018-1-15

在 MATLAB Online 中打开

1 个投票

Interp1 is defined for functions that have a SINGLE y value for any x. Since you are trying to call it for frequency as a function of phase, that clearly fails to be true. interp1 simply is not applicable to multi-valued relationships.

You have two simple options.

1. Just find the pairs of points that cross your line of interest. Use Linear interpolation, to locate the intersection. With so few points on a relationship that is so ill-defined, linear interpolation is the best you can ask for.

2. Download a code like Doug Schwarz's intersections code. Find it on the file exchange . It will generate all three intersections. Again, it will do only linear interpolation. You pass it two curves as sets of points. One of the curves will be simply the (red) straight line at a constant phase, given two frequencies.

[fint,phaseint] = intersections(f,phase,[.1 1000],[-135 -135])
fint =
         0.91
         1.75
          190
phaseint =
  -135
  -135
  -135

Although, since you are plotting this on a log axis, a log-linear interpolant might be better.

[fint,phaseint] = intersections(log10(f),phase,log10([.1 1000]),[-135 -135])
fint =
         -0.1
     0.083333
          2.1
phaseint =
  -135
  -135
  -135
fint = 10.^fint
fint =
      0.79433
       1.2115
       125.89