functioninterpolate

Interpolate linearly in a vector

Extends from Modelica.Icons.Function (Icon for functions).

Information

Syntax

// Real    x[:], y[:], xi, yi;
// Integer iLast, iNew;
        yi = Vectors.interpolate(x,y,xi);
(yi, iNew) = Vectors.interpolate(x,y,xi,iLast=1);

Description

The function call "Vectors.interpolate(x,y,xi)" interpolates linearly in vectors (x,y) and returns the value yi that corresponds to xi. Vector x[:] must consist of monotonically increasing values. If xi < x[1] or > x[end], then extrapolation takes places through the first or last two x[:] values, respectively. If the x and y vectors have length 1, then always y[1] is returned. The search for the interval x[iNew] ≤ xi < x[iNew+1] starts at the optional input argument "iLast". The index "iNew" is returned as output argument. The usage of "iLast" and "iNew" is useful to increase the efficiency of the call, if many interpolations take place. If x has two or more identical values then interpolation utilizes the x-value with the largest index.

Example

  Real x1[:] = { 0,  2,  4,  6,  8, 10};
  Real x2[:] = { 1,  2,  3,  3,  4,  5};
  Real y[:]  = {10, 20, 30, 40, 50, 60};
algorithm
  (yi, iNew) := Vectors.interpolate(x1,y,5);  // yi = 35, iNew=3
  (yi, iNew) := Vectors.interpolate(x2,y,4);  // yi = 50, iNew=5
  (yi, iNew) := Vectors.interpolate(x2,y,3);  // yi = 40, iNew=4

Inputs

TypeNameDefaultDescription
Real[:]xAbscissa table vector (strict monotonically increasing values required)
Real[size(x, 1)]yOrdinate table vector
RealxiDesired abscissa value
IntegeriLast1Index used in last search

Outputs

TypeNameDefaultDescription
RealyiOrdinate value corresponding to xi
IntegeriNew1xi is in the interval x[iNew] <= xi < x[iNew+1]