1*0 empty double row vector

X = [-14  -12   -10. -6.  -2  2.  6.  8   10   12   14   16  18]; 
Y = [-0.8 -1.1 -.88 -.44  0  .44 .88 1.1 1.32 1.54 1.76 1.98 1.7];
y1 = intrp(-20,X,Y)

out = -0.8000

y1 = -0.8000

y2 = intrp(-11,X,Y)

out = -0.9900

y2 = -0.9900

y3 = intrp(9,X,Y)

out = 1.2100

y3 = 1.2100

y4 = intrp(19,X,Y)

out =

  1×0 empty double row vector


y4 =

  1×0 empty double row vector

function out = intrp(in,X,Y)
% interpolate a 2-d function
%  in: input value of x
%  X: input vector
%  Y: output vector
%  out: output value of Y=f(X) for X = in
X = [-14  -12   -10. -6.  -2  2.  6.  8   10   12   14   16  18];
Y = [-0.8 -1.1 -.88 -.44  0  .44 .88 1.1 1.32 1.54 1.76 1.98 1.7];
if (in < -14)
    loc = find(in < X,1);
    out = Y(loc)
elseif (in > 18)
    loc = find(in < X,1);
    out = Y(loc-1)
else
    loc = find(in < X,1);
    out = Y(loc-1)+(Y(loc)-Y(loc-1))/(X(loc)-X(loc-1))*(in-X(loc-1))
end
end

Answer 1

[Walter Roberson]

19 < X is never true, so find() is going to return empty.

What result were you hoping for in the case where the input value is greater than all of the X values?

Question: why are you passing X into your function but then ignoring the input X and re-assigning values to X inside the function?

[Chirag]

I was hoping to get y=f(X)=1.7 for any values that are greater than X=18. But, seems I am lost in translation somewhere.

[Walter Roberson]

I suspect you want an algorithm closer to:

in value less than or equal to X(1) should return Y(1)
in value greater than or equal to X(end) should return Y(end)
otherwise do your calculation

[Chirag]

Yes. Something like that.