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

回答(1 个)

Walter Roberson
Walter Roberson 2023-3-17

0 个投票

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?

3 个评论

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.
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
Yes. Something like that.

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