pchip extrapolation?

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Tim
Tim 2011-12-19
Hi All Trying to use pchip to process sensor data from raw digital values. Seems to work fine when the raw values are within the range of calibration values used us inputs to pchip. However, when raw values are smaller than those values used to calibrate the sensor the output is unexpected. Using the example for pchip from Mlab help:
x = -3:3; y = [-1 -1 -1 0 1 1 1]; t = -3:.01:3; p = pchip(x,y,t);
pchip (x,y,-4); ans = -1 (as expected)
However, if change y to:
y = [-1 -0.98 -0.94 0 0.94 0.98 1];
pchip (x,y,-4); ans = -0.9933
Why do answers for x values <-3 increase? I would expect them to linearly decrease i.e. pchip (x,y,-4); ans = -1.02
Thanks in advance

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采纳的回答

Robert Cumming
Robert Cumming 2011-12-19
your extrapolating quite far (compared to the source data), plot the data and you will see why the numbers are not what "you expect" - extrapolation must always be treated with care.
x = -3:3;
y = [-1 -1 -1 0 1 1 1];
t = -3:.01:3;
plot ( [-5:0.01:5], pchip(x,y,[-5:0.01:5]), 'r' );
hold on;
y = [-1 -0.98 -0.94 0 0.94 0.98 1];
plot ( [-5:0.01:5], pchip(x,y,[-5:0.01:5]), 'k' );
grid on
  2 个评论
Walter Roberson
Walter Roberson 2011-12-19
"Prediction is very difficult, especially if it's about the future."
(Nils Bohr)
Robert Cumming
Robert Cumming 2011-12-19
Indeed...
maybe I should have put "extreme" care...

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更多回答（1 个）

Andrew Newell
Andrew Newell 2011-12-19
Why would you expect the output to be linear when the input isn't? Look at the end points:
y = [-1 -0.98 -0.94 0 0.94 0.98 1];
diff(y)
ans =
0.0200 0.0400 0.9400 0.9400 0.0400 0.0200
If you change y to
y = [-1.02 -0.98 -0.94 0 0.94 0.98 1.02];
then you do get linear extrapolation: pchip(x,y,-4) is equal to -1.06.
EDIT: moreover, the extrapolation remains linear for larger absolute values.

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