pchip extrapolation?

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Tim
Tim il 19 Dic 2011
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

Risposta accettata

Robert Cumming
Robert Cumming il 19 Dic 2011
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 Commenti
Walter Roberson
Walter Roberson il 19 Dic 2011
"Prediction is very difficult, especially if it's about the future."
(Nils Bohr)
Robert Cumming
Robert Cumming il 19 Dic 2011
Indeed...
maybe I should have put "extreme" care...

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Più risposte (1)

Andrew Newell
Andrew Newell il 19 Dic 2011
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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