EXTRAPVAL must be a scalar, error while using interp3.
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Hi I am using interp3 to extrapolate data at Tj = 35 0r any value within the range Tj', with ids = 49, Vdd = 500. The data is as follows:
Tj = [25:25:150];Ids= 35;Vdd = 600; E=[0.4 0.378 0.36 0.365 0.37 0.4]*10^-3;
I am usinfg the following command:
interp3(Tj,Ids,Vdd,E,Tj',49,500,'linear','extrap')
and get the following error:
EXTRAPVAL must be a scalar.
Please help.
Thanks
Sneha
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Antworten (2)
Walter Roberson
am 19 Sep. 2018
interp2() and interp3() always do extrapolation unless you specify a scalar value to be used to replace any attempt to query outside the defined range.
interp1() extrapolates by default for 'pchip', 'spline', or 'makima' but not the other methods, and needs 'extrap' to be specified for the other methods; you can also specify an extrapolation scalar value to replace any attempt to query outside the defined range.
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Walter Roberson
am 20 Sep. 2018
Figure 26 by itself does not give enough information for that. I doubt that all of the graphs together give enough information for that.
Figure 26 has as input one Ids and multiple Tj. There are no other similar figures with different Ids.
Therefore to estimate Eon at different Ids, based upon the information given in that graph, you would need to have a relationship between Ids and Tj, along the lines of varying Ids would permit you to predict a different effective Tj, which you would then look up to get Eon.
Unfortunately, it looks to me as if even if you synthesize all of the information from the graphs together, there is not enough information to predict change of Tj with varying Ids. Tj is not the output (Y axis) on any of the graphs, so you cannot find Tj as being "caused" by any change to Ids. Ids is, though, the Y axis on several of the graphs, so hypothetically you might be able to find a combination of steps that relates Ids as temperature changes... but I doubt it.
KSSV
am 19 Sep. 2018
YOu should specify the value of extrapval. Then the value specified will be assigned to all the values which lie out side the domain.
extrapval = 1 ;
interp3(Tj,Ids,Vdd,E,Tj',49,500,'linear',extrapval)
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