EXTRAPVAL must be a scalar, error while using interp3.

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Sneha Thakur
Sneha Thakur 2018 年 9 月 19 日
コメント済み: Walter Roberson 2018 年 9 月 20 日
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

回答 (2 件)

Walter Roberson
Walter Roberson 2018 年 9 月 19 日
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.
  6 件のコメント
Sneha Thakur
Sneha Thakur 2018 年 9 月 19 日
Hi Walter
Thank you so much for your time.
I tried using it, but it leads to error:
_Error using griddedInterpolant_
Interpolation requires at least two sample points in each dimension.
Which is obvious as, Ids and Vdd have only value. But this is what is available from the graph in attached datasheet (Fig. 26). I want to use Fig. 26 to get Eon values at a different Ids and Vdd, as Tj is varied.
I am still trying to find the exact tool/command I can use for this. I dont think interp is going to work or I have to understand it much better.
I really appreciate your time. Please do let me know if you have something that can be of help to me.
Thanks
Sneha
Walter Roberson
Walter Roberson 2018 年 9 月 20 日
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.

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KSSV
KSSV 2018 年 9 月 19 日
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)
  2 件のコメント
Sneha Thakur
Sneha Thakur 2018 年 9 月 19 日
Hi KSSV
Thanks for the reply.
But i do not want to assign just any value to the extrap. I want Matlab to calculate the value based on graph equation. I want an answer almost true to the solution.
Is there any other method I can do this with?
Thanks
Sneha
Sneha Thakur
Sneha Thakur 2018 年 9 月 19 日
Hi KSSV
Can you please look into my other equation too,if possible.
It is the same problem, But I guess with datasheet attached and looking at the graphs you would be able to help me.
Thanks
Sneha

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