Hello all,
I have datapoints, expressed in a x-vector and a y-vector.
Just by plotting x and y the points are already pretty smooth. The gradient between the first x-points is very big, so I would like to smooth the curve even more. How do I do it the best way? Interpolation between points is the answer, but how to I do it in mathcad?
I attached a mathcad prime 3.0 sheet where I try to do some tests, but without success so far.
The graph in the screenshot is already pretty accurate, but thats because I have enough datapoints.
In a worst case, where I cannot obtain enough datapoints, I would need some kind of interpolation...
The optimum would be even to reconstruct a function out datapoints. Is this possible?
I could do further analysis by having a function f(x)
The datapoints were obtained from this graph for thermal resistance over a copper area:
Best regards!
ACCEPTED SOLUTION
Accepted Solutions
Here are two other models.
Both have the x-axis as horizontal asymptote and use just three parameters.
For playing around with different models I suggest you download the trial/evaluation version of "CurveExpert Professional". You can testdrive it (for I think 30 days) in full mode and then it falls back to a very useful mode with limited functionality which is always good enough for trying different fits for a given data set. CurveExpert will try to fit about a hundred predefined function models to your data and you may also define your own models. It sorts them for best fit and most of the times finds good guess values to come to a fitting result itself. In Prime you have to play around with the guesses yourself to get genfit do its job. Needless to say - the models above were found using CurveExpert Pro.
The sheet you attached does not have much value without the data!
I guess that what you are looking for is not interpolation but rather some kind of regression.
Maybe some type of hyperbola or a more elaborate rational function type.
You may try "genfit" after you have settled for a specific type of regression function type.
Of course it would help if you know the type of function you can expect.
Here is a quick try using some few data optically sampled from the picture you provided.
For my convenience, I implemented it in MC 15, but it should work the same way in Prime.
Here is my 10 ct:
(I grabbed the data from the two vectors on the first page of your file, and assigned that to x and y, exported them using WRITEPRN and then imported it into Mathcad 11)
Success!
Luc
Ok, I had to redigitize the graph since (as Werner pointed out) you didn't send your data.
Attached is a Prime Express file and a pdf of it.
I would expect that a log/log plot would present a straight line; it doesn't (perhaps an artifact of digitization.) At any rate, a second order polynomial fit of the log of X and Y provides a reasonable curve.
(The method of getting there is due to having only Express, so polyfit needs to be invented.)
I just noticed that you evaluated the two vectors anyway and as we can see the last result when we open a Prime sheet, I can copy the data myself and assign it to the variables.
I don't know what happened in your attempt to duplicate my first approach. But when I deleted the region "r(s):=f(s,a,b,c,d)" and retyped it from scratch, things worked OK. So you may give that a try. Maybe the reason is Primes infamous autolabel "feature" - don't know.
Here is what I see after a changed the limits on the axis accordingly:
If x approaches infinity, r(x) appraoches zero - so the x-axis is the horizontal asymptote.
In my second appoach I realize that the new algorithm in Prime seems to be even more sensitive with respect to the guess values that the one on Mathcad. You found out about that yourself.
After changing the axis limits I guess this is the better fit of the two.
The graph approaches the value of c (approx. 10) if x goes to infinity.
According you attempt to duplicate Luc's contribution:
1) You must put the constraints and the minerr command into a solve block
2) You must use vector indices! The ones you get by pressing [
When Luc writes xn1, both the n and also the 1 are vector indices!
This graph approaches minus infinity when x goes to infinity
You may consider an easier approach without using the vector n , like
or you also can use "genfit" to do the job:
The values are a little bit different because in Lucs original approach only five data points (specified by the vector n) where considered, but now all 12 data points are taken in account.
Here are two other models.
Both have the x-axis as horizontal asymptote and use just three parameters.
For playing around with different models I suggest you download the trial/evaluation version of "CurveExpert Professional". You can testdrive it (for I think 30 days) in full mode and then it falls back to a very useful mode with limited functionality which is always good enough for trying different fits for a given data set. CurveExpert will try to fit about a hundred predefined function models to your data and you may also define your own models. It sorts them for best fit and most of the times finds good guess values to come to a fitting result itself. In Prime you have to play around with the guesses yourself to get genfit do its job. Needless to say - the models above were found using CurveExpert Pro.
To change plot scales, you can select and change the first, the second and the last value along each of the axes.
See the picture above, The red arrows point to the y-axis parameters, the green are for the x-axis.
Zoomed in on the bottom y-value.
If you change the bottom y-value to 0, and the second one (which is now 0) to 20, and the top one to 200, you get an axis from 0 to 200 with ticks every 20.
It is great to see how intuitive this was designed...
Success!
Luc
@xyz123 wrote:
Hello all,
could you please help again to interpolate smoothly the following datapoints?
That depends on how you would define "smooth interpolation" and what you would need the interpolation function for. For many cases a simple linear interpolation like the one shown in the next picture would suffice and be as good and suitable as any spline interpolation.
It may also be that you are looking for a regression function. If thats the case it would help to know more about the data, what process it stems from and what kind of function its assumed to be underlying.
Here are two different examples for a regression and you always may fit your own function type to the data using Mathcads genfit function as shown in the last example.
