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How to interpolate intermediate values?

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Jacky Jo
Jacky Jo on 21 Jun 2018
Commented: Jacky Jo on 21 Jun 2018
I have an array with 110 values. let say:
M1_allvalues = [1,2,10,-1,-2,..,-10, 1, 2..........,10]
I simply want to make the array to a size of 3600 values in it, by interpolating the values in between each array element. There would be approximately 32-33 values between each element to achieve 3600 values array. For example:
Between 1 and 2 in the given array some 32 values, then the beginning would be:
newArray = [1, 1.03, 1.06, 1.09........,1.97, 2,......, 3,........, 110]
How do I do that? I was thinking of this:
for i=1:length(M1_allvalues) - 1
newArray(i,1)= M1_allvalues(i,1): (32-33 vales): M1_allvalues(i + 1,1);
end
could you me some idea?
  2 Comments
Walter Roberson
Walter Roberson on 21 Jun 2018
Is it required that the existing elements all appear in the output exactly? If equal spacing were used then some elements might only be approximated.
Jacky Jo
Jacky Jo on 21 Jun 2018
Edited: Jacky Jo on 21 Jun 2018
@Walter Roberson Yes. Because they represent a curve. If we consider only the start and end the curve will not be thereafter. In other words, upsampling may be. The array look like as follows. For instance, I would like to have between -0.716515302 and -0.738227694 I would like to have 32/33 values. The similar way for all. Do we have any inbuild function for that?
M1_allvalues = [
-0.716515302
-0.738227694
-0.734134713
-0.795537839
-0.721594973
-0.733554246
-0.633938598
-0.646957164
-0.639987555
-0.650192438
-0.69239733
-0.844869671
-0.891111276
-0.97870424
-0.97912141
-0.87953858
-0.87995575
-0.98037292
-0.98079009
-0.98120726
-0.98162443
-0.9820416
-0.982458771
-0.982875941
-0.983293111
1.014964305
0.98042719
0.974544314
0.688186533
0.658870952
0.595105523
0.390863794
0.199207123
0.215257305
0.10254977
-0.051655171
-0.234441557
-0.306198555
-0.304774991
-0.420750606
-0.559915604
-0.507888071
-0.64621695
-1.025155038
-0.515321439
-0.672936192
-0.620544176
-0.660977732
-0.673011345
-0.785923311
-0.607349152
-0.580663475
-0.63031181
-0.765443108
-0.662154214
-0.646247998
-0.628825981
-0.625081084
-0.618565838
-0.661108932
-0.645737598
-0.606520318
-0.639491009
-0.653300752
-0.670253078
-0.659396223
-0.778644332
-0.870010559
-0.939977168
-0.941245761
-0.942514355
-0.943782948
-0.945051542
-0.946320136
-0.947588729
-0.948857323
-0.950125916
-0.95139451
-0.952663103
-0.953931697
1.102059236
0.800832245
0.878517853
0.771312906
0.682760904
0.524194329
0.845499176
0.780904075
0.185449607
0.08183194
0.005542497
-0.129033531
-0.230487187
-0.316896749
-0.413027888
-0.507425914
-0.599394836
-0.616839434
-0.63656488
-0.669927747
-0.668568359
-0.646474955
-0.646278772
-0.582780115
-0.735585371
-0.772032042
-0.757837306
-0.734415844
-0.740307607
-0.74551905 ]

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Accepted Answer

KSSV
KSSV on 21 Jun 2018
Edited: KSSV on 21 Jun 2018
% M1_allvalues = [1,2,3,....,110] ;
iwant = linspace(min(M1_allvalues),max(M1_allvalues ),3600) ;
  3 Comments
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KSSV
KSSV on 21 Jun 2018
Let A be your data.
N = round(3600/length(A)) ;
iwant = zeros(N,length(A)-1) ;
for i = 1:length(A)-1
iwant(:,i) = linspace(A(i),A(i+1),N) ;
end
iwant = iwant(:) ;
Jacky Jo
Jacky Jo on 21 Jun 2018
Yes.. That works.... Thanks a lot

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