MATLAB: I converted log to linear in part c and it’s giving me error

MATLAB

C_ = log10(C);
figure(2);
semilogy(T,C,'-');
xlabel('time');
ylabel('concentration(mg/L)');
%PART C
t = [17,19,21,23 ]; 
c = [8.1,4.3,2.3,1.1];% series of data points 
cnew = log10(c);
y_1 = interp1(t,c,25,'linear','extrap')
y_2 = interp1(t,c,27,'linear','extrap')
y_3= interp1(t,c,29,'linear','extrap')
c_linear = 10^(cnew)

Best Answer

Replace line:

c_linear = 10^(cnew)

with :

c_linear = 10.^(cnew)

Read about MATLAB element by element operations. https://in.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html

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MATLAB: Interpt 1 is giving me error

The function name is interp1 not interpt1

MATLAB: How to use inerpolation option with end values? If I use interp1 command it’s giving NaN as the output if input is out of the range. I want to use end values if input is out of the range

>> X = [1,2,3];
>> Y = [12,5000,12000];
>> interp1(X,Y,4,'nearest','extrap')
ans =  12000
>> interp1(X,Y,4,'linear','extrap')
ans = 19000

But note that you cannot pick a method like linear, spline, etc, and also limit the extrapolation to the end values:

>> interp1(X,Y,[0,1.5,4],'linear','extrap')  % linear -> not end values
ans =
   -4976    2506   19000
>> interp1(X,Y,[0,1.5,4],'nearest','extrap') % nearest -> not linear interpolated
ans =
      12    5000   12000

But you can easily select the interpolation method and limit to the end values, by simply using max and min:

>> max(Y(1),min(Y(end),interp1(X,Y,[0,1.5,4],'linear','extrap')))
ans =
      12    2506   12000

If you only need to extrapolate in one direction then you could set the extrapolation value:

>> interp1(X,Y,[1.5,4],'linear',Y(end))
ans =
    2506   12000

Best Answer

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MATLAB: Interpt 1 is giving me error

MATLAB: How to use inerpolation option with end values? If I use interp1 command it’s giving NaN as the output if input is out of the range. I want to use end values if input is out of the range

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