1-d curve equation

I have two arrays:

1_time ->

t=(0:15);

2_measurements ->

data1=[  0  7  20  28  35  48  57  61  72  82  96  98  111  121  129  144  ];

I have to get the estimation at t=20 ( linear method ) ,and the 'minimum & maximum' difference between the linear equation line and the data1 curve .

My method was :

1 Making a plot ->

plot(t,data1);

2 Getting the estimated at t=20 by using the numerical panel (tool-> basic fitting)

3 And finally i get the 'minimum & maximum' difference between the linear equation line and the data1 curve , by plotting the residuals

Is there any code to get what i want and not by using plotting .. ect ? I am a newbie when it come to matlab

[Image Analyst]

This is a 1-D curve, not a 2-D curve. For each element of your dependent variable, "data1", there is exactly one independent variable, "t", that specifies it, not two. data1 is a 1-D vector, not a 2-D matrix.

[MRT gang]

Question is updated ! thank you .

Answer 1

[Star Strider]

If you are confident of a linear fit, just calculate the extrapolated ‘data1’ value at ‘t=20’:

t=(0:15);
data1=[  0  7  20  28  35  48  57  61  72  82  96  98  111  121  129  144  ];

B = [t'  ones(size(t'))]\data1';                                % Estimate Coefficient Of Linear Fit
data1_fit = [t'  ones(size(t'))]*B;                             % Create Regression Line To Plot

data1_20 = [20  1]*B                                            % Extrapolate Fit To ‘t=20’

figure(1)
plot(t, data1, 'bp')
hold on
plot(t, data1_fit, '-g')
plot(20, data1_20, '*r')
hold off
grid
axis([0  21    -10  200])

This code plots the data and fit using the mldivide,\ operator, and plots and calculates the extrapolated value:

data1_20 =

         186.43

[MRT gang]

Really appreciate your help . Any idea about the second half of what i wanted ? thanks

[Star Strider]

My pleasure.

I believe I did both. The plot is figure(1), and the extrapolation is this line:

data1_20 = [20  1]*B                                            % Extrapolate Fit To ‘t=20’

producing:

data1_20 =

           186.43

I didn’t see part 3. about the residuals and residual plot until I went looking for it just now. Add this to the end of my previous code, and I believe we’ve covered everything:

resid = data1 - data1_fit';
resid_extr = [min(resid) max(resid);  min(abs(resid))  max(abs(resid))];
logc_vct = [abs(resid) == resid_extr(2,1);  abs(resid) == resid_extr(2,2)];

figure(2)
stem(t, resid)
hold on
plot(t(logc_vct(1,:)), abs(resid(logc_vct(1,:))), '*r',  t(logc_vct(2,:)), abs(resid(logc_vct(2,:))), '*r') 
hold off
grid
xlabel('Time')
ylabel('Residual')
axis([-1  16    ylim])

The ‘resid_extr’ matrix are the minima and maxima of the residuals and the absolute values of the residuals. I plotted all and highlighted the minimum and maximum absolute values. I used logical vectors rather than the find function in the ‘logc_vct’ array, just for variety. It doesn’t make any difference in the behaviour of the code.

[MRT gang]

thank you very much ! i appreciate your help

[Star Strider]

My pleasure!

Answer 2

[Azzi Abdelmalek]

out=interp1(t,data1,20,'linear','extrap')

[MRT gang]

the result is not the same when using my method ( plot->basic fitting>numerical estimation )