Predicting the value at y(t=8) using my model

Hi,

I am completeing some LLS analysis and need to predict the value of my model after 8 seconds.

% Problem 3

clc

close all

% Load Data

t = LLS_Data3(:,1);

y = LLS_Data3(:,2);

% Plot the data

figure;

plot(t,y,'.')

% Apply LLS

AL = [t,ones(size(x))]; %

thetaL = inv(AL.'*AL)*(AL.')*y;

yML = thetaL(1)*t + thetaL(2); % Based on equation

% Plot Error over time

figure(1)

errorL = y - yML;

plot(t,errorL, 'x')

xlabel('Time')

ylabel('Error')

hold on

yline(0, 'k')

hold off

This is my current code but am unsure how to use the model to predict my Y value (yML) at t = 8 seconds. The given data set runs for 10 seconds and there are currently 7 values given throughout the 10 second period.

Any help would be appreciated,

Thank you

[Sam Chak]

@Matt Boyles: The given data set runs for 10 seconds and there are currently 7 values given throughout the 10 second period.

Are you suggesting that there are only 7 data points over the entire 10-second period?

Are you looking to predict the output at exactly t = 8 seconds using the LLS model?

[Matt Boyles]

Yes exactly, thanks

Answer 1

[Torsten]

thetaL(1)*8 + thetaL(2)

[Matt Boyles]

Thankyou,

What about for a more complex equation like

yM7 = theta7(1)*t.^7 + theta7(2)*t.^6 + theta7(3)*t.^5 + theta7(4)*t.^4 + theta7(5)*t.^3 + theta7(6)*t.^2 + theta7(7)*t + theta7(8); %

Is it the same premis that I would just multiply each t value by 8? it doesnt seem correct to me.

Thanks for your help!!

[Torsten]

If you have a function yM7(t) and you want to find its value at t=8, how do you do that ? You insert 8 for t, don't you ?

Answer 2

[Sam Chak]

Hi @Matt Boyles

You can try something like the following to estimate the output. However, there is no guarantee that the estimation at sec is accurate, as shown in the following example.

subplot(2, 1, 1)
nPts1 = 1001;       % number of points
x1   = linspace(0, 10, nPts1);
y1   = sin(4*pi/10*x1) + sin(6*pi/10*x1);
plot(x1, y1, 'linewidth', 1.5), grid on
xlabel('x')

subplot(2, 1, 2)
nPts2 = 7;          % number of points
x2    = linspace(0, 10, nPts2);
y2    = sin(4*pi/10*x2) + sin(6*pi/10*x2);
% plot(x2, y2, 'rp')
xlabel('x')

f = fit(x2', y2', 'poly6')

f = 
     Linear model Poly6:
     f(x) = p1*x^6 + p2*x^5 + p3*x^4 + p4*x^3 + p5*x^2 + 
                    p6*x + p7
     Coefficients:
       p1 =  -5.586e-17
       p2 =    0.005051
       p3 =     -0.1263
       p4 =       1.099
       p5 =      -3.858
       p6 =       4.443
       p7 =   2.993e-13

plot(f, x2, y2), grid on

actual_t8 = sin(4*pi/10*8) + sin(6*pi/10*8)

actual_t8 = 7.7716e-16

estim_t8  = f(8)

estim_t8 = -0.3861

[Matt Boyles]

Thank you!