Regression Fit a SinWave to a dataset?
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Ok..I'm almost there. Yesterday I received some help but still have questions (I'm an ultra beginner).
I love baseball and have a data set with the amount of runs scored by the Pittsburgh Pirates for their last 501 games. The data set is sampled by 1. I completed a spectral analysis on that data set and a 3.5678 game cycle is identified in the data set. I would like to know exactly how to: 1.) generate the sin/cos wave with a period of 3.5678 games 2.) fit the newly created sin/cos wave to the Score data set.
Any code or other insight will be MOST appreciated.
1 Kommentar
Daniel Shub
am 1 Mai 2012
http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer
http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers
Akzeptierte Antwort
Wayne King
am 1 Mai 2012
Something is probably off in your spectral analysis in identifying the frequency. You have 501 points, which means the spacing between frequencies is 1/501. The closest you can actually get to a period of 3.5678 is 0.2794 cycles/game. So let me use that
y = score(:); %make sure score is a column vector
X = ones(501,3);
n = 0:500;
X(:,2) = cos(2*pi*0.2794*n)';
X(:,3) = sin(2*pi*0.2794*n)';
betahat = X\y;
Fit is
yhat = betahat(1)+betahat(2)*cos(2*pi*0.2794*n)+betahat(3)*sin(2*pi*0.2794*n);
If you have the Statistics Toolbox
[b,bint,r,rint,stats] = regress(y,X);
returns useful information in stats for example:
the R-square statistic, the F statistic and p value for the full model, and an estimate of the error variance.
If you have the Statistics Toolbox and R2012a, you can use
mdl = LinearModel.fit(X,y)
which outputs a table. You can get the coefficients with:
mdl.Coefficients.Estimate
2 Kommentare
Weitere Antworten (2)
Wayne King
am 1 Mai 2012
Because when you do a spectral analysis, the frequency spacing in your case is 1/501. So you only get frequency estimates at 0, 1/501, 2/501,3/501, 4/501. So I was assuming that your spectral analysis really revealed a peak at the closest Nyquist frequency (a multiple of 1/501).
If you want, you can use 0.2803 which is 1/3.5678 for the frequency. That should not change the results much I would think.
2 Kommentare
Dr. Seis
am 1 Mai 2012
He gave you the hint... it is a multiple of 1/501
140/501 = 0.2794...
He chose this because ABS(140/501 - 1/3.5678) < ABS(141/501 - 1/3.5678), which means 140/501 is closer to the frequency you are after.
Wayne King
am 1 Mai 2012
You are not using what I gave you. You are using
%betahat = X/y;
Look at the example I gave you:
y = score(:); %make sure score is a column vector
X = ones(501,3);
n = 0:500;
X(:,2) = cos(2*pi*0.2794*n)';
X(:,3) = sin(2*pi*0.2794*n)';
betahat = X\y;
It's very important which slash you use.
2 Kommentare
Siehe auch
Kategorien
Mehr zu Conway's Game of Life finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!