least squares with seasonal component in matlab - matlab

Question

I was reading a paper which looked at investigating trends in monthly wind speed data for the past 20 years or so. The paper uses a number of different statistical approaches, which I am trying to replicate here.
The first method used is a simple linear regression model of the form
$$ y(t) = a_{1}t + b_{1} $$
where $a_{1}$ and $b_{1}$ can be determined by standard least squares.
Then they specify that some of the potential error in the linear regression model can be removed explicitly by accounting for the seasonal signal by fitting a model of the form:
$$ y(t) = a_{2}t + b_{2}\sin\left(\frac{2\pi}{12t} + c_{2}\right) + d_{2}$$
where coefficients $a_{2}$, $b_{2}$, $c_{2}$, and $d_{2}$ can be determined by least squares. They then go on to specify that this model was also tested with additional harmonic components of 3, 4, and 6 months.
Using the following data as an example:
% 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960
y = [112 115 145 171 196 204 242 284 315 340 360 417 % Jan
118 126 150 180 196 188 233 277 301 318 342 391 % Feb
132 141 178 193 236 235 267 317 356 362 406 419 % Mar
129 135 163 181 235 227 269 313 348 348 396 461 % Apr
121 125 172 183 229 234 270 318 355 363 420 472 % May
135 149 178 218 243 264 315 374 422 435 472 535 % Jun
148 170 199 230 264 302 364 413 465 491 548 622 % Jul
148 170 199 242 272 293 347 405 467 505 559 606 % Aug
136 158 184 209 237 259 312 355 404 404 463 508 % Sep
119 133 162 191 211 229 274 306 347 359 407 461 % Oct
104 114 146 172 180 203 237 271 305 310 362 390 % Nov
118 140 166 194 201 229 278 306 336 337 405 432 ]; % Dec
time = datestr(datenum(yr(:),mo(:),1));
jday = datenum(time,'dd-mmm-yyyy');
y2 = reshape(y,[],1);
plot(jday,y2)
Can anyone demonstrate how the model above can be written in matlab?

Answer 1

Notice that your model is actually linear, we can use a trigonometric identity to show that. To use a nonlinear model use nlinfit.
Using your data I wrote the following script to compute and compare the different methods:
(you can comment out the opts.RobustWgtFun = 'bisquare'; line to see that it's exactly like the linear fit with the 12 periodicity)
% y = [112 115 ...
y2 = reshape(y,[],1);
t=(1:144).';
% trend
T = [ones(size(t)) t];
B=T\y2;
y_trend = T*B;
% least squeare, using linear fit and the 12 periodicity only
T = [ones(size(t)) t sin(2*pi*t/12) cos(2*pi*t/12)];
B=T\y2;
y_sincos = T*B;
% least squeare, using linear fit and 3,4,6,12 periodicities
addharmonics = [3 4 6];
T = [T bsxfun(#(h,t)sin(2*pi*t/h),addharmonics,t) bsxfun(#(h,t)cos(2*pi*t/h),addharmonics,t)];
B=T\y2;
y_sincos2 = T*B;
% least squeare with bisquare weights,
% using non-linear model of a linear fit and the 12 periodicity only
opts = statset('nlinfit');
opts.RobustWgtFun = 'bisquare';
b0 = [1;1;0;1];
modelfun = #(b,x) b(1)*x+b(2)*sin((b(3)+x)*2*pi/12)+b(4);
b = nlinfit(t,y2,modelfun,b0,opts);
% plot a comparison
figure
plot(t,y2,t,y_trend,t,modelfun(b,t),t,y_sincos,t,y_sincos2)
legend('Original','Trend','bisquare weight - 12 periodicity only', ...
'least square - 12 periodicity only','least square - 3,4,6,12 periodicities', ...
'Location','NorthWest');
xlim(minmax(t'));