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Unformatted text preview: 1 EAS6939 HW#1 Name: Taiki Matsumura UFID: 65358317 Date: 1/13/2010 1. To solve the standard deviation of the 11 noise, I use the following formula, N N n 1 Where, n is number of the samples, N is the value of ith samples of noise, and N is the mean of noise. I can obtain by the value of noise n 11, N 0.138 Then I obtain the standard deviation of the noise, . . It can be seen that is fairly closed to the specific standard deviation (=0.1). 2. For linear polynomial approximation with noise, I use following vector form X Where y 0.0687 0.0955 0.2558 0.4731 0.3290 0.5473 0.5695 0.5277 0.6926 0.9035 1.0225 , X 1 1 0.1 1 0.2 1 0.3 1 0.4 1 0.5 1 0.6 1 0.7 1 0.8 1 0.9 1 1 , b b So that I can obtain X T X 11 5.5 5.5 3.85 , X T 5.3478 3.7087 Now I write the normal equation, X T X X T , 11b 5.5b 5.3478 5.5b 3.85b 3.7087 And solve it to obtain b 0.0158 and b 0.9407 . So that the leastsquare fit is y 0.0158 0.9407x Sum of the squares of the error is calculated as SS T X T 3.6552 3.5733 0.0819 2 Finally related error measures, such as rms error e , standard deviation , coefficient of multiple determination R , adjusted coefficient of multiple determination R , PRESS rms error e PrESS , and rms error by calculated analytically (by exact integration) e _ are calculated as followings....
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 Spring '08
 PETERIFJU

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