solutions8 - ponding values: 1 = 1 . 9323 1- 2 = 1 . 8967 1...

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Matlab Solutions Time Series (2DD23) Week 8 Solution to the Matlab Exercise ( D + C sin 2 t + φ 2 )) sin 1 t + ϕ 1 ) (1) Use: 2 cos A cos B = cos ( A + B ) + cos ( A - B ) (see exercise 2.3, page 26. So sin A sin B = cos ( A - π 2 ) cos ( B - π 2 ) = 1 2 cos ( A + B - π) + 1 2 cos ( A - B ) = - 1 2 cos ( A + B ) + 1 2 cos ( A - B ) Now we can rewrite formula 1. ( D + C sin 2 t + φ 2 )) sin 1 t ϕ 1 ) = D sin ( ω 1 t + ϕ 1 ) + C sin 2 t + φ 2 ) sin 1 t + ϕ 1 ) = ··· + C ± 1 2 cos ( 1 - ω 2 ) t + ϕ 1 - ϕ 2 ) - 1 2 cos ( 1 + ω 2 ) t + ϕ 1 + ϕ 2 ) ² So the peaks are at frequencies ω 1 , ω 1 - ω 2 and ω 1 + ω 2 . Use searchpeaks to find the corres-
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Unformatted text preview: ponding values: 1 = 1 . 9323 1- 2 = 1 . 8967 1 + 2 = 1 . 9674 This means that 1 = 1 . 9323 , so the length of one period is 1 1 . 9323 = . 5175 day = 12 . 42 hours. Furthermore 2 = 1 . 9323-1 . 8967 , so the length of one period is 1 . 0356 = 28 . 87 day. 1...
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