Test1_THans - Take Home Part of Test I The Answers (Using...

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Take Home Part of Test I The Answers (Using Mathcad) Problems I & II (combined): The period: p 3 The function: First define f0(t) to be the given function for t between 0 and p, and zero elsewhere. f0 t ( ) t 0 t 1 < if 1 1 t 2 < if 3 t 2 t 3 < if 0 otherwise This "shift by np" functions uses MathCad's mod function to compute the x between 0 and p such that t=x+np for some integer n. shift_by_np t ( ) if mod t p , ( ) 0 mod t p , ( ) , p mod t p , ( ) , ( ) So f(t) is the composition of f0 with the "shift by np" function: f t ( ) f0 shift_by_np t ( ) ( ) Maximum number of terms in the Fourier partial sum : M 25 Compute the Fourier Coefficients: A 0 1 p 0 p t f t ( ) d . Note: I've set q k = 2 pw k since 2 k appears in so many of the following computations. k 1 M .. w k k p q k 2 p . w k . a k 2 p 0 p t f t ( ) cos q k t . . d . b k 2 p 0 p t f t ( ) sin q k t . . d .
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FS N t , ( ) A 0 1 N k a k cos q k t . . b k sin q k t . . = b k 0.073 0.067 0.026 0.013 0.015 9.297 10 3 . 6.131 10 3 . 7.152 10 3 . 5.083 10 3 . 3.71 10 3 . 4.331 10 3 . 3.309 10 3 . 2.557 10 3 . 2.972 10 3 . 2.371 10 3 . 4.604 10 15 . 2.198 10 3 . 1.807 10 3 . 1.486 10 3 . 1.709 10 3 . 1.437 10 3 . 1.203 10 3 . 1.378 10 3 . 1.18 10 3 . 1.002 10 3 . a k 0.39 0.117 0.018 0.034 0.023 7.441 10 3 . 0.013 0.01 4.227 10 3 . 7.189 10 3 . 6.048 10 3 . 2.846 10 3 . 4.561 10 3 . 4.014 10 3 . 2.056 10 3 . 3.21 10 3 . 2.91 10 3 . 1.628 10 3 . 2.646 10 3 . 2.271 10 3 . 1.488 10 3 . 1.933 10 3 . 1.818 10 3 . 1.096 10 3 . 1.549 10 3 . The Fourier Coefficients: k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 A 0 0.722 =
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Graphs of the Function and Various Partial Sums of the Fourier Series : The Function and the 4 th Partial Sum: f t ( ) FS 4 t , ( ) t 1 0 1 2
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This note was uploaded on 11/07/2011 for the course MA 460 taught by Professor Staff during the Fall '11 term at University of Alabama in Huntsville.

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Test1_THans - Take Home Part of Test I The Answers (Using...

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