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homework2 - 1 1/2 P2.9.l w-21r/2- a 2dt—l 9 o 0 2 31/2...

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Unformatted text preview: . \. / . 1 1/2 P2.9.l w -21r/2-«; a - 2dt—l; \\ _ 9! o 0 2- 31/2 1/2 1/2 an - ~3- __{/22 cos mrt dt - 4 { cosmrt dt - 2%; 1/2 bn - 22- f 2 sin mrt dt - 0 (integral of odd fn. over symmetric interval). -1/2 ° sin(n1r/2) Thus, the series is: f(t) - l + 22 W cos n1rt. 14-1 b) Expanding, we have: f(t) - 1 + %cos 1rt - {licos 3m: + é—"cos 5M: - This result has the same harmonics (i.e., multiples of we) as Example 2.5.1 but: 1) Has an average value of 1, in contrast to 0 in Example 2.5.1. 2) Expands in cosines rather than sines as a result of the even symmetry, in contrast to the odd (folded) symmetry in Example 2.5.1. Note from a comparison with Drill Problem 2.5.1 that time scaling affects “’01 but does not affect the Fourier series coefficients. <:;2:11.1 U ing Eqs. (2.49), (2.52) in Eq. (2.69), we have: T{f2(t) dt — a3 + 2n;1%(an-jbn)(an+jbn) - a3 + % Z (a§+b§). n-l {(3,181) a) e‘zcos[1r(2-1)] =- -0.l35; b) e'3u(t-3) — 0.05u(t-3); c) (4)2 + 4 - 20; d) [(2)2 + 41/2 - 4; e) (3)3/3 + 4(3) - 21; ...
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