Question

The sketch of the following periodic function f (t) given in one period f(t) t2 -1, 0s t s 2 is given as follows f(t) 2 -1 We proceed as follows to find the Fourier series representation of f (t) (Note:Jt2 cos at dt = 2t as at + (a--)sina:Jt2 sin at dt = 2t sin at + sin at. Г t2 sin at dt-tsi. )cos at.) Please scroll to the bottom of page for END of question a) The period of this function, 2p, is (5 marks) Please choose one: a)O 4 b)o 8 c)o2 d)O None of the above

Answer 1

Given that f(t) = t^2/2 -1, 0 < equal to t < equal to 2 (a) the period of this function f(t) is 2 (b) the fourier series is f(t) = a_0 /2 + sigma^infinity _n-1(cos n pi t) + sigma^infinity_n-1 b_nsin(n pi t) a_0 = 1/1 integral _0^2 f(t) dt = integral_0^2 (t^2/2 - 1)dt = (t^3/6-t)_0^2 = 8/6 -2 = -4/6 = -2/3 (c) a_n = 1/1 integral _0^2 f(t) cos n pi t dt = integral _0^2 (t^2/2-1) cos n pi t dt = 1/2 integral _0^2 t^2 cos n pi t dt - integral _0^2 cos n pi t dt 1/2 [2t cos n pi t/n^2 pi^2 +(t^2/n pi - 2/ n^3 pi^3) sin n pi t]_0^2 - (sin n pi t/ n pi)_0^2 = 1/2 [4/n^2 pi^2]-(0) = 2/n^2 pi ^2 (d) b_n = 1/1 integral _0^2 f(t) sin n pi t dt = integral _0^2 (t^2/2-1) sin n pi t dt