Question

The function f(t) is defined by -3t+6, 0<t<4 f(t) = -3, 4 < t < 5. Let f (t) denote the periodic extension of f(t), with period 5. Evaluate f (-2.3), f (O), f (7.5), f (9.2) and state the value to which the Fourier series of f (t), FS(t), converges for each of the following values: t = 0,t = 2.8, t = 4, t = 5. Enter all your answers correct to one decimal place. • A: Enter f(-2.3): • B: Enter f (0): • C: Enter f (7.5): D: Enter f (9.2): • E: Enter FS(O):

Also :
FS (2.8)
FS (4)
FS (5)

Answer 1

Solution: Given function is f(t) = -3t+6, Oct 24 -3, uc tcs Let i(t) denote the periodic extension of f(t) with periods. Then f(t) = f(t) for octes with f (t+sn) = f(t) Now -2.1 - 1.5 (A) † (-2.3) - (-2.3+5) - (2.1)=f(2.1) --3(2.7) +6 = (B) Ể (0) = (0+5) = f(5) = 4(5) = -3 (c) € (7-5) = f (1.5-5) = f (2-5)= f(2.5) = -3(2.5)+6 = a f (9.2) = (9.2-5) = f(4.2) = f(4.2) = -3 (E) FS (0) = [ f(0+0) + f(0-0) J= { [f(0+0) + f(s-o)] = < (6-3)= ( 3 [:Ojs beint of dis- continuity (F) 65 (28) Ê (2.8) = f(2.8) = -3(2-8) +6=1-2-4 (6) F5(4) = £ [ f(4+0) + f(4-o)] = { [f(4to) + f(4-0) ] = 1 / 2 [-3-6]= [2 / 3 [:4 is a point of discontinuity (H) Fts FS (5) FS (0) - 2