Question

Find a Fourier series expansion of the periodic function 0 -T -asts 2 - f(t) = 6 cost T <<- 2 2 0 I SISE 2 f(t) = f (t +21) Select one: a f(t)= 12 12 5 (-1)** cos nt 1 2n-1 b. f(t) = 12.12 F(-1)** cos 2nt T 4n-1 C 6 12 =+ 125 (-1) C05 211 472-1 6 12 (-1) * cosm d

Answer 1

0 f(t) -tets- 6 cost - 1/2 st sa ㅈ 0 X 2 sts Fourier series expansion of f(t) is f(t) = he + Ź Can cos nt + by sin nt) n=1 KIN Pou to guide = { atau séunitate foute Szerst dt T 2 costat = 12 7 = 12. [st] 12. (sin Q-smo) 12 - • 7/2 6 cost cosnt dt and an -IK X T: ft) = 0 in -A <t<- À I and -7/2 Asten → An = À S?6cost ess nt at 6 cost cosnt dt picost cosnt is even function an = À l {cos (n+1) t + C05 (n-1)t} de ㅈ wal

sin(n+1) + 6 ㅈ sin (n-1) t Na 7 an + N+I M-1 o 6 ㅈ .sin (n+1) sin (n-1) + 2 | M+! n-1 케이 (n-1) Sin{n+1) 졸 + (h+i) sin h-)졸 +) (n-1 N - Sar) [simg 8 + c6 sing] + tory [ snea-coasing 이k 어즈 1 (n-1) cos + (n+) ( - is 6. ㅈ 101) 600 (-452 3 nT = -2005 k 2 2 다 므05 용 1 다 12 ㅈ cos na 47 1 12 (-)) 2 ㅠ 4n즈 1 ㅈ And bn = 1 / 4 ( f(t) sinnt at - T - 2 6 costsinnt at ㅈ 2

lon - o To cost sinnt is odd function] .: The required Fourier series is ft) 6 ㅈ + Σ 12 T 40² n=1 M (-)nt cos nt} > ft) = 4 + 1 osnt) 7 fe - ht c.yntless nt 12 = 4n21 T 1=1

Thank you.