## Answers

Given Periodic signal Time period T = 3 W = 2 pi/T = 2 pi/3 We know the c_n = 1/T integral^T_0 x (t) e^-i n w_0 t dt c_n = 1/3 [integral^3_0 t e^-in 2 pi/3 t dt] c_n = 1/3 [integral^1_0 t e^-in 2 pi/3 t dt] By integrating by parts c_n = 1/3 [t e^-j n 2 pi/3 t/-j n 2 pi/3 - e^-j n 2 pi/3 t/(-j n 2 pi/3)^2]^1_0 c_n = 1/3 [e^-j n 2 pi/3/-j n 2 pi/3 - e^-j n 2 pi/3/(-j n 2 pi/3)^2 - [0 - 1/(-j n 2 pi/3)^2]] c_n = 1/3 [-e^j n 2 pi/3/j n 2 pi/3 + e^-j n 2 pi/3/(n 2 pi/3)^2 - 1/(n 2 pi/3)^2] \r\n c_n = 1/3 3/2 pi n j e^-j n 2 pi/3 + 3/4 pi^2 n^2 e^-j n 2 pi/3 - 3/4 pi^2 n^2 c_n = j e^-j n 2 pi/3/2 pi n + 3/4 pi^2 n^2 e^-j n 2 pi/3 - 3/4 pi^2 n^2 c_n

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