Question

2 Find the esponential Fourier-series expansion of the periodic signal x(1 + 2π) = x(1)

Answer 1

Given x(t) = t^2 - pi < t lessthanorequalto pi and x (2 pi + t) = x(t) implies x(t) is a periodic signal with period of T = 2 pi Exponential Fourier series is represented by x(t) = sigma^infinity_n = -infinity C_n e^j n omega_0 t where C_n - 1/T integral_T x(t) e^- j n omega_0 t dt calculation of C_n: C_n = 1/2 pi integral^pi_pi t^2 e^- j n omega_0 t dt since x(t) is even signal C_n = 1/2 pi times 2 integral^pi_0 t^r e^- j n omega_0 t \r\n C_n = 1/2 pi times 2 [t^2 e^- j n omega_0 t/- j n omega_0 + 2 t e^- j n omega_0 t/+ n^2 omega^2_0 + 2 e^-j n omega_0 t/j n^3 omega^3_0]^pi_t = 0 C_n = 1/ pi [pi^2 e^-j n omega_0 pi/- j n omega_0 + 2 pi e^-j n omega_0 pi/n^2 omega^2_0 + 2 e^- j n omega pi/j n^3 omega^3_0] - 1/pi