Question

The trigonometric Fourier series of the signal f(t) derived in the lecture notes as e-a, 0 < t T, with T π was n=1 where, 16n 1-e 2 a0 = π (1-e-2), a,- 41-e 2 2 and bn - Show that, f(t) = (1-e-2) +- COSL2n n=1

Answer 1

Given a_0 = 2/pi(1 - e^-pi/2), a_n = 4/pi(1 - e^-pi/2)/(1 + 1 bn^2) b_n = 1b_n(1 - e^-pi/2)/pi(1 + 1b n^2) f(t) = 2/pi(1 - e^-pi/2) + sigma^infinity_n = 1 4/pi(1 - e^-pi/2)/(1 + 1b n^2) cos n omega_0 t + 1bn(1 - e^-pi/2)/n(1 + 1b n^2) sin n omega_0 t = 2/pi(1 - e^-pi/2) + 4/pi(1 - e^-pi/2) sigma^infinity_n = 1 1/1 + 1b n^2 cosn omega_0 t + 4n/1 + 1b n^2 sin n omega_0 t Let 1/a + 1 b n^2 = a and 4n/1 + 1b n^2 = b we know a cos theta + b sin theta = squareroot a^2 + b^2(a/squareroot a^2 + b^2 cos theta + b/squareroot a^2 + b^2 sin theta) = squareroot a^2 + b^2 cos (theta - tan^-1(b/a) using this formula we can write squareroot a^2 + b^2 = squareroot 1/(1 + 1 b n^2) + (4n/1 b n^2)^2 = squareroot (1 + 1bn^2)/(1 + 1bn^2)^2 = 1/squareroot 1 + 1bn^2 tan^-1 b/a = tan^-1 4n/1 + 1b n^2/1/1 + 1bn^2 = tan^-1 4n