Answer to Question #267362 in Differential Equations for ron

Question

Find fourier transform of

f(t)={ e^2it, -1<t<1

0,otherwise

Answer 1

$e^{2it}=cos2t+isin2t$

$f(t) = a_0 +∑_{n=1}^{\infin}(a_n cos (nw_0t)+b_n sin (nw_0t))$

$a_0=\frac{1}{T}\int^{t_0+T}_{t_0}f(t)dt$

$a_n=\frac{2}{T}\int^{t_0+T}_{t_0}f(t)cos(nw_0t)dt$

$b_n=\frac{2}{T}\int^{t_0+T}_{t_0}f(t)sin(nw_0t)dt$

we have:

$T=2,w_0=2\pi/T=\pi$

then:

$a_0=\frac{1}{2}\int^{1}_{-1}(cos2t+isin2t)dt=\frac{1}{4}(sin2t-icos2t)|^1_{-1}=$

$=\frac{1}{2}sin2$

$a_n=\int^{1}_{-1}(cos2t+isin2t)cos(n\pi t)dt=\frac{2\pi ncos2sin(\pi n)-4sin2cos(\pi n)}{\pi^2n^2-4}=\frac{(-1)^{n-1}4sin2}{\pi^2n^2-4}$

$b_n=\int^{1}_{-1}(cos2t+isin2t)sin(n\pi t)dt=\frac{4cos2sin(\pi n)-2\pi nsin2cos(\pi n)}{\pi^2n^2-4}i=\frac{(-1)^{n-1}2sin2}{\pi^2n^2-4}i$

$f(t) = \frac{1}{2}sin2+∑_{n=1}^{\infin}(\frac{(-1)^{n-1}4sin2}{\pi^2n^2-4} cos (n\pi t)+i\frac{(-1)^{n-1}2sin2}{\pi^2n^2-4} sin (n\pi t))$