Question #102649

Determine the Fourier transform of the function


f(t)={sin3t -pie<=t<=+pie



=0 otherwise

Expert's answer

Ft[f]=12π∫−∞+∞f(t)e−iwtdt=12π∫−πeπesin3te−iwtdt=F_t[f]=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}f(t)e^{-iwt}dt=\frac{1}{\sqrt{2\pi}}\int\limits_{-\pi e}^{\pi e}sin3te^{-iwt}dt=

=12πe−iwt3cos⁡3t−iwsin⁡3tw2−9−πeπe=i2π(sin⁡πe(w+3)w+3−sin⁡πe(w−3)w−3)=\frac{1}{\sqrt{2\pi}}\frac{e^{-iwt}3\cos3t-iw\sin3t}{w^2-9}_{-\pi e}^{\pi e}=\frac{i}{\sqrt{2\pi}}\left( \frac{\sin\pi e(w+3)}{w+3}-\frac{\sin\pi e(w-3)}{w-3}\right)


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