Question #102649
Determine the Fourier transform of the function

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

=0 otherwise
1
Expert's answer
2020-02-21T10:09:39-0500

Ft[f]=12π+f(t)eiwtdt=12ππeπesin3teiwtdt=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=

=12πeiwt3cos3tiwsin3tw29πeπe=i2π(sinπe(w+3)w+3sinπe(w3)w3)=\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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