The inverse Fourier transform of \(X\left( {j\omega } \right) = \
A. –t rect (2t + 4) + t rect (2t - 4)
B. t rect (2t + 4) - t rect (2t - 4)
C. –t rect (2(t + 4)) + t rect (2(t – 4))
D. t rect (2(t + 4)) - t rect (2(t – 4))
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Right Answer is: C
SOLUTION
\(s\left( {j\omega } \right) = \frac{{2\sin \left( {2\omega / 4 } \right)}}{\omega }\mathop \leftrightarrow \limits^{FT} s\left( t \right) = rect\left( {2t} \right) = \left\{ {\begin{array}{*{20}{c}} 1&,&{\left| t \right| \le 2}\\ 0&,&{otherwise} \end{array}} \right.\)
\({s_1}\left( {j\omega } \right) = 2\sin \left( {4\omega } \right)s\left( {j\omega } \right)\mathop \leftrightarrow \limits^{FT} {s_1}\left( t \right) = - j\;s\left( {t + 4} \right) + j\;s\left( {t - 4} \right)\)
\(x\left( {j\omega } \right) = \frac{d}{{d\omega }}{s_1}\left( {j\omega } \right)\mathop \leftrightarrow \limits^{FT} x\left( t \right) = - j\;t\;{s_1}\left( t \right)\)
\(x\left( t \right) = - t\;rect\;\left( {2\left( {t + 4} \right)} \right) + t\;rect\;\left( {2\left( {t - 4} \right)} \right)\)