Question

Which of the following is the Fourier transform of $x\left(t\right)=\{\begin{array}{cc}2;& -1\le t<0\\ 1;& 0\le t<1\\ 0;& elsewhere\end{array}$

• A. $X\left(\omega \right)=\frac{2\sin \omega }{\omega }+\frac{j}{\omega }-\frac{j{e}^{j\omega }}{\omega }$
• B. $X\left(\omega \right)=\frac{2\cos \omega }{\omega }-\frac{1}{\omega }+\frac{j{e}^{j\omega }}{\omega }$
• C. $X\left(\omega \right)=\frac{2\sin \omega }{\omega }+\frac{j}{\omega }+\frac{e^{j\omega }}{\omega }$
• D. $X\left(\omega \right)=\frac{2\cos \omega }{\omega }+\frac{j}{\omega }+\frac{e^{j\omega }}{\omega }$

Answer 1

The correct option is A $X\left(\omega \right)=\frac{2\sin \omega }{\omega }+\frac{j}{\omega }-\frac{j{e}^{j\omega }}{\omega }$


$2rect(t+\frac{1}{2})\stackrel{F.T}{⟷}2Sa\left(\frac{\omega }{2}\right)e^{\frac{j\omega }{2}}$

$rect(t-\frac{1}{2})\stackrel{F.T}{⟷}Sa\left(\frac{\omega }{2}\right)e^{-\frac{j\omega }{2}}$

$x\left(t\right)\stackrel{F.T}{⟷}X\left(\omega \right)$

$X\left(\omega \right)=2\frac{sin\frac{\omega }{2}}{\omega /2}{e}^{\frac{j\omega }{2}}+\frac{sin\frac{\omega }{2}}{\omega /2}{e}^{-\frac{j\omega }{2}}$

$=4\frac{sin\frac{\omega }{2}}{\omega }{e}^{\frac{j\omega }{2}}+2\frac{sin\frac{\omega }{2}}{\omega }{e}^{-\frac{j\omega }{2}}$

$=4\frac{sin\frac{\omega }{2}(cos\frac{\omega }{2}+jsin\frac{\omega }{2})}{\omega }+2\frac{sin\frac{\omega }{2}(cos\frac{\omega }{2}-jsin\frac{\omega }{2})}{\omega }$

$X\left(\omega \right)=\frac{2sin\omega +j4si{n}^2\frac{\omega }{2}+sin\omega -j2si{n}^2\frac{\omega }{2}}{\omega }$

$=\frac{3sin\omega +j2si{n}^2\frac{\omega }{2}}{\omega }=\frac{3sin\omega +j(1-cos\omega )}{\omega }$

$=\frac{3sin\omega }{\omega }+\frac{j}{\omega }-\frac{jcos\omega }{\omega }$

$=\frac{2sin\omega }{\omega }+\frac{j}{\omega }-\frac{(jcos\omega -sin\omega )}{\omega }$

$=\frac{2sin\omega }{\omega }+\frac{j}{\omega }-\frac{j(jcos\omega -sin\omega )}{j\omega }$

$=\frac{2sin\omega }{\omega }+\frac{j}{\omega }-\frac{j{e}^{j\omega }}{\omega }$