Question

$\text{The Fourier transform of the complex sinusoidal pulse shown below is}z\left(t\right)=\{\begin{array}{cc}{e}^{j10t};& |t|<\pi \\ 0;& |t|>\pi \end{array}$

• A. $\frac{1}{\omega -10}sin\left(\right(\omega -10\left)\pi \right)$
• B. $\frac{2}{\omega -10}sin\left(\right(\omega -10\left)\pi \right)$
• C. $\frac{2}{\omega }sin\left(\right(\omega -10\left)\pi \right)$
• D. $\frac{1}{2\omega }sin\left(\right(\omega -10\left)\pi \right)$

Answer 1

The correct option is B $\frac{2}{\omega -10}sin\left(\right(\omega -10\left)\pi \right)$

$\text{Given, sinusoidal pulse}z\left(t\right)=\{\begin{array}{cc}{e}^{j10t};& |t|<\pi \\ 0;& |t|>\pi \end{array}\text{We may express z(t) as the product of a complex sinusoid}{e}^{j10t}\text{and a rectangluar pulse x(t)}\text{Let}x\left(t\right)=\{\begin{array}{cc}1;& |t|<\pi \\ 0;& |t|>\pi \end{array}\text{Fourier transform of x(t) is X(j}\omega )\therefore X(j\omega )={\int }_{-\infty }^{\infty }x(t)e^{-j\omega t}dt={\int }_{-\pi }^{\pi }1.e^{-j\omega t}dt={\left[\frac{e^{-j\omega t}}{-j\omega }\right]}_{-\pi }^{\pi }=-\frac{1}{j\omega }[e^{-j\pi \omega }-e^{+j\omega \pi }]=\frac{e^{j\omega \pi }-e^{-j\omega \pi }}{j\omega }=\frac{2}{\omega }\left[\frac{e^{j\omega \pi }-e^{-j\omega \pi }}{2j}\right]\therefore X(j\omega )=\frac{2}{\omega }sin(\omega \pi \left)\text{By using frequency shifting property of Fourier transform, we get,}z\right(t)=e^{j10t}.x(t\left)\stackrel{FT}{⟷}X\right(j(\omega -10)\left)z\right(t\left)\stackrel{FT}{⟷}\frac{2}{\omega -10}sin\right((\omega -10)\pi )$