Question

the Fourier Transform of the signal $x(t) = e^{-t}u(t - 1)$ is Select one: \begin{itemize} \item a. $X(\omega) = \frac{1}{(j\omega + 1)e}$ \item b. $X(\omega) = \frac{e^{-j\omega}}{(j\omega + 1)e}$ \item c. $X(\omega) = \frac{e^{-j\omega}}{(j\omega + 1)}$ \item d. $X(\omega) = \frac{1}{(j\omega + 1)}$ \end{itemize}

          the Fourier Transform of the signal $x(t) = e^{-t}u(t - 1)$ is
Select one:
\begin{itemize}
    \item a. $X(\omega) = \frac{1}{(j\omega + 1)e}$
    \item b. $X(\omega) = \frac{e^{-j\omega}}{(j\omega + 1)e}$
    \item c. $X(\omega) = \frac{e^{-j\omega}}{(j\omega + 1)}$
    \item d. $X(\omega) = \frac{1}{(j\omega + 1)}$
\end{itemize}

the Fourier Transform of the signal x(t) = e^-tu(t - 1) is
Select one:

* a. X(ω) = (1)/((jω + 1)e)

* b. X(ω) = (e^-jω)/((jω + 1)e)

* c. X(ω) = (e^-jω)/((jω + 1))

* d. X(ω) = (1)/((jω + 1))

Added by Connor E.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sachin Rao

Step 1

The given signal x(t) is defined as x(t) = e^(-t)u(t - 1), where u(t) is the unit step function. The unit step function u(t) is defined as u(t) = 1 for t ≥ 0 and u(t) = 0 for t < 0. Therefore, the signal x(t) is only non-zero for t ≥ 1. Show more…

Show all steps