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 P: 138 1. The problem statement, all variables and given/known data Find the Fourier Transform of the following function: $$y(t) = \left( \begin{array}{cc} 0,& \ \ t<1 \\1-e^{-(t-1)},& \ \ 1 < t < 5 \\e^{-(t-5)}-e^{-(t-1)},& \ \ t \geq 5 \end{array}$$ 2. Relevant equations I employed the following transforms in my attempt at a solution: $$x(t-t_0) \longleftrightarrow e^{-j\omega t_0}X(j\omega)$$ $$u(t) \longleftrightarrow \frac{1}{j\omega}+\pi\delta(\omega)$$ $$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j\omega} \ \ \ (\mbox{Real}(a)>0)$$ 3. The attempt at a solution First, I rewrote the piecewise function using the unit step function: $$y(t) = u(t-1)-u(t-1)e^{-(t-1)} + u(t-5)e^{-(t-5)} - u(t-5)$$ Next I used the transforms listed above to get the following: $$Y(j\omega) = e^{-j\omega}\left(\frac{1}{j\omega}+\pi\delta(\omega)\right) - e^{-j\omega}\left(\frac{1}{1+j\omega}\right) + e^{-j5\omega}\left(\frac{1}{1+j\omega}\right) - e^{-j5\omega}\left(\frac{1}{j\omega}+\pi\delta(\omega)\right)$$ After some algebra and application of Euler's formula I got: $$Y(j\omega) = 2e^{-j3\omega}sin(2\omega)\left[\frac{1}{\omega} + j\pi\delta(\omega)-j\right]$$ But the answer should be: $$Y(j\omega) = \frac{2e^{-j3\omega}sin(2\omega)}{w(1+j\omega)}$$ Did I do something wrong? Or is there some way to turn $$\frac{1}{\omega} + j\pi\delta(\omega)-j$$ into $$\frac{1}{\omega(1+j\omega)}$$ ?? Thanks!