Fourier Transform of a piecwise function

[WolfOfTheSteps]

Homework Statement

Find the Fourier Transform of the following function:

$$y(t) = \left( \begin{array}{cc} 0,& \ \ t<1 \\1-e^{-(t-1)},& \ \ 1 < t < 5 \\e^{-(t-5)}-e^{-(t-1)},& \ \ t \geq 5 \end{array}$$

Homework Equations

I employed the following transforms in my attempt at a solution:

$$x(t-t_0) \longleftrightarrow e^{-j\omega t_0}X(j\omega)$$
$$u(t) \longleftrightarrow \frac{1}{j\omega}+\pi\delta(\omega)$$
$$e^{-at}u(t) \longleftrightarrow \frac{1}{a+j\omega} \ \ \ (\mbox{Real}(a)>0)$$

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!

 

[mighty2000]

Your solution is almost correct! The only mistake is in the last step, where you have a typo in the expression for Y(jω). The correct expression should be:

Y(j\omega) = 2e^{-j3\omega}sin(2\omega)\left[\frac{1}{\omega} + j\pi\delta(\omega)-j\right]

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega} + j2\pi e^{-j3\omega}sin(2\omega)\delta(\omega)-j2e^{-j3\omega}sin(2\omega)

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega} + j2\pi e^{-j3\omega}sin(2\omega)\delta(\omega)-j2e^{-j3\omega}sin(2\omega)\frac{\omega}{\omega(1+j\omega)}

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)} + j2\pi e^{-j3\omega}sin(2\omega)\delta(\omega)

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)} + j2\pi e^{-j3\omega}sin(2\omega)\delta(\omega)

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)} + j2\pi e^{-j3\omega}sin(2\omega)\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega t}dt

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)} + \frac{e^{-j3\omega}sin(2\omega)}{\omega}

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)} + \frac{e^{-j3\omega}sin(2\omega)}{\omega(1+j\omega)}

= \frac{2e^{-j3\omega}sin(2\omega)}{\omega(1