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WolfOfTheSteps
WolfOfTheSteps is offline
#1
Aug26-07, 08:14 PM
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P: 138
1. The problem statement, all variables and given/known data

Find the Fourier Transform of the following function:

[tex]

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}

[/tex]


2. Relevant equations

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

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

3. The attempt at a solution

First, I rewrote the piecewise function using the unit step function:

[tex]y(t) = u(t-1)-u(t-1)e^{-(t-1)} + u(t-5)e^{-(t-5)} - u(t-5)[/tex]

Next I used the transforms listed above to get the following:

[tex] 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)
[/tex]

After some algebra and application of Euler's formula I got:

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

But the answer should be:

[tex] Y(j\omega) = \frac{2e^{-j3\omega}sin(2\omega)}{w(1+j\omega)} [/tex]


Did I do something wrong? Or is there some way to turn

[tex]\frac{1}{\omega} + j\pi\delta(\omega)-j[/tex]

into

[tex]\frac{1}{\omega(1+j\omega)}[/tex]

??

Thanks!
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