View Single Post
Aug26-07, 08:14 PM
WolfOfTheSteps's Avatar
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:

[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)

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]




Phys.Org News Partner Science news on
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds