View Single Post
WolfOfTheSteps
#1
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:

[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!
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100