MHB Aidan's question via email about Fourier Transforms (2)

  • Thread starter Thread starter Prove It
  • Start date Start date
  • Tags Tags
    Email Fourier
Click For Summary
SUMMARY

The Fourier Transform of the function $\displaystyle 3\,H\left( t - 1 \right) \mathrm{e}^{-2\,t}$ is calculated using the Second Shift Theorem. By substituting $v = t - 1$, the expression simplifies to $\displaystyle 3\,\mathrm{e}^{-2 - \mathrm{i}\,\omega} \left( \frac{1}{2 + \mathrm{i}\,\omega} \right)$. The Heaviside unit step function, $H(t)$, is utilized, and the Fourier transform of $H(t)e^{-at}$ is confirmed as $\frac{1}{a+i\omega}$. This discussion employs the non-unitary angular frequency convention for Fourier transforms.

PREREQUISITES
  • Understanding of Fourier Transforms
  • Familiarity with the Heaviside unit step function
  • Knowledge of the Second Shift Theorem
  • Basic algebraic manipulation of exponential functions
NEXT STEPS
  • Study the properties of the Second Shift Theorem in Fourier analysis
  • Learn about the applications of the Heaviside function in signal processing
  • Explore the derivation of Fourier transforms for exponential functions
  • Investigate the implications of using non-unitary angular frequency in Fourier transforms
USEFUL FOR

Students and professionals in electrical engineering, applied mathematics, and signal processing who are working with Fourier analysis and time-domain transformations.

Prove It
Gold Member
MHB
Messages
1,434
Reaction score
20
Find the Fourier Transform of $\displaystyle 3\,H\left( t - 1 \right) \mathrm{e}^{-2\,t} $.

In order to use the Second Shift Theorem, the function needs to be entirely of the form $\displaystyle f\left( t - 1 \right) $. To do this let $\displaystyle v = t - 1 \implies t = v + 1 $, then

$\displaystyle \begin{align*}
\mathrm{e}^{-2\,t} &= \mathrm{e}^{-2 \, \left( v + 1 \right) } \\
&= \mathrm{e}^{-2\,v - 2 } \\
&= \mathrm{e}^{-2\,\left( t - 1 \right) - 2 } \\
&= \mathrm{e}^{-2\,\left( t - 1 \right) } \,\mathrm{e}^{-2}
\end{align*} $

And so

$\displaystyle \begin{align*} \mathcal{F}\,\left\{ 3\,H\left( t - 1 \right) \mathrm{e}^{-2\,t} \right\} &= 3\,\mathrm{e}^{-2}\,\mathcal{F}\,\left\{ H\left( t - 1 \right) \mathrm{e}^{-2\,\left( t - 1 \right) } \right\} \\
&= 3\,\mathrm{e}^{-2}\,\mathrm{e}^{-\mathrm{i}\,\omega} \,\mathcal{F}\,\left\{ H\left( t \right) \mathrm{e}^{-2\,t} \right\} \\ &= 3\,\mathrm{e}^{-2 - \mathrm{i}\,\omega} \left( \frac{1}{2 + \mathrm{i}\,\omega } \right) \end{align*} $
 
Mathematics news on Phys.org
  • Like
Likes Greg Bernhardt

Similar threads

  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 1 ·
Replies
1
Views
11K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 4 ·
Replies
4
Views
11K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
10K