Question

What is the Fourier transform of $e^{-3(t-2)}u(t-2)$? $\frac{1}{3+j\omega}$ $\frac{1}{-3+j\omega}$ $\frac{1}{3+j\omega}e^{-j2\omega}$ $\frac{1}{3+j\omega}e^{j2\omega}$

Name: what is the fourier transform of e 3t 2ut 2 frac13jomega frac1 3jomega frac13jomegae j2omega frac13jomegaej2omega 88828
Uploaded: 2022-05-06T05:48:59-08:00
Duration: 3 min 38 s
Channel: Shaiju T
Description: what is the fourier transform of e 3t 2ut 2 frac13jomega frac1 3jomega frac13jomegae j2omega frac13jomegaej2omega 88828

          What is the Fourier transform of $e^{-3(t-2)}u(t-2)$?
$\frac{1}{3+j\omega}$
$\frac{1}{-3+j\omega}$
$\frac{1}{3+j\omega}e^{-j2\omega}$
$\frac{1}{3+j\omega}e^{j2\omega}$

$what is the fourier transform of e 3t 2ut 2 frac13jomega frac1 3jomega frac13jomegae j2omega frac13jomegaej2omega 88828$

Added by Kathy T.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Shaiju T

05/06/2022

Step 1

The Fourier transform of $x(t)$ is given by: $X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt = \int_{-\infty}^{\infty} e^{-3(t-2)}u(t-2)e^{-j\omega t}dt$ Since $u(t-2) = 0$ for $t < 2$ and $u(t-2) = 1$ for $t \ge 2$, we have: $X(j\omega) = Show more…

Show all steps

Thanks for your feedback!

What is the Fourier transform of $e^{-3(t-2)}u(t-2)$? $\frac{1}{3+j\omega}$ $\frac{1}{-3+j\omega}$ $\frac{1}{3+j\omega}e^{-j2\omega}$ $\frac{1}{3+j\omega}e^{j2\omega}$