How to take the fourier transform of a function?

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To find the Fourier transform of x(t) = e^(-t) sin(t) for t >= 0, the correct integral is ∫x(t)e^(-j2πt)dt, with bounds from 0 to ∞. The sine function can be expressed using exponential notation as sin(t) = (1/2j)(e^(jt) - e^(-jt)). Although the integration may seem complex, it's manageable by recognizing that the limit as t approaches infinity of e^(-(a + jb)t) equals zero when a > 0. The discussion emphasizes the importance of focusing on the correct limits and the exponential decay in the integral. Ultimately, the integral can be solved without resorting to convolution methods.
XcKyle93
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Homework Statement


Find the Fourier transform of x(t) = e-t sin(t), t >=0.

We're barely 3 weeks into my signals course, and my professor has already introduced the Fourier transform. I barely understand what it means, but I just want to get through this problem set.

Homework Equations


I honestly don't know. I know it's an improper integral with bounds -∞ and ∞, that is
∫x(t)e-j2∏tdt

The Attempt at a Solution


I get something VERY LONG which does not seem right.
 
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How can sin(t) expressed in exponential notation?

BTW your integral is missing dt. It is dt, right?
 
Yes, that is correct.

I know that sin(t) = (1/2) * j * (e-t - et), so x(t) = (1/2)*j*(et(-j-1) - et(j-1)). But then I'd have something really painful to integrate, right?
 
You're right, can't do it that way. Will come back to you in a short while.
 
Looks like convolution in the frequency domain is the way to go. You remember the convolution theorem?

So you'll need F(ω) for f(t) = exp(-at) and f(t) = sin(ωt), both for t > 0 and both = 0 for t < 0. The first one is easy; let me know how you're managing with the second ... remember when you take F(ω) to integrate from 0 to ∞, not -∞ to +∞.

EDIT: never mind convolution. You can do the integral.

The integral is not as bad as you think. Key point is that lim t→∞ {e-(a + jb)t} = 0 providing a > 0 which it is in your case.
 
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