Calculate the Fourier Transform

AI Thread Summary
The discussion focuses on calculating the Fourier Transform of the function x(t) = e^(-|t|) cos(2t). The first integral from -∞ to 0 is straightforward, yielding a specific result, while the second integral from 0 to ∞ raises questions about handling infinite boundaries. Participants note that the exponential decay in both integrals approaches zero at their respective limits, simplifying the calculations. The function is identified as even, indicating that both integrals yield equal results. Understanding the symmetry of the function aids in solving the problem effectively.
rht1369
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Homework Statement



calculate the Fourier Transform of the following function:

Homework Equations



x(t) = e-|t| cos(2t)

The Attempt at a Solution



0-∞ et ((e2jt + e-2jt) / 2) e-jωt + ∫0 e-t ((e2jt + e-2jt) / 2) e-jωt

The first integral is easy to calculate and equals: (1/2) * (1/(1+(2-ω)j) + 1/(1+(-2-ω)j))
But how about the second integral?
 
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You did the first integral OK. Why do you think the second integral is harder than the first one?

But if you sketch a graph of x(t), you might notice something interesting that saves you some work...
 
AlephZero said:
You did the first integral OK. Why do you think the second integral is harder than the first one?

But if you sketch a graph of x(t), you might notice something interesting that saves you some work...

My problem is with the infinite boundaries. The first integral is simple since is done from -∞ to 0 and it makes the limit of exponential part will be zero But I am confused about the second one since its boundaries are from 0 to ∞ and I can get it how the limit of exponential part of this function will be zero too!

I sketched the graph. Do you mean that the function is asymmetric? so the doing the integration for one side will be equal to the another?
 
rht1369 said:
But I am confused about the second one since its boundaries are from 0 to ∞ and I can get it how the limit of exponential part of this function will be zero too!

The real part of the second integral is e^{-t} which goes to 0 as t goes to +\infty. That is similar to the first integral, where e^{+t} goes to 0 as t goes to -\infty.

Do you mean that the function is asymmetric? so the doing the integration for one side will be equal to the another?

It would be better to call it "an even function" not "asymmetric", but you got the point that the two integrals are equal.
 

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