Calculate the Fourier Transform

[rht1369]

Homework Statement

calculate the Fourier Transform of the following function:

Homework Equations

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

The Attempt at a Solution

∫⁰_-∞ e^t ((e^2jt + e^-2jt) / 2) e^-jωt + ∫^∞₀ e^-t ((e^2jt + 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?

 

[AlephZero]

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...

 

[rht1369]

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?

 

[AlephZero]

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.

 

[DeeNos]

I would suggest using the properties of the Fourier Transform to simplify the second integral. Since the function is even, we can rewrite the integral as 2∫0∞ e-t cos(2t) e-jωt. We can then use the property of shifting to rewrite this as 2∫0∞ e-(t+jω) cos(2(t+jω)) dt. Using Euler's formula, we can further simplify this to 2∫0∞ e-(t+jω) (e2j(t+jω) + e-2j(t+jω))/2 dt. Finally, using the property of scaling, we can rewrite this as 2∫0∞ e-(t+jω) (e2jt + e-2jt)/2 e2jωt dt. This integral can then be evaluated using the same method as the first integral, resulting in the final Fourier Transform of x(t) as (1/2) * (1/(1+(2-ω)j) + 1/(1+(-2-ω)j)).