Fourier Transform with two functions

[kpou]

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Homework Statement

Find the Fourier Transform of

y = exp(^{}-at)sin(\omega_{}0t) for t ≥ 0
and = 0 for t < 0

Find the amplitudes C(\omega, S(\omega), and energy spectrum \Phi' for \omega > 0 if the term that peaks at negative frequency can be disregarded for pos frequency.

Homework Equations

Y(\omega) = C(\omega) - iS(\omega)
\Phi' = C^2(\omega) + iS^2(\omega)
C(\omega)= \int( y(t)cosw(\omegat)dt from -∞ -> ∞
S(\omega)= \int ( y(t)sin(\omegat)dt from -∞ -> ∞

The Attempt at a Solution

I have page after page of trying to simplify the algebra down with no luck. In my text it writes "y" in the equation with no function of ( t ) or (\omega) for most every other equation I see contains either of those. Is there something different about y?

It looks like I have to take the Fourier of two functions exp(^{}-at) and sin(\omega_{}0t) over t = 0 -> \infty

I try exp(-at)[/itex]sin\omega_{}0tcos(\omegat) using sin(bx)=(exp(ibx)-exp(-ibx)/2i

Am I missing something? Are there any algebraic tricks I may be missing? Thanks !

C(\omega)= \int( y(t)cosw(\omegat)dt from 0 -> ∞ Since y(t) = 0 for negative t

= \int exp(-at)sin(ω₀)t)cos(ω₀t)dt
= \int exp(-at)[ (1/2i) ( exp(iω₀t) - exp(-iω₀t) ) (1/2) ( exp(iωt) + exp(-iωt))] dt

 

[fzero]

In this case, it's much easier to compute Y(\omega) directly and determine C(\omega), S(\omega) as the real and imaginary parts. You will want to use the identity

\sin z = \frac{e^{iz} - e^{-iz}}{2i}.