Fourier Transform with two functions

[kpou]

$\geq$

Homework Statement

Find the Fourier Transform of

y = exp($^{}-at$)sin($\omega$$_{}0$t) 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($\omega$t)dt from -∞ -> ∞
S($\omega$)= $\int$ ( y(t)sin($\omega$t)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$$_{}0$t) over t = 0 -> $\infty$

I try exp(-at)[/itex]sin$\omega$$_{}0$tcos($\omega$t) 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($\omega$t)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

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