Question

. use the ft analysis equation to compute the fourier transforms of: (a) e−2(t 2)u(t) (b) δ(t 3) −δ(t −3)

Answer 1

It looks like you're asked to find the Fourier transforms of (a) $e^{-2(t+2)}u(t)$ and (b) $\delta(t+3)-\delta(t-3)$.

(a) We have by definition of the Fourier transform,

$\mathcal F \left\{e^{-2(t+2)}u(t)\right\} = \displaystyle \int_{-\infty}^\infty e^{-2(t+2)} u(t) e^{-i\,2\pi\xi t} \, dt \\\\ ~~~~~~~~ = \int_0^\infty e^{-2(t+2)} e^{-i\,2\pi\xi t} \, dt \\\\ ~~~~~~~~ = \frac1{e^4} \int_0^\infty e^{-(2+i\,2\pi\xi) t} \, dt$

Substitute $s=(2+i\,2\pi\xi)t$ and $ds = (2+i\,2\pi\xi) \, dt$.

$\mathcal F \left\{e^{-2(t+2)}u(t)\right\} = \displaystyle \frac1{e^4(2+i\,2\pi\xi)} \int_0^\infty e^{-s} \, ds = \boxed{\frac1{2e^4(1+i\,\pi\xi)}}$

This may or may not agree with whatever solution you expect, depending on the definition you are using for the transform. For the record, I'm using the definition

$\hat f(\xi) = \displaystyle \mathcal F\left\{f(x)\right\} = \int_{-\infty}^\infty f(x) e^{-i\,2\pi\xi x} \, dx$

(b) The Fourier transform is

$\mathcal F\left\{\delta (t + 3) - \delta (t - 3)\right\} = \int_{-\infty}^\infty (\delta(t+3)-\delta(t-3)) e^{-i\,2\pi\xi t} \, dt \\\\ ~~~~~~~~ = e^{-i\,2\pi\xi(-3)} - e^{-i\,2\pi\xi(3)} \\\\ ~~~~~~~~ = e^{i\,6\pi\xi} - e^{-i\,6\pi\xi} \\\\ ~~~~~~~~ = 2 \sinh(i\,6\pi\xi) \\\\ ~~~~~~~~ = \boxed{2i\sin(6\pi\xi)}$

where we use the property of the delta function that

$\displaystyle \int_{-\infty}^\infty f(x) \delta(x-x_0) \, dx = f(x_0)$