Fourier Transform of shifted unit step function

Question

Find the Fourier transform of $$x(t) = (t-2) e^{-t} u(t-2)$$ I got $e^{-t}u(t-2) \to (e^{-2j2\pi f})/j2\pi f$ but i'm not sure how to combine with the term $(t-2)$. Any suggestions? Thanks in advance.

Answer 1

First, notice that $$\mathcal{F}\{te^{-t}u(t)\}=\frac{1}{(1+j\omega)^2}$$

Then we can see that

$$\begin{align}x(t) &= (t-2) e^{-t} u(t-2)\\ &=\frac{e^{2}}{e^{2}}\left( (t-2) e^{-t} u(t-2)\right)\\ &=\frac{1}{e^{2}}\left((t-2) e^{-(t-2)} u(t-2)\right)\end{align}$$

From$$\mathcal{F}\{f(t-t_0)\}=F(j\omega)e^{-j\omega t_0}$$ (for $t_0=2$) and the linearity of the FT we can conclude that

$$\mathcal{F}\{x(t)\}=\frac{1}{e^{2}}\frac{1}{(1+j\omega)^2}e^{-2j\omega}=\frac{e^{-2(1+j\omega)}}{(1+j\omega)^2}$$