I try to compute the Fourier transformation of $$h(t) = e^{-2(t-1)}u(t-1)$$ I was looking at the general rule $$e^{-at}u(t) \Rightarrow \frac{1}{a+j\omega}$$ which would lead me to the following
$$h(t) = e^2e^{-2t}u(t-1)$$ $$H(\omega) = e^2\frac{1}{2 + i\omega}e^{-i\omega} = \frac{e^{2-i\omega}}{2+i\omega}$$
I was wondering if that is correct? Also is there any good way to plot its magnitude?
Let $$\mathscr{F}\{h(t)\} = H(\omega)$$ It's easy to prove $$\mathscr{F}\{h(t-t_0)\} = e^{-j\omega t_0}H(\omega)$$ As you said $$\mathscr{F}\{e^{-at}u(t)\} = \frac{1}{a+j\omega} \ ,\ \ \ \ \ \Re\{a\} \gt 0 $$ Put $a = 2$ $$\mathscr{F}\{e^{-2t}u(t)\} = \frac{1}{2+j\omega} $$ Then $t_0 = 1$ and we got $$\mathscr{F}\{e^{-2(t-1)}u(t-1)\} = \frac{e^{-j\omega }}{2+j\omega}$$
This could be easily verified by WA.
Edit: The answers are the same $$ \frac{-j e^{-j \omega}}{\omega - 2 j} = \frac{j}{j}\frac{-j e^{-j \omega}}{\omega - 2 j} = \frac{e^{-j\omega }}{2+j\omega}$$
And the magnitude is
$$|\frac{e^{-j\omega }}{2+j\omega}| = \frac{|e^{-j\omega }|}{|2+j\omega|} = \frac{1}{\sqrt{4 + \omega^2}}$$ Because $$|e^{-j\omega }| = |j\sin(-\omega) + \cos(-\omega)| = \sqrt{\sin^2(\omega) + \cos^2(\omega)} = 1$$ And $$|2+j\omega| = \sqrt{2^2 + \omega^2}$$