Consider $$x(t) = \text{rect} \left(\frac{t}{2}-\frac{3}{2}\right)$$
From FT-Table, we know $$\mathcal{F}\big(x(at)\big) = \frac{1}{|a|}X\left(j\frac{\omega}{a}\right)$$ and $$\mathcal{F}\big(x(t-t_0)\big) = X\left(j\omega\right)e^{-j\omega t_0}$$
Now I am confused the FT of the top function should be $$X(j\omega)=2\text{sinc}\left(\frac{\omega}{\pi}\right)e^{-j\omega \frac{3}{2}}$$ or $$X(j\omega)=2\text{sinc}\left(\frac{\omega}{\pi}\right)e^{-j\omega 3}$$
Can anyone please provide me with a strict way to identify the correct answer?
By definition, take$$X(\omega)=\int_{-\infty}^{\infty} \operatorname{rect}\left(\frac{t}{\tau}\right) e^{-j \omega t} d t$$ $\text { Since rect }(t / \tau)=1 \text { for }-\tau / 2<t<\tau / 2 \text { and } 0 \text { otherwise, we have } $ $$X(\omega)=\int_{-\tau / 2}^{\tau / 2} e^{-j \omega t} d t=-\frac{1}{j \omega}\left(e^{-j \omega r / 2}-e^{j \omega r / 2}\right)=\frac{2 \sin \left(\frac{\omega \tau}{2}\right)}{\omega}=\tau \frac{\sin \left(\frac{\omega \tau}{2}\right)}{\left(\frac{\omega \tau}{2}\right)}=\tau \operatorname{sinc}\left(\frac{\omega \tau}{2}\right).$$ We take $\tau=2$, thus $X(\omega)=2\text{sinc}(\omega).$ Subsequently, the translation rule you give us $$ 2\text{sinc}(\omega)e^{- 3j \omega },$$ since we translate by 3. You take another definition of rect that gives us the $1/\pi$?
Also:
I presume $X=\mathcal F(x),$ so be aware the transformation rules read $$\mathcal F(f(at))=\frac1{|a|}\hat f(\frac \omega{|a|})$$ and $$\mathcal F(f(t-t_0))= \hat f(\omega )e^{-jt_0\omega},$$ where $\hat f=\mathcal F(f)$, thus the complex number $j$ is not within $f$.