I am working on a convolution question where $x_1$ is the sinc function and $x_2$ is: $$x_2(t)=\frac{\sin(2\pi (t-1))}{\pi(t-1)}$$ I have attempted to solve this using the properties of the fourier transform by saying the F.T. of the sinc function is $1$, and the time shift property gives me $e^{jw(-1)}$ as the phase shift element.
I am unsure how to handle the $2$ scaling factor in the argument of the sine function. My book gives the solution as $e^{-2w}$, but I don't know how they eliminate the $j$ from the time shift either.
At this point, I would really appreciate an explanation of the fourier transform of the general case, where: $$x_N(t) = A\frac{\sin(M\pi(t-N))}{\pi(t-N)}$$
because I am consistently confused with these, and the book's explanations are no help to me understanding what's going on.
Let $x_N$ be the function given by $$x_N(t)=A\frac{\sin(M\pi(t-N))}{\pi(t-N)}$$ The Fourier Transform of $x_N$ is $$\begin{align} X_N(\omega)&=\mathscr{F}\{x_N\}(\omega)\\\\ &=\int_{-\infty}^\infty x_N(t)e^{i\omega t}\,dt\\\\ &=Ae^{iN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{i\omega t}\,dt\tag1 \end{align}$$ Enforcinng the substitution $t\mapsto t/M\pi$ reveals $$\begin{align} X_N(\omega)&=\frac{Ae^{iN\omega }}\pi\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{i(\omega/M\pi) t}\,dt\\\\ &=\frac{Ae^{iN\omega }}2\left(\text{sgn}(\omega/M\pi +1)-\text{sgn}(\omega/M\pi -1)\right)\\\\ &=\begin{cases}Ae^{iN\omega}&,|\omega|<M\pi\\\\0&,\text{elsewhere}\end{cases} \end{align}$$ where we used the Fourier Transform of the sinc function, $\text{sinc}(t)=\frac{\sin(t)}{t}$
$$\mathscr{\text{sinc}}(\omega)=\begin{cases}\pi&,|\omega|<1\\\\0&,\text{elsewhere}\end{cases}$$