Update: Fourier Transform of a shifted and scaled $\operatorname{sinc}$ signal

Question

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 $$\begin{align} X_N(j\omega)&=\mathscr{F}\{x_N\}(j\omega)\\\\ &=\int_{-\infty}^\infty x_N(t)e^{-j\omega t}\,dt\\\\ &=Ae^{-jN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{-j\omega t}\,dt\tag1 \end{align}$$ Enforcing the substitution $t \rightarrow t/M\pi$ reveals $$\begin{align} X_N(j\omega )&=\frac{Ae^{-jN\omega }}\pi\int_{-\infty}^\infty \frac{\sin(t)}{t}e^{-j(\omega/M\pi) t}\,dt\\\\ &=\frac{Ae^{-jN\omega }}2\left(\text{sgn}(-\omega/M\pi +1)-\text{sgn}(-\omega/M\pi -1)\right)\\\\ &=\begin{cases}Ae^{-jN\omega}&,|\omega|<M\pi\\\\0&,\text{elsewhere}\end{cases} \end{align}$$ where we used the Fourier Transform of the sinc function, $\operatorname{sinc}(t)=\frac{\sin(t)}{t}$

$$\mathscr{\operatorname{rect}}(\omega)=\begin{cases}\pi&,|\omega|<1\\\\0&,\text{elsewhere}\end{cases}$$

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Let $\displaystyle x(t):=\frac{\sin(2\pi(t-1))}{\pi(t-1)}$. Then we have $A=1,M=2, N=1$ so we get : $$ X(j\omega)=\begin{cases} e^{-j\omega}&\text{if $|\omega|<2\pi$}\\ 0&\text{if otherwise}\end{cases} $$

Is this example correct? because in my book it says the answer is $e^{2\omega}$

I would hope for someone to assist me in determining if this is correct or not. I have spent entire day on this problem. I would much appreciate any help and thank you :)

Answer 1

HINT:

Instead of having someone else confirm your result, it would be more instructive to confirm it yourself by computing the inverse Fourier transform of the expression for $X(j\omega)$ that you obtained:

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega\tag{1}$$

In the definition $(1)$ it is assumed that in the forward transform you use a negative exponent and no scaling constant, i.e.,

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{2}$$

After having derived and confirmed the result yourself, it would be useful to remember the formula for the impulse response of an ideal low pass filter with cut-off $\omega_c$. Combined with the time shifting property of the Fourier transform it is straightforward to immediately write down the transform of your original function $x_N(t)$.