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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, $\operatorname{sinc}(t)=\frac{\sin(t)}{t}$

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


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 :)

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  • $\begingroup$ Are you sure that you use the definition of the Fourier transform with a positive sign in the exponent? In signal processing it's much more common to have a negative exponent (and a positive exponent in the inverse transform). $\endgroup$
    – Matt L.
    Mar 23 at 10:34
  • $\begingroup$ With that being said I believe that only $N$ changes. @MattL. $\endgroup$
    – SPARSE
    Mar 23 at 10:36
  • $\begingroup$ It's no big deal really, just convention. You just gotta be clear about which definition you use because the result will change accordingly. Just check the definition used in your book and follow it. $\endgroup$
    – Matt L.
    Mar 23 at 10:38
  • $\begingroup$ Here I followed the definition of positive exponent for the $e^{j\omega t}$ term. The book did not present which method but I am guessing like the majority of problems related to Fourier I think they used $e^{-j\omega t}$ so I will try to rederive it using the correct convention and recheck my answer. @MattL. $\endgroup$
    – SPARSE
    Mar 23 at 10:41
  • $\begingroup$ I have updated my derivation but still unsure of my answer. @MattL. $\endgroup$
    – SPARSE
    Mar 23 at 13:44
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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 $x_N(t)$.

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