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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 $$\begin{align} X_N(j\omega)&=\mathscr{F}\{x_N\}(j\omega)\\\\ &=\int_{-\infty}^\infty x_N(t)e^{-j\omega t}\,dt\\\\ &=Ae^{-jN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{-j\omega t}\,dt\tag1 \end{align}$$ Enforcinng the substitution $t\mapsto t/M\pi$ 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, $\text{sinc}(t)=\frac{\sin(t)}{t}$

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

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 :)

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 :)

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 $$\begin{align} X_N(j\omega)&=\mathscr{F}\{x_N\}(j\omega)\\\\ &=\int_{-\infty}^\infty x_N(t)e^{-j\omega t}\,dt\\\\ &=Ae^{-jN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{-j\omega t}\,dt\tag1 \end{align}$$ Enforcinng the substitution $t\mapsto t/M\pi$ 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, $\text{sinc}(t)=\frac{\sin(t)}{t}$

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

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 any help I would much appreciate it :)

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 $$\begin{align} X_N(j\omega)&=\mathscr{F}\{x_N\}(j\omega)\\\\ &=\int_{-\infty}^\infty x_N(t)e^{-j\omega t}\,dt\\\\ &=Ae^{-jN\omega }\int_{-\infty}^\infty \frac{\sin(M\pi t)}{\pi t}e^{-j\omega t}\,dt\tag1 \end{align}$$ Enforcinng the substitution $t\mapsto t/M\pi$ 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, $\text{sinc}(t)=\frac{\sin(t)}{t}$

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

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 :)