I'm preparing for an exam in the signals and systems class I'm taking. One of the practice exams has a problem that requires you to take the Fourier transform of $\text{sinc}(4t)$.
From a table of Fourier transform pairs I found: $\dfrac{\omega_b}{\pi}\text{sinc}\left(\dfrac{\omega_b t}{\pi} \right)\Rightarrow \text{rect}(\omega/2\omega_b)$.
Using this I tried to match the given $\text{sinc}(4t)$ by rewriting it as $\dfrac14\dfrac{4\pi}{\pi}\text{sinc}\left(\dfrac{4\pi t}{\pi} \right)$. From this I get that $\omega_b = 4\pi$ and thus the Fourier transform should yield $\dfrac{1}{4}\text{rect}\left(\omega/8\pi \right)$.
But, in the exam solutions they show the Fourier transform to yield $\dfrac{\pi}{4}\text{rect}(\omega/8)$.
Any ideas where I'm going wrong?
Best Answer
Because this is for signal analysis class, I will assume that $\newcommand{\Res}{\operatorname{Res}}\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\sinc}{\operatorname{sinc}}\sinc(t)=\frac{\sin(\pi t)}{\pi t}$.
Since $\sinc(4t)$ is an even function, we have $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}e^{-2\pi i\,xt}\,\mathrm{d}t &=\int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}\cos(2\pi xt)\,\mathrm{d}t\\ &=\int_{-\infty}^\infty\frac{\sin(t(4\pi+2\pi x))+\sin(t(4\pi-2\pi x))}{8\pi t}\mathrm{d}t\tag{1} \end{align} $$ First note that $\int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t$ is odd in $k$. For $k>0$, let's compute $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t &=\int_{-\infty}^\infty\frac{e^{i\pi kt}-e^{-i\pi kt}}{2\pi it}\mathrm{d}t\\ &=\int_{\gamma^+}\frac{e^{i\pi kt}}{2\pi it}\mathrm{d}t-\int_{\gamma^-}\frac{e^{-i\pi kt}}{2\pi it}\mathrm{d}t\\ &=2\pi i\Res\left(\frac{e^{i\pi kt}}{2\pi it},0\right)-0\\ &=2\pi i\cdot\frac{1}{2\pi i}\\ &=1\tag{2} \end{align} $$ where both $\gamma^+$ and $\gamma^-$ go from $-R-i$ to $R-i$ and then $\gamma^+$ circles back counter-clockwise to $-R-i$ and $\gamma^-$ circles back clockwise to $-R-i$ and $R\to\infty$.
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Equation $(2)$ and its oddness in $k$ says that $$ \int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t=\sgn(k)\tag{3} $$ Combining $(1)$ and $(3)$ yields $$ \int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}e^{-2\pi i\,xt}\,\mathrm{d}t=\frac{\sgn(4+2x)+\sgn(4-2x)}{8}\tag{4} $$ which is $\frac14$ for $x\in\left(-2,2\right)$, $\frac18$ for $x\in\left\{-2,2\right\}$, and $0$ elsewhere.