fourier analysislaplace transform
I want to solve the Laplace transform of
$$\frac{\sin(4(t-1))}{\pi(t-1)} $$
I know that Fourier transform of a sinc funcion is rect function. However I found Laplace transform as $\cot^{-1}(s)$. What is the reason of this difference?
My Solution
Secondly, How can I use the Laplace transform properties to solve this question?
Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\ &=& \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega - \int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;. \end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' + \int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.
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.
Best Answer
HINT:
Let $F(s)$ be the Laplace Transform of $f(t)=\frac{\sin(t)}{t}u(t)$ where $u(t)$ is the unit step function (Heaviside Function). Then, we have that
$$F(s)=\int_0^\infty \frac{\sin(t)}{t}e^{-st}\,dt \tag1$$
Differentiating $(1)$ yields
$$F'(s)=-\int_0^\infty \sin(t)e^{-st}\,dt=-\frac1{s^2+1}\tag 2$$
Use $F(0)=\pi/2$ along with Laplace Transform relationships for time shifting and scaling.
Can you finish now?
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