Fourier transform of $\frac{\sin(6\pi t)}{t}$

Question

I have to calculate the fourier transform of this function in time domain: $\frac{\sin(6\pi t)}{t}$. First I tough to use the definition of $\operatorname{sinc}$ function as $\operatorname{sinc}(t)=\frac{\sin(2\pi t)}{t}$, but I can't get the same result of my professor: $\pi \operatorname{rect}(f/6)$. How can I get from the function to its fourier transform in this case? Just to make me understand better, I'd like to proceed in the way you proceed to calculate derivatives: you know some examples and from that you derivate everything. Thankyou!

Answer 1

The function $f(t)=\sin(6\pi t)/t$ is square integrable on $\mathbb{R}$. So the Fourier transform is $$ \hat{f}(s)= \lim_{R\rightarrow \infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{\sin(6\pi t)}{t}e^{-ist}dt. $$ Because $\sin(6\pi t)/t$ is even and $e^{-ist}=\cos(st)-i\sin(st)$ has odd imaginary part, then the integral for the Fourier transform reduces to a real part only. The integral of $\sin(6\pi t)\cos(st)/t$ is even in $t$; so the Fourier transform reduces to $$ \hat{f}(s)=\lim_{R\rightarrow\infty}\frac{2}{\sqrt{2\pi}}\int_{0}^{R}\frac{1}{t}\sin(6\pi t)\cos(st)dt \\ =\lim_{R\rightarrow\infty}\frac{2}{\sqrt{2\pi}}\int_{0}^{R}\frac{1}{t}\{\sin(6\pi t+st)+\sin(6\pi t-st)\}dt \\ =\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{0}^{R}\frac{\sin(t(6\pi+s))}{t}+\frac{\sin(t(6\pi-s))}{t}dt \\ =\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{0}^{R(6\pi+s)}\frac{\sin(u)}{u}du+\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{0}^{R(6\pi-s)}\frac{\sin(u)}{u}du. $$ The final result comes from noticing that $$ \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{0}^{Rv}\frac{\sin(u)}{u}du = \left\{\begin{array}{cc} \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{\sin(u)}{u}du, & v > 0 \\ 0 & v = 0 \\ -\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{\sin(u)}{du}, & v < 0. \end{array}\right. $$ All you have to do is break your problem up into cases where $s < -6\pi$, $-6\pi < s < 6\pi$, $6\pi < s$. Of course you'll need to know $$ \int_{0}^{\infty}\frac{\sin(u)}{u}du = \frac{\pi}{2}. $$