Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
You can't automatically use any table of Fourier transforms, since not everybody uses the same conventions (=definitions). Could you post the definition(s) you use in class and the ones that came with the table? –  Gerben Mar 16 '11 at 21:47
    
What exactly do you mean be conventions. As far as the transform pair goes the one I posted in the question is what is listed in our book. –  blcArmadillo Mar 16 '11 at 22:03
5  
There are two definitions of the sinc function in use; either $\sin x/x$ or $\sin \pi x/\pi x$. Could it be that the exam and the transform table differ in that regard? –  Hans Lundmark Mar 16 '11 at 22:32
2  
Also, there are two normalizations for the Fourier Transform; $\int_{\mathbb{R}}f(t)e^{-2\pi ixt}\mathrm{d}t$ and $\int_{\mathbb{R}}f(t)e^{-ixt}\mathrm{d}t$. –  robjohn May 8 '12 at 21:35

2 Answers 2

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 $R\to\infty$.

$\hspace{4cm}$enter image description here

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.

share|improve this answer

I think u can follow the steps in this post:

http://eagle.lamost.org/?p=38679

,simply change z to 4t in the Sinc funtion.

share|improve this answer
2  
Generally we prefer (though it is not a hard rule) that links be accompanied by a brief description of what the link links to. In your case, it may help to explain that (a) you are linking to a blog post and (b) your link uses the residue theorem, and so is very similar to robjohn's answer. –  Willie Wong May 9 '12 at 7:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.