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The Fourier transform table we are given, lists Time function Vs Fourier transform in terms of f. But I want to have w(simple omega) instead of f. Is there a method to do the translation easily?

EG: I have rect(t/T) <==> Tsinc(fT). in the table. From that
I need to get (W/pi)*sinc(Wt) <==> rect(w/2W) 
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2 Answers 2

up vote 1 down vote accepted

Let $$X(f) = \int_{-\infty}^{\infty} x(t) \exp(-i 2\pi ft)\,\mathrm dt$$ denote the Fourier transform of $x(t)$ which relationship we denote as $$x(t) \leftrightarrow X(f).$$ An alternative definition defines the Fourier transform of $x(t)$ as $\hat{X}(\omega)$ where $$\begin{align*} \hat{X}(\omega) &= \int_{-\infty}^{\infty} x(t) \exp(-i\omega t)\,\mathrm dt\\ &= \int_{-\infty}^{\infty} x(t) \exp(-i 2\pi (\omega/2\pi) t)\,\mathrm dt\\ &= X(\omega/2\pi). \end{align*}$$ Thus the "method to do the translation easily" that you are looking for is as follows.

If we know $X(f)$, the Fourier transform of $x(t)$ with respect to the frequency variable $f$, then we can find $\hat{X}(\omega)$, the Fourier transform of $x(t)$ with respect to the radian frequency variable $\omega$ simply by substituting $\omega/2\pi$ for $f$ everywhere in $X(f)$.

As an application to your example, you know the Fourier transform pair $$\text{rect}(t/T) \leftrightarrow T\text{sinc}(fT)$$ as a transform pair. Here, the sinc function is defined as $$\text{sinc}(\alpha) = \begin{cases} \frac{\sin(\pi\alpha)}{\pi\alpha}, & \alpha \neq 0,\\ 1, & \alpha = 0,\end{cases} ~~\text{but see my comment later}$$ Now, the duality theorem of Fourier transforms says if $x(t) \leftrightarrow X(f)$, then $X(t) \leftrightarrow x(-f)$ which applies to the transform pair listed above gives

$$T\text{sinc}(tT) \leftrightarrow \text{rect}(f/T)$$

where we have used the fact that $\text{rect}(\cdot)$ is an even function of its argument. replacing the constant $T$ by $W/\pi$ and replacing $f$ by $\omega/2\pi$, the radian frequency transform pair is

$$\frac{W}{\pi}\text{sinc}(Wt/\pi) \leftrightarrow \text{rect}(\omega/2W)$$

This is not quite the answer you need to arrive at, and so I will let you in on a secret. Most people who tend to use radian frequencies to define Fourier transforms tend to define the sinc function slightly differently from the definition I gave above. I will use Sinc for this other definition which is $$\text{Sinc}(\alpha) = \begin{cases} \frac{\sin(\alpha)}{\alpha}, & \alpha \neq 0,\\ 1, & \alpha = 0.\end{cases} ~~\text{note the difference between sinc and Sinc}$$ Thus we have $$\frac{W}{\pi}\text{sinc}(Wt/\pi) = \frac{W}{\pi}\times\frac{\sin(Wt)}{Wt} = \frac{W}{\pi}\text{Sinc}(Wt)$$ and so your transform pair actually is

$$\frac{W}{\pi}\text{Sinc}(Wt) \leftrightarrow \text{rect}(\omega/2W)$$

which is what you have been struggling to show. In other words, you need to check the definition of sinc in addition to scaling the frequency variable when you go from $f$ to $\omega$ in your Fourier transforms.

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It is news to me that the normalized sinc function (vs. Sinc) is not what "most people" use. But you answer the OP's confusion and mine, so thanks. –  daniel Dec 3 '12 at 3:22
    
@daniel My preference is to use Fourier transforms with respect to $f$ because of symmetry (no $1/2\pi$ factors) and simplicity (e.g. $\text{rect}(t)\leftrightarrow \text{sinc}(f)$) with $\text{sinc}(x)=\frac{\sin\pi(x)}{\pi x}$ because $\text{sinc}(n)=0$ for $n\neq 0$ is easier for me to remember, but there are many people (especially mathematicians and people doing digital signal processing) who always use $\omega$ and Sinc instead. –  Dilip Sarwate Dec 3 '12 at 15:32
    
Hmm. But for example Wiki's tables used normalized sinc with notation $\omega.$ Is sinc(f) standard? And a belated +1. –  daniel Dec 4 '12 at 0:19
1  
@daniel I have commonly seen the normalized sinc used with $f$. The unnormalized Sinc is usually found with $\omega$ but I am sure that counterexamples can be found for assertions "normalized with $f$ always" and "unnormalized with $\omega$ always". Wikipedia's tables list Fourier transform pairs for three different versions of the Fourier transform and it is not surprising that the same definition of sinc is used in the table. To switch to different definitions would just add confusion. –  Dilip Sarwate Dec 4 '12 at 2:03

This is too long for a comment. Looks as though there is small error in the problem as written?

You have $$\text{rect}(t/T) \leftrightarrow T \text{sinc} (fT)$$ but shouldn't this (from [1] below) be $$ \text{rect}(t/T) \leftrightarrow T \text{sinc} (\omega T)?$$

From tables we have the pairs $$(1)\hspace{10mm}f(at) \leftrightarrow \frac{1}{|a|}\hat{f}(\frac{\omega}{a}) $$ and

$$(2)\hspace{10mm}\frac{1}{|a|}\text{rect}(\frac{\omega}{2\pi a}) \leftrightarrow \text{sinc} (at)$$

You have $$\frac{W}{\pi}\text{sinc}(Wt) \leftrightarrow \text{rect}(\frac{\omega}{2W})$$

If we start with your $\text{rect}\frac{\omega}{2W}$ then from (2) we anticipate a constant $a = W/\pi$ and which would give us $\frac{\pi}{W}\text{rect}\frac{\omega}{2W}.$ Then there must be a second constant $b = W/\pi$ such that $b \frac{\pi}{W}\text{rect}\frac{\omega}{2W} = \text{rect}\frac{\omega}{2W}.$ Then the corresponding element of the transform pair would be $$b~ \text{sinc} (at) = \frac{W}{\pi}\text{sinc}\frac{Wt}{\pi} $$

I do not see how to get rid of the $\pi$ in the denominator of the last expression.

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See my answer for a different view on the matter. $$\begin{align*}\mathcal F[\text{rect}(t/T)&=\int_{-T/2}^{T/2}\exp(-i2\pi ft)\,\mathrm dt\\&=\left.\frac{\exp(-i2\pi ft}{-i2\pi f} \right|_{-T/2}^{T/2}\\&=T\frac{\exp(\pi fT)-\exp(-\pi fT)}{(2i)(\pi f T}\\&=T\frac{\sin(\pi fT)}{\pi fT}\\&= T\cdot\text{sinc}(fT)\end{align*}$$ where $\text{sinc}(\alpha)$ is defined as $\frac{\sin(\pi\alpha)}{\pi\alpha}$ and not as $\frac{\sin(\alpha)}{\alpha}$. –  Dilip Sarwate Dec 3 '12 at 2:07

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