Fourier transform of sinc function

Question

Let us consider the Fourier transform of $\mathrm{sinc}$ function. As I know it is equal to a rectangular function in frequency domain and I want to get it myself, I know there is a lot of material about this, but I want to learn it by myself. We have $\mathrm{sinc}$ function whhich is defined as $$ \mathrm{sinc}(\omega_0\,t) = \sin(\omega_0\,t)/(\omega_0\,t). $$ Its Fourier transform $$ \int_{\Bbb R} \sin(\omega_0\,t) \,e^{-i\,\omega\,t}/(\omega_0\,t)\,\mathrm dt $$ can be represented as $$ \int_{\Bbb R} \sin(\omega_0\,t)\,(\cos(\omega\,t) - i \,\sin(\omega\,t))/(\omega_0\,t)\,\mathrm dt. $$ Because we can distribute in brackets and consider that integral of difference is equal differences of integrals, we get $$ \int_{\Bbb R} \sin(\omega_0 \,t) \cos(\omega\,t)/(\omega_o\,t)\,\mathrm dt - \int_{\Bbb R} \sin(\omega_0\,t) \sin(\omega\,t)/(\omega_o\,t)\,\mathrm dt $$

but first product is zero, right? Because sine and cosine are orthogonal, so how could continue? Please help me.

Answer 1

In terms of deriving the Fourier Transform, I will make some use of techniques highlighted in http://www.claysturner.com/dsp/FTofSync.pdf

Let us start with your expression

$$\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)\cos(\omega t)}{(\omega_ot)}dt-\int_{-\infty}^{\infty}\frac{j\sin(\omega_0t)\sin(\omega t)}{(\omega_ot)}dt$$

Examining the integrals, the term $\omega_0t$ in the denominator makes evaluating the integral more involved.

We can use some relevant trigonometric identities so that we can express $$\sin(\omega_0t)\cos(\omega t)=\frac{1}{2}[\sin((\omega+\omega_0)t)-\sin((\omega-\omega_0)t)]$$

$$\sin(\omega_0t)\sin(\omega t)=\frac{1}{2}[\cos((\omega-\omega_0)t)-\cos((\omega+\omega_0)t)]$$

To deal with the awkward $\omega_0t$ term, we can use the following identity to convert the single integral into a double integral that is far nicer to evaluate: $$\frac{1}{\omega_0t}=\int_0^{\infty}e^{-\omega_0ts}ds$$

Thus the integral to evaluate is $$\int_0^{\infty}\int_{-\infty}^{\infty}\sin(\omega_0t)\cos(\omega t)e^{-\omega_0tx}dtdx-j\int_0^{\infty}\int_{-\infty}^{\infty}\sin(\omega_0t)\sin(\omega t)e^{-\omega_0tx}dtdx$$

which expands to $$\frac{1}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\sin((\omega+\omega_0)t)e^{-\omega_0tx}dtdx-\frac{1}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\sin((\omega-\omega_0)t)e^{-\omega_0tx}dtdx\\-\frac{j}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\cos((\omega-\omega_0)t)e^{-\omega_0tx}dtdx+\frac{j}{2}\int_0^{\infty}\int_{-\infty}^{\infty}\cos((\omega+\omega_0)t)e^{-\omega_0tx}dtdx$$

We can exploit Fubini's theorem to rewrite the integral as

$$\frac{1}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\sin((\omega+\omega_0)t)e^{-\omega_0xt}dt\right]dx-\frac{1}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\sin((\omega-\omega_0)t)e^{-\omega_0xt}dt\right]dx\\-\frac{j}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\cos((\omega-\omega_0)t)e^{-\omega_0xt}dt\right]dx+\frac{j}{2}\int_{-\infty}^{\infty}\left[\int_0^{\infty}\cos((\omega+\omega_0)t)e^{-\omega_0xt}dt\right]dx$$

We shall use the following integral identities to calculate the above integrals :- $$\int_0^{\infty}\sin(at)e^{-st}dt=\frac{a}{a^2+s^2}$$ $$\int_0^{\infty}\cos(at)e^{-st}dt=\frac{s}{a^2+s^2}$$

This results in $$\frac{1}{2}\int_{-\infty}^{\infty}\left[\frac{(\omega+\omega_0)}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx-\frac{1}{2}\int_{-\infty}^{\infty}\left[\frac{(\omega-\omega_0)}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx\\-\frac{j}{2}\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx+\frac{j}{2}\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx\text{ (Eq. 1)}$$

To evaluate the real component integrals in (Eq. $1$), we use the following result:- $$\int_{-\infty}^{\infty}\frac{a}{a^2+s^s}ds=\frac{|a|}{a}\int_{-\infty}^{\infty}\frac{1}{1+y^2}dy=sgn(a)\left[\arctan y\right]^{\infty}_{-\infty}=sgn(a)\pi$$

This leads to $$\int_{-\infty}^{\infty}\left[\frac{(\omega+\omega_0)}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx=sgn(\omega+\omega_0)\pi$$ $$\int_{-\infty}^{\infty}\left[\frac{(\omega-\omega_0)}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx=sgn(\omega-\omega_0)\pi$$

As regards the imaginary component integrals, note that the numerator is simply a constant times the derivative of the denominator, so we have

$$\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx=\frac{1}{2\omega_0}\left[\ln\{(\omega-\omega_0)^2+\omega_0^2x^2\}\right]_{-\infty}^{\infty}$$

$$\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx=\frac{1}{2\omega_0}\left[\ln\{(\omega+\omega_0)^2+\omega_0^2x^2\}\right]_{-\infty}^{\infty}$$

Combining the imaginary component integrals as per the integral in (Eq. $1$) we wish to evaluate and noting that they have opposite signs, we have $$-\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]dx+\int_{-\infty}^{\infty}\left[\frac{\omega_0x}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]dx\\=\lim_{x\rightarrow\infty}\frac{1}{2\omega_0}ln\left[\frac{(\omega+\omega_0)^2+\omega_0^2x^2}{(\omega-\omega_0)^2+\omega_0^2x^2}\right]+\lim_{x\rightarrow -\infty}\frac{1}{2\omega_0}ln\left[\frac{(\omega-\omega_0)^2+\omega_0^2x^2}{(\omega+\omega_0)^2+\omega_0^2x^2}\right]=\frac{2}{\omega_0}\ln(1)=0$$ Thus the imaginary terms in the integral cancel out, leading to the integral in (Eq. $1$) being a real result, as follows $$\frac{\pi}{2}\left[sgn(\omega+\omega_0)-sgn(\omega-\omega_0)\right]$$

Putting everything together we have: $$\int_{-\infty}^{\infty}\frac{\sin(\omega_0t)\cos(\omega t)}{(\omega_ot)}dt-\int_{-\infty}^{\infty}\frac{j\sin(\omega_0t)\sin(\omega t)}{(\omega_ot)}dt=\frac{\pi}{2}\left[sgn(\omega+\omega_0)-sgn(\omega-\omega_0)\right]$$

The result is a rectangular function that starts from frequency $-\omega_0$ and ends at frequency $\omega_0$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\fermi\pars{t} \equiv {\sin\pars{\omega_{0}t} \over \omega_{0}t} =\int_{-\infty}^{\infty}\hat{\fermi}\pars{\omega}\expo{-\ic\omega t}\, {\dd\omega \over 2\pi}\,,\qquad\qquad\hat{\fermi}\pars{\omega}:\ {\large ?}}$

\begin{align} \color{#c00000}{\hat{\fermi}\pars{\omega}} &= \int_{-\infty}^{\infty} {\rm sinc}\pars{\omega_{0}t} \expo{\ic\omega t}\,\dd t ={1 \over \verts{\omega_{0}}}\int_{-\infty}^{\infty}{\sin\pars{t} \over t}\, \expo{\ic\omega t/\verts{\omega_{0}}}\,\dd t \\[3mm]&={1 \over \verts{\omega_{0}}}\int_{-\infty}^{\infty} \pars{\half\int_{-1}^{1}\expo{\ic\nu t}\,\dd\nu} \expo{\ic\omega t/\verts{\omega_{0}}}\,\dd t =\color{#c00000}{{1 \over 2\verts{\omega_{0}}}\int_{-1}^{1}\dd\nu\ \overbrace{\int_{-\infty}^{\infty} \expo{\ic\pars{\nu + \omega/\verts{\omega_{0}}}t}\,\dd t} ^{\ds{2\pi\,\delta\pars{\nu + {\omega \over \verts{\omega_{0}}}}}}} \end{align} where $\ds{\delta\pars{x}}$ is the Dirac Delta 'Function'.

Then \begin{align} \color{#c00000}{\hat{\fermi}\pars{\omega}}&={\pi \over \verts{\omega_{0}}}\, \Theta\pars{1 - \verts{\omega \over \verts{\omega_{0}}}} \end{align} where $\ds{\Theta\pars{x}}$ is the Heaviside Step Function.

$$\color{#00f}{\large% \hat{\fermi}\pars{\omega}={\pi \over \verts{\omega_{0}}}\, \Theta\pars{\verts{\omega_{0}} - \verts{\omega}}} $$

Answer 3

This simple method is missing. Since for $\omega_0\geq 0$, $$ \int_{\Bbb R} \mathbf{1}_{[-\omega_0,\omega_0]}(\omega) \,e^{i\,\omega\,t}\,\mathrm d \omega = \int_{-\omega_0}^{\omega_0} \,e^{i\,\omega\,t}\,\mathrm d \omega \\ = \frac{e^{i\, \omega_0\,t} - e^{-i\,\omega_0\,t}}{i\, t} \\ = 2\,\omega_0\,\mathrm{sinc}(\omega_0\,t) $$ one deduces by the Fourier inversion theorem that $$ \int_{\Bbb R} \mathrm{sinc}(\omega_0\,t) \,e^{-i\,\omega\,t}\,\mathrm d t = \frac{\pi}{\omega_0}\,\mathbf{1}_{[-\omega_0,\omega_0]}(\omega). $$