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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.

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  • $\begingroup$ any help ?i could not continue $\endgroup$ Commented Apr 2, 2014 at 18:54

4 Answers 4

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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 $\omega_0$.

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  • $\begingroup$ You are welcome - the website I've linked in the answer proved a great resource. $\endgroup$ Commented Apr 8, 2014 at 17:40
  • $\begingroup$ a bit deep,but i get point $\endgroup$ Commented Apr 8, 2014 at 17:40
  • $\begingroup$ I agree it's quite involved, but I thought deriving the result from first principles would be an interesting exercise. $\endgroup$ Commented Apr 8, 2014 at 17:41
  • $\begingroup$ interesting is Fubini’s theorem by the way $\endgroup$ Commented Apr 8, 2014 at 17:42
  • $\begingroup$ what does it says if you could explain? $\endgroup$ Commented Apr 8, 2014 at 17:43
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\fermi\pars{t} \equiv {\sin\pars{\omega_{0}t} \over \omega_{0}t} =\int_{-\infty}^{\infty}\hat{\fermi}\pars{\omega}\expo{-\ic\omega t}\, {\dd\omega \over 2\pi}\,,\qquad\qquad\hat{\fermi}\pars{\omega}:\ {\large ?}}$

\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.

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

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  • $\begingroup$ @datodatuashvili You're welcome. I enjoyed it. $\endgroup$ Commented Apr 9, 2014 at 7:58
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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 $$ \int_{\Bbb R} \mathrm{sinc}(\omega_0\,t) \,e^{-i\,\omega\,t}\,\mathrm d t = \frac{\pi}{\omega_0}\,\mathbf{1}_{[-\omega_0,\omega_0]}(\omega). $$

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We know that the Fourier transform of Sinc(z) is,

\begin{equation*} \int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-i \omega z} dz\end{equation*}

and

\begin{equation*} \int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-i \omega z} dz = \int_{-\infty}^{+\infty} {{e^{iz}-e^{-iz}}\over{2iz}} e^{-i \omega z} dz. \end{equation*}

So,

(1) \begin{equation*} \int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-i \omega z} dz={1 \over {2i}} \int_{-\infty}^{+\infty} {{e^{i(1-\omega)z}}\over{z}} dz + {1 \over {2i}} \int_{+\infty}^{-\infty} {{e^{-i(1+\omega)z}}\over{z}} dz. \end{equation*}

Let us consider the first item, when \omega < 1, namely (1-\omega)>0, we can choose the path below to do the contour integration.

enter image description here

Using the method of complex residues, we take the contour with no singular point, separating the path into four parts, namely A, B, C and D shown as the red letters in the figure. Thus,

(2) \begin{equation*} \begin{split} \oint {{e^{i(1-\omega)z}}\over{z}} dz=\lim_{\substack{{\rho \to 0}\\{R\to\infty}}} \int_{-R}^{-\rho}{{e^{i(1-\omega)z}}\over{z}} dz+ \lim_{\rho \to 0} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz +\lim_{\substack{{\rho \to 0}\\{R\to\infty}}} \int^{R}_{\rho}{{e^{i(1-\omega)z}}\over{z}} dz\\ +\lim_{R \to \infty} \int_{D}{{e^{i(1-\omega)z}}\over{z}} dz = 2\pi i {\mathrm{Res}} f(z)=0. \end{split} \end{equation*}

Also if (1-\omega)>0, namely ~\omega < 1, and when ~{R \to \infty} and ~{\rho \to 0}, it is easy to prove using Jordan’s lemma, that the fourth term is equal to zero. Thus we can the rewrite the rest items as,

\begin{equation*} \int_{-\infty}^{0_{-}}{{e^{i(1-\omega)z}}\over{z}} dz+\int^{+\infty}_{0_{+}}{{e^{i(1-\omega)z}}\over{z}} dz =\int_{-\infty}^{+\infty}{{e^{i(1-\omega)z}}\over{z}} dz =-\lim_{\rho \to 0} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz.\end{equation*}

In order to calculate the value of the item on the right side, I use z=\rho e^{i\theta}, so dz=i \rho e^{i\theta}d\theta, then we have,

\begin{equation*} \lim_{\rho \to 0} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz=\lim_{\rho \to 0}\int_{\pi}^0 \exp[i (1-\omega) \rho e^{i \theta}] i d\theta=\int_{\pi}^0id\theta=i(0-\pi)=-i \pi.\end{equation*}

So, in the end, we get the value of the first integration item shown in the Eq.(1),

(3) \begin{equation*} \int_{-\infty}^{+\infty}{{e^{i(1-\omega)z}}\over{z}} dz = i\pi,\quad \omega<1. \end{equation*}

When \omega>1, according to Jordan’s lemma, we can not get the fourth item in the Eq.(2) to be zero. So we have to choose another path for finalising the integration. So I chose the path below to deal with such a situation,

enter image description here

It is not hard to calculate the same way as the previous one,

(4) \begin{equation*} \begin{split} \oint {{e^{i(1-\omega)z}}\over{z}} dz=\lim_{\substack{{\rho \to 0}\\{R\to\infty}}} \int_{R}^{\rho}{{e^{i(1-\omega)z}}\over{z}} dz+ \lim_{\rho \to 0} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz +\lim_{\substack{{\rho \to 0}\\{R\to\infty}}} \int^{-R}_{-\rho}{{e^{i(1-\omega)z}}\over{z}} dz\\ +\lim_{R \to \infty}\int_{D}{{e^{i(1-\omega)z}}\over{z}} dz = 2\pi i {\mathrm{Res}} f(z)=0, \end{split} \end{equation*}

and here again I use the same trick, z=R e^{i\theta}, so dz=i R e^{i\theta}d\theta. Because \omega>1,

\begin{equation*}\lim_{R \to \infty} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz=\lim_{R \to \infty}\int_{\pi}^0 \exp[i (1-\omega) R e^{i \theta}] i d\theta=\int_{\pi}^0 0 \times id\theta=0,\end{equation*}

\begin{equation*} \lim_{\rho \to 0} \int_{B}{{e^{i(1-\omega)z}}\over{z}} dz=\lim_{\rho \to 0}\int^{\pi}_0 \exp[i (1-\omega) \rho e^{i \theta}] i d\theta=\int^{\pi}_0 i d\theta=i(\pi-0)=i \pi.\end{equation*}

So, in the end,

(5) \begin{equation*} \int_{-\infty}^{+\infty}{{e^{i(1-\omega)z}}\over{z}} dz = i\pi,\quad \omega>1. \end{equation*}

Now, let’s move on to the second part in the Eq.(1). Following the same steps, using contour above real axis I get,

(6) \begin{equation*} \int_{+\infty}^{-\infty}{{e^{-i(1+\omega)z}}\over{z}} dz = \lim_{\rho \to 0}\int_{\pi}^0 \exp[-i (1+\omega) \rho e^{i \theta}] i d\theta = -i\pi,\quad \omega< -1; \end{equation*}

and using contour blow real axis I get,

(7) \begin{equation*} \int_{+\infty}^{-\infty}{{e^{-i(1+\omega)z}}\over{z}} dz = \lim_{\rho \to 0}\int^{\pi}_0 \exp[-i(1+\omega) \rho e^{i \theta}] i d\theta = i\pi,\quad \omega>-1. \end{equation*}

Finally, we can write the answer as, if -1< \omega<1

\begin{equation*}\int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-i \omega z} dz={1 \over {2i}} \int_{-\infty}^{+\infty} {{e^{i(1-\omega)z}}\over{z}} dz + {1 \over {2i}} \int_{+\infty}^{-\infty} {{e^{-i(1+\omega)z}}\over{z}} dz={{\pi}\over2}+{{\pi}\over2}=\pi;\end{equation*}

and if \omega>1 and \omega<-1

\begin{equation*}\int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-i \omega z} dz={1 \over {2i}} \int_{-\infty}^{+\infty} {{e^{i(1-\omega)z}}\over{z}} dz + {1 \over {2i}} \int_{+\infty}^{-\infty} {{e^{-i(1+\omega)z}}\over{z}} dz={{\pi}\over2}-{{\pi}\over2}=0.\end{equation*}

(Note: if we define the Fourier Transform as \begin{equation*}\int_{-\infty}^{+\infty} {{\sin(z)}\over{z}} e^{-2\pi i \omega z} dz \end{equation*}, Then the result could be a little different.)

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  • $\begingroup$ Note that $\lim_{R \to \infty}\exp[i (1-\omega) R e^{i \theta}] = 0$ for $\omega \gt1$ is not true. $\endgroup$
    – S.H.W
    Commented Jun 8, 2020 at 23:55

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