Author Topic: HA6 problem 4  (Read 500 times)

Victor Ivrii

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HA6 problem 4
« on: March 05, 2015, 08:37:57 AM »
Find Fourier transforms of

a.  $f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\
   & 0 && |x|\ge a;\end{aligned}\right.$

b.  $f(x)=\left\{\begin{aligned} & x && |x|\le a,\\
  & 0 && |x|\ge a;\end{aligned}\right.$

c.  Using (a) calculate $\int_{-\infty}^\infty \frac{\sin  (x)}{x}\,dx$.

Yiyun Liu

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Re: HA6 problem 4
« Reply #1 on: March 05, 2015, 09:00:43 PM »
\begin{gathered}
  part(a): \hfill \\
  \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty  {f(x){e^{ - i\omega x}}dx}  \hfill \\
        = \frac{1}{{2\pi }}\int\limits_{ - a}^a {{e^{ - i\omega x}}dx}  \hfill \\
   = \frac{1}{{2\pi }}(\frac{{{e^{i\omega a}} - {e^{ - i\omega a}}}}{{i\omega }}) \hfill \\
   = \frac{{\sin (\omega a)}}{{\pi \omega }} \hfill \\
   \hfill \\
  part(b): \hfill \\
   \hfill \\
  take  f(x) = xg(x),where,  g(x) = 1 \hfill \\
  hence,    \hat f(\omega ) = i\frac{{d\hat g(\omega )}}{{d\omega }} = i\frac{{d(\frac{{\sin (\omega a)}}{{\pi \omega }})}}{{d\omega }} = i(\frac{{a\cos (a\omega )}}{{\pi \omega }} - \frac{{\sin (a\omega )}}{{\pi {\omega ^2}}}) \hfill \\
   \hfill \\
  part(c): \hfill \\
   \hfill \\
   IFT: \hfill \\
  f(x) = \int\limits_{ - \infty }^\infty  { \hat f(\omega ){e^{i\omega x}}d\omega  = } \int\limits_{ - \infty }^\infty  {\frac{{\sin (\omega a)}}{{\pi \omega }}} {e^{i\omega x}}d\omega  \hfill \\
  thus,\int\limits_{ - \infty }^\infty  {\frac{{\sin (xa)}}{x}{e^{i\omega x}}} dx = \pi ,      |\omega | \leqslant a,if  a = 1,and     \omega  = 0. \hfill \\
\end{gathered} \]

Victor Ivrii

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Re: HA6 problem 4
« Reply #2 on: March 08, 2015, 08:45:27 AM »
In many papers some answers like $=0$ if this and $=\pi$ if that for (c ) was given. It is a very simple question: calculate $I:=\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx$. We got that
$$
\frac{1}{\pi}\int _{-\infty}^\infty \frac{\sin(ak)}{k }\underbracket{e^{ikx}}\,dk= \left\{\begin{aligned}&1 &&|x|<a\\ &0 &&|x|\ge a\end{aligned}\right.$$
We do not need underlined factor, so we take $x=0$ and arrive to
$$
\int_{-\infty}^\infty \frac{\sin(a k)}{k}\,dk= \pi.
$$
So taking $a=1$ we have $I=\pi$.

Remark. Above integral does not depend on $a\ne 0$, indeed one can see it changing variables $k_{new}=a k$.
« Last Edit: March 08, 2015, 08:47:39 AM by Victor Ivrii »