Author Topic: HA6 problem 3  (Read 414 times)

Victor Ivrii

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HA6 problem 3
« on: March 05, 2015, 08:36:46 AM »
Let $\alpha>0$. Based on Fourier transform of $e^{-\alpha x^2/2}$ find Fourier transforms of

a.  $e^{-\alpha x^2/2}\cos (\beta x)$, $e^{-\alpha x^2/2}\sin (\beta x)$;

b.  $ x e^{-\alpha x^2/2}\cos (\beta x)$, $x e^{-\alpha x^2/2}\sin (\beta x)$.

Yiyun Liu

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Re: HA6 problem 3
« Reply #1 on: March 05, 2015, 09:07:14 PM »
\[\begin{gathered}
  part(a): \hfill \\
   \hfill \\
  f(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}} \hfill \\
  \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty  {{e^{\frac{{ - \alpha {x^2}}}{2}}}} {e^{ - i\omega x}}dx = \frac{1}{{\sqrt {2\pi \alpha } }}{e^{\frac{{ - {\omega ^2}}}{{2\alpha }}}} \hfill \\
  g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) = \frac{1}{{2i}}f(x)({e^{i\beta x}} - {e^{ - i\beta x}}) \hfill \\
  thus,\hat g(\omega ) = \frac{1}{{2i}}\left[ {(\hat f(\omega  - \beta ) - (\hat f(\omega  + \beta )} \right] \hfill \\
   = \frac{1}{{2i\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega  - \beta )}^2}}}{{2\alpha }}}} - {e^{\frac{{ - {{(\omega  + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\
  similar, \hfill \\
  g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) = \frac{1}{2}f(x)({e^{i\beta x}} + {e^{ - i\beta x}}) \hfill \\
  \hat g(x) = \frac{1}{2}\left[ {(\hat f(\omega  - \beta ) + (\hat f(\omega  + \beta )} \right] = \frac{1}{{2\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega  - \beta )}^2}}}{{2\alpha }}}} + {e^{\frac{{ - {{(\omega  + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\
   \hfill \\
  part(b); \hfill \\
   \hfill \\
  f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) \hfill \\
  let,f(x) = xg(x),g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x),hence, \hfill \\
  \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{i}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega  - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega  - \beta )}^2}}}{{2\alpha }}}}} - \frac{{(\omega  + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega  + \beta )}^2}}}{{2\alpha }}}}) \hfill \\
  likewise, \hfill \\
  f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) \hfill \\
  f(x) = xg(x),where,g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x),hence, \hfill \\
  \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{1}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega  - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega  - \beta )}^2}}}{{2\alpha }}}}} + \frac{{(\omega  + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega  + \beta )}^2}}}{{2\alpha }}}}) \hfill \\
   \hfill \\
\end{gathered} \]

Victor Ivrii

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Re: HA6 problem 3
« Reply #2 on: March 07, 2015, 05:13:12 AM »
Yiyun, you should use MathJax only in math mode (not for a text).