### Author Topic: HA6 problem 1  (Read 1629 times)

#### Victor Ivrii

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##### HA6 problem 1
« on: March 05, 2015, 08:34:49 AM »
Let $\alpha >0$. Find Fourier transforms of

a.  $e^{-\alpha|x|}$;

b.  $e^{-\alpha|x|}\cos(\beta x)$, $e^{-\alpha|x|}\sin (\beta x)$ with $\beta >0$;

c.  $x e^{-\alpha|x|}$ with $\beta >0$;

d.  $xe^{-\alpha|x|}\cos(\beta x)$, $x e^{-\alpha|x|}\sin (\beta x)$ with $\beta >0$.

#### Mark Nunez

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##### Re: HA6 problem 1
« Reply #1 on: March 05, 2015, 09:22:36 PM »

#### Jessica Chen

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##### Re: HA6 problem 1
« Reply #2 on: March 05, 2015, 10:15:48 PM »
a.
\begin{align}
\hat{f}(\omega) &=\ \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x}dx\\
&=\  \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x -i\omega x} +\int_{0}^{\infty} e^{-\alpha x -i\omega x}]dx\\
&= \frac{1}{2\pi}[\int_{-\infty}^{0} (\cos (\omega x)-i\sin (\omega x))e^{\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x))e^{-\alpha x } +\int_{0}^{\infty} (\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+i\sin (\omega x)+\cos (\omega x)-i\sin (\omega x))e^{-\alpha x}]dx\\
&= \frac{1}{2\pi}[\int_{0}^{\infty} (\cos (\omega x)+\cos (\omega x))e^{-\alpha x}]dx\\
&= \frac{2}{2\pi}\int_{0}^{\infty}\cos(\omega x)e^{-\alpha x} dx\\
&= \frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}
\end{align}

b.
1)
Let $f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\cos(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2}e^{i\beta x}+\frac{1}{2}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2}e^{-\alpha |x|}e^{i\beta x}+\frac{1}{2}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2}\hat{f}(\omega - \beta) + \frac{1}{2}\hat{f}(\omega + \beta)$.

From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}

2)
Let $f(\omega) = e^{-\alpha |x|}$, and let
\begin{align}
g(x) &=\ e^{-\alpha |x|}\sin(\beta x)\\
&=\ e^{-\alpha |x|}(\frac{1}{2i}e^{i\beta x}-\frac{1}{2i}e^{-i\beta x})\: \mbox{ (By Trig identity)}\\
&=\ \frac{1}{2i}e^{-\alpha |x|}e^{i\beta x}-\frac{1}{2i}e^{-\alpha |x|}e^{-i\beta x}
\end{align}
Which satisfies the form $\hat{g}(\omega) = \frac{1}{2i}\hat{f}(\omega - \beta) - \frac{1}{2i}\hat{f}(\omega + \beta)$.

From part a we have found $\hat{f}\frac{\alpha}{\pi (\alpha ^2 +\omega ^2)}$
Thus\begin{align}
\hat{g}(\omega) = \frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2)})-\frac{1}{2i}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})
\end{align}

c.
Let $f(\omega) = e^{-\alpha |x|}$, and let
$g(x) = xe^{-\alpha |x|}$
By the theorem $\hat{g}(\omega) = i\hat{f'}(\omega)$, We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\
&=\  i \frac{d}{d\omega }(\frac{\alpha }{\pi (\alpha ^2 +\omega ^2)})\\

&=\ i \frac{\alpha}{\pi}\frac{-2\omega}{(\alpha ^2 +\omega ^2)^2}\\
\end{align}

d.
Similar to part c),
Let $f(\omega) = e^{-\alpha |x|}cos(\beta x)$, and let $g(x) = xe^{-\alpha |x|}cos(\beta x)$

By the theorem $\hat{g}(\omega) = i\hat{f'}(\omega)$, where
$\hat{f}(\omega) = \frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})$

We have
\begin{align}
\hat{g}(\omega) &=\ i \frac{d\hat{f}(\omega)}{d\omega}\\

&=\ i \frac{d}{d\omega }\hat{f}(\omega)\\
& =\ i \frac{d}{d\omega }(\frac{\alpha}{\pi (\alpha ^2 +(\omega - \beta) ^2})+\frac{1}{2}(\frac{\alpha}{\pi (\alpha ^2 +(\omega + \beta) ^2)})\\
&=\ i\frac{\alpha}{\pi}[(\frac{-(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2})+\frac{-(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]\\
&=\ \frac{-\alpha i}{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}+\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]
\end{align}

Similar to above, we can get $g(x) = xe^{-\alpha |x|}sin(\beta x)$ by applying part b.2 formula and use the method we used in part c, then we get:
$\hat{g}(\omega) = \frac{-\alpha }{\pi}[\frac{\omega - \beta}{(\alpha^2 + (\omega - \beta)^2)^2}-\frac{\omega + \beta}{(\alpha^2 + (\omega + \beta)^2)^2}]$
« Last Edit: March 05, 2015, 11:29:12 PM by Jessica Chen »