### Author Topic: TT2-P5  (Read 2882 times)

#### Victor Ivrii

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##### TT2-P5
« on: November 18, 2015, 08:43:06 PM »
Find Fourier transforms of the  function
\begin{equation*}
f(x)= \left\{\begin{aligned}
&\cos (x) &&|x|<\frac{\pi}{2},\\
&0 &&|x|>\frac{\pi}{2}
\end{aligned}\right.
\end{equation*}
and write this function as a Fourier integral.

#### Xi Yue Wang

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##### Re: TT2-P5
« Reply #1 on: November 18, 2015, 10:46:46 PM »
To get Fourier transform,
$$\hat{f}(k) = \frac{1}{2\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(x)e^{-ikx} dx$$
Let $g(x) = 1, f(x) = g(x)cos(x) = g(x)\frac{e^{ix}+e^{-ix}}{2} = \frac{1}{2}[g(x)e^{ix} + g(x)e^{-ix}] = \hat{f}(k) = \frac{1}{2}[\hat{g}(k-1) + \hat{g}(k+1)]$
$$\hat{g}(k) = \frac{1}{2\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} e^{-ikx} dx\\=\frac{1}{2\pi}\frac{e^{\frac{-ik\pi}{2}}-e^{\frac{ik\pi}{2}}}{-ik}\\=\frac{1}{2\pi}\frac{e^{\frac{ik\pi}{2}}-e^{\frac{-ik\pi}{2}}}{ik}\\=\frac{1}{k\pi}\sin(\frac{k\pi}{2})$$
Then, $$\hat{f}(k) = \frac{1}{2}[\hat{g}(k-1) + \hat{g}(k+1)]\\=\frac{1}{2}[\frac{1}{(k-1)\pi}\sin(\frac{(k-1)\pi}{2}) + \frac{1}{(k+1)\pi}\sin(\frac{(k+1)\pi}{2})]\\=\frac{1}{2}[\frac{1}{(k-1)\pi}\sin(\frac{(k\pi-\pi)}{2}) + \frac{1}{(k+1)\pi}\sin(\frac{(k\pi+\pi)}{2}]\\=\frac{1}{2}[\frac{1}{(1-k)\pi}\cos(\frac{k\pi}{2})+\frac{1}{(1+k)\pi}\cos(\frac{k\pi}{2})]\\=\frac{2}{2\pi(1-k^2)}\cos(\frac{k\pi}{2}) \\= \frac{1}{\pi(1-k^2)}\cos(\frac{k\pi}{2})$$
Hence, we write $f(x)$ as a Fourier integral.
$$f(x) = \int_{-\infty}^{\infty} \frac{1}{\pi(1-k^2)}\cos(\frac{k\pi}{2})e^{ikx} dk$$
« Last Edit: November 19, 2015, 12:00:11 AM by Xi Yue Wang »

#### Emily Deibert

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##### Re: TT2-P5
« Reply #2 on: November 18, 2015, 11:12:03 PM »
I think you accidentally added a factor of $\frac{1}{2}$ in your answer.

#### Bruce Wu

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##### Re: TT2-P5
« Reply #3 on: November 18, 2015, 11:20:20 PM »
I agree with Emily. I got $$\frac{\cos(\pi k/2)}{\pi (1-k^2)}$$

#### Emily Deibert

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##### Re: TT2-P5
« Reply #4 on: November 18, 2015, 11:22:32 PM »
Indeed, I got the same. Thank you for commenting Fei Fan Wu.

#### Xi Yue Wang

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##### Re: TT2-P5
« Reply #5 on: November 18, 2015, 11:43:28 PM »
Oh, Yes! I found where I miss a 2. Thank you Emily and Fei Fan!