APM346-2018S > Term Test 2

TT2--P5

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Victor Ivrii:
Find Fourier transforms of the  function
\begin{equation}
f(x)=\cos^2(x)e^{-|x|}
\end{equation}
and write this function as a Fourier integral.

Tristan Fraser:
We first note that $ \cos^2 x  = (\frac{(e^{ix} + e^{-ix})}{2})^2  = \frac{2 + e^{2ix} + e^{-2ix}}{4} $

From there, we take the fourier transform:  $$\hat{f(}k) = \frac{1}{2\pi} \int_{-\infty }^{\infty } e^{-ikx} e^{-|x|} ( \frac{2 + e^{2ix} + e^{-2ix}}{4}) = \frac{1}{2\pi }\left ( \int_{0}^{\infty } \frac{2e^{-x(1+ki)} + e^{-x(1+ki - 2i)} + e^{-x(1+ki + 2i))}}{4} dx +\int_{-\infty }^{0} \frac{2e^{x(1- ki)} + e^{x(1- ki - 2i) }+ e^{x(1- ki + 2i))}}{4} dx\right )$$

We integrate that thing, and note that by evaluating at the bounds of $0,\pm \infty $, only the terms evaluated at 0 can be kept, netting us:

$$\hat{f(}k) =  \frac{1}{2\pi}( \frac{1}{2(1+ki)} + \frac{1}{4(1+ki - 2i)} + \frac{1}{4(1+ki+2i)} + \frac{1}{2(1-ki)} + \frac{1}{4(1-ki-2i)} + \frac{1}{4(1-ki+2i)}) $$

While this expression may be ugly, what is useful about it is the fact that these are fractions of complex numbers and their conjugates (noteably first and fourth terms, 2nd and 6th, 3rd and 5th) so we'd get the following fraction

$$\hat{f(}k) =  \frac{1}{2\pi}( \frac{2}{4(1+ k^2)} + \frac{2}{16(1+ (k-2)^2)} + \frac{2}{16(1+ (k+2)^2)} ) $$

Thus, we can plug this function into the IFT:

$$ f(x) =  \int_{-\infty}^{\infty} \hat{f}(k) e^{ikx} dk  = \int_{-\infty}^{\infty} \frac{1}{4\pi}( \frac{1}{(1+ k^2)} + \frac{1}{4(1+ (k-2)^2)} + \frac{1}{4(1+ (k+2)^2)}) e^{ikx} dk  $$

Leaving the function as a Fourier integral.

Jingxuan Zhang:
I suggest somewhat different numerical factor. Not much different.

Victor Ivrii:

--- Quote from: Jingxuan Zhang on March 23, 2018, 08:00:54 PM ---I suggest somewhat different numerical factor. Not much different.

--- End quote ---
Please correct.

Jingxuan Zhang:
\begin{equation}\begin{split}
\hat{f}(\omega)&=\frac{1}{2\pi}\int_{-\infty}^\infty (\frac{1}{4}e^{2ix}+\frac{1}{4}e^{-2ix}+\frac{1}{2})e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega-2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{4}e^{-|x|}e^{-i(\omega+2) x}\,dx+\frac{1}{2\pi}\int_{-\infty}^\infty \frac{1}{2}e^{-|x|}e^{-i\omega x}\,dx\\
&=\frac{1}{4\pi(1+(x-2)^2)}+\frac{1}{4\pi(1+(x+2)^2)}+\frac{1}{2\pi(1+x^2)}
\end{split}\end{equation}

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