Fourier transform

The Fourier transform is a mathematical operation that produces a function in one domain by integrating a function from another domain.

As mentioned in the previous article, a Fourier series represents a function $f(x)$ within a specific interval, for example ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ . Outside this interval, the function is assumed to repeat itself. Therefore, a non-periodic function cannot be accurately represented by a Fourier series. If we want a mathematical tool to represent both periodic and non-periodic functions, we need a more generalised version of the Fourier series that is defined on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ , as $L\rightarrow\infty$ . To express this mathematically, let ${k_n=\frac{2\pi n}{L}}$ and substitute eq113 for this new interval in eq112 to give

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{L}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\;\;\;\;\;\;\;\;114$

Since ${\Delta k_n=\frac{2\pi \Delta n}{L}}$ and $\Delta n=1$ ,

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{L}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\Delta n=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ik_nx}dx\biggr\]e^{ik_nx}\Delta k_n$

As ${\lim_{L\rightarrow\infty}\Delta k_n=\lim_{L\rightarrow\infty}\frac{2\pi\Delta n}{L}\rightarrow 0}$ , the discrete sum in the above equation becomes a continuous integral:

$f(x)=\sum_{n=-\infty}^{\infty}\biggr\[\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\biggr\]e^{ikx}dk\;\;\;\;\;\;\;\;115$

where we have replaced the discrete variable $k_n$ with the continuous variable $k$ .

Question

Show that $\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$ in eq115 is equal to the function $g(k)$ .

Answer

The integral $\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$ is essentially a function of $k$ , as there is a corresponding integral value (output) for every value of $k$ (input).

Therefore,

$f(x)=\int_{-\infty}^{\infty}g(k)e^{ikx}dk\;\;\;where\;\;\;g(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\;\;\;\;\;\;\;\;117$

$g(k)$ is known as the Fourier transform of $f(x)$ , and vice versa. The two equations in eq117 are regarded as a Fourier transform pair, which allow one function to be transformed into the other function. We can also say that $g(k)$ is the inverse Fourier transform of $f(x)$ , and vice versa.

Question

i) Show that $h(x)=f(x)g(x)$ is an even function if both $f(x)$ and $g(x)$ are even functions.

ii) Show that $h(x)=f(x)g(x)$ is an odd function if $f(x)$ is an even function and $g(x)$ is an odd function or vice versa.

iii) If $f(x)$ is an odd function, show that $\int_{-\infty}^{\infty}f(x)dx=0$ , assuming that the integral is carried out over a symmetric interval.

iv) Is $\int_{-\infty}^{\infty}q(x)dx=0$ if $q(x)$ is an even function?

Answer

i) An even function and an odd function are defined as $f(-x)=f(x)$ and $f(-x)=-f(x)$ respectively. So, $h(-x)=f(-x)g(-x)=f(x)g(x)=h(x)$ .

ii) $h(-x)=f(-x)g(-x)=-f(x)g(x)=-h(x)$ .

iii) Using the Cauchy principal value, ${\int_{-\infty}^{\infty}f(x)dx=\lim_{a\rightarrow\infty,b\rightarrow 0}\biggr\[\int_{-a}^bf(x)dx+\int_b^af(x)dx\biggr\]=\int_{-\infty}^0f(x)dx+\int_0^{\infty}f(x)dx}$ . Let $x=-x$ , $dx=-dx$ for $\int_{-\infty}^0f(x)dx$ . So,

$\int_{-\infty}^{\infty}f(x)dx=-\int_{\infty}^{0}f(-x)dx+\int_{0}^{\infty}f(x)dx=0$

iv) No. With $x=-x$ , $dx=-dx$ for $\int_{-\infty}^0f(x)dx$ ,

$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^0f(x)dx+\int_0^{\infty}f(x)dx=2\int_{0}^{\infty}f(x)dx$

Special forms of eq117 arise according to whether $f(x)$ is even or odd. From eq117,

$g(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)coskxdx-\frac{i}{2\pi}\int_{-\infty}^{\infty}f(x)sinkxdx$

$coskx$ is an even function, while $sinkx$ is an odd function. So, if $f(x)$ is even, the first integrand is an even function, while the second integrand is an odd function. As shown in parts iii and iv the Q&A above,

$g(k)=\frac{1}{\pi}\int_{0}^{\infty}f(x)coskxdx\;\;\;\;\;\;\;\;118$

It follows that $g(k)$ is also an even function because $g(-k)=\frac{1}{\pi}\int_{0}^{\infty}f(x)cos(-kx)dx=\frac{1}{\pi}\int_{0}^{\infty}f(x)coskxdx=g(k)$ . From eq117,

$f(x)=\int_{-\infty}^{\infty}g(k)e^{ikx}dx=\int_{-\infty}^{\infty}g(k)coskxdk+i\int_{-\infty}^{\infty}g(k)sinkxdk=2\int_{0}^{\infty}g(k)coskxdk\;\;\;\;\;\;\;\;119$
Eq118 and eq119 are known as a pair of Fourier cosine transforms, which has applications in Fourier-transform infrared spectroscopy.

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