Derive Fourier Transform from Fourier Series

Derive from Fourier Series

Fourier Series

See Fourier Series (Real) for details, we will directly give its form: The Fourier series expansion of the function $f$ is

$$f(x)\sim \frac{a_0}{2}+\sum _{n=1}^{\infty }\left(a_n\cos \frac{n\pi x}{l}+b_n\sin \frac{n\pi x}{l}\right)$$

where

$$\begin{array}{rl}{a}_n& =\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \frac{n\pi x}{l}\mathrm{d}x,n=0,1,2,\cdots \\ b_n& =\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \frac{n\pi x}{l}\mathrm{d}x,n=1,2,\cdots \end{array}$$

Complex Form of Fourier Series

Using Euler's formula, we can derive

$$\cos \theta =\frac{{\mathrm{e}}^{\mathrm{i}\theta }+{\mathrm{e}}^{-\mathrm{i}\theta }}{2},\sin \theta =-\mathrm{i}\left(\frac{{\mathrm{e}}^{\mathrm{i}\theta }-{\mathrm{e}}^{-\mathrm{i}\theta }}{2}\right)$$

Substituting them into the above series expression, assuming the series converges, let $\omega =\frac{2\pi }{T}=\frac{2\pi }{2l}=\frac{\pi }{l}$, we have

$$\begin{array}{rl}f(x)& =\frac{a_0}{2}+\sum _{n=1}^{\infty }\left(a_n\frac{{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}+{\mathrm{e}}^{-\mathrm{i}n{\omega }_{0}x}}{2}-\mathrm{i}{b}_n\frac{{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}-{\mathrm{e}}^{-\mathrm{i}n{\omega }_{0}x}}{2}\right)\\ & =\frac{a_0}{2}+\sum _{n=1}^{\infty }\frac{1}{2}(a_n-\mathrm{i}{b}_n){\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}+\sum _{n=1}^{\infty }\frac{1}{2}(a_n+\mathrm{i}{b}_n){\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\\ & =\frac{a_0}{2}+\sum _{n=1}^{\infty }\frac{1}{2}(a_n-\mathrm{i}{b}_n){\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}+\sum _{n=-\infty }^{-1}\frac{1}{2}(a_{-n}+\mathrm{i}{b}_{-n}){\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\\ & =\sum _{-\infty }^{\infty }{d}_n{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\end{array}$$

where

$$d_n=\{\begin{array}{ll}\frac{1}{2}(a_n-\mathrm{i}{b}_{-n}),& n>0\\ \frac{1}{2}{a}_0,& n=0\\ \frac{1}{2}(a_{-n}-\mathrm{i}{b}_{-n}),& n<0\end{array}$$

It can be verified that for any value of $n$, $d_n$ can be expressed as

$$d_n=\frac{1}{2l}{\int }_0^{2l}{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x=\frac{1}{T}{\int }_0^T{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x$$

where ${\omega }_0=\frac{2\pi }{T}$ is called the fundamental frequency, and thus we obtain the complete expression of the complex exponential form Fourier series expansion

$$f(x)=\sum _{n=-\infty }^{+\infty }\left[\frac{1}{T}{\int }_0^T{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x\right]{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}$$

Fourier Transform

Fourier series expansion is for periodic functions, but in reality, most signals are non-periodic. Non-periodic functions can be viewed as periodic functions with $T\to +\infty$, and when $T=+\infty$, the fundamental frequency ${\omega }_0$ becomes the differential $\mathrm{d}\omega$, so the summation needs to be converted to an integral

$$\begin{array}{rl}f(x)& =\sum _{n=-\infty }^{+\infty }\left[\frac{1}{T}{\int }_0^T{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x\right]{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\\ & =\sum _{n=-\infty }^{+\infty }\left[\frac{{\omega }_0}{2\pi }{\int }_0^T{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x\right]{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\\ & =\sum _{n=-\infty }^{+\infty }\left[\omega {\int }_0^T{\mathrm{e}}^{-in{\omega }_{0}x}f(x)\mathrm{d}x\right]{\mathrm{e}}^{\mathrm{i}n{\omega }_{0}x}\end{array}$$

Here we take $\omega =\frac{{\omega }_0}{2\pi }$, let $\omega =\mathrm{d}\omega \to 0$ then we have

$$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{e}^{-i2\pi \omega x}\cdot f(x)\mathrm{d}x\cdot e^{i2\pi \omega x}\mathrm{d}\omega$$

where

$$F(\omega )={\int }_{-\infty }^{+\infty }{e}^{-i2\pi \omega x}\cdot f(x)\mathrm{d}x$$

is called the Fourier transform, and

$$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }F(\omega )\cdot e^{i2\pi \omega x}\mathrm{d}\omega$$

is called the inverse Fourier transform