Fourier series and the Fourier transform (a)

Clever mathematicians always like to see the world from another perspective, the Fourier transform method gives people a look at and understand the world from a different angle.

A function of polynomial expansion in higher mathematics where most people have learned the Taylor expansion, which is a specific function expansion method for the power series, there are important applications in calculus. The Fourier expansion is a function that meet specific criteria developed on a set of orthogonal trigonometric series, we can define the set of trigonometric series:

$\left \{sin(0x),cos(0x),sin(x),cos(x),sin(2x),cos(2x)...sin(nx),cos(nx)\right\}$

and

$\ Int _ {- \ the} ^ {\} the sin (nx) DX = 0$

$\ Int _ {- \ pi} ^ {\ pi} cos (nx) dx = 0$

$\ _ You {- \ pi} ^ {\ pi} sin (nx) cos (mx) dx = 0$

The above three formulas described orthogonality of the set of trigonometric lines.

More simply, it is to be considered a specific function can be superimposed comes from the numerous sine function. Since the trigonometric functions have periodicity, we can first consider the period $T$ as a function $f(t)$ of Fourier expansion.

Based on the face of the popular argument Fourier series, you can write:

$f(t)=A_{0}+\sum_{i=1}^{\infty }A_{i}sin(\frac{2i\pi t}{T})+\varphi _{i})$

And using the angle difference equation high school expands to:

$f(t)=A_{0}+\sum_{i=1}^{\infty }A_{i} \left \{sin(\frac{2i\pi t}{T})cos\varphi_{i}+cos(\frac{2i\pi t}{T})sin\varphi _{i} \right \}$

And so $\frac{a_{0}}{2}=A_{0}$ , $a_{i}=A_{i}sin(\varphi_{i} )$ , $b_{i}=A_{i}cos(\varphi_{i})$ , $\omega =\frac{2\pi}{T}$ end up:

$f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}cos(i\omega t)+b_{i}sin(i\omega t)\right \}$

This is the period of $2\pi$ the Fourier series expansion of a function. We now find $\frac{a_{0}}{2}$ , $a_{i}$ , $b_{i}$ :

Type of $f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}cos(i\omega t)+b_{i}sin(i\omega t)\right \}$ integration of:

$\int_{-T}^{T}f(t)dt=\int_{-T}^{T}\frac{a_{0}}{2}dt+\sum_{i=1}^{\infty }\left \{\int_{-T}^{T}a_{i}cos(i\omega t)dt+\int_{-T}^{T}b_{i}sin(i\omega t)dt\right \}$

The orthogonality of trigonometric lines:

$\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)dt=T{\frac{a_{0}}{2}}$

$a_{0}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)dt$

Formula $f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}cos(i\omega t)+b_{i}sin(i\omega t)\right \}$ multiplies $sin(i\omega t)$ and integration of:

$\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt=\int_{\frac{-T}{2}}^{\frac{T}{2}}\frac{a_{0}}{2}sin(i\omega t)dt+\sum_{i=1}^{\infty }\left \{\int_{\frac{-T}{2}}^{\frac{T}{2}}a_{i}sin(i\omega t)cos(i\omega t)dt+\int_{\frac{-T}{2}}^{\frac{T}{2}}b_{i}sin^2(i\omega t)dt\right \}$

The orthogonality of trigonometric lines:

$\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt=0+\sum_{i=1}^{\infty }\left \{0+\int_{\frac{-T}{2}}^{\frac{T}{2}}b_{i}sin^2(i\omega t)dt\right \}$

Using the formula in descending $sin^2x=\frac{1+cos2x}{2}, cos^2x=\frac{1-cos2x}{2}$ get:

$\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt=\frac{1}{2}\sum_{i=1}^{\infty }\left \{\int_{\frac{-T}{2}}^{\frac{T}{2}}b_{i}(1-cos2x)dt\right \}$

$\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt=\frac{T}{2}bi$

$b_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt$

Similarly get:

$a_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)cos(i\omega t)dt$

Finishing the above process, we get triangular form of the Fourier series expansion is:

$f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}cos(i\omega t)+b_{i}sin(i\omega t)\right \}$

$a_{0}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)dt$

$a_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)cos(i\omega t)dt$

$b_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt$

Now we will need to expand the function of further generalization, and now wish to set $f(t)$ is not a periodic function, we can think, in fact, non-periodic function can be viewed as an infinite cycle that is $T\rightarrow \infty$ a periodic function. We need to lead Euler formula as a ready formula:

$e ^ {jx} = cosx + jsinx$ ， $j=\sqrt{-1}$

$cosx = \ frac {e ^ {jx} + e ^ {- jx}} {2}$

$sinx = \ frac {e ^ {jx} -e ^ {- jx}} {2j}$

Formula $f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}cos(i\omega t)+b_{i}sin(i\omega t)\right \}$ can be rewritten as:

$f(t)=\frac{a_{0}}{2}+\sum_{i=1}^{\infty }\left \{a_{i}\frac{e^{i\omega t}+e^{-i\omega t}}{2}+b_{i}\frac{e^{i\omega t}-e^{-i\omega t}}{2j}\right \}$

Order was:

$f(t)=\frac{a_{0}}{2}+\frac{1}{2}\sum_{i=1}^{\infty }\left \{e^{i\omega t}(a_{i}-jb_{i})+e^{-i\omega t}(a_{i}+jb_{i})\right \}$

For $a_{i}-jb_{i}$ there are:

$a_{i}-jb_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)cos(i\omega t)dt-j\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)sin(i\omega t)dt$

$=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)\left \{cos(i\omega t)-jsin(i\omega t)\right \}dt$

$=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-ij\omega t}dt$

Similarly there are:

$a_{i}+jb_{i}=\frac{2}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{ij\omega t}dt$

Note that when $i=0$ the time $a_{i}-jb_{i}=a_{i}+jb_{i}=a_{0}$ , and when $i<0$ the time $a_{i}-jb_{i}=a_{i}+jb_{i}$

and so:

$f(t)=\sum_{-\infty }^{\infty }e^{i\omega t}\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-ij\omega t}dt$

Order $c_{i}=\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-ij\omega t}dt$ , $\omega_{i}=i\omega$ are:

$f(t)=}\sum_{-\infty }^{\infty }c_{i}e^{j\omega_{i}t}$

For non-periodic functions, the form of:

$f(t)=\lim_{T\rightarrow \infty }\sum_{-\infty }^{\infty }e^{j\omega_{i}t}\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega_{i}t}dt$

When the cycle of infinite angular frequency $\omega$ by discrete variables as a continuous variable, it might be $\frac{1}{T}$ the writing $\frac{\omega}{2\pi}$ , the final form of the Fourier series as follows:

$f(t)=\frac{1}{2\pi}\int_{-\infty }^{\infty }\left [ \int_{-\infty }^{\infty }f(t)e^{-j\omega_{i}t}dt \right ]e^{j\omega_{i}t}d\omega$

Fourier series and the Fourier transform (a)

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