Integral transformation formula derivation and examples

If in a $Ｔ$ $T$ function cycle $f(x)$ in

$\bigg[-\frac{T}{2}, \frac{T}{2}\bigg]$

Satisfies the Dirichlet condition, namely:

1. Except for a limited number of discontinuities of the first type, continuous everywhere

2. Segmented monotonic, the number of monotonic intervals is limited

Then $f(x)$ the fourier series is expressed as:

$f(t)\approx \frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[a_ncos(n\omega_0t) + b_nsin(n\omega_0t)\bigg]$

$\bigg[-\frac{T}{2}, \frac{T}{2}\bigg]$

The upper converges everywhere, and converges at $f(x)$ the continuous point of $f(t)$ , where,

$\omega_0 =\frac{2\pi}{T}$

Integrate both sides of the above formula:

$\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)dt=\int_{-\frac{T}{2}}^{\frac{T}{2}}\bigg(\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[a_ncos(n\omega_0t) + b_nsin(n\omega_0t)\bigg]\bigg) dt=\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{a_0}{2}dt\\=\frac{a_0}{2}\cdot T$

and so:

$a_0=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)dt$

for $\\ (m=1,2,3,\cdots)$

$\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)cos(m\omega_0t)dt=\int_{-\frac{T}{2}}^{\frac{T}{2}}\bigg(\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[a_ncos(n\omega_0t) + b_nsin(n\omega_0t)\bigg]\bigg)cos(m\omega_0t) dt\\=0+a_n\int_{-\frac{T}{2}}^{\frac{T}{2}}cos^2(n\omega_0t)dt=a_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1+cos(2n\omega_0t)}{2}dt=a_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1}{2}dt +a_n\int_{-\frac{T}{2}}^{\frac{T}{2}} \frac{cos(2n\omega_0 t)}{2}dt\\=a_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1}{2}dt +a_n\int_{-\frac{T}{2}}^{\frac{T}{2}} \frac{cos(2n\omega_0 t)}{4n\omega_0}d(2n\omega_0 t)=\frac{a_n}{2}\bigg|_{-\frac{T}{2}}^{\frac{T}{2}}+\frac{a_n}{4n\omega_0}sin(2n\omega_0t)\bigg|_{-\frac{T}{2}}^{\frac{T}{2}}\\=a_n\cdot \frac{T}{2}+0$

and so:

$\\a_n=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)cos(n\omega_0 t)dt \quad(n=1,2, 3,\cdots)$

$\\ \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)sin(m\omega_0t)dt=\int_{-\frac{T}{2}}^{\frac{T}{2}}\bigg(\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[a_ncos(n\omega_0t) + b_nsin(n\omega_0t)\bigg]\bigg)sin(m\omega_0t) dt\\=0+b_n\int_{-\frac{T}{2}}^{\frac{T}{2}}sin^2(n\omega_0t)dt=b_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1-cos(2n\omega_0t)}{2}dt=b_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1}{2}dt -b_n\int_{-\frac{T}{2}}^{\frac{T}{2}} \frac{cos(2n\omega_0 t)}{2}dt\\=b_n\int_{-\frac{T}{2}}^{\frac{T}{2}}\frac{1}{2}dt - b_n\int_{-\frac{T}{2}}^{\frac{T}{2}} \frac{cos(2n\omega_0 t)}{4n\omega_0}d(2n\omega_0 t)=\frac{b_n}{2}\bigg|_{-\frac{T}{2}}^{\frac{T}{2}}-\frac{b_n}{4n\omega_0}sin(2n\omega_0t)\bigg|_{-\frac{T}{2}}^{\frac{T}{2}}\\=b_n\cdot \frac{T}{2}+0$

and so:

$\\b_n=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)sin(n\omega_0 t)dt \quad(n=1,2, 3,\cdots)$

In summary:

$a_0=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)dt$

$\\a_n=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)cos(n\omega_0 t)dt \quad(n=1,2, 3,\cdots)$

$\\b_n=\frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)sin(n\omega_0 t)dt \quad(n=1,2, 3,\cdots)$

In the field of electronic communication, Euler’s formula is often used:

$cos(t)=\frac{e^{it}+e^{-it}}{2}$

$sin(t)=\frac{e^{it}-e^{-it}}{2i}$

and so:

$\\f(t)\approx \frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[a_ncos(n\omega_0t) + b_nsin(n\omega_0 t)\bigg]\\=\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[ \frac{a_n}{2}(e^{in\omega_0 t}+e^{-in\omega_0 t})-i\cdot \frac{b_n}{2}(e^{in\omega_0 t}-e^{-in\omega_0 t})\bigg]\\=\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[ \frac{a_n-ib_n}{2}e^{in\omega_0 t}+\frac{a_n+ib_n}{2}e^{-in\omega_0 t}\bigg]$

make:

$c_0=\frac{a_0}{2}$

$c_n=\frac{a_n-ib_n}{2}$

$c_{-n}=\frac{a_n+ib_n}{2}$

Get the complex exponential form of the fourier series:

$\\f(t)\approx\frac{a_0}{2}+\sum_{n=1}^{\infty }\bigg[ \frac{a_n-ib_n}{2}e^{in\omega_0 t}+\frac{a_n+ib_n}{2}e^{-in\omega_0 t}\bigg]\\= c_0+\sum_{n=1}^{\infty }\bigg(c_ne^{in\omega_0 t} + c_{-n}e^{-in\omega_0 t}\bigg)=\sum_{n=-\infty}^{\infty}C_ne^{in\omega_0 t}$

Here:

$C_0=\frac{a_0}{2}=\frac{1}{2}\cdot \frac{2}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)dt=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)dt$

$\\C_n=\frac{a_n-ib_n}{2}=\frac{1}{2}\cdot \frac{2}{T}\bigg[\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)cos(n\omega_0 t)dt-i\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)sin(n\omega _0 t)dt\bigg] \\=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)\bigg[cos(n\omega_0 t)-isin(n\omega_0 t)\bigg]dt=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{-in\omega_0 t}dt \quad (n=1,2, 3, \cdots)$

Similarly:

$C_{-n}=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{in\omega_0 t}dt \quad (n=1,2, 3, \cdots)$

The above is $C_0, C_{-n},C_n$ written in a unified form:

$C_{n}=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{-in\omega_0 t}dt \quad (n=0,\pm1,\pm2, \pm3, \cdots)$

make

$\omega _n=n\omega_0 \quad (n=0,\pm1,\pm2, \pm3, \cdots)$

Combining the above formulas, we can get:

$\mathbf{f(t)\approx\frac{1}{T}\sum_{n=-\infty }^{\infty}\bigg[ \int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{-i\omega _nt}dt\bigg]e^{i\omega _n t}}$

After splitting, the Fourier series form is obtained:

$\\\mathbf{ F(n )=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt \quad (n \in Z)}$

$\\\mathbf{f(t)=\sum_{n=-\infty}^{\infty }F(n)\cdot e^{in\omega_0 t}\quad (n \in Z)}$

The Fourier series derives the Fourier transform of non-periodic signals:

$\\f(t)=\sum_{n=-\infty}^{\infty }F(n)\cdot e^{in\omega_0 t}=\sum_{n=-\infty}^{\infty }\bigg[\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}\\=\frac{1}{T}\sum_{n=-\infty}^{\infty }\bigg[\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}$

When $T\rightarrow \infty$ , the period signal becomes aperiodic signal, because $\omega_0 =\frac{2\pi}{T}$ , of Fourier series:

$\\ \lim_{T \to \infty }f(t)=\lim_{T \to \infty }\frac{1}{T}\sum_{n=-\infty}^{\infty }\bigg[\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}\\=\lim_{T \to \infty }\frac{\omega_0}{2\pi}\sum_{n=-\infty}^{\infty }\bigg[\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}\\=\lim_{T \to \infty }\frac{1}{2\pi}\sum_{n=-\infty}^{\infty }\bigg[\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}\cdot \omega_0$

At that $T\rightarrow \infty$ time, $\Delta \omega =n\omega_0 - (n-1)\omega_0=\omega_0\rightarrow 0$

According to the differential element method of calculus, the accumulation of the outside can be regarded as finding the base $\omega_0$ and the height

$\bigg[\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-in\omega_0 t}dt\bigg]\cdot e^{in\omega_0 t}$

The area of the graph:

and so:

$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt \qquad (-\infty<\omega< \infty )$

$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega \qquad (-\infty<\omega< \infty )$

An example from Fourier series to Fourier transform:

The analytical formula of this function is:

$f(x)=\left\{\begin{matrix} 0 \qquad x>-\frac{T}{2}\ and \ x < -\tau \\ 1 \qquad \ \ x >-\tau \ and \ x < \tau\\ 0 \qquad \ \ \ \ x > \tau \ and \ x < \frac{T}{2} \end{matrix}\right.$

$\\ a_0=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)dx = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(x)dx =\frac{1}{T}\int_{-\tau}^{\tau}1\cdot dx = \frac{1}{T}\int_{-\tau}^{\tau}1\cdot dx\\=\frac{1}{T}\bigg|^{\tau}_{-\tau}=\frac{2\tau}{T}$

$\\ b_n =0$

$\\ a_n =\frac{2}{T}\int_{-\frac{T}{2}}^{-\frac{T}{2}}f(x)cos(n\cdot \frac{2\pi}{T}\cdot x)dx=\frac{2}{T}\int_{-\tau}^{\tau}1\cdot cos(n\cdot \frac{2\pi}{T}\cdot x)dx \\=\frac{T}{2n\pi}\cdot \frac{2}{T}\int_{-\tau}^{\tau}1\cdot cos(n\cdot \frac{2\pi}{T}\cdot x)d(n\cdot \frac{2\pi}{T}\cdot x)=\frac{1}{n\pi}sin(n\cdot \frac{2\pi}{T}\cdot x)\bigg|_{-\tau}^{\tau}=\frac{2sin(\frac{2n\pi \tau}{T})}{n\pi}$

The function graph is:

python code:


# -*- coding: utf-8 -*-
"""
Created on Mon Feb  1 13:57:21 2021
@author: czl
"""
from pylab import *
 
x = mgrid[-20:20:0.01]
 
def fourier_wave():
    a0 = 3/16
    s=a0
    
    for n in range(1,1000,1):
        bn = 0
        an = 2*sin((2*n*pi*1.5/16))/(n*pi)
        s0 = an*cos(n*x*(2*pi/16))+bn*sin(n*x*(2*pi/16))
        s=s+s0
        
    plot(x,s,'orange',linewidth=0.6)
    title('fourier_transform')
    show()    
 
fourier_wave()

The Fourier transform coefficients in complex exponential form are:

$f (n \ omega_0) = \ frac {a_n-ib_n} {2} = \ frac {a_n} {2} = \ frac {sin (\ frac {2n \ pi \ tau} {T})} {n \ pi }$

$f(n\cdot \frac{2\pi}{T})=\frac{sin(\frac{2n\pi \tau}{T})}{n\pi}$

Density spectrum:

$T\cdot f(n\cdot \frac{2\pi}{T})=2\tau \cdot \frac{sin(\frac{2n\pi \tau}{T})}{\frac{2n\pi\tau}{T}}$

When $T->\infty$ the time:

$F (\ omega) = T \ cdot f (n \ cdot \ frac {2 \ pi} {T}) = 2 \ tau \ cdot \ frac {sin (\ omega \ tau)} {\ omega \ tau}$

The following diagram shows is that when $T->\infty$ , the spectral density represented by the signal.

The meaning of negative frequency here is the direction of rotation of the unit circle, not the concept of "negative" in the ordinary sense.

Integral transformation formula derivation and examples

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