傅里叶级数与傅里叶变换_Part3_周期为2L的函数展开为傅里叶级数

参考链接：

DR_CAN老师的原视频

0、复习Part2的内容

参考链接：傅里叶级数与傅里叶变换_Part2_周期为2Π的函数展开为傅里叶级数

对于周期为 T=2πT = 2\piT=2π周期函数，即f(x)=f(x+2π)f\left( x \right) = f\left( {x + 2\pi } \right)f(x)=f(x+2π) ，它的傅里叶级数展开形式如下：

f(x)=a02+∑n=1∞ancos⁡nx+∑n=1∞bnsin⁡nxf\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} + \sum\limits_{n = 1}^\infty {{b_n}\sin nx}f(x)=2a0​​+n=1∑∞​an​cosnx+n=1∑∞​bn​sinnx

其中，

a0=1π∫−ππf(x)dxan=1π∫−ππf(x)cos⁡nxdxbn=1π∫−ππf(x)sin⁡nxdx\begin{array}{l} {a_0} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)dx} \\\\ {a_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)\cos nxdx} \\\\ {b_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {f\left( x \right)\sin nxdx} \end{array}a0​=π1​∫−ππ​f(x)dxan​=π1​∫−ππ​f(x)cosnxdxbn​=π1​∫−ππ​f(x)sinnxdx​

1、周期为2L的函数展开为傅里叶级数

对于，f(t)=f(t+2L)f\left( t \right) = f\left( {t + 2L} \right)f(t)=f(t+2L) ，周期为T=2LT = 2LT=2L的函数

如果想用运用Part2掌握的以T=2πT = 2\piT=2π为周期的f(t)=f(t+2π)f\left( t \right) = f\left( {t + 2\pi} \right)f(t)=f(t+2π)的傅里级数展开公式，需要做一个换元。

令 x=πLt⇒t=Lπxx = \frac{\pi }{L}t \Rightarrow t = \frac{L}{\pi }xx=Lπ​t⇒t=πL​x

f(t)=f(πLx)≜g(x)f\left( t \right) = f\left( {\frac{\pi }{L}x} \right) \triangleq g\left( x \right)f(t)=f(Lπ​x)≜g(x)

通过换元，我们就把周期为T=2LT = 2LT=2L的函数，f(t)=f(t+2L)f\left( t \right) = f\left( {t + 2L} \right)f(t)=f(t+2L) 变成了周期为T=2πT = 2\piT=2π的函数g(t)=g(t+2π)g\left( t \right) =g\left( {t + 2\pi} \right)g(t)=g(t+2π) 。

根据Part2的内容我们知道

g(x)=a02+∑n=1∞ancos⁡nx+∑n=1∞bnsin⁡nxg\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} + \sum\limits_{n = 1}^\infty {{b_n}\sin nx}g(x)=2a0​​+n=1∑∞​an​cosnx+n=1∑∞​bn​sinnx

其中，

a0=1π∫−ππg(x)dxan=1π∫−ππg(x)cos⁡nxdxbn=1π∫−ππg(x)sin⁡nxdx\begin{array}{l} {a_0} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)dx} \\\\ {a_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)\cos nxdx} \\\\ {b_n} = \frac{1}{\pi }\int_{ - \pi }^\pi {g\left( x \right)\sin nxdx} \end{array}a0​=π1​∫−ππ​g(x)dxan​=π1​∫−ππ​g(x)cosnxdxbn​=π1​∫−ππ​g(x)sinnxdx​

👇

现在要做的，就是把x=πLtx = \frac{\pi }{L}tx=Lπ​t带进去

cos⁡nx=cos⁡nπLt,sin⁡nx=sin⁡nπLt\cos nx = \cos \frac{{n\pi }}{L}t,\sin nx = \sin \frac{{n\pi }}{L}tcosnx=cosLnπ​t,sinnx=sinLnπ​t

g(x)=f(t)g\left( x \right) = f\left( t \right)g(x)=f(t)

∫−ππdx=∫−LLdπLt=πL∫−LLdt\int_{ - \pi }^\pi {dx} = \int_{ - L}^L {d\frac{\pi }{L}t} = \frac{\pi }{L}\int_{ - L}^L {dt}∫−ππ​dx=∫−LL​dLπ​t=Lπ​∫−LL​dt

1π∫−ππdx=1π⋅(πL∫−LLdt)=1L∫−LLdt\frac{1}{\pi }\int_{ - \pi }^\pi {dx} = \frac{1}{\pi } \cdot \left( {\frac{\pi }{L}\int_{ - L}^L {dt} } \right) = \frac{1}{L}\int_{ - L}^L {dt}π1​∫−ππ​dx=π1​⋅(Lπ​∫−LL​dt)=L1​∫−LL​dt

讲上述算出的式子带入到g(x)g\left( x\right)g(x)中。

f(t)=g(x)=a02+∑n=1∞ancos⁡nπLt+∑n=1∞bnsin⁡nπLtf\left( t \right) = g\left( x\right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos \pmb{\frac{{n\pi }}{L}t}} + \sum\limits_{n = 1}^\infty {{b_n}\sin \pmb{\frac{{n\pi }}{L}t}}f(t)=g(x)=2a0​​+n=1∑∞​an​cosLnπ​tLnπ​t+n=1∑∞​bn​sinLnπ​tLnπ​t

其中

a0=1L∫−LLf(t)dtan=1L∫−LLf(t)cos⁡nπLtdtbn=1L∫−LLf(t)sin⁡nπLtdt\begin{array}{l} {a_0} =\pmb{ \frac{1}{L}}\int_{ - L}^L {f\left( t \right)dt} \\\\ {a_n} =\pmb{ \frac{1}{L}}\int_{ - L}^L {f\left( t \right)\cos \pmb{\frac{{n\pi }}{L}tdt}} \\\\ {b_n} = \pmb{\frac{1}{L}}\int_{ - L}^L {f\left( t \right)\sin \pmb{\frac{{n\pi }}{L}tdt}} \end{array}a0​=L1​L1​∫−LL​f(t)dtan​=L1​L1​∫−LL​f(t)cosLnπ​tdtLnπ​tdtbn​=L1​L1​∫−LL​f(t)sinLnπ​tdtLnπ​tdt​

在工程当中，由于时间是t≥0t \ge 0t≥0的，所以ttt是从000开始的，假设周期为 T=2LT=2LT=2L，ω≜πL=2π2L=2πT\omega \triangleq \frac{\pi }{L} = \frac{{2\pi }}{{2L}} = \frac{{2\pi }}{T}ω≜Lπ​=2L2π​=T2π​ ，这个ω\omegaω本质上就是角频率。

再来看积分，因为−L-L−L ~ LLL是一个周期， 也000 ~ 2L2L2L是一个周期，因此 ∫−LLdt→∫02Ldt→∫0Tdt\int_{ - L}^L {dt} \to \int_0^{2L} {dt} \to \int_0^T {dt}∫−LL​dt→∫02L​dt→∫0T​dt，把上述的符号带入到周期为T=2LT = 2LT=2L的傅里叶级数展开公式当中，就可以得到。

f(t)=a02+∑n=1∞ancos⁡nωt+∑n=1∞bnsin⁡nωtf\left( t \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos n\omega t} + \sum\limits_{n = 1}^\infty {{b_n}\sin n\omega t}f(t)=2a0​​+n=1∑∞​an​cosnωt+n=1∑∞​bn​sinnωt

a0=2T∫0Tf(t)dtan=2T∫0Tf(t)cos⁡nωtdtbn=2T∫0Tf(t)sin⁡nωtdt\begin{array}{l} {a_0} = \frac{2}{T}\int_0^T {f\left( t \right)dt} \\\\ {a_n} = \frac{2}{T}\int_0^T {f\left( t \right)\cos n\omega tdt} \\\\ {b_n} = \frac{2}{T}\int_0^T {f\left( t \right)\sin n\omega tdt} \end{array}a0​=T2​∫0T​f(t)dtan​=T2​∫0T​f(t)cosnωtdtbn​=T2​∫0T​f(t)sinnωtdt​

伏笔： 考虑，当T→∞T \to \inftyT→∞时，f(t)f\left( t \right)f(t) 不再为周期函数，那时候 f(t)f\left( t \right)f(t)该给如何展开呢？ 这就是傅里叶变换啦。

2、例子

把下图这个函数，傅里叶展开一下

T=20,ω=2πT=2π20=110πT = 20,\omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{20}} = \frac{1}{{10}}\piT=20,ω=T2π​=202π​=101​π

a0=2T∫0Tf(t)dt=2T∫0T27dt+2T∫T2T3dt=7+3=10{a_0} = \frac{2}{T}\int_0^T {f\left( t \right)dt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7dt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3dt} = 7 + 3 = 10a0​=T2​∫0T​f(t)dt=T2​∫02T​​7dt+T2​∫2T​T​3dt=7+3=10

an=2T∫0Tf(t)cos⁡nωtdt=2T∫0T27cos⁡nπ10tdt+2T∫T2T3cos⁡nπ10dt=110⋅(70nπsin⁡nπ10t∣010)+110⋅(30nπsin⁡nπ10t∣1020)=110⋅(0+0)=0\begin{aligned} {a_n} &= \frac{2}{T}\int_0^T {f\left( t \right)\cos n\omega tdt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7\cos \frac{{n\pi }}{{10}}tdt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3\cos \frac{{n\pi }}{{10}}dt} \\ & = \frac{1}{{10}} \cdot \left( {\frac{{70}}{{n\pi }}\sin \frac{{n\pi }}{{10}}t\left| {_0^{10}} \right.} \right) + \frac{1}{{10}} \cdot \left( {\frac{{30}}{{n\pi }}\sin \frac{{n\pi }}{{10}}t\left| {_{10}^{20}} \right.} \right) = \frac{1}{{10}} \cdot \left( {0 + 0} \right) = 0 \end{aligned}an​​=T2​∫0T​f(t)cosnωtdt=T2​∫02T​​7cos10nπ​tdt+T2​∫2T​T​3cos10nπ​dt=101​⋅(nπ70​sin10nπ​t∣∣​010​)+101​⋅(nπ30​sin10nπ​t∣∣​1020​)=101​⋅(0+0)=0​

bn=2T∫0Tf(t)sin⁡nωtdt=2T∫0T27sin⁡nπ10tdt+2T∫T2T3sin⁡nπ10dt=110⋅(−70nπcos⁡nπ10t∣010)+110⋅(−30nπcos⁡nπ10t∣1020)=−7nπ(cos⁡nπ−1)−3nπ(cos⁡2nπ−cos⁡nπ)\begin{aligned} {b_n} &= \frac{2}{T}\int_0^T {f\left( t \right)\sin n\omega tdt} = \frac{2}{T}\int_0^{\frac{T}{2}} {7\sin \frac{{n\pi }}{{10}}tdt} + \frac{2}{T}\int_{\frac{T}{2}}^T {3\sin \frac{{n\pi }}{{10}}dt} \\ &= \frac{1}{{10}} \cdot \left( { - \frac{{70}}{{n\pi }}\cos \frac{{n\pi }}{{10}}t\left| {_0^{10}} \right.} \right) + \frac{1}{{10}} \cdot \left( { - \frac{{30}}{{n\pi }}\cos \frac{{n\pi }}{{10}}t\left| {_{10}^{20}} \right.} \right)\\ & = - \frac{7}{{n\pi }}\left( {\cos n\pi - 1} \right) - \frac{3}{{n\pi }}\left( {\cos 2n\pi - \cos n\pi } \right) \end{aligned}bn​​=T2​∫0T​f(t)sinnωtdt=T2​∫02T​​7sin10nπ​tdt+T2​∫2T​T​3sin10nπ​dt=101​⋅(−nπ70​cos10nπ​t∣∣​010​)+101​⋅(−nπ30​cos10nπ​t∣∣​1020​)=−nπ7​(cosnπ−1)−nπ3​(cos2nπ−cosnπ)​

当 nnn为偶数时，cos⁡nπ=cos⁡2nπ=1\cos n\pi = \cos 2n\pi = 1cosnπ=cos2nπ=1

bn=−7nπ(1−1)+−3nπ(1−1)=0{b_n} = - \frac{7}{{n\pi }}\left( {1 - 1} \right) + - \frac{3}{{n\pi }}\left( {1 - 1} \right) = 0bn​=−nπ7​(1−1)+−nπ3​(1−1)=0

当 nnn为奇数时，cos⁡nπ=−1,cos⁡2nπ=1\cos n\pi = - 1,\cos 2n\pi = 1cosnπ=−1,cos2nπ=1

bn=−7nπ(−1−1)−3nπ[1−(−1)]=8nπ{b_n} = - \frac{7}{{n\pi }}\left( { - 1 - 1} \right) - \frac{3}{{n\pi }}\left[ {1 - \left( { - 1} \right)} \right] = \frac{8}{{n\pi }}bn​=−nπ7​(−1−1)−nπ3​[1−(−1)]=nπ8​

所以

f(t)=102+∑n=1∞0⋅cos⁡π10nt+∑n=1∞bnsin⁡π10nt=5+∑n=1∞8nπ⋅sin⁡nπ10t,n=1,3,5,7,⋯\begin{aligned} f\left( t \right) &= \frac{{10}}{2} + \sum\limits_{n = 1}^\infty {0 \cdot \cos \frac{\pi }{{10}}nt} + \sum\limits_{n = 1}^\infty {{b_n}\sin \frac{\pi }{{10}}nt} \\ &= 5 + \sum\limits_{n = 1}^\infty {\frac{8}{{n\pi }} \cdot \sin \frac{{n\pi }}{{10}}} t,n = 1,3,5,7, \cdots \end{aligned}f(t)​=210​+n=1∑∞​0⋅cos10π​nt+n=1∑∞​bn​sin10π​nt=5+n=1∑∞​nπ8​⋅sin10nπ​t,n=1,3,5,7,⋯​

下面给出MatLAB仿真结果

MatLAB代码如下:

close all; clear all; clc;dt = 0.1;t = 0:dt:40;ft = zeros(length(t),1);for i=1:length(t)if (t(i) >= 0 && t(i) <= 10) || (t(i) >= 20 && t(i) <= 30)ft(i) = 7;elseft(i) = 3;endendfigure(1);plot(t,ft,'r-','LineWidth',2);grid on;hold on;axis([0,40,0,10]);FourierSerier0 = 5 * ones(length(t),1);% n = 1FourierSerier1 = (8/(1*pi)) * sin((1*pi/10)*t);% n = 3FourierSerier3 = (8/(3*pi)) * sin((3*pi/10)*t);% n = 5FourierSerier5 = (8/(5*pi)) * sin((5*pi/10)*t);% n = 7FourierSerier7 = (8/(7*pi)) * sin((7*pi/10)*t);% n = 9FourierSerier9 = (8/(9*pi)) * sin((9*pi/10)*t);% n = 11FourierSerier11 = (8/(11*pi)) * sin((11*pi/10)*t);Fs5 = FourierSerier0 + FourierSerier1 + FourierSerier3 + FourierSerier5;Fs5 = Fs5';Fs11 = FourierSerier0 + FourierSerier1 + FourierSerier3 + FourierSerier5 + FourierSerier7 + FourierSerier9 + FourierSerier11;Fs11 = Fs11';plot(t,Fs5,'b-','LineWidth',1.5); hold on;plot(t,Fs11,'k-','LineWidth',2.5);hold on;legend('f(t)','f(t)傅里叶级数（n取到5）',' f(t)傅里叶级数（n取到11）');

系列学习链接：欢迎大家点赞、收藏、留言讨论。

傅里叶级数与傅里叶变换_Part0_欧拉公式证明+三角函数和差公式证明

傅里叶级数与傅里叶变换_Part1_三角函数系的正交性

傅里叶级数与傅里叶变换_Part2_周期为2Π的函数展开为傅里叶级数

傅里叶级数与傅里叶变换_Part3_周期为2L的函数展开为傅里叶级数

傅里叶级数与傅里叶变换_Part4_傅里叶级数的复数形式

傅里叶级数与傅里叶变换_Part5_傅里叶级数推导傅里叶变换

傅里叶级数与傅里叶变换_Part6_离散傅里叶变换推导

傅里叶级数与傅里叶变换_Part7_离散傅里叶变换的性质