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_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 = 2 π T = 2\pi T=2π周期函数,即 f ( x ) = f ( x + 2 π ) 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 ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ n x + ∑ n = 1 ∞ b n sin ⁡ n x f\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=1ancosnx+n=1bnsinnx
其中,
a 0 = 1 π ∫ − π π f ( x ) d x a n = 1 π ∫ − π π f ( x ) cos ⁡ n x d x b n = 1 π ∫ − π π f ( x ) sin ⁡ n x d x \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=π1ππf(x)dxan=π1ππf(x)cosnxdxbn=π1ππf(x)sinnxdx

1、周期为2L的函数展开为傅里叶级数

对于, f ( t ) = f ( t + 2 L ) f\left( t \right) = f\left( {t + 2L} \right) f(t)=f(t+2L) ,周期为 T = 2 L T = 2L T=2L的函数
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_第1张图片
如果想用运用Part2掌握的以 T = 2 π T = 2\pi T=2π为周期的 f ( t ) = f ( t + 2 π ) 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。

x = π L t ⇒ t = L π x x = \frac{\pi }{L}t \Rightarrow t = \frac{L}{\pi }x x=Lπtt=πLx

t t t x x x
2 L 2L 2L x = π L t = π L ⋅ 2 L = 2 π x = \frac{\pi }{L}t = \frac{\pi }{L} \cdot 2L = 2\pi x=Lπt=Lπ2L=2π
4 L 4L 4L 4 π 4\pi 4π
0 0 0 0 0 0

f ( t ) = f ( π L x ) ≜ g ( x ) f\left( t \right) = f\left( {\frac{\pi }{L}x} \right) \triangleq g\left( x \right) f(t)=f(Lπx)g(x)
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_第2张图片
通过换元,我们就把周期为 T = 2 L T = 2L T=2L的函数, f ( t ) = f ( t + 2 L ) f\left( t \right) = f\left( {t + 2L} \right) f(t)=f(t+2L) 变成了周期为 T = 2 π T = 2\pi T=2π的函数 g ( t ) = g ( t + 2 π ) 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 ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ n x + ∑ n = 1 ∞ b n sin ⁡ n x g\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=1ancosnx+n=1bnsinnx
其中,
a 0 = 1 π ∫ − π π g ( x ) d x a n = 1 π ∫ − π π g ( x ) cos ⁡ n x d x b n = 1 π ∫ − π π g ( x ) sin ⁡ n x d x \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=π1ππg(x)dxan=π1ππg(x)cosnxdxbn=π1ππg(x)sinnxdx

现在要做的,就是把 x = π L t x = \frac{\pi }{L}t x=Lπt带进去

cos ⁡ n x = cos ⁡ n π L t , sin ⁡ n x = sin ⁡ n π L t \cos nx = \cos \frac{{n\pi }}{L}t,\sin nx = \sin \frac{{n\pi }}{L}t cosnx=cosLt,sinnx=sinLt

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

x x x t t t
− π -\pi π t = L π x = L π ⋅ − π = − L t = \frac{L}{\pi }x = \frac{L}{\pi } \cdot - \pi = - L t=πLx=πLπ=L
π \pi π L L L

∫ − π π d x = ∫ − L L d π L t = π L ∫ − L L d t \int_{ - \pi }^\pi {dx} = \int_{ - L}^L {d\frac{\pi }{L}t} = \frac{\pi }{L}\int_{ - L}^L {dt} ππdx=LLdLπt=LπLLdt

1 π ∫ − π π d x = 1 π ⋅ ( π L ∫ − L L d t ) = 1 L ∫ − L L d t \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} π1ππdx=π1(LπLLdt)=L1LLdt

讲上述算出的式子带入到 g ( x ) g\left( x\right) g(x)中。

f ( t ) = g ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ n π L t + ∑ n = 1 ∞ b n sin ⁡ n π L t f\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=1ancosLtLt+n=1bnsinLtLt
其中
a 0 = 1 L ∫ − L L f ( t ) d t a n = 1 L ∫ − L L f ( t ) cos ⁡ n π L t d t b n = 1 L ∫ − L L f ( t ) sin ⁡ n π L t d t \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=L1L1LLf(t)dtan=L1L1LLf(t)cosLtdtLtdtbn=L1L1LLf(t)sinLtdtLtdt

在工程当中,由于时间是 t ≥ 0 t \ge 0 t0的,所以 t t t是从 0 0 0开始的,假设周期为 T = 2 L T=2L T=2L ω ≜ π L = 2 π 2 L = 2 π T \omega \triangleq \frac{\pi }{L} = \frac{{2\pi }}{{2L}} = \frac{{2\pi }}{T} ωLπ=2L2π=T2π ,这个 ω \omega ω本质上就是角频率。

再来看积分,因为 − L -L L ~ L L L是一个周期, 也 0 0 0 ~ 2 L 2L 2L是一个周期,因此 ∫ − L L d t → ∫ 0 2 L d t → ∫ 0 T d t \int_{ - L}^L {dt} \to \int_0^{2L} {dt} \to \int_0^T {dt} LLdt02Ldt0Tdt,把上述的符号带入到周期为 T = 2 L T = 2L T=2L的傅里叶级数展开公式当中,就可以得到。

f ( t ) = a 0 2 + ∑ n = 1 ∞ a n cos ⁡ n ω t + ∑ n = 1 ∞ b n sin ⁡ n ω t f\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=1ancost+n=1bnsint

a 0 = 2 T ∫ 0 T f ( t ) d t a n = 2 T ∫ 0 T f ( t ) cos ⁡ n ω t d t b n = 2 T ∫ 0 T f ( t ) sin ⁡ n ω t d t \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=T20Tf(t)dtan=T20Tf(t)costdtbn=T20Tf(t)sintdt

伏笔: 考虑,当 T → ∞ T \to \infty T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
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_第3张图片
T = 20 , ω = 2 π T = 2 π 20 = 1 10 π T = 20,\omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{20}} = \frac{1}{{10}}\pi T=20,ω=T2π=202π=101π

a 0 = 2 T ∫ 0 T f ( t ) d t = 2 T ∫ 0 T 2 7 d t + 2 T ∫ T 2 T 3 d t = 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 = 10 a0=T20Tf(t)dt=T202T7dt+T22TT3dt=7+3=10

a n = 2 T ∫ 0 T f ( t ) cos ⁡ n ω t d t = 2 T ∫ 0 T 2 7 cos ⁡ n π 10 t d t + 2 T ∫ T 2 T 3 cos ⁡ n π 10 d t = 1 10 ⋅ ( 70 n π sin ⁡ n π 10 t ∣ 0 10 ) + 1 10 ⋅ ( 30 n π sin ⁡ n π 10 t ∣ 10 20 ) = 1 10 ⋅ ( 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=T20Tf(t)costdt=T202T7cos10tdt+T22TT3cos10dt=101(70sin10t 010)+101(30sin10t 1020)=101(0+0)=0

b n = 2 T ∫ 0 T f ( t ) sin ⁡ n ω t d t = 2 T ∫ 0 T 2 7 sin ⁡ n π 10 t d t + 2 T ∫ T 2 T 3 sin ⁡ n π 10 d t = 1 10 ⋅ ( − 70 n π cos ⁡ n π 10 t ∣ 0 10 ) + 1 10 ⋅ ( − 30 n π cos ⁡ n π 10 t ∣ 10 20 ) = − 7 n π ( cos ⁡ n π − 1 ) − 3 n π ( cos ⁡ 2 n π − cos ⁡ n π ) \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=T20Tf(t)sintdt=T202T7sin10tdt+T22TT3sin10dt=101(70cos10t 010)+101(30cos10t 1020)=7(cos1)3(cos2cos)

n n n为偶数时, cos ⁡ n π = cos ⁡ 2 n π = 1 \cos n\pi = \cos 2n\pi = 1 cos=cos2=1

b n = − 7 n π ( 1 − 1 ) + − 3 n π ( 1 − 1 ) = 0 {b_n} = - \frac{7}{{n\pi }}\left( {1 - 1} \right) + - \frac{3}{{n\pi }}\left( {1 - 1} \right) = 0 bn=7(11)+3(11)=0

n n n为奇数时, cos ⁡ n π = − 1 , cos ⁡ 2 n π = 1 \cos n\pi = - 1,\cos 2n\pi = 1 cos=1,cos2=1

b n = − 7 n π ( − 1 − 1 ) − 3 n π [ 1 − ( − 1 ) ] = 8 n π {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=7(11)3[1(1)]=8

所以
f ( t ) = 10 2 + ∑ n = 1 ∞ 0 ⋅ cos ⁡ π 10 n t + ∑ n = 1 ∞ b n sin ⁡ π 10 n t = 5 + ∑ n = 1 ∞ 8 n π ⋅ sin ⁡ n π 10 t , 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=10cos10πnt+n=1bnsin10πnt=5+n=18sin10t,n=1,3,5,7,

下面给出MatLAB仿真结果
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_第4张图片
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;
    else
        ft(i) = 3;
    end
end
figure(1);plot(t,ft,'r-','LineWidth',2);grid on;hold on;axis([0,40,0,10]);
 
FourierSerier0 = 5 * ones(length(t),1);
% n = 1
FourierSerier1 = (8/(1*pi)) * sin((1*pi/10)*t);
% n = 3
FourierSerier3 = (8/(3*pi)) * sin((3*pi/10)*t);
% n = 5
FourierSerier5 = (8/(5*pi)) * sin((5*pi/10)*t);
% n = 7
FourierSerier7 = (8/(7*pi)) * sin((7*pi/10)*t);
% n = 9
FourierSerier9 = (8/(9*pi)) * sin((9*pi/10)*t);
 
% n = 11
FourierSerier11 = (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)');

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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

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