信号与系统-函数的正交分解
函数的正交分解
一个函数族 { 1 , c o s ( ω 0 t ) , c o s ( 2 ω 0 t ) , … … , c o s ( n ω 0 t ) , s i n ( ω 0 t ) , s i n ( 2 ω 0 t ) , … … , s i n ( n ω 0 t ) } \{1,cos(\omega_0t),cos(2\omega_0t),……,cos(n\omega_0t),sin(\omega_0t),sin(2\omega_0t),……,sin(n\omega_0t)\} {1,cos(ω0t),cos(2ω0t),……,cos(nω0t),sin(ω0t),sin(2ω0t),……,sin(nω0t)}任取两个，在一个周期 T 0 = 2 π ω 0 T_0 = \frac{2\pi}{\omega_0} T0=ω02π上积分为0.
内积
若一个运算 < . > <.> <.>满足以下4个性质
- 交换律 $
- 齐次性 $ <\lambda x,y>= \lambda
- 叠加性 $
- 非负性 $
把此运算叫做内积
定义：$
定义：$
正交基函数
若函数族 e 1 , e 2 , … … , e N , … … = { e k } k = 1 ∼ + ∞ {e_1,e_2,……,e_N,……}=\{e_k\}_{k=1 \sim +\infty} e1,e2,……,eN,……={ek}k=1∼+∞，满足
- ∀ k \forall k ∀k有 $
- ∀ k 1 ≠ k 2 \forall k_1 \ne k_2 ∀k1=k2，有$
则称 { e k } k = 1 ∼ + ∞ \{e_k\}_{k=1 \sim +\infty} {ek}k=1∼+∞是正交函数族
其内积定义为 < e k 1 ( t ) , e k 2 ( t ) > = ∫ T 0 e k 1 ( t ) e k 2 ( t ) ‾ d t
标准正交基函数
若函数族 { e k } k = 1 ∼ + ∞ \{e_k\}_{k=1 \sim +\infty} {ek}k=1∼+∞ 满足
- ∀ k \forall k ∀k有 $
- ∀ k 1 ≠ k 2 \forall k_1 \ne k_2 ∀k1=k2，有$
把正交基变为标准正交叫做正交基的标准化
从正交基的角度理解傅里叶级数(复数形式)
{ x ( t ) = ∑ k = − ∞ + ∞ a k e j k ω 0 t a k = 1 T 0 ∫ 0 T 0 x ( t ) e − j k ω 0 t d t \left \{ \begin{array}{c} x(t)= \sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t}\\ a_k=\frac{1}{T_0}\int_{0}^{T_0}x(t)e^{-jk\omega_0t}dt \end{array} \right. {x(t)=∑k=−∞+∞akejkω0tak=T01∫0T0x(t)e−jkω0tdt
证明 < e j k ω 0 t , e j k ω 0 t > < e^{jk\omega_0t}, e^{jk\omega_0t}> <ejkω0t,ejkω0t>为一组正交基函数
< e j k ω 0 t , e j k ω 0 t > = ∫ T 0 e j k ω 0 t e j k ω 0 t ‾ d t = T 0 < e^{jk\omega_0t}, e^{jk\omega_0t} > = \int _{T_0} e^{jk\omega_0t} \overline {e^{jk\omega_0t}} dt = T_0 <ejkω0t,ejkω0t>=∫T0ejkω0tejkω0tdt=T0
若 k 1 ≠ k 2 k_1 \ne k_2 k1=k2，则
< e j k ω 0 t , e j k ω 0 t > = ∫ T 0 e j k 1 ω 0 t e j k 2 ω 0 t ‾ d t = ∫ T 0 e j ( k 1 − k 2 ) ω 0 t d t = 1 j ( k 1 − k 2 ) ω 0 e j ( k 1 − k 2 ) ω 0 t ∣ t = 0 T 0 = 0 \begin{aligned} < e^{jk\omega_0t}, e^{jk\omega_0t} > &= \int _{T_0} e^{jk_1\omega_0t} \overline {e^{jk_2\omega_0t}} dt \\ &= \int _{T_0} e^{j(k_1-k_2)\omega_0t}dt \\ &= \frac{1}{j(k_1-k_2) \omega_0} e^{j(k_1-k_2)\omega_0t} | ^{T_0} _{t=0} = 0\\ \end{aligned} <ejkω0t,ejkω0t>=∫T0ejk1ω0tejk2ω0tdt=∫T0ej(k1−k2)ω0tdt=j(k1−k2)ω01ej(k1−k2)ω0t∣t=0T0=0
则得证 < e j k ω 0 t , e j k ω 0 t > < e^{jk\omega_0t}, e^{jk\omega_0t}> <ejkω0t,ejkω0t>为一组正交基函数
x ( t ) = ∑ k = − ∞ + ∞ a k e j k ω 0 t x(t)= \sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t} x(t)=∑k=−∞+∞akejkω0t
则
< x ( t ) , e j k ω 0 t > = < ∑ u = − ∞ + ∞ a u e j u ω 0 t , e j k ω 0 t > = ∑ u = − ∞ + ∞ a u < e j u ω 0 t , e j k ω 0 t > = T 0 a k \begin{aligned} < x(t),e^{jk\omega_0t} > &= <\sum_{u=-\infty}^{+\infty}a_ue^{ju\omega_0t} , e^{jk\omega_0t}> \\ &= \sum_{u=-\infty}^{+\infty}a_u < e^{ju\omega_0t},e^{jk\omega_0t} > \\ &= T_0 a_k \end{aligned} <x(t),ejkω0t>=<u=−∞∑+∞auejuω0t,ejkω0t>=u=−∞∑+∞au<ejuω0t,ejkω0t>=T0ak
a k = 1 T 0 < x ( t ) , e j k ω 0 t > = 1 T 0 ∫ T 0 x ( t ) e j k ω 0 t ‾ d t = 1 T 0 ∫ T 0 x ( t ) e − j k ω 0 t d t \begin{aligned} a_k &= \frac{1}{T_0} < x(t),e^{jk\omega_0t} > \\ &= \frac{1}{T_0} \int _{T_0} x(t) \overline {e^{jk\omega_0t}} dt\\ &= \frac{1}{T_0} \int _{T_0} x(t) {e^{-jk\omega_0t}}dt \end{aligned} ak=T01<x(t),ejkω0t>=T01∫T0x(t)ejkω0tdt=T01∫T0x(t)e−jkω0tdt
参考文献
- https://www.bilibili.com/video/BV1g94y1Q76G?p=17&vd_source=b3a15f5d8887594220b3104779be9fc3