傅里叶级数的目的、形式与物理意义
傅里叶级数的目的
傅里叶级数的三种形式
标准三角展开式
f ( t ) = a 0 + ∑ n = 1 ∞ ( a n cos ( n ω 0 t ) + b n sin ( n ω 0 t ) ) f(t)=a_0+\sum_{n=1}^{\infty} (a_n\cos(n\omega_0 t)+b_n\sin(n\omega_0 t)) f ( t ) = a 0 + n = 1 ∑ ∞ ( a n cos ( n ω 0 t ) + b n sin ( n ω 0 t )) 其中
a 0 = 1 T ∫ T f ( t ) d t a n = 2 T ∫ T f ( t ) cos ( n ω 0 t ) d t b n = 2 T ∫ T f ( t ) sin ( n ω 0 t ) d t a_0=\frac{1}{T}\int_T f(t)\,dt\\
a_n=\frac{2}{T}\int_T f(t)\cos(n\omega_0 t)\,dt\\
b_n=\frac{2}{T}\int_T f(t)\sin(n\omega_0 t)\,dt a 0 = T 1 ∫ T f ( t ) d t a n = T 2 ∫ T f ( t ) cos ( n ω 0 t ) d t b n = T 2 ∫ T f ( t ) sin ( n ω 0 t ) d t 合并三角展开式
f ( t ) = A 0 + ∑ n = 1 ∞ A n cos ( n ω 0 t − θ n ) f(t)=A_0+\sum_{n=1}^{\infty} A_n\cos(n\omega_0 t-\theta_n)
f ( t ) = A 0 + n = 1 ∑ ∞ A n cos ( n ω 0 t − θ n ) 其中
A 0 = a 0 = 1 T ∫ T f ( t ) d t A n = a n 2 + b n 2 θ n = tan − 1 b n a n A_0=a_0=\frac{1}{T}\int_T f(t)\,dt\\
A_n=\sqrt{a_n^2+b_n^2}\\
\theta_n=\tan^{-1}\frac{b_n}{a_n} A 0 = a 0 = T 1 ∫ T f ( t ) d t A n = a n 2 + b n 2 θ n = tan − 1 a n b n 指数展开式
由欧拉公式
cos ( n ω 0 t ) = e j n ω 0 t + e − j n ω 0 t 2 \cos(n\omega_0t)=\frac{e^{jn\omega_0t}+e^{-jn\omega_0t}}{2} cos ( n ω 0 t ) = 2 e j n ω 0 t + e − j n ω 0 t sin ( n ω 0 t ) = e j n ω 0 t − e − j n ω 0 t 2 j \sin(n\omega_0t)=\frac{e^{jn\omega_0t}-e^{-jn\omega_0t}}{2j} sin ( n ω 0 t ) = 2 j e j n ω 0 t − e − j n ω 0 t 代入标准三角展开式
f ( t ) = a 0 + ∑ n = 1 ∞ ( a n cos ( n ω 0 t ) + b n sin ( n ω 0 t ) ) f(t)=a_0+\sum_{n=1}^{\infty}(a_n\cos(n\omega_0t)+b_n\sin(n\omega_0t)) f ( t ) = a 0 + n = 1 ∑ ∞ ( a n cos ( n ω 0 t ) + b n sin ( n ω 0 t )) 得
f ( t ) = a 0 + ∑ n = 1 ∞ [ a n e j n ω 0 t + e − j n ω 0 t 2 + b n e j n ω 0 t − e − j n ω 0 t 2 j ] f(t)=a_0+\sum_{n=1}^{\infty}\left[a_n\frac{e^{jn\omega_0t}+e^{-jn\omega_0t}}{2}+b_n\frac{e^{jn\omega_0t}-e^{-jn\omega_0t}}{2j}\right] f ( t ) = a 0 + n = 1 ∑ ∞ [ a n 2 e j n ω 0 t + e − j n ω 0 t + b n 2 j e j n ω 0 t − e − j n ω 0 t ] 整理得
f ( t ) = a 0 + ∑ n = 1 ∞ [ a n − j b n 2 e j n ω 0 t + a n + j b n 2 e − j n ω 0 t ] f(t)=a_0+\sum_{n=1}^{\infty}\left[\frac{a_n-jb_n}{2}e^{jn\omega_0t}+\frac{a_n+jb_n}{2}e^{-jn\omega_0t}\right] f ( t ) = a 0 + n = 1 ∑ ∞ [ 2 a n − j b n e j n ω 0 t + 2 a n + j b n e − j n ω 0 t ] 令
C 0 = a 0 C n = a n − j b n 2 C − n = a n + j b n 2 C_0=a_0 \qquad C_n=\frac{a_n-jb_n}{2} \qquad C_{-n}=\frac{a_n+jb_n}{2} C 0 = a 0 C n = 2 a n − j b n C − n = 2 a n + j b n 于是
f ( t ) = ∑ n = − ∞ ∞ C n e j n ω 0 t f(t)=\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_0t} f ( t ) = n = − ∞ ∑ ∞ C n e j n ω 0 t 其中
C n = 1 T ∫ T f ( t ) e − j n ω 0 t d t C_n=\frac{1}{T}\int_Tf(t)e^{-jn\omega_0t}\,dt C n = T 1 ∫ T f ( t ) e − j n ω 0 t d t 利用周期函数的对称性计算傅里叶级数
偶对称
奇对称
半波对称
四分之一波对称
傅里叶级数在电路分析中的应用
电路分析中的合并三角形式
周期性电源的电路解题
周期性电源的功率计算
傅里叶级数计算平均功率
设电压和电流的傅里叶级数分别为
v ( t ) = V D C + ∑ m = 1 ∞ V m cos ( m ω 0 t − θ V m ) v(t)=V_{DC}+\sum_{m=1}^{\infty}V_m\cos(m\omega_0t-\theta_{Vm}) v ( t ) = V D C + m = 1 ∑ ∞ V m cos ( m ω 0 t − θ V m ) i ( t ) = I D C + ∑ n = 1 ∞ I n cos ( n ω 0 t − θ I n ) i(t)=I_{DC}+\sum_{n=1}^{\infty}I_n\cos(n\omega_0t-\theta_{In}) i ( t ) = I D C + n = 1 ∑ ∞ I n cos ( n ω 0 t − θ I n ) 平均功率定义为
P = 1 T ∫ t 0 t 0 + T v ( t ) i ( t ) d t P=\frac{1}{T}\int_{t_0}^{t_0+T}v(t)i(t)\,dt P = T 1 ∫ t 0 t 0 + T v ( t ) i ( t ) d t 将 v(t) 和 i(t) 代入：
P = 1 T ∫ t 0 t 0 + T [ V D C + ∑ m = 1 ∞ V m cos ( m ω 0 t − θ V m ) ] [ I D C + ∑ n = 1 ∞ I n cos ( n ω 0 t − θ I n ) ] P=\frac{1}{T}\int_{t_0}^{t_0+T}
\left[V_{DC}+\sum_{m=1}^{\infty}V_m\cos(m\omega_0t-\theta_{Vm})\right]\left[I_{DC}+\sum_{n=1}^{\infty}I_n\cos(n\omega_0t-\theta_{In})\right] P = T 1 ∫ t 0 t 0 + T [ V D C + m = 1 ∑ ∞ V m cos ( m ω 0 t − θ V m ) ] [ I D C + n = 1 ∑ ∞ I n cos ( n ω 0 t − θ I n ) ] 展开：
P = 1 T ∫ t 0 t 0 + T [ V D C I D C + ∑ m = 1 ∞ ∑ n = 1 ∞ V m I n cos ( m ω 0 t − θ V m ) cos ( n ω 0 t − θ I n ) + V D C ∑ n = 1 ∞ I n cos ( n ω 0 t − θ I n ) + I D C ∑ n = 1 ∞ V n cos ( n ω 0 t − θ V n ) ] d t \begin{aligned}
P={}&\frac{1}{T}\int_{t_0}^{t_0+T}
\Bigg[
V_{DC}I_{DC}+\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
V_mI_n
\cos(m\omega_0t-\theta_{Vm})
\cos(n\omega_0t-\theta_{In})\\
&+V_{DC}\sum_{n=1}^{\infty}I_n\cos(n\omega_0t-\theta_{In})
+I_{DC}\sum_{n=1}^{\infty}V_n\cos(n\omega_0t-\theta_{Vn})
\Bigg]dt
\end{aligned} P = T 1 ∫ t 0 t 0 + T [ V D C I D C + m = 1 ∑ ∞ n = 1 ∑ ∞ V m I n cos ( m ω 0 t − θ V m ) cos ( n ω 0 t − θ I n ) + V D C n = 1 ∑ ∞ I n cos ( n ω 0 t − θ I n ) + I D C n = 1 ∑ ∞ V n cos ( n ω 0 t − θ V n ) ] d t 由于
三角函数的正交性
在一个完整周期 (T) 内，三角函数满足以下正交关系：
∫ t 0 t 0 + T sin ( m ω 0 t ) d t = 0 for all m \int_{t_0}^{t_0+T}\sin(m\omega_0t)\,dt=0
\qquad\text{for all }m ∫ t 0 t 0 + T sin ( m ω 0 t ) d t = 0 for all m ∫ t 0 t 0 + T cos ( m ω 0 t ) d t = 0 for all m \int_{t_0}^{t_0+T}\cos(m\omega_0t)\,dt=0
\qquad\text{for all }m ∫ t 0 t 0 + T cos ( m ω 0 t ) d t = 0 for all m ∫ t 0 t 0 + T cos ( m ω 0 t ) sin ( n ω 0 t ) d t = 0 for all m , n \int_{t_0}^{t_0+T}
\cos(m\omega_0t)\sin(n\omega_0t)\,dt=0
\qquad\text{for all }m,n ∫ t 0 t 0 + T cos ( m ω 0 t ) sin ( n ω 0 t ) d t = 0 for all m , n ∫ t 0 t 0 + T sin ( m ω 0 t ) sin ( n ω 0 t ) d t = { 0 , m ≠ n T 2 , m = n \int_{t_0}^{t_0+T}
\sin(m\omega_0t)\sin(n\omega_0t)\,dt
=
\begin{cases}
0,&m\ne n\\
\dfrac{T}{2},&m=n
\end{cases} ∫ t 0 t 0 + T sin ( m ω 0 t ) sin ( n ω 0 t ) d t = ⎩ ⎨ ⎧ 0 , 2 T , m = n m = n ∫ t 0 t 0 + T cos ( m ω 0 t ) cos ( n ω 0 t ) d t = { 0 , m ≠ n T 2 , m = n \int_{t_0}^{t_0+T}
\cos(m\omega_0t)\cos(n\omega_0t)\,dt
=
\begin{cases}
0,&m\ne n\\
\dfrac{T}{2},&m=n
\end{cases} ∫ t 0 t 0 + T cos ( m ω 0 t ) cos ( n ω 0 t ) d t = ⎩ ⎨ ⎧ 0 , 2 T , m = n m = n 因此
P = 1 T ∫ t 0 t 0 + T [ V D C I D C + ∑ n = 1 ∞ V n I n cos ( n ω 0 t − θ V n ) cos ( n ω 0 t − θ I n ) ] d t
P=\frac{1}{T}\int_{t_0}^{t_0+T}
\Bigg[
V_{DC}I_{DC}+\sum_{n=1}^{\infty}
V_nI_n
\cos(n\omega_0t-\theta_{Vn})
\cos(n\omega_0t-\theta_{In})
\Bigg]dt P = T 1 ∫ t 0 t 0 + T [ V D C I D C + n = 1 ∑ ∞ V n I n cos ( n ω 0 t − θ V n ) cos ( n ω 0 t − θ I n ) ] d t 根据积化和差公式
P = V D C I D C + 1 T ∑ n = 1 ∞ V n I n 2 ∫ t 0 t 0 + T [ c o s ( θ V n − θ I n ) + c o s ( 2 n ω 0 t − θ V n − θ I n ) ] d t \begin{aligned}
P
={}&V_{DC}I_{DC}
+\frac{1}{T}\sum_{n=1}^{\infty}
\frac{V_n I_n}{2}
\int_{t_0}^{t_0+T}
\Big[
cos(\theta_{Vn}-\theta_{In})+cos(2n\omega_0t-\theta_{Vn}-\theta_{In})
\Big]dt
\end{aligned} P = V D C I D C + T 1 n = 1 ∑ ∞ 2 V n I n ∫ t 0 t 0 + T [ cos ( θ V n − θ I n ) + cos ( 2 n ω 0 t − θ V n − θ I n ) ] d t 其中
∫ t 0 t 0 + T c o s ( 2 n ω 0 t − θ V n − θ I n ) d t = 0 \int_{t_0}^{t_0+T}cos(2n\omega_0t-\theta_{Vn}-\theta_{In})\,dt=0 ∫ t 0 t 0 + T cos ( 2 n ω 0 t − θ V n − θ I n ) d t = 0 因此得到
P = V D C I D C + 1 T ∑ n = 1 ∞ V n I n 2 ∫ t 0 t 0 + T c o s ( θ V n − θ I n ) d t = V D C I D C + ∑ n = 1 ∞ V n I n 2 c o s ( θ V n − θ I n ) \begin{aligned}
P&=V_{DC}I_{DC}+\frac{1}{T}\sum_{n=1}^{\infty}\frac{V_n I_n}{2}\int_{t_0}^{t_0+T}cos(\theta_{Vn}-\theta_{In})\,dt\\
&=V_{DC}I_{DC}+\sum_{n=1}^{\infty}\frac{V_n I_n}{2}cos(\theta_{Vn}-\theta_{In})
\end{aligned} P = V D C I D C + T 1 n = 1 ∑ ∞ 2 V n I n ∫ t 0 t 0 + T cos ( θ V n − θ I n ) d t = V D C I D C + n = 1 ∑ ∞ 2 V n I n cos ( θ V n − θ I n ) P = V D C I D C + 1 2 ∑ n = 1 ∞ V n I n cos ( θ V n − θ I n ) \boxed{
P
=
V_{DC}I_{DC}
+
\frac{1}{2}
\sum_{n=1}^{\infty}
V_nI_n
\cos(\theta_{Vn}-\theta_{In})
} P = V D C I D C + 2 1 n = 1 ∑ ∞ V n I n cos ( θ V n − θ I n ) 这就是含有直流分量和各次谐波的周期信号的平均功率公式。
方均根值与等效直流电计算平均功率