4 傅里叶级数的复数形式

傅里叶级数的复数形式

欧拉公式
结论

f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n c o s n w t + b n s i n n w t ] f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_ncosnwt+b_nsinnwt] f(t)=2a0+∑n=1∞[ancosnwt+bnsinnwt]

a 0 = 2 T ∫ 0 T f ( t ) d t a_0=\frac{2}{T}\int_{0}^T f(t) dt a0=T2∫0Tf(t)dt

a n = 2 T ∫ 0 T f ( t ) c o s n w t d t a_n=\frac{2}{T}\int_{0}^{T}f(t)cosnwtdt an=T2∫0Tf(t)cosnwtdt

b n = 2 T ∫ 0 T f ( t ) s i n n w t d t b_n=\frac{2}{T}\int_{0}^{T}f(t)sinnwtdt bn=T2∫0Tf(t)sinnwtdt

欧拉公式

e i θ = c o s θ + i s i n θ e^{i \theta}=cos \theta + i sin \theta eiθ=cosθ+isinθ

c o s θ = 1 2 ( e i θ + e − i θ ) cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta}) cosθ=21(eiθ+e−iθ)

s i n θ = − 1 2 ( e i θ − e − i θ ) sin\theta=-\frac{1}{2}(e^{i\theta}-e^{-i\theta}) sinθ=−21(eiθ−e−iθ)

f ( t ) = a 0 2 + ∑ n = 1 ∞ [ a n c o s n w t + b n s i n n w t ] = a 0 2 + ∑ n = 1 ∞ [ a n 1 2 ( e i n w t + e − i n w t ) − 1 2 i b n ( e i n w t − e − i n w t ) ] = a 0 2 + ∑ n = 1 ∞ [ a n − i b n 2 e i n w t + a n + i b n 2 e − i n w t ] = a 0 2 + ∑ n = 1 ∞ a n − i b n 2 e i n w t + ∑ n = 1 ∞ a n + i b n 2 e − i n w t = ∑ n = 0 0 a 0 2 e i n w t + ∑ n = 1 ∞ a n − i b n 2 e i n w t + ∑ n = − ∞ − 1 a − n + i b − n 2 e i n w t = ∑ − ∞ ∞ C n e i n w t \begin{aligned} f(t) &=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_ncosnwt+b_nsinnwt] \\ &=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_n \frac{1}{2}(e^{inwt}+e^{-inwt})-\frac{1}{2}ib_n(e^{inwt}-e^{-inwt})] \\ &=\frac{a_0}{2}+\sum_{n=1}^{\infty}[\frac{a_n - ib_n}{2}e^{inwt}+\frac{a_n+ib_n}{2}e^{-inwt}] \\ &=\frac{a_0}{2}+\sum_{n=1}^{\infty}\frac{a_n - ib_n}{2}e^{inwt}+\sum_{n=1}^{\infty}\frac{a_n+ib_n}{2}e^{-inwt} \\ &=\sum_{n=0}^0\frac{a_0}{2}e^{inwt}+\sum_{n=1}^{\infty}\frac{a_n - ib_n}{2}e^{inwt} + \sum_{n=-\infty}^{-1}\frac{a_{-n}+ib_{-n}}{2}e^{inwt} \\ &=\sum_{-\infty}^{\infty}Cne^{inwt} \end{aligned} f(t)=2a0+n=1∑∞[ancosnwt+bnsinnwt]=2a0+n=1∑∞[an21(einwt+e−inwt)−21ibn(einwt−e−inwt)]=2a0+n=1∑∞[2an−ibneinwt+2an+ibne−inwt]=2a0+n=1∑∞2an−ibneinwt+n=1∑∞2an+ibne−inwt=n=0∑02a0einwt+n=1∑∞2an−ibneinwt+n=−∞∑−12a−n+ib−neinwt=−∞∑∞Cneinwt
C n = { a 0 2 , n = 0 a n − i b n 2 , n > 0 a − n + i b − n 2 , n < 0 Cn= \begin{cases} \frac{a_0}{2}, & \text {$n=0$} \\ \frac{a_n - ib_n}{2}, & \text{$n > 0$} \\ \frac{a_{-n}+ib_{-n}}{2}, & \text{$n < 0$} \\ \end{cases} Cn=⎩⎪⎨⎪⎧2a0,2an−ibn,2a−n+ib−n,n=0n>0n<0
当 n = 0 n=0 n=0
C n = a 0 2 = 1 2 ⋅ 2 T ∫ 0 T f ( t ) d t = 1 T ∫ 0 T f ( t ) d t Cn=\frac{a_0}{2}=\frac{1}{2} \cdot \frac{2}{T} \int_{0}^T f(t)dt=\frac{1}{T}\int_0^Tf(t)dt Cn=2a0=21⋅T2∫0Tf(t)dt=T1∫0Tf(t)dt
当 n > 0 n>0 n>0
C n = 1 2 ( 2 T ∫ 0 T f ( t ) c o s n w t d t − i 2 T ∫ 0 T f ( t ) s i n n w t d t ) = 1 T ∫ 0 T f ( t ) ( c o s n w t − i s i n n w t ) d t = 1 T ∫ 0 T f ( t ) e − i n w t d t \begin{aligned} Cn &=\frac{1}{2}(\frac{2}{T}\int_{0}^{T}f(t)cosnwtdt - i\frac{2}{T}\int_{0}^{T}f(t)sinnwtdt) \\ &=\frac{1}{T}\int_{0}^{T}f(t)(cosnwt-isinnwt)dt \\ &=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt \end{aligned} Cn=21(T2∫0Tf(t)cosnwtdt−iT2∫0Tf(t)sinnwtdt)=T1∫0Tf(t)(cosnwt−isinnwt)dt=T1∫0Tf(t)e−inwtdt
c o s n w t − i s i n n w t = c o s ( − n w t ) + i s i n ( − n w t ) = e − i n w t cosnwt-isinnwt=cos(-nwt)+isin(-nwt)=e^{-inwt} cosnwt−isinnwt=cos(−nwt)+isin(−nwt)=e−inwt

当 n < 0 n<0 n<0
C n = 1 2 ( 2 T ∫ 0 T f ( t ) c o s ( − n ) w t d t + i 2 T ∫ 0 T f ( t ) s i n ( − n ) w t d t ) = 1 T ∫ 0 T f ( t ) ( c o s n w t − i s i n n w t ) d t = 1 T ∫ 0 T f ( t ) e − i n w t d t \begin{aligned} Cn &=\frac{1}{2}(\frac{2}{T}\int_{0}^{T}f(t)cos(-n)wtdt + i\frac{2}{T}\int_{0}^{T}f(t)sin(-n)wtdt) \\ &=\frac{1}{T}\int_{0}^{T}f(t)(cosnwt-isinnwt)dt \\ &=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt \end{aligned} Cn=21(T2∫0Tf(t)cos(−n)wtdt+iT2∫0Tf(t)sin(−n)wtdt)=T1∫0Tf(t)(cosnwt−isinnwt)dt=T1∫0Tf(t)e−inwtdt

当 n = 0 n=0 n=0
1 T ∫ 0 T f ( t ) e − i n w t d t = 1 T ∫ 0 T f ( t ) d t \frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt=\frac{1}{T}\int_0^Tf(t)dt T1∫0Tf(t)e−inwtdt=T1∫0Tf(t)dt

结论

f ( t ) = f ( t + T ) f(t) = f(t+T) f(t)=f(t+T)

f ( t ) = ∑ − ∞ ∞ C n e i n w t f(t) = \sum_{-\infty}^{\infty}Cne^{inwt} f(t)=∑−∞∞Cneinwt

C n = 1 T ∫ 0 T f ( t ) e − i n w t d t Cn=\frac{1}{T}\int_{0}^{T}f(t)e^{-inwt}dt Cn=T1∫0Tf(t)e−inwtdt

原视频:
https://www.bilibili.com/video/av35047004/?spm_id_from=333.788.videocard.0