根据函数的奇偶性可推断出函数的傅里叶级数的那些值?
奇函数 an=0 bn为积分2倍
偶函数 an为积分2倍 bn=0
因为an是乘以cos(偶函数) bn是乘以sin(奇函数)
奇函数x偶函数=奇函数 积分为0
偶函数x偶函数=偶函数 积分为2倍
这里的积分2倍是上面意思呢?
比如一个函数是这样
f ( t ) = ∫ − T T f ( t ) d t f(t)=\int_{-T}^{T}f(t)dt\\ f(t)=∫−TTf(t)dt
那么就可以写成
f ( t ) = 2 ∫ 0 T f ( t ) d t f(t)=2\int_{0}^{T}f(t)dt f(t)=2∫0Tf(t)dt
三角函数形式
a 0 = 2 × 2 T ∫ 0 τ 2 1 d t = 2 τ T a_{0}=2\times\frac{2}{T} \int_{0}^{\frac{\tau}{2}}1d t=\frac{2\tau}{T} a0=2×T2∫02τ1dt=T2τ
a 0 2 = τ T \frac{a_0}{2}=\frac{\tau}{T} 2a0=Tτ
a n = 2 × 2 T ∫ 0 τ 2 f ( t ) c o s ( n w t ) d t = 4 T [ 1 n w s i n ( n w t ) ∣ 0 τ 2 ] = 4 T 1 n w s i n n w τ 2 a_n=2\times\frac{2}{T}\int_{0}^{\frac{\tau}{2}}f(t)cos(nwt)dt\\ ~\\ =\frac{4}{T}[\frac{1}{nw}sin(nwt)|_0^\frac{\tau}{2}]\\ ~\\ =\frac{4}{T}\frac{1}{nw}sin\frac{nw\tau}{2} an=2×T2∫02τf(t)cos(nwt)dt =T4[nw1sin(nwt)∣02τ] =T4nw1sin2nwτ
b n = 0 b_n=0 bn=0
∴ f ( t ) = τ T + ∑ n = 1 ∞ 4 T 1 n w s i n n w τ 2 c o s n w T \therefore f(t)=\frac{\tau}{T}+\sum_{n=1}^{\infty}\frac{4}{T}\frac{1}{nw}sin\frac{nw\tau}{2}cosnwT ∴f(t)=Tτ+n=1∑∞T4nw1sin2nwτcosnwT
指数形式
F n = 1 T ∫ − T 2 T 2 f ( t ) e − i n w t d t = 1 T ∫ − τ 2 τ 2 e − i n w t d t = 1 T e − i n w t − i n w ∣ − τ 2 τ 2 ∵ sin ( θ ) = e i θ − e − i θ 2 i ∴ F n = 2 T s i n ( n w τ 2 ) n w = τ T S a ( n w τ 2 ) , n = 0 , ± 1 , ± 2 , ⋯ F_n=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-inwt}dt\\ ~\\ =\frac{1}{T}\int_{-\frac{\tau}{2}}^{\frac{\tau}{2}}e^{-inwt}dt\\ ~\\ =\frac{1}{T}~\frac{e^{-inwt}}{-inw}|_{-\frac{\tau}{2}}^{\frac{\tau}{2}}\\ ~\\ \because \sin (\theta)=\frac{e^{i \theta}-e^{-i \theta}}{2 i}\\ ~\\ \therefore F_n=\frac{2}{T}\frac{sin(nw\frac{\tau}{2})}{nw}\\ ~\\ =\frac{\tau}{T}Sa(\frac{nw\tau}{2}) ,n=0,\pm1,\pm2,\cdots Fn=T1∫−2T2Tf(t)e−inwtdt =T1∫−2τ2τe−inwtdt =T1 −inwe−inwt∣−2τ2τ ∵sin(θ)=2ieiθ−e−iθ ∴Fn=T2nwsin(nw2τ) =TτSa(2nwτ),n=0,±1,±2,⋯
∴ f ( t ) = ∑ − ∞ ∞ τ T S a ( n w τ 2 ) e i n w t \therefore f(t) =\sum_{-\infty}^{\infty}\frac{\tau}{T}Sa(\frac{nw\tau}{2})e^{inwt} ∴f(t)=−∞∑∞TτSa(2nwτ)einwt
在这之前,我们先要了解单位冲击函数这个概念
{ δ ( t ) = 0 , t ≠ 0 ∫ − ∞ ∞ δ ( t ) d t = 1 \left\{\begin{array}{cc} \delta(t)=0, & t \neq 0 \\ \int_{-\infty}^{\infty} \delta(t) d t=1 \end{array}\right. {δ(t)=0,∫−∞∞δ(t)dt=1t=0
对应图像:
这个函数可以理解为一个矩形,长很小,宽很大,面积为1
它是对强度极大、作用时间极短暂且积分有限的一类理想化数学模型。
f ( t ) δ ( t ) = f ( 0 ) δ ( t ) f(t) \delta(t)=f(0) \delta(t) f(t)δ(t)=f(0)δ(t)
∫ − ∞ ∞ f ( t ) δ ( t ) d t = f ( 0 ) \int_{-\infty}^{\infty} f(t) \delta(t) \mathrm{d} t=f(0) ∫−∞∞f(t)δ(t)dt=f(0)
f ( t ) δ ( t − a ) = f ( a ) δ ( t − a ) f(t) \delta(t-a)=f(\mathrm{a}) \delta(t-a) f(t)δ(t−a)=f(a)δ(t−a)
∫ − ∞ ∞ f ( t ) δ ( t − a ) d t = f ( a ) \int_{-\infty}^{\infty} f(t) \delta(t-a) \mathrm{d} t=f(a) ∫−∞∞f(t)δ(t−a)dt=f(a)
试对下式求其傅里叶变换:
f ( t ) = ( 10 t 2 + 9 t + 2 ) δ ( t ) f(t)=\left(10 t^{2}+9 t+2\right) \delta(t) f(t)=(10t2+9t+2)δ(t)
答 : f ( t ) = 2 δ ( t ) ∵ ∫ − ∞ ∞ δ ( t ) e − i ω t d t = 1 ∴ F n = ∫ − ∞ ∞ 2 δ ( t ) e − i w t d t = 2 答:f(t)=2\delta(t)\\ ~\\ \because \int_{-\infty}^{\infty} \delta(t) \mathrm{e}^{-i \omega t} \mathrm{d} t=1 \\ ~\\ \therefore F_n=\int_{-\infty}^{\infty}2\delta(t)e^{-iwt}dt\\ =2 答:f(t)=2δ(t) ∵∫−∞∞δ(t)e−iωtdt=1 ∴Fn=∫−∞∞2δ(t)e−iwtdt=2