傅里叶变换公式推导

作者是个没有系统学习过高等数学的高中OIer,推导过程可能不严谨,有问题的地方望大家纠正一下qwq
因为作者之前没学过有关知识,是边看文章边推边写出来的,所以推导思路根原文章基本一致,原文章在此:【Math】复数表示个傅里叶变换


傅里叶级数公式:

f ( t ) = A 0 + ∑ n = 1 ∞ [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] , { A 0 = 1 2 π ∫ − π π f ( t ) d t a n = 1 2 π ∫ − π π c o s ( n ω t ) f ( t ) d t b n = 1 2 π ∫ − π π s i n ( n ω t ) f ( t ) d t f(t) = A_0 + \sum_{n=1}^{\infty}\Big[ a_ncos(n\omega t) + b_nsin(n\omega t) \Big] , \begin{cases} A_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(t)dt \\ a_n = \frac{1}{2\pi} \int_{-\pi}^{\pi}cos(n\omega t)f(t)dt \\ b_n = \frac{1}{2\pi} \int_{-\pi}^{\pi}sin(n\omega t)f(t)dt \end{cases} f(t)=A0​+n=1∑∞​[an​cos(nωt)+bn​sin(nωt)],⎩⎪⎨⎪⎧​A0​=2π1​∫−ππ​f(t)dtan​=2π1​∫−ππ​cos(nωt)f(t)dtbn​=2π1​∫−ππ​sin(nωt)f(t)dt​
  推导:首先我们把一个周期函数用一堆正弦函数来表达:
f ( t ) = A 0 + ∑ n = 1 ∞ s i n ( n ω t + φ n ) = A 0 + ∑ n = 1 ∞ [ A n s i n ( n ω t ) c o s φ n + A n c o s ( n ω t ) s i n φ n ] 和 差 角 公 式 = A 0 + ∑ n = 1 ∞ [ A n s i n φ n c o s ( n ω t ) + A n c o s φ n s i n ( n ω t ) ] 令    a n = A n s i n φ n , b n = A n c o s φ n 得 : f ( t ) = A 0 + ∑ n = 1 ∞ [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] 两 边 同 时 取 积 分 得 到 : ∫ − π π f ( t ) d t = ∫ − π π A 0 d t + ∫ − π π ∑ n = 1 ∞ [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] d t \begin{aligned} f(t) = &A_0 + \sum_{n=1}^{\infty} sin(n\omega t + \varphi_n) \\ = & A_0 + \sum_{n=1}^{\infty} \Big[A_nsin(n\omega t)cos \varphi_n + A_n cos(n\omega t)sin\varphi_n\Big] \qquad 和差角公式\\ = & A_0 + \sum_{n=1}^{\infty} \Big[A_n sin\varphi_ncos(n\omega t) + A_n cos \varphi_nsin(n\omega t)\Big] \\ 令 \; a_n = &A_nsin\varphi_n ,b_n = A_ncos\varphi_n 得:\\ f(t) = &A_0 + \sum_{n = 1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)] \\ 两边同时取&积分得到: \\ \int_{-\pi}^{\pi}f(t)dt = & \int_{-\pi}^{\pi}A_0dt + \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt \end{aligned} f(t)===令an​=f(t)=两边同时取∫−ππ​f(t)dt=​A0​+n=1∑∞​sin(nωt+φn​)A0​+n=1∑∞​[An​sin(nωt)cosφn​+An​cos(nωt)sinφn​]和差角公式A0​+n=1∑∞​[An​sinφn​cos(nωt)+An​cosφn​sin(nωt)]An​sinφn​,bn​=An​cosφn​得:A0​+n=1∑∞​[an​cos(nωt)+bn​sin(nωt)]积分得到:∫−ππ​A0​dt+∫−ππ​n=1∑∞​[an​cos(nωt)+bn​sin(nωt)]dt​
  我们先把这一项算出来:
∫ − π π ∑ n = 1 ∞ [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] d t \begin{aligned} \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt \end{aligned} ∫−ππ​n=1∑∞​[an​cos(nωt)+bn​sin(nωt)]dt​
  要计算这一项,只需要知道求和里面的东西等于几,也就是:
∫ − π π [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] d t = a n n ω s i n ( n ω t ) − b n n ω c o s ( n ω t ) ∣ − π π = [ a n n ω s i n ( n ω π ) − b n n ω c o s ( n ω π ) ] − [ a n n ω s i n ( − n ω π ) − b n n ω c o s ( − n ω π ) ] = a n n ω s i n ( n ω π ) − a n n ω s i n ( − n ω π ) + b n n ω c o s ( − n ω π ) − b n n ω c o s ( n ω π ) = a n n ω [ s i n ( n ω π ) − s i n ( − n ω π ) ] + b n n ω [ c o s ( − n ω π ) − c o s ( n ω π ) ] \begin{aligned} &\int_{-\pi}^{\pi}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt \\ = & \frac{a_n}{n\omega}sin(n\omega t) - \frac{b_n}{n\omega}cos(n\omega t) \bigg|_{-\pi}^{\pi} \\ = & \Big[ \frac{a_n}{n\omega}sin(n\omega \pi) - \frac{b_n}{n\omega}cos(n\omega \pi) \Big] - \Big[ \frac{a_n}{n\omega}sin(-n\omega \pi) - \frac{b_n}{n\omega}cos(-n\omega \pi) \Big] \\ = & \frac{a_n}{n\omega}sin(n\omega \pi) - \frac{a_n}{n\omega}sin(-n\omega \pi) + \frac{b_n}{n\omega}cos(-n\omega \pi) - \frac{b_n}{n\omega}cos(n\omega \pi) \\ = & \frac{a_n}{n\omega}[sin(n\omega \pi) - sin(-n\omega\pi)] + \frac{b_n}{n\omega}[cos(-n\omega \pi) - cos(n\omega\pi)] \end{aligned} ====​∫−ππ​[an​cos(nωt)+bn​sin(nωt)]dtnωan​​sin(nωt)−nωbn​​cos(nωt)∣∣∣∣​−ππ​[nωan​​sin(nωπ)−nωbn​​cos(nωπ)]−[nωan​​sin(−nωπ)−nωbn​​cos(−nωπ)]nωan​​sin(nωπ)−nωan​​sin(−nωπ)+nωbn​​cos(−nωπ)−nωbn​​cos(nωπ)nωan​​[sin(nωπ)−sin(−nωπ)]+nωbn​​[cos(−nωπ)−cos(nωπ)]​
  因为 s i n k π = 0 sink\pi = 0 sinkπ=0 且 c o s x = c o x ( − x ) cosx = cox(-x) cosx=cox(−x),所以上面那一项总体就是 0。即:
∫ − π π [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] d t = 0 \int_{-\pi}^{\pi}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt = 0 ∫−ππ​[an​cos(nωt)+bn​sin(nωt)]dt=0
  所以原式可以化为:
∫ − π π f ( t ) d t = ∫ − π π A 0 d t + ∫ − π π ∑ n = 1 ∞ [ a n c o s ( n ω t ) + b n s i n ( n ω t ) ] d t = A 0 t ∣ − π π + 0 = [ π − ( − π ) ] A 0 = 2 π A 0 ∴ A 0 = 1 2 π ∫ − π π f ( t ) d t \begin{aligned} \int_{-\pi}^{\pi}f(t)dt = & \int_{-\pi}^{\pi}A_0dt + \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt \\ = & A_0t \bigg|_{-\pi}^{\pi} + 0 = [\pi - (-\pi)]A_0 = 2\pi A_0 \\ \therefore A_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}&f(t)dt \end{aligned} ∫−ππ​f(t)dt==∴A0​=2π1​∫−ππ​​∫−ππ​A0​dt+∫−ππ​n=1∑∞​[an​cos(nωt)+bn​sin(nωt)]dtA0​t∣∣∣∣​−ππ​+0=[π−(−π)]A0​=2πA0​f(t)dt​
  然后我们来求 a n a_n an​ 和 b n b_n bn​。等式两边同时乘上 c o s ( k ω t ) cos(k\omega t) cos(kωt)得到了:
f ( t ) c o s ( k ω t ) = A 0 c o s ( k ω t ) + ∑ n = 1 ∞ [ a n c o s ( n ω t ) c o s ( k ω t ) + b n s i n ( n ω t ) c o s ( k ω t ) ] \begin{aligned} f(t)cos(k\omega t) = &A_0cos(k\omega t) + \sum_{n = 1}^{\infty}[a_ncos(n\omega t)cos(k\omega t) + b_nsin(n\omega t)cos(k\omega t)]\\ \end{aligned} f(t)cos(kωt)=​A0​cos(kωt)+n=1∑∞​[an​cos(nωt)cos(kωt)+bn​sin(nωt)cos(kωt)]​
  两边同时积分得到了:
∫ − π π f ( t ) c o s ( k ω t ) d t = A 0 ∫ − π π c o s ( k ω t ) d t + ∑ n = 1 ∞ [ a n ∫ − π π c o s ( n ω t ) c o s ( k ω t ) d t + b n ∫ − π π s i n ( n ω t ) c o s ( k ω t ) d t ] \begin{aligned} \int_{-\pi}^{\pi} f(t)cos(k\omega t)dt = &A_0\int_{-\pi}^{\pi}cos(k\omega t)dt +\\ \sum_{n = 1}^{\infty}[a_n\int_{-\pi}^{\pi}&cos(n\omega t)cos(k\omega t)dt + b_n\int_{-\pi}^{\pi}sin(n\omega t)cos(k\omega t)dt]\\ \end{aligned} ∫−ππ​f(t)cos(kωt)dt=n=1∑∞​[an​∫−ππ​​A0​∫−ππ​cos(kωt)dt+cos(nωt)cos(kωt)dt+bn​∫−ππ​sin(nωt)cos(kωt)dt]​
  又因为这些项都是 0:
A 0 ∫ − π π c o s ( k ω t ) d t = 0 b n ∫ − π π s i n ( n ω t ) c o s ( k ω t ) d t = 0 \begin{aligned} A_0\int_{-\pi}^{\pi}cos(k\omega t)dt = 0\\ b_n\int_{-\pi}^{\pi}sin(n\omega t)cos(k\omega t)dt = 0 \end{aligned} A0​∫−ππ​cos(kωt)dt=0bn​∫−ππ​sin(nωt)cos(kωt)dt=0​
  所以:

∫ − π π f ( t ) c o s ( k ω t ) d t = a n ∑ n = 1 ∞ ∫ − π π c o s ( n ω t ) c o s ( k ω t ) d t = a n ∫ − π π c o s 2 ( n ω t ) d t = a n 2 ∫ − π π [ 1 + c o s ( 2 n ω t ) ] d t 半 角 公 式 = a n 2 [ ∫ − π π 1 d t + ∫ − π π c o s ( 2 n ω t ) d t ] \begin{aligned} \int_{-\pi}^{\pi} f(t)cos(k\omega t)dt = &a_n\sum_{n = 1}^{\infty}\int_{-\pi}^{\pi}cos(n\omega t)cos(k\omega t)dt \\ = &a_n \int_{-\pi}^{\pi}cos^2(n\omega t)dt \\ = & \frac{a_n}{2} \int_{-\pi}^{\pi} [1 + cos(2n\omega t)]dt \qquad 半角公式\\ = & \frac{a_n}{2} \Big[\int_{-\pi}^{\pi}1dt + \int_{-\pi}^{\pi}cos(2n\omega t)dt \Big] \end{aligned} ∫−ππ​f(t)cos(kωt)dt====​an​n=1∑∞​∫−ππ​cos(nωt)cos(kωt)dtan​∫−ππ​cos2(nωt)dt2an​​∫−ππ​[1+cos(2nωt)]dt半角公式2an​​[∫−ππ​1dt+∫−ππ​cos(2nωt)dt]​
  又因为:
∫ − π π c o s ( 2 n ω t ) d t = 0 \int_{-\pi}^{\pi}cos(2n\omega t)dt = 0 ∫−ππ​cos(2nωt)dt=0
  所以:
∫ − π π f ( t ) c o s ( k ω t ) d t = a n ∑ n = 1 ∞ ∫ − π π c o s ( n ω t ) c o s ( k ω t ) d t = a n 2 ∫ − π π 1 d t = a n 2 × t ∣ − π π = a n 2 [ π − ( − π ) ] = π a n ∴ a n = 1 π ∫ − π π f ( t ) c o s ( n ω t ) d t ( n = k ) \begin{aligned} &\int_{-\pi}^{\pi} f(t)cos(k\omega t)dt = a_n\sum_{n = 1}^{\infty}\int_{-\pi}^{\pi}cos(n\omega t)cos(k\omega t)dt \\ = & \frac{a_n}{2} \int_{-\pi}^{\pi}1dt = \frac{a_n}{2} \times t \bigg|_{-\pi}^{\pi} = \frac{a_n}{2} [\pi - (-\pi)] = \pi a_n\\ \therefore &a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)cos(n\omega t)dt \qquad \qquad \qquad \qquad (n=k) \end{aligned} =∴​∫−ππ​f(t)cos(kωt)dt=an​n=1∑∞​∫−ππ​cos(nωt)cos(kωt)dt2an​​∫−ππ​1dt=2an​​×t∣∣∣∣​−ππ​=2an​​[π−(−π)]=πan​an​=π1​∫−ππ​f(t)cos(nωt)dt(n=k)​
  计算 b n b_n bn​ 的方法和计算 a n a_n an​ 的基本相同,可以得到:
b n = 1 π ∫ − π π f ( t ) s i n ( k ω t ) d t ( n = k ) b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)sin(k\omega t)dt \qquad \qquad \qquad \qquad \quad (n = k) bn​=π1​∫−ππ​f(t)sin(kωt)dt(n=k)

傅里叶指数形式:

f ( t ) = 1 T ∑ n = − ∞ + ∞ ∫ t 0 t o + T e − i n ω t d t    ⋅    e i n ω t f(t) = \frac{1}{T}\sum_{n = -\infty}^{+\infty}\int_{t_0}^{t_o+T}e^{-in\omega t}dt \; \cdot \; e^{in\omega t} f(t)=T1​n=−∞∑+∞​∫t0​to​+T​e−inωtdt⋅einωt
  推导如下:
  由欧拉公式( e i x = c o s x + i s i n x e^{ix} = cosx + isinx eix=cosx+isinx)我们可以推出:
c o s x = e i x + e − i x 2 s i n x = − i ⋅ e i x − e − i x 2 cosx = \frac{e^{ix} + e^{-ix}}{2} \qquad sinx = -i \cdot \frac{e^{ix} - e^{-ix}}{2} cosx=2eix+e−ix​sinx=−i⋅2eix−e−ix​
  将两式带入傅里叶级数得到:
f ( t ) = A 0 + ∑ n = 1 ∞ [ a n e i n ω t + e − i n ω t 2 + b n ⋅ ( − i ) e i n ω t − e − i n ω t 2 ] = A 0 + ∑ n = 1 ∞ [ a n − i b n 2 e i n ω t + a n + i b n 2 e − i n ω t ] \begin{aligned} f(t) = &A_0 + \sum_{n = 1}^{\infty}[a_n\frac{e^{in\omega t} + e^{-in\omega t}}{2} + b_n \cdot (-i)\frac{e^{in\omega t} - e^{-in\omega t}}{2} ]\\ = &A_0 + \sum_{n = 1}^{\infty}[\frac{a_n-ib_n}{2}e^{in\omega t } + \frac{a_n + ib_n}{2}e^{-in\omega t}]\\ \end{aligned} f(t)==​A0​+n=1∑∞​[an​2einωt+e−inωt​+bn​⋅(−i)2einωt−e−inωt​]A0​+n=1∑∞​[2an​−ibn​​einωt+2an​+ibn​​e−inωt]​
  又因为:
a n = 1 π ∫ − π π f ( t ) c o s ( n ω t ) d t ( n = k ) b n = 1 π ∫ − π π f ( t ) s i n ( k ω t ) d t ( n = k ) \begin{aligned} a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)cos(n\omega t)dt \qquad \qquad \qquad \qquad (n=k)\\ b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)sin(k\omega t)dt \qquad \qquad \qquad \qquad \quad (n = k) \end{aligned} an​=π1​∫−ππ​f(t)cos(nωt)dt(n=k)bn​=π1​∫−ππ​f(t)sin(kωt)dt(n=k)​
  所以:
a n − i b n 2 = 1 2 ⋅ [ 1 π ∫ − π π f ( t ) c o s ( n ω t ) d t − i ⋅ 1 π ∫ − π π f ( t ) s i n ( k ω t ) d t ] = 1 2 ⋅ 2 T [ ∫ t 0 t 0 + T c o s ( n ω t ) f ( t ) d t − i ∫ t 0 t 0 + T s i n ( n ω t ) f ( t ) d t ] = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) [ c o s ( n ω t ) − i s i n ( n ω t ) ] d t = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) [ e i n ω t + e − i n ω t 2 − i ⋅ ( − i ) e i n ω t − e − i n ω t 2 ] d t = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) e − i n ω t d t \begin{aligned} \frac{a_n - ib_n}{2} = &\frac{1}{2} \cdot \Bigg[ \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)cos(n\omega t)dt - i \cdot \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)sin(k\omega t)dt \Bigg] \\ = &\frac{1}{2} \cdot \frac{2}{T}\Bigg[ \int_{t_0}^{t_0+T}cos(n\omega t)f(t)dt - i \int_{t_0}^{t_0+T}sin(n\omega t)f(t)dt \Bigg]\\ = & \frac{1}{T} \cdot \int_{t_0}^{t_0+T}f(t)[cos(n\omega t) - isin(n\omega t)]dt \\ = & \frac{1}{T} \cdot \int_{t_0}^{t_0+T}f(t)\Big[\frac{e^{in\omega t} + e^{-in\omega t}}{2} - i \cdot (-i)\frac{e^{in\omega t} - e^{-in\omega t}}{2}\Big]dt \\ = & \frac{1}{T} \cdot \int _{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \end{aligned} 2an​−ibn​​=====​21​⋅[π1​∫−ππ​f(t)cos(nωt)dt−i⋅π1​∫−ππ​f(t)sin(kωt)dt]21​⋅T2​[∫t0​t0​+T​cos(nωt)f(t)dt−i∫t0​t0​+T​sin(nωt)f(t)dt]T1​⋅∫t0​t0​+T​f(t)[cos(nωt)−isin(nωt)]dtT1​⋅∫t0​t0​+T​f(t)[2einωt+e−inωt​−i⋅(−i)2einωt−e−inωt​]dtT1​⋅∫t0​t0​+T​f(t)e−inωtdt​
a n + i b n 2 = 1 2 ⋅ [ 1 π ∫ − π π f ( t ) c o s ( n ω t ) d t + i ⋅ 1 π ∫ − π π f ( t ) s i n ( k ω t ) d t ] = 1 2 ⋅ 2 T [ ∫ t 0 t 0 + T c o s ( n ω t ) f ( t ) d t + i ∫ t 0 t 0 + T s i n ( n ω t ) f ( t ) d t ] = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) [ c o s ( n ω t ) + i s i n ( n ω t ) ] d t = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) [ e i n ω t + e − i n ω t 2 + i ⋅ ( − i ) e i n ω t − e − i n ω t 2 ] d t = 1 T ⋅ ∫ t 0 t 0 + T f ( t ) e i n ω t d t \begin{aligned} \frac{a_n + ib_n}{2} = &\frac{1}{2} \cdot \Bigg[ \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)cos(n\omega t)dt + i \cdot \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)sin(k\omega t)dt \Bigg] \\ = &\frac{1}{2} \cdot \frac{2}{T}\Bigg[ \int_{t_0}^{t_0+T}cos(n\omega t)f(t)dt + i \int_{t_0}^{t_0+T}sin(n\omega t)f(t)dt \Bigg]\\ = & \frac{1}{T} \cdot \int_{t_0}^{t_0+T}f(t)[cos(n\omega t) + isin(n\omega t)]dt \\ = & \frac{1}{T} \cdot \int_{t_0}^{t_0+T}f(t)\Big[\frac{e^{in\omega t} + e^{-in\omega t}}{2} + i \cdot (-i)\frac{e^{in\omega t} - e^{-in\omega t}}{2}\Big]dt \\ = & \frac{1}{T} \cdot \int _{t_0}^{t_0+T}f(t)e^{in\omega t}dt \end{aligned} 2an​+ibn​​=====​21​⋅[π1​∫−ππ​f(t)cos(nωt)dt+i⋅π1​∫−ππ​f(t)sin(kωt)dt]21​⋅T2​[∫t0​t0​+T​cos(nωt)f(t)dt+i∫t0​t0​+T​sin(nωt)f(t)dt]T1​⋅∫t0​t0​+T​f(t)[cos(nωt)+isin(nωt)]dtT1​⋅∫t0​t0​+T​f(t)[2einωt+e−inωt​+i⋅(−i)2einωt−e−inωt​]dtT1​⋅∫t0​t0​+T​f(t)einωtdt​
  然后再带回去:
f ( t ) = A 0 + ∑ n = 1 ∞ [ a n − i b n 2 e i n ω t + a n + i b n 2 e − i n ω t ] = 1 T ∫ t 0 t 0 + T f ( t ) d t + ∑ n = 1 ∞ [ 1 T ⋅ ∫ t 0 t 0 + T f ( t ) e − i n ω t d t ⋅ e i n ω t + 1 T ⋅ ∫ t 0 t 0 + T f ( t ) e i n ω t d t ⋅ e − i n ω t ] = 1 T ∫ t 0 t 0 + T f ( t ) d t + 1 T ∑ n = 1 ∞ ∫ t 0 t 0 + T f ( t ) e − i n ω t d t ⋅ e i n ω t + 1 T ∑ n = − ∞ − 1 ∫ t 0 t 0 + T f ( t ) e − i n ω t d t ⋅ e i n ω t = 1 T ∑ n = − ∞ + ∞ ∫ t 0 t 0 + T f ( t ) e − i n ω t d t ⋅ e i n ω t \begin{aligned} f(t) = &A_0 + \sum_{n = 1}^{\infty}[\frac{a_n-ib_n}{2}e^{in\omega t } + \frac{a_n + ib_n}{2}e^{-in\omega t}]\\ = & \frac{1}{T}\int_{t_0}^{t_0+T}f(t)dt + \sum_{n=1}^{\infty}\Big[ \frac{1}{T} \cdot \int _{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \cdot e^{in\omega t} + \frac{1}{T} \cdot \int _{t_0}^{t_0+T}f(t)e^{in\omega t}dt \cdot e^{-in\omega t} \Big] \\ = &\frac{1}{T}\int_{t_0}^{t_0+T}f(t)dt + \frac{1}{T}\sum_{n=1}^{\infty}\int_{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \cdot e^{in\omega t} + \frac{1}{T}\sum_{n = -\infty}^{-1}\int_{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \cdot e^{in\omega t}\\ = & \frac{1}{T}\sum_{n=-\infty}^{+\infty}\int_{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \cdot e^{in \omega t} \end{aligned} f(t)====​A0​+n=1∑∞​[2an​−ibn​​einωt+2an​+ibn​​e−inωt]T1​∫t0​t0​+T​f(t)dt+n=1∑∞​[T1​⋅∫t0​t0​+T​f(t)e−inωtdt⋅einωt+T1​⋅∫t0​t0​+T​f(t)einωtdt⋅e−inωt]T1​∫t0​t0​+T​f(t)dt+T1​n=1∑∞​∫t0​t0​+T​f(t)e−inωtdt⋅einωt+T1​n=−∞∑−1​∫t0​t0​+T​f(t)e−inωtdt⋅einωtT1​n=−∞∑+∞​∫t0​t0​+T​f(t)e−inωtdt⋅einωt​

傅里叶变换和傅里叶逆变换

  傅里叶变换:
F ( ω x ) = ∫ − ∞ + ∞ f ( t ) s − i ω x t d t \begin{aligned} F(\omega_x) = \int_{-\infty}^{+\infty}f(t)s^{-i\omega_x t}dt \end{aligned} F(ωx​)=∫−∞+∞​f(t)s−iωx​tdt​
  傅里叶逆变换:
f ( t ) = 1 2 π ∫ − ∞ + ∞ F ( ω x ) e i ω x t d ω \begin{aligned} f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega_x)e^{i\omega_x t}d\omega \end{aligned} f(t)=2π1​∫−∞+∞​F(ωx​)eiωx​tdω​
  其中 F ( ω ) F(\omega) F(ω) 叫做 f ( t ) f(t) f(t) 的像函数, f ( t ) f(t) f(t) 叫做 F ( ω ) F(\omega) F(ω) 的像原函数。 F ( ω ) F(\omega) F(ω) 是 f ( t ) f(t) f(t) 的像。 f ( t ) f(t) f(t) 是 F ( ω ) F(\omega) F(ω) 原像。
  推导过程如下:
  令:
ω x = n ω F ( ω x ) = ∫ − ∞ + ∞ f ( t ) s − i ω x t d t \omega_x = n\omega \qquad F(\omega_x) = \int_{-\infty}^{+\infty}f(t)s^{-i\omega_x t}dt ωx​=nωF(ωx​)=∫−∞+∞​f(t)s−iωx​tdt
  又因为:
f ( t ) = 1 T ∑ n = − ∞ + ∞ ∫ t 0 t 0 + T f ( t ) e − i n ω t d t ⋅ e i n ω t f(t) = \frac{1}{T}\sum_{n=-\infty}^{+\infty}\int_{t_0}^{t_0+T}f(t)e^{-in\omega t}dt \cdot e^{in \omega t} f(t)=T1​n=−∞∑+∞​∫t0​t0​+T​f(t)e−inωtdt⋅einωt
  所以:
f ( t ) = 1 T ∑ − ∞ + ∞ ∫ − ∞ + ∞ f ( t ) e − i ω x t d t ⋅ e i ω x t = 1 T ∑ − ∞ + ∞ [ F ( ω x ) e i ω x t ] = 1 2 π ∫ − ∞ + ∞ F ( ω x ) e i ω x t d ω x \begin{aligned} f(t) = &\frac{1}{T}\sum_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(t)e^{-i\omega_xt}dt \cdot e^{i\omega_xt} \\ = &\frac{1}{T} \sum_{-\infty}^{+\infty}[F(\omega_x)e^{i\omega_xt}] = \frac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega_x)e^{i\omega_xt}d\omega_x \end{aligned} f(t)==​T1​−∞∑+∞​∫−∞+∞​f(t)e−iωx​tdt⋅eiωx​tT1​−∞∑+∞​[F(ωx​)eiωx​t]=2π1​∫−∞+∞​F(ωx​)eiωx​tdωx​​

离散傅里叶变换

  离散傅里叶变换:
F ( n ) = ∑ n = 0 N f ( t ) e i 2 π n N t F(n) = \sum_{n = 0}^{N}f(t)e^{i\frac{2\pi n}{N}t} F(n)=n=0∑N​f(t)eiN2πn​t
  逆离散傅里叶变换:
f ( t ) = ∑ n = 0 N 1 N F ( n ) e i 2 π n N t f(t) = \sum_{n=0}^{N}\frac{1}{N}F(n)e^{i\frac{2\pi n}{N}t} f(t)=n=0∑N​N1​F(n)eiN2πn​t

上一篇:从傅里叶级数到傅里叶变换


下一篇:简述linux下的statm