标签：cos frac 推导 变换 傅里叶 int dt pi omega
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：
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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∑∞[ancos(nωt)+bnsin(nωt)]，⎩⎪⎨⎪⎧A0=2π1∫−ππf(t)dtan=2π1∫−ππcos(nωt)f(t)dtbn=2π1∫−ππsin(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：
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\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∫−ππf(t)dt=A0+n=1∑∞sin(nωt+φn)A0+n=1∑∞[Ansin(nωt)cosφn+Ancos(nωt)sinφn]和差角公式A0+n=1∑∞[Ansinφncos(nωt)+Ancosφnsin(nωt)]Ansinφn，bn=Ancosφn得：A0+n=1∑∞[ancos(nωt)+bnsin(nωt)]积分得到：∫−ππA0dt+∫−ππn=1∑∞[ancos(nωt)+bnsin(nωt)]dt
我们先把这一项算出来：
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\begin{aligned} \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt \end{aligned}
∫−ππn=1∑∞[ancos(nωt)+bnsin(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：
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\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[ancos(nωt)+bnsin(nωt)]dtnωansin(nωt)−nωbncos(nωt)∣∣∣∣−ππ[nωansin(nωπ)−nωbncos(nωπ)]−[nωansin(−nωπ)−nωbncos(−nωπ)]nωansin(nωπ)−nωansin(−nωπ)+nωbncos(−nωπ)−nωbncos(nωπ)nωan[sin(nωπ)−sin(−nωπ)]+nωbn[cos(−nωπ)−cos(nωπ)]
因为
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sink\pi = 0
sinkπ=0 且
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cosx = cox(-x)
cosx=cox(−x)，所以上面那一项总体就是 0。即：
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\int_{-\pi}^{\pi}[a_ncos(n\omega t) + b_nsin(n\omega t)]dt = 0
∫−ππ[ancos(nωt)+bnsin(nωt)]dt=0
所以原式可以化为：
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\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}
∫−ππf(t)dt==∴A0=2π1∫−ππ∫−ππA0dt+∫−ππn=1∑∞[ancos(nωt)+bnsin(nωt)]dtA0t∣∣∣∣−ππ+0=[π−(−π)]A0=2πA0f(t)dt
然后我们来求
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cos(k\omega t)
cos(kωt)得到了：
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\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)=A0cos(kωt)+n=1∑∞[ancos(nωt)cos(kωt)+bnsin(nωt)cos(kωt)]
两边同时积分得到了：
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\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}
∫−ππf(t)cos(kωt)dt=n=1∑∞[an∫−ππA0∫−ππcos(kωt)dt+cos(nωt)cos(kωt)dt+bn∫−ππsin(nωt)cos(kωt)dt]
又因为这些项都是 0：
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\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∫−ππcos(kωt)dt=0bn∫−ππsin(nωt)cos(kωt)dt=0
所以：
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\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}
∫−ππf(t)cos(kωt)dt====ann=1∑∞∫−ππcos(nωt)cos(kωt)dtan∫−ππcos2(nωt)dt2an∫−ππ[1+cos(2nωt)]dt半角公式2an[∫−ππ1dt+∫−ππcos(2nωt)dt]
又因为：
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\int_{-\pi}^{\pi}cos(2n\omega t)dt = 0
∫−ππcos(2nωt)dt=0
所以：
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\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}
=∴∫−ππf(t)cos(kωt)dt=ann=1∑∞∫−ππcos(nωt)cos(kωt)dt2an∫−ππ1dt=2an×t∣∣∣∣−ππ=2an[π−(−π)]=πanan=π1∫−ππf(t)cos(nωt)dt(n=k)
计算
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bn 的方法和计算
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b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)sin(k\omega t)dt \qquad \qquad \qquad \qquad \quad (n = k)
bn=π1∫−ππf(t)sin(kωt)dt(n=k)
傅里叶指数形式：
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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)=T1n=−∞∑+∞∫t0to+Te−inωtdt⋅einωt
推导如下：
由欧拉公式(
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eix=cosx+isinx)我们可以推出：
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cosx = \frac{e^{ix} + e^{-ix}}{2} \qquad sinx = -i \cdot \frac{e^{ix} - e^{-ix}}{2}
cosx=2eix+e−ixsinx=−i⋅2eix−e−ix
将两式带入傅里叶级数得到：
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\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∑∞[an2einωt+e−inωt+bn⋅(−i)2einωt−e−inωt]A0+n=1∑∞[2an−ibneinωt+2an+ibne−inωt]
又因为：
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\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∫−ππf(t)cos(nωt)dt(n=k)bn=π1∫−ππf(t)sin(kωt)dt(n=k)
所以：
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\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∫−ππf(t)cos(nωt)dt−i⋅π1∫−ππf(t)sin(kωt)dt]21⋅T2[∫t0t0+Tcos(nωt)f(t)dt−i∫t0t0+Tsin(nωt)f(t)dt]T1⋅∫t0t0+Tf(t)[cos(nωt)−isin(nωt)]dtT1⋅∫t0t0+Tf(t)[2einωt+e−inωt−i⋅(−i)2einωt−e−inωt]dtT1⋅∫t0t0+Tf(t)e−inωtdt
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\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∫−ππf(t)cos(nωt)dt+i⋅π1∫−ππf(t)sin(kωt)dt]21⋅T2[∫t0t0+Tcos(nωt)f(t)dt+i∫t0t0+Tsin(nωt)f(t)dt]T1⋅∫t0t0+Tf(t)[cos(nωt)+isin(nωt)]dtT1⋅∫t0t0+Tf(t)[2einωt+e−inωt+i⋅(−i)2einωt−e−inωt]dtT1⋅∫t0t0+Tf(t)einωtdt
然后再带回去：
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\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−ibneinωt+2an+ibne−inωt]T1∫t0t0+Tf(t)dt+n=1∑∞[T1⋅∫t0t0+Tf(t)e−inωtdt⋅einωt+T1⋅∫t0t0+Tf(t)einωtdt⋅e−inωt]T1∫t0t0+Tf(t)dt+T1n=1∑∞∫t0t0+Tf(t)e−inωtdt⋅einωt+T1n=−∞∑−1∫t0t0+Tf(t)e−inωtdt⋅einωtT1n=−∞∑+∞∫t0t0+Tf(t)e−inωtdt⋅einωt
傅里叶变换和傅里叶逆变换
傅里叶变换：
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\begin{aligned} F(\omega_x) = \int_{-\infty}^{+\infty}f(t)s^{-i\omega_x t}dt \end{aligned}
F(ωx)=∫−∞+∞f(t)s−iωxtdt
傅里叶逆变换：
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\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)eiωxtdω
其中
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F(\omega)
F(ω) 叫做
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f(t) 的像函数，
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F(ω) 的像原函数。
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推导过程如下：
令：
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\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−iωxtdt
又因为：
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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)=T1n=−∞∑+∞∫t0t0+Tf(t)e−inωtdt⋅einωt
所以：
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\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−iωxtdt⋅eiωxtT1−∞∑+∞[F(ωx)eiωxt]=2π1∫−∞+∞F(ωx)eiωxtdωx
离散傅里叶变换
离散傅里叶变换：
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F(n) = \sum_{n = 0}^{N}f(t)e^{i\frac{2\pi n}{N}t}
F(n)=n=0∑Nf(t)eiN2πnt
逆离散傅里叶变换：
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f(t) = \sum_{n=0}^{N}\frac{1}{N}F(n)e^{i\frac{2\pi n}{N}t}
f(t)=n=0∑NN1F(n)eiN2πnt
标签：cos,frac,推导,变换,傅里叶,int,dt,pi,omega 来源： https://blog.csdn.net/ID246783/article/details/120848291
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