Mathematics Yoshio Sep 8th, 2023 at 8:00 PM 8 0
泰勒展開式（Taylor Expansion）
泰勒級數 Taylor SeriesTaylor Series
A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function \(f(x)\) about a point \(x=a\) is given by
\[f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n\]
where \(n!\) denotes the factorial of \(f(n)\). In the more compact sigma notation, this can be written as
\[\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n\]
In mathematics, a Power Series is an infinite series of the form
\[f(x)=a_0+a_1(x\;-\;c)+a_2{(x\;-\;c)}^2+...+a_n{(x\;-\;c)}^n=\sum_{n=0}^\infty a_n(x\;-\;c)^n\]
When the center of the series is equal to zero, it is called by Maclaurin Series and takes the simpler form
\[f(x)=a_0+a_1x+a_2x^2+...+a_nx^n=\sum_{n=0}^\infty a_nx^n\]
A polynomial in a single indeterminate x can always be written in the form
\[f(x) = c_0 + c_1 x + c_2 x^2 + ...+ c_k x^k+...=\sum_{k=0}^\infty c_k x^k\]
\(f^{'}(x)=\frac{df(x)}{dx}=c_1+c_2\cdot2\cdot x+c_3\cdot3\cdot x^2+c_4\cdot4\cdot x^3+...\)
\(f^{''}(x)=\frac{df^{'}(x)}{dx}=c_2\cdot2\cdot1+c_3\cdot3\cdot2\cdot x+c_4\cdot4\cdot3\cdot x^2+...\)
\(f^{'''}(x)=\frac{df^{''}(x)}{dx}=c_3\cdot3\cdot2\cdot1+c_4\cdot4\cdot3\cdot2\cdot x+...\)
\(f^{(k)}(x)=\frac{df^{(k-1)}(x)}{dx}=c_k\cdot k!+c_{k+1}\cdot(k+1)!\cdot x+...\)
\(c_k = \frac{f^{(k)}(0)}{k!}\)
\[f(x)=f(0)+\frac{f'(0)}{1!}x+...+\frac{f^{(k)}(0)}{k!}x^k+...=\sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k\]
\[f(x) = c_0 + c_1 x + c_2 x^2 + ...+ c_k x^k+...=\sum_{k=0}^\infty c_k x^k\]
When the center of the series moves to \(a\)
\[f(x)=f(a)+\frac{f^{'}(a)}{1!}(x-a)+...+\frac{f^{(k)}(a)}{k!}(x-a)^k+...=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k\]
For Complex Function
\[f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n,\;\left|z-z_0\right|<{R}\]
\[a_n=\frac{f^{(n)}(z_0)}{n!},\;n=0,1,2,...\]
Maclaurin series
Most Maclaurin series expressible in terms of elementary functions can be determined through the composition and combination of the following functions:
| Function | Maclaurin Series | Interval of Convergence |
|---|---|---|
| \(\dfrac{1}{1-x}\) | \(\displaystyle \sum_{k=0}^{\infty} x^k\) | \(-1 < x < 1\) |
| \(e^x\) | \(\displaystyle \sum_{k=0}^{\infty} \dfrac{x^k}{k!}\) | \(-\infty < x < \infty\) |
| \(\ln(1+x)\) | \(\displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}x^k}{k}\) | \(-1 < x \leqslant 1\) |
| \(\sin x\) | \(\displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k+1}}{(2k+1)!}\) | \(-\infty < x < \infty\) |
| \(\cos x\) | \(\displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k x^{2k}}{(2k)!}\) | \(-\infty < x < \infty\) |
Laurent series
In the mathematics, the Laurent series of a complex function \(f(z)\) is a representation of that function as a power series which includes terms of negative degree.
Suppose that \(f(z)\) is analytic on the annulus
\[A: r_1 < |z - z_0|\;{<}\;r_2\]
Then \(f(z)\) can be expressed as a series
\[f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\sum_{n=1}^\infty\frac{b_n}{(z-z_0)^n}\]
The coefficients have the formulus
\[a_n=\frac1{2\pi i}\oint_C\frac{f(z)dz}{(z-z_0)^{n+1}},\;n=0,\;1,\;2,\;...\]
\[b_n=\frac1{2\pi i}\oint_C(z-z_0)^{n-1}f(z)dz,\;n=1,\;2,\;3,\;...\]
where \(C\) is any circle \(|z-z_0|=r\) inside the annulus
Furthermore
The series \(\sum_{n = 0}^{\infty} a_n (z - z_0)^n\) converges to an analytic function for \(|z - z_0| {<} r_2\).
It is called the analytic or regular part of the Laurent series. [positive powers] (Taylor series)
The series \(\sum_{n = 1}^{\infty} \dfrac{b_n}{(z - z_0)^n}\) converges to an analytic function for \(|z - z_0| {>} r_1\).
It is called the singular or principal part of the Laurent series. [negative powers]
\[f(z)=...+\frac{a_{-m}}{{(z-z_0)}^m}+\frac{a_{-m+1}}{{(z-z_0)}^{m-1}}+...+\frac{a_{-2}}{{(z-z_0)}^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+...\]
\[f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n\]
\[a_n=\frac1{2\mathrm{πi}}\oint_C\frac{f(z)}{{(z-z_0)}^{n+1}}dz,\;z_o\;is\;not\;at\;ROC\]
\[f(z)=\sum_{n=-\infty}^\infty(\frac1{2\mathrm{πi}}\oint_C\frac{f(w)}{{(w-z_0)}^{k+1}}dw)(z-z_0)^n\]
Cauchy's integral formula
\[f(a)=\frac1{2\mathrm{πi}}\oint_C\frac{f(z)}{z-a}dz\]
Ex
\(let\;C\;be\;the\;contour\;described\;by\;\left|z-1\right|=4,\;To\;find\;\oint_C\frac{e^{3z}}{z-\mathrm{πi}}dz\;around\;the\;contour\;C\)
\(f(a)=\frac1{2\mathrm{πi}}\oint_C\frac{f(z)}{z-a}dz\)
Sol
\(\oint_C\frac{f(z)}{z-a}dz=2\mathrm{πi}\cdot f(a)=2\mathrm{πi}e^{3\mathrm{πi}}\)
\(=2\mathrm{πi}(\cos3\mathrm\pi+\mathrm{isin}3\mathrm\pi)=-2\mathrm{πi}\)
Ex
To find \(\frac1{2\mathrm{πi}}\oint_C\frac{\mathrm{cosπ}z}{z^2-1}dz,\;C\;is\;the\;area\;(-2-i,\;2-i,\;2+i,-2+i)\)
Sol
\(\frac1{2\mathrm{πi}}\oint_C\frac{\mathrm{cosπ}z}{(z+1)(z-1)}dz=\frac1{2\mathrm{πi}}\cdot\frac12(\oint_C\frac{\mathrm{cosπ}z}{(z-1)}dz-\oint_C\frac{\mathrm{cosπ}z}{(z+1)}dz)\)
\(=\frac1{2\mathrm{πi}}\cdot\frac12\mathrm{cosπ}-\frac1{2\mathrm{πi}}\cdot\frac12\cos(-\mathrm\pi)=0\)
Pf
\(\oint_C\frac{f(z)}{z-a}dz=\oint_\Gamma\frac{f(z)}{z-a}dz\)
\(z=a+\varepsilon e^{i\theta}\)
\(dz=\varepsilon ie^{i\theta}d\theta\)
\(=\int_0^{2\mathrm\pi}\frac{f(a+\varepsilon e^{i\theta})}{\varepsilon e^{i\theta}}\varepsilon ie^{i\theta}d\theta\)
\(=i\int_0^{2\mathrm\pi}f(a+\varepsilon e^{i\theta})d\theta\)
\(\lim_{\varepsilon\rightarrow0}\oint_C\frac{f(z)}{z-a}dz=\lim_{\varepsilon\rightarrow0}i\int_0^{2\mathrm\pi}f(a+\varepsilon e^{i\theta})d\theta\)
\(=i\int_0^{2\mathrm\pi}f(a)d\theta\)
\(=2\mathrm{πi}\cdot\mathrm f(\mathrm a)\)
\(\therefore f(a)=\frac1{2\mathrm{πi}}\oint_C\frac{f(z)}{z-a}dz\)
Cauchy's differentiation formula
\[f^{(n)}(a)=\frac{n!}{2\mathrm{πi}}\oint_C\frac{f(z)}{{(z-a)}^{n+1}}dz\]
Ex
\(let\;C\;be\;the\;contour\;described\;by\;\left|z\right|=2,\;To\;find\;\oint_C\frac{e^{iz}}{z^3}dz\;around\;the\;contour\;C\)
\(f^{(n)}(a)=\frac{n!}{2\mathrm{πi}}\oint_C\frac{f(z)}{{(z-a)}^{n+1}}dz\)
\(\oint_C\frac{e^{iz}}{z^3}dz={\left.\frac{2\mathrm{πi}}{2!}f''(z)\right|}_{z=0}\)
\(=\frac{2\mathrm{πi}}{2!}\cdot i\cdot i\cdot{\left.e^{iz}\right|}_{z=0}=-\mathrm{πi}\)
傅立葉級數（Fourier Series）
Fourier SeriesFourier series is used to describe a periodic signal in terms of cosine and sine waves. In other other words, it allows us to model any arbitrary periodic signal with a combination of sines and cosines.
Consider expressing the Fourier series as
\[f(x)=\sum_{n=-\infty}^{+\infty}(a_n\cos\;nx+b_n\sin\;nx)\]
where \[a_n=\frac1{2\pi}\int_{-\pi}^{+\pi}f(t)cos\;nxt\;dt\]
and \[b_n=\frac1{2\pi}\int_{-\pi}^{+\pi}f(t)\sin\;nxt\;dt\]
The usual Fourier series involving sines and cosines is obtained by taking \(f_{1}(x) = cosx\) and \(f_2(x)=sinx\). Since these functions form a complete orthogonal system over \([-π,π]\), the Fourier series of a function \(f(x)\) is given by
where
and \(n=1, 2, 3, ....\) Note that the coefficient of the constant term \(a_0\) has been written in a special form compared to the general form for a generalized Fourier series in order to preserve symmetry with the definitions of \(a_n\) and \(b_n\).
Note that \(cosϕ\) is an even function, i.e. \(cos(−ϕ)=cosϕ\), and therefore \(a_{-n} = a_n\). Similarly, \(sinϕ\) is an odd function, i.e. \(sin(-ϕ)=-sinϕ\), and therefore \(b_{-n} =-b_n\) . Thus, we can combine the terms with the positive and negative \(n\) of the same absolute value, which will result in doubled coefficients \(2a_n\) and \(2b_n\) for \(n > 0\), while the terms with \(n < 0\) will be dropped. Since the terms with \(n = 0\) do not have a pair, they will not be doubled, so \(a_0\) will remain the same, and \(b_0\) can be omitted since \(b_0 = 0\).
\(f(x)=\sum_{k=0}^\infty(A_k\cos\frac{2\pi kx}T+B_k\sin\frac{2\pi kx}T)\)
\(=A_0+\sum_{k=1}^\infty(A_k\cos\frac{2\pi kx}T+B_k\sin\frac{2\pi kx}T)\)
\(A_0:\;DC\;component\)
\(\frac{2\pi}T:\;fundamental\;component\)
\(\cos(\omega x),\;\cos(\frac{2\pi x}T),\;\cos(2\pi fx)\)
\(f:Hz,\;\omega=2\pi f=\frac{2\pi}T\)
\(\int_Tf(x)dx=\int_TA_0dx+\sum_{k=1}^\infty\int_TA_k\cos\frac{2\pi kx}Tdx+\sum_{k=1}^\infty\int_TB_k\sin\frac{2\pi kx}Tdx\)
\(=A_0T\)
\(\therefore A_0=\frac1T\int_Tf(t)dt\) average in one period
\(\int_Tf(x)\cos\frac{2\pi mx}Tdx,\;m=1,2,3,..\)
\(=\int_TA_0\cos\frac{2\pi mx}Tdx+\sum_{k=1}^\infty\int_TA_k\cos\frac{2\pi kx}T\cos\frac{2\pi mx}Tdx+\sum_{k=1}^\infty\int_TB_k\sin\frac{2\pi kx}T\cos\frac{2\pi mx}Tdx\)
\(\cos\frac{2\pi kx}T\cos\frac{2\pi mx}T=\frac12\lbrack\cancel{\cos\frac{2\pi}T(k+m)x}+\cos\frac{2\pi}T(k-m)x\rbrack\)
\(\sin\frac{2\pi kx}T\cos\frac{2\pi mx}T=\frac12\lbrack\cancel{\sin\frac{2\pi}T(k+m)x}+\cancel{\sin\frac{2\pi}T(k-m)x}\rbrack\)
\(\Rightarrow\int_Tf(x)\cos\frac{2\pi mx}Tdx=\int_TA_m\cos^2\frac{2\pi mx}Tdx=\frac12A_mT\)
\(A_m=\frac2T\int_Tf(x)\cos\frac{2\pi mx}Tdx\)
\(A_k=\frac2T\int_Tf(x)\cos\frac{2\pi kx}Tdx\)
Similarly, we have \(B_k=\frac2T\int_Tf(x)\sin\frac{2\pi kx}Tdx\)
For a periodic function, which satisfies Dirichlet condition, it can be expressed by a Fourier series.
Dirichlet conditions:
1. The number of discontinuous points is finite in one period.
2. The number of maximum and minimum points is finite in one period. \(\sin(\frac1x)\)
3. The function is absolutely integrable in one period.
Although each of the infinite number of sinusoidal functions is continuous, their sum may be discontinuous.
\(\int_Tf_T(x)\sin m\omega_0xdx,\;m=1,2,3,..\)
\(=\int_TA_0\sin m\omega_0xdx+\sum_{k=1}^\infty\int_TA_k\cos k\omega_0x\cdot\sin m\omega_0xdx+\sum_{k=1}^\infty\int_TB_k\sin k\omega_0x\cdot\sin m\omega_0xdx\)
\(\int_TA_0\sin m\omega_0xdx=0\)
\(\int_T\cos k\omega_0x\cdot\sin m\omega_0xdx=\frac12\int_T\lbrack\sin(k+m)\omega_0x-s\mathrm{in}(k-m)\omega_0x\rbrack dx=0\)
\(\int_T\sin k\omega_0x\cdot\sin m\omega_0xdx=\frac12\int_T\lbrack\cos(k-m)\omega_0x-\cos(k+m)\omega_0x\rbrack dx\)
\(=\frac12\int_T\lbrack\cos(k-m)\omega_0x=\left\{\begin{array}{l}\frac T2,\;k=m\\0,\;k\neq m\end{array}\right.\)
\(f(x)=A_0+\sum_{k=1}^\infty(A_k\cos\frac{2\pi kx}T+B_k\sin\frac{2\pi kx}T)\)
where \(A_0=\frac1T\int_Tf(x)dx\)
\(A_k=\frac2T\int_Tf(x)\cos\frac{2\pi kx}Tdx\)
\(B_k=\frac2T\int_Tf(x)sin\frac{2\pi kx}Tdx\)
Fourier Series
\(f_T(t)=A_0+\sum_{k=1}^\infty A_k\cos k\omega_0t+\sum_{k=1}^\infty B_k\sin k\omega_0t,\;-\infty\;<\;t\;<\;\infty\)
\(A_0=\frac1T\int_Tf(t)dt\)
\(A_k=\frac2T\int_Tf(t)\cos k\omega_0t\;dt\)
\(B_k=\frac2T\int_Tf(t)\sin k\omega_0t\;dt\)
\(\omega_0=\frac{2\pi}T\)
\(\{\omega_0,\;2\omega_0,\;3\omega_0,...\}\)
\(If\;f(x)\;is\;an\;even\;function,\;f(x)=A_0+\sum_{k=1}^\infty A_k\cos k\omega_0x\)
\(If\;f(x)\;is\;an\;odd\;function,\;f(x)=\sum_{k=1}^\infty A_k\sin k\omega_0x\)
Finite Duration Function
\(f(t)=f_T(t),\;a\;\leq\;t\;<\;b\)
\(f(t)=A_0+\sum_{k=1}^\infty(A_k\cos k\omega_0t+B_k\sin k\omega_0t),\;a\;\leq\;t\;<\;b\)
A finite duration function can not be uniquely expressed by a Fourier Series.
\(\cos k\omega_0x=\frac12(e^{jk\omega_0x}+e^{-jk\omega_0x})\)
\(\sin k\omega_0x=\frac1{2j}(e^{jk\omega_0x}-e^{-jk\omega_0x})\)
\(f_T(x)=A_0+\sum_{k=1}^\infty\lbrack\frac{A_k}2(e^{jk\omega_0x}+e^{-jk\omega_0x})+\frac{B_k}{2j}(e^{jk\omega_0x}-e^{-jk\omega_0x})\rbrack\)
\(f_T(x)=A_0+\sum_{k=1}^\infty(\frac{A_k}2+\frac{B_k}{2j})e^{jk\omega_0x}+\sum_{k=1}^\infty(\frac{A_k}2-\frac{B_k}{2j})e^{-jk\omega_0x}\)
\(f_T(x)=A_0+\sum_{k=1}^\infty(\frac{A_k}2+\frac{B_k}{2j})e^{jk\omega_0x}+\sum_{m=-1}^{-\infty}(\frac{A_{-m}}2-\frac{B_{-m}}{2j})e^{jm\omega_0x}\)
\(=A_0+\sum_{k=1}^\infty C_ke^{jk\omega_0x}+\sum_{m=-1}^{-\infty}C_{-m}e^{jm\omega_0x}\)
\(C_k=\frac12(A_k-jB_k),\;C_{-m}=\frac12(A_{-m}+jB_{-m})\)
\(f_T(x)=C_0+\sum_{k=1}^\infty(C_ke^{jk\omega_0x}+C_{-k}e^{-jk\omega_0x})=\sum_{k=-\infty}^\infty C_ke^{jk\omega_0x}\)
\(where\;C_0=A_0\)
\(C_k=\left\{\begin{array}{l}\frac12(A_k-jB_k),\;k>0\\\frac12(A_{-k}+jB_{-k}),\;k<0\end{array}\right.\)
\(C_k=\frac12(A_k-jB_k)=\vert C_k\vert e^{j\phi_k}\)
\(where\;\vert C_k\vert=\vert C_{-k}\vert=\frac12\sqrt{A_k^2+B_k^2},\)
\(\phi_k=-\phi_{-k}=-\tan^{-1}(\frac{B_k}{A_k})\;and\;C_{-k}=C_k^\ast\)
magnitude: even
phase: odd
\(f_T(t)=\sum_{k=-\infty}^\infty C_ke^{jk\omega_0t}\)
\(\int_Tf_T(t)e^{-jm\omega_0t}dt,\;(m:given)\)
\(=\int_T\sum_{k=-\infty}^\infty C_ke^{jk\omega_0t}e^{-jm\omega_0t}dt\)
\(=\sum_{k=-\infty}^\infty C_k\int_Te^{j(k-m)\omega_0t}dt\)
\(=C_m\int_Tdt=TC_m\)
\(\Rightarrow C_k=\frac1T\int_Tf_T(t)e^{-jk\omega_0t}dt,\;k\in\mathcal Z\)
\(C_k^\ast=\frac1T\int_Tf_T(t)e^{jk\omega_0t}dt,\;k\in\mathcal Z\)
\(=\frac1T\int_Tf_T(t)e^{-j(-k)\omega_0t}dt\)
\(=C_{-k}\)
\(C_k=\vert C_k\vert e^{j\phi_k}=\vert C_k\vert\cos\phi_k+j\vert C_k\vert\sin\phi_k\)
[Ex] \(f(x)=\left\{\begin{array}{l}2,\;-1\leq x<1\\0,\;-2.5\leq x<-1,\;1\leq x<2.5\end{array}\right.\)
\(f(t)=A_0+\sum_{k=1}^\infty A_k\cos k\omega_0t+\sum_{k=1}^\infty B_k\sin k\omega_0t,\;T=5\)
\(=\sum_{k=-\infty}^\infty C_ke^{jk\omega_0x}\)
\(A_0=\frac15\int_{-1}^12dt=\frac45\)
\(A_k=\frac25\int_{-1}^12\cos k\omega_0tdt=\frac25\int_{-1}^12\cos k\frac{2\pi}5tdt\)
\(=\frac45\frac5{2k\pi}\left.\sin k\frac{2\pi}5\right|_{-1}^1=\frac4{k\pi}\sin\frac{2k\pi}5\)
\(B_k=\frac25\int_{-1}^12\sin k\frac{2\pi}5tdt\)
\(=\frac45\frac5{2k\pi}\left.cosk\frac{2\pi}5\right|_{-1}^1=0\)
\(C_k=\frac15\int_{-1}^12\cdot e^{-jk\omega_0t}dt\)
\(=\frac25\frac1{-jk\omega_0}\left.e^{-jk\omega_0t}\right|_{-1}^1\)
\(=\frac{2j}{5k}\cdot\frac5{2\pi}(e^{-jk\omega_0}-e^{jk\omega_0})\)
\(=\frac j{k\pi}\cdot-2j\cdot\sin k\omega_0=\frac2{k\pi}\sin k\frac{2\pi}5\)
\(C_k^\ast=C_{-k}=\frac2{k\pi}\sin k\frac{2\pi}5\)
\(\vert C_k\vert=\vert\frac2{k\pi}\sin k\frac{2\pi}5\vert\)
\(\phi_k=\left\{\begin{array}{l}0,\;C_k\;>\;0\\\pi,\;C_k\;<\;0\end{array}\right.\)
\(\phi_{-k}=\left\{\begin{array}{l}0,\;C_{-k}\;>\;0\\-\pi,\;C_{-k}\;<\;0\end{array}\right.\)
\(C_k=\frac2{k\pi}\sin k\frac{2\pi}5=\frac{\sin\frac{2k\pi}5}{\frac{2k\pi}5\cdot{\frac54}}=\frac45\sin\frac{2k}5\pi=(0.8)\sin(0.4k\pi)\)
The Power of periodic function
by mean-square value \(P=\frac1T\int_Tf_T^2(x)dx\)
\(P=\frac1T\int_T(\sum_{k=-\infty}^\infty C_ke^{jk\omega_0x}\sum_{n=-\infty}^\infty C_ke^{jn\omega_0x})dx\)
\(=\frac1T\int_T(\sum_{k=-\infty}^\infty C_ke^{jk\omega_0x}\sum_{m=-\infty}^\infty C_{-m}e^{-jm\omega_0x})dx\)
\(=\frac1T\int_T(\sum_{k=-\infty}^\infty\sum_{m=-\infty}^\infty C_kC_{-m}e^{j(k-m)\omega_0x})dx\)
\(=\sum_{k=-\infty}^\infty\sum_{m=-\infty}^\infty\frac1T\int_TC_kC_{-m}e^{j(k-m)\omega_0x}dx\)
\(\int_TC_kC_{-m}e^{j(k-m)\omega_0x}dx=0\;for\;k\neq m\)
\(P=\sum_{k=-\infty}^\infty\frac1T\int_TC_kC_{-k}dx\)
\(=\sum_{k=-\infty}^\infty C_kC_{-k}=\sum_{k=-\infty}^\infty C_kC_k^\ast=\sum_{k=-\infty}^\infty\vert C_k\vert^2\)
Finally, we obtain the Parseval’s theorem as
\(P=\frac1T\int_Tf_T^2(x)dx=\sum_{k=-\infty}^\infty\vert C_k\vert^2\)
Parseval's Theorem
\(\int_{-\infty }^{\infty }s^{2}(t)dt=\int_{-\infty }^{\infty }\left ( \left | S(f) \right | \right )^{2}df \nonumber\)
\(f{(x)}\sim\frac{a_0}2+\sum_{n=1}^\infty{(a_n\cos nx+b_n\sin nx)}\)
\(\frac1\pi\int_{-\pi}^\pi{\vert f^2{(x)}\vert}{}{\mathrm d}x=\frac{{a_0}^2}2+\sum_{n=1}^\infty{({a_n}^2+{b_n}^2)}\)
\(f{(x)}=\sum_{n=-\infty}^\infty c_ne^{jnx},\;where\;c_n=\frac1{2\pi}\int_{-\pi}^\pi f{(t)}e^{-jnt}{}{\mathrm d}t\)
Then \(\frac1{2\pi}\int_{-\pi}^\pi{\vert f{(x)}\vert}^2{}{\mathrm d}x=\sum_{n=-\infty}^\infty{\vert c_n\vert}^2\)
[Ex] \(f{(x)}={\{\begin{array}{l}0,\;-\pi\;\leq\;x\;\leq0\\1,\;0\;<\;x\;\leq\;\pi\end{array}}.\)
\(a_0=\frac1\pi\int\limits_{-\pi}^\pi f{(x)}dx=\frac1\pi\int\limits_0^\pi1dx=\frac1\pi⋅\pi=1.\)
\(for\;n\neq0\)
\(a_n=\frac1\pi\int\limits_{-\pi}^\pi f{(x)}\cos\;nxdx=\frac1\pi\int\limits_0^\pi1⋅\cos\;nxdx=\frac1\pi{\lbrack{{(\frac{\sin\;nx}n)}\vert}_0^\pi\rbrack}=\frac1{\pi n}⋅0=0,\)
\(b_n=\frac1\pi\int\limits_{-\pi}^\pi f{(x)}\sin\;nxdx=\frac1\pi\int\limits_0^\pi1⋅\sin\;nxdx=\frac1\pi{\lbrack{{(-\frac{\cos\;nx}n)}\vert}_0^\pi\rbrack}=-\frac1{\pi n}⋅{(\cos\;n\pi-\cos\;0)}=\frac{1-\cos\;n\pi}{\pi n}.\)
Since \(\cos\;n\pi={(-1)}^n,\)
\(b_n=\frac{1-{(-1)}^n}{\pi n}.\)
\(f{(x)}=\frac12+\sum_{n=1}^\infty\frac{1-{(-1)}^n}{\pi n}\sin\;nx.\)
\(f{(x)}=\frac12+\frac{1-{(-1)}}\pi\sin\;x+\frac{1-{(-1)}^2}{2\pi}\sin\;2x+\frac{1-{(-1)}^3}{3\pi}\sin\;3x+\frac{1-{(-1)}^4}{4\pi}\sin\;4x+\frac{1-{(-1)}^5}{5\pi}\sin\;5x+\dots\)
\(=\frac12+\frac2\pi\sin\;x+\frac2{3\pi}\sin\;3x+\frac2{5\pi}\sin\;5x+\dots\)
傅立葉轉換（Fourier Transform）
Fourier TransformThe 1D Fourier transform is:
\(f_T(t)=\sum_{k=-\infty}^\infty C_ke^{jk\omega_0t}\)
\(C_k=\frac1T\int_Tf_T(x)e^{-jk\omega_0x}dx\)
\(\therefore f_T(t)=\sum_{k=-\infty}^\infty\frac1T\int_{-\frac T2}^\frac T2f_T(x)e^{-jk\omega_0x}dx\cdot e^{jk\omega_0t},\;\omega_0=\frac{2\pi}T\)
\(\omega_0=\frac{2\pi}T=\triangle\omega\rightarrow0\)
\(f_T(t)=\sum_{k=-\infty}^\infty\frac{\triangle\omega}{2\pi}\int_{-\infty}^\infty f_T(x)e^{-j\omega x}dx\cdot e^{jk\triangle\omega t}\)
\(\int_{-\infty}^\infty f_T(x)e^{-j\omega x}dx=F(\omega)\)
\(f_T(t)=\int_{-\infty}^\infty\frac1{2\pi}F(\omega)e^{j\omega t}d\omega\)
\(\left\{\begin{array}{l}f(t)=\frac1{2\pi}\int_{-\infty}^\infty F(\omega)e^{j\omega t}d\omega\\F(\omega)=\int_{-\infty}^\infty f(t)e^{-j\omega t}dt\end{array}\right.\)
\(Fourier\;transform\;of\;f(t):\;F(\omega)\)
\(ℱ\{f(t)\}=F(\omega)\)
\(ℱ^{-1}\{F(\omega)\}=f(t)\)
The 3D Fourier transform is:
\(\left\{\begin{array}{l}f(x)=\frac1{{(2\pi)}^3}\int_{-\infty}^\infty F(\omega)e^{j\omega x}d^3\omega=ℱ^{-1}\lbrack F(\omega)\rbrack\\F(\omega)=\int_{-\infty}^\infty f(x)e^{-j\omega x}d^3x=ℱ\lbrack f(x)\rbrack\end{array}\right.\)
Consider a spacetime plane-wave \(e^{ikx}=e^{i(\omega t-kx)}\)
\(f(x)=\frac1{{(2\pi)}^4}\int_{-\infty}^\infty F(k)e^{-jkx}d^4\omega=\frac1{{(2\pi)}^4}\int_{-\infty}^\infty F(k)e^{-j(\omega t-kx)}d^4\omega\)
\(F(k)=\int_{-\infty}^\infty f(x)e^{jkx}d^4x=\int_{-\infty}^\infty f(x)e^{j(\omega t-kx)}d^4x\)
Multi-Dimensional Fourier Transform
\(F(k_1,k_2,\dots,k_d)=\int_{-\infty}^\infty dx_1\;e^{-ik_1x_1}\;\int_{-\infty}^\infty dx_2\;e^{-ik_2x_2}\,\cdots\,\int_{-\infty}^\infty dx_d\;e^{-ik_dx_d}\,f(x_1,x_2,\dots,x_N)\)
\(F(\vec k)=\int\;f(\vec x)e^{-i\,\vec k\cdot\vec x}d^dx\)
\(f(\vec x)=\int\;F(\vec k)e^{i\,\vec k\cdot\vec x}\frac{d^dk}{(2\pi)^d}\)
\(\delta^d(\vec x-\vec x')=\int e^{\left[i\vec k\cdot\left(\vec x-\vec x'\right)\right]}\frac{d^dk}{(2\pi)^d}\)
\(\int\,f(\vec x)\delta^d(\vec x-\vec x')d^dx=f(\vec x')\)
The Fourier transform (FT) is a transform that converts a function into a form that describes the frequencies present in the original function. The output of the transform is a complex-valued function of frequency. The Fourier transform can also be generalized to functions of several variables on Euclidean space, sending a function of 3-dimensional 'position space' to a function of 3-dimensional momentum (or a function of space and time to a function of 4-momentum).
傅立葉轉換介紹The Fourier transform of the expression \(f = f(x)\) with respect to the variable \(x\) at the point \(w\) is
\[F{(w)}=ℱ\lbrack f(x)\rbrack=c\int\limits_{-\infty}^\infty f{(x)}e^{iswx}dx\]
\(c\) and \(s\) are parameters of the Fourier transform. The fourier function uses \(c = 1, s = -1\).
\[F(m,n)=ℱ\lbrack f(x,y)\rbrack=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}{f(x,y)e^{-j\frac{2\pi mx}M}}e^{-j\frac{2\pi ny}N}\]
where \(0\leq m\leq M-1\) and \(0\leq n\leq N-1\)
Properties of Continuous-Time Fourier Transform (CTFT)
Fourier Transform
The Fourier transform of a continuous-time function \(x(t)\) is defined as,
\[X(\omega)=ℱ\lbrack x(t)\rbrack=\int_{-\infty}^\infty x(t)e^{-j\omega t}dt\]
Inverse Fourier Transform
The inverse Fourier transform of a continuous-time function is defined as,
\[x(t)=ℱ^{-1}\lbrack X(\omega)\rbrack=\frac1{2\pi}\int_{-\infty}^\infty X(\omega)e^{j\omega t}d\omega\]
Short-Time Fourier Transform (STFT)
Short-time Fourier transform (STFT), is a Fourier-related transform used to determine the sinusoidal frequency and phase content of local sections of a signal as it changes over time.
STFT is a method of analysis used for analyzing non-stationary signals. It extracts several frames of signals with a window that moves with time. If the time window is sufficiently narrow, each frame extracted can be viewed as stationary so that Fourier transform can be used. With the window moving along the time axis, the relation between the variance of frequency and time can be identified.
Fourier Transform (FT)
Time domain signals transfer to Frequency domain signals
\[F{(\omega)}=\int\limits_{-\infty}^\infty f{(t)}e^{-i\omega t}dt\]
Short-Time Fourier Transform (STFT)
\[F{(\tau,\omega)}=\int\limits_{-\infty}^\infty f{(t)}w{(t-\tau)}e^{-i\omega t}dt\]
where \(\omega (t)\) represents the sliding window that emphasizes local frequency components within it. The size of the window is related to the time resolution and frequency resolution of STFT. The shorter the window, the higher the time resolution. However, this is usually accompanied by poor frequency resolution. For a long window, the frequency resolution is high, but the time resolution is low. This phenomenon reflects Heisenberg's uncertainty principle.
Different window shapes produce different results, and Origin provides up to nine different windows for addressing specific needs: Rectangle, Welch, Triangular, Bartlett, Hanning, Hamming, Blackman, Gaussian and Kaiser.
The window function, is commonly a Hann window or Gaussian window centered around zero.
Discrete-time Fourier transform (DTFT)
離散時間傅立葉轉換
The discrete-time Fourier transform (DTFT) is analogous to a Fourier series, except instead of starting with a periodic function of time and producing discrete sequence over frequency, it starts with a discrete sequence in time and produces a periodic function in frequency. It is the transform-pair relationship between a discrete-time signal and its continuous-frequency transform. The utility of this frequency domain function is rooted in the Poisson summation formula.
The DTFT is often used to analyze samples of a continuous function. The term discrete-time refers to the fact that the transform operates on discrete data, often samples whose interval has units of time. From uniformly spaced samples it produces a function of frequency that is a periodic summation of the continuous Fourier transform of the original continuous function.
Periodic Function in Frequency
Inverse Discrete-Time Fourier Transform (IDTFT)
The inverse discrete-time Fourier transform (IDTFT) is defined as the process of finding the discrete-time sequence \(x(n)\) from its frequency response \(X(ω)\).
\(\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\: \frac{1}{2\pi}\int_{-\pi}^{\pi}\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\mathit{e}^{\mathit{j\omega n}}\:\mathit{d\omega}}\)
\(\)
Discrete Fourier transform (DFT)
In mathematics, the discrete Fourier transform (DFT) converts a finite sequence of equally-spaced samples of a function into a same-length sequence of equally-spaced samples of the discrete-time Fourier transform (DTFT), which is a complex-valued function of frequency.
The discrete Fourier transform (DFT) is a method for converting a sequence of \(N\) complex numbers \(x_0, x_1, ..., x_{N-1}\)
\[ X_k = \sum_{n=0}^{N-1} x_n e^{-2\pi i kn/N}, \]
for \( 0 \le k \le N-1.\)
The \(x_i\) are thought of as the values of a function, or signal, at equally spaced times \(t=0,1,…,N−1\). he output \(X_k\) is a complex number which encodes the amplitude and phase of a sinusoidal wave with frequency \(\frac kN\) cycles per time unit. (This comes from Euler's formula: \(e^{2\pi ikn/N}=\cos(2\pi kn/N)+i\sin(2\pi kn/N)\)) The effect of computing the \(X_k\) is to find the coefficients of an approximation of the signal by a linear combination of such waves. Since each wave has an integer number of cycles per \(N\) time units, the approximation will be periodic with period \(N\). This approximation is given by the inverse Fourier transform
\[x_n=\frac1N\;\sum_{k=0}^{N-1}X_ke^{2\mathrm{πikn}/\mathrm N}\]
It is a consequence of orthogonality with respect to the complex dot product.
The DFT is useful in many applications, including the simple signal spectral analysis outlined above. Knowing how a signal can be expressed as a combination of waves allows for manipulation of that signal and comparisons of different signals:
Digital files (jpg, mp3, etc.) can be shrunk by eliminating contributions from the least important waves in the combination.
Different sound files can be compared by comparing the coefficients \(X_k\) of the DFT.
Radio waves can be filtered to avoid "noise" and listen to the important components of the signal.
DFT example A
Find the DFT of \((x_0,x_1,x_2,x_3) = (0,1,0,0).\)
In this case, \[ X_k = \sum_{n=0}^3 x_n e^{-2\pi i kn/4} = e^{-2\pi i k/4}. \] So \( X_0 = 1, X_1 = -i, X_2 = -1, X_3 = i.\) The answer is \( (1,-i,-1,i).\)
Proceeding as in the previous example, this gives the expansion \[ x_n = 1 - i e^{2\pi i n/4} - e^{4\pi i n/4} + i e^{6\pi i n/4}. \] Converting the complex coefficients to complex exponentials gives \[ \begin{align} x_n &= 1 + e^{6\pi i/4} e^{2\pi i n/4} + e^{4\pi i/4} e^{4\pi i n/4} + e^{2\pi i/4} e^{6 \pi i n/4} \\ &= 1 + e^{2\pi i(n+3)/4} + e^{2\pi i (2n+2)/4} + e^{2\pi i (3n+1)/4}. \end{align} \] This is an example of phase shifting occurring in the sum. Taking the real parts of both sides gives a sum of cosine waves: \[ x_n = 1 + \cos(2\pi n/4 + 3\pi/2) + \cos(4\pi n/4 + \pi) + \cos(6\pi n/4 + \pi/2), \] where the addition of \(3\pi/2, \pi, \pi/2\) has the effect of shifting the waves forward by \( 270^\circ, 180^\circ, 90^\circ,\) respectively.
Discrete Fourier Transform is with real and imaginary parts - \(ReX(i)\) cosine wave amplitudes and \(ImX(i)\) sine wave amplitudes.
DFT example B
Determine the 8-point DFT of the sequence x(n)={1,1,1,1,1,1,0,0}
\[X(\mathrm{k})=\sum_{n=0}^{N-1} x(\mathrm{n}) \mathrm{e}^{-j 2 \pi k n / N} k=0,1, \ldots N-1\]
for \(N = 8\)
\[X(\mathrm{k})=\sum_{n=0}^7 x(\mathrm{n}) \mathrm{e}^{-j \pi k n / 4} k=0,1,2 \ldots . . N-1\\\]
for \(N = 0\)
\(\begin{aligned} X(0) & =\sum_{n=0}^7 x(n) \\\\ \mathrm{X}(0) & =\mathrm{x}(0)+\mathrm{x}(1)+\mathrm{x}(2)+\mathrm{x}(3)+\mathrm{x}(4)+\mathrm{x}(5)+\mathrm{x}(6)+\mathrm{x}(7) \\\\ & =1+1+1+1+1+1+0+0 \\\\ & =6 \end{aligned}\)
for \(N = 1\)
\(\begin{aligned} X(1) & =\sum_{n=0}^7 x(n) \mathrm{e}^{-j \pi n / 4} \\\\ \mathrm{X}(1) & =\mathrm{x}(0)+\mathrm{x}(1) \mathrm{e}^{-\mathrm{j} \pi / 4}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} \pi / 2}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 3 \pi / 4}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 5 \pi / 4}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 3 \pi / 2}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 7 \pi / 4} \\\\ & =1+0.707-\mathrm{j} 0.707-\mathrm{j}-0.707-\mathrm{j} 0.707-1-0.707+\mathrm{j} 0.707 \quad=-0.707-\mathrm{j} 1.707\\ \end{aligned}\)
for \(N = 2\)
\begin{aligned} X(2) & =\sum_{n=0}^7 x(n) e^{-j \pi n / 2} \\\\ X(2) & =x(0)+\mathrm{x}(1) \mathrm{e}^{-\mathrm{j} \pi / 2}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} \pi}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 3 \pi / 2}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} 2 \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 5 \pi / 2}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 3 \pi}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 7 \pi / 2} \\\\ & =1-\mathrm{j}-1+\mathrm{j}+1-\mathrm{j} \\\\ & =1-\mathrm{j}\\ \end{aligned}
for \(N = 3\)
\(\begin{aligned} X(3) & =\sum_{n=0}^7 x(n) e^{-j 3 \pi n / 4} \\\\ X(3) & =x(0)+x(1) e^{-j 3 \pi / 4}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} \pi / 2}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 9 \pi / 4}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} 3 \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 15 \pi / 4}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 9 / / 4}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 21 \pi / 4} \\\\ & =1-0.707-\mathrm{j} 0.707+\mathrm{j}+0.707-\mathrm{j} 0.707-1+0.707+\mathrm{j} 0.707 \\\\ & =0.707+\mathrm{j} 0.293\\ \end{aligned}\)
for \(N = 4\)
\(\begin{aligned} X(4) & =\sum_{n=0}^7 x(n) \mathrm{e}^{-j \pi n}\\ \\ X(4) & =\mathrm{x}(0)+\mathrm{x}(1) \mathrm{e}^{-j \pi}+\mathrm{x}(2) \mathrm{e}^{-j \pi 2}+\mathrm{x}(3) \mathrm{e}^{-j \pi 3}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} \pi 4}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} \pi 5}+\mathrm{x}(6) \mathrm{e}^{-j \pi 6}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} \pi 7} \\\\ & =1-1+1-1+1-1=0 \end{aligned}\\\)
for \(N = 5\)
\(\begin{aligned} X(5) & =\sum_{n=0}^7 x(n) e^{-j 5 \pi n / 4} \\\\ X(5) & =x(0)+\mathrm{x}(1) \mathrm{e}^{-\mathrm{j} 5 \pi / 4}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} 5 \pi / 2}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 5 \pi n / 4}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} 5 \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 25 \pi / 4}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 15 \pi / 2}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 35 \pi / 4} \\\\ & =1-0.707+\mathrm{j} 0.707-\mathrm{j}+0.707+\mathrm{j} 0.707-1+0.707-\mathrm{j} 0.707 \\\\ & =0.707-\mathrm{j} 0.293\\ \end{aligned}\)
for \(N = 6\)
\(\begin{aligned} X(6) & =\sum_{n=0}^7 x(n) e^{-j 3 \pi n / 2} \\\\ \mathrm{X}(6) & =\mathrm{x}(0)+\mathrm{x}(1) \mathrm{e}^{-\mathrm{j} 3 \pi / 2}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} 3 \pi}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 9 \pi / 2}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} 6 \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 15 \pi}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 9 \pi}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 21 \pi / 2} \\\\ & =1+\mathrm{j}-1-\mathrm{j}+1+\mathrm{j} \Rightarrow \quad=1+\mathrm{j}\\ \end{aligned}\)
for \(N = 7\)
\(\begin{aligned} \mathrm{X}(7) & =\mathrm{x}(0)+\mathrm{x}(1) \mathrm{e}^{-\mathrm{j} 7 \pi / 4}+\mathrm{x}(2) \mathrm{e}^{-\mathrm{j} 7 \pi / 2}+\mathrm{x}(3) \mathrm{e}^{-\mathrm{j} 21 \pi / 4}+\mathrm{x}(4) \mathrm{e}^{-\mathrm{j} 7 \pi}+\mathrm{x}(5) \mathrm{e}^{-\mathrm{j} 35 \pi / 4}+\mathrm{x}(6) \mathrm{e}^{-\mathrm{j} 21 \pi / 2}+\mathrm{x}(7) \mathrm{e}^{-\mathrm{j} 49 \pi / 4} \\\\ & =1+0.707+\mathrm{j} 0.707+\mathrm{j}-0.707+\mathrm{j} 0.707-1-0.707-\mathrm{j} 0.707 \\\\ = & -0.707+\mathrm{j} 1.707 \\\\ & \mathrm{X}(\mathrm{K})=\{6,0.707-\mathrm{j} 1.707,1-\mathrm{j}, 0.707+\mathrm{j} 0.293,0,0.707-\mathrm{j} 0.293,1+\mathrm{j},-0.707+\mathrm{j} 1.707\}\\ \end{aligned}\)
Fast Fourier Transform (FFT)
