111年北科大光電碩士班-工程數學詳解

 國立臺北科技大學 l11學 年度碩 士班招生考試

系所組別 :2401、 2402、 2403光 電工程系碩士班
第一節 工程數學 試題

解答：$$\delta =\begin{bmatrix}0 & 1/2 & 0\\ 1/2& 0 & 0\\ 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}0 & 1 & -1\\ 0& 1 & 1\\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix}1 & 0 & 0\\ 0& 1/2 & 0\\ 0 & 0 & -1/2 \end{bmatrix} \begin{bmatrix}0 & 0& 1\\ 1/2& 1/2 & 0\\ -1/2 & 1/2 & 0 \end{bmatrix} \equiv PDP^{-1} \\ \Rightarrow \cos \delta =P\begin{bmatrix}\cos 1 & 0 & 0\\ 0& \cos(1/2) & 0\\ 0 & 0 & \cos(1/2) \end{bmatrix}P^{-1} = \begin{bmatrix}\cos (1/2) & 0 & 0\\ 0& \cos(1/2) & 0\\ 0 & 0 & \cos 1 \end{bmatrix} \\ \Rightarrow \sin \delta =P\begin{bmatrix}\sin 1 & 0 & 0\\ 0& \sin(1/2) & 0\\ 0 & 0 & -\sin(1/2) \end{bmatrix}P^{-1} = \begin{bmatrix}0& \sin(1/2) & 0\\ \sin(1/2) & 0 & 0\\ 0 & 0 & \sin 1 \end{bmatrix}\\ 因此e^{i\delta} =\cos \delta +i\sin \delta=\begin{bmatrix}\cos (1/2) & 0 & 0\\ 0& \cos(1/2) & 0\\ 0 & 0 & \cos 1 \end{bmatrix}+\begin{bmatrix}0& i\sin(1/2) & 0\\ i\sin(1/2) & 0 & 0\\ 0 & 0 & i\sin 1 \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix}\cos(1/2)& i\sin(1/2) & 0\\ i\sin(1/2) & \cos(1/2) & 0\\ 0 & 0 & i\sin 1 +\cos 1\end{bmatrix}}$$

解答：$$L\{y''-3y'+2y \}=L\{4t-8\} \Rightarrow (s^2-3s+2)Y(s)-2s-1={4\over s^2}-{8\over s} \\ \Rightarrow Y(s)={4\over s^2(s-2)(s-1)}-{8\over s(s-2)(s-1)}+ {2s+1\over (s-2)(s-1)} \\ = \left( {3\over s}+ {2\over s^2}+ {1\over s-2}-{4\over s-1}\right)-\left( {4\over s}+{4\over s-2}-{8\over s-1}\right) +\left({5\over s-2} -{3\over s-1} \right) \\=-{1\over s}+ {2\over s^2}+{2\over s-2}+{1\over s-1} \Rightarrow y(t)=L^{-1}\{Y(s)\} \\ \Rightarrow \bbox[red, 2pt]{y(t)=-1+2t+2e^{2t} +e^t}$$

解答：$$y''+4y=0 \Rightarrow y_h=c_1\cos(2x) +c_2\sin(2x)\\ 令\cases{y_1= \cos(2x)\\ y_2=\sin(2x)} \Rightarrow W=\begin{vmatrix} y_1& y_2 \\ y_1' & y_2'\end{vmatrix} =\begin{vmatrix} \cos(2x)& \sin(2x) \\ -2\sin(2x) & 2\cos(2x) \end{vmatrix} =2 \\ \Rightarrow y_p =-\cos(2x) \int{ \sin(2x)5 x e^{-x}\over 2}dx +\sin(2x) \int{ \cos(2x) 5xe^{-x}\over 2}dx \\ =-{5\over 2} \cos(2x)\left({1\over 25}e^{-x}((3-5x)\sin(2x) -2(5x+2)\cos(2x)) \right)\\ \qquad +{5\over 2}\sin(2x)\left( {1\over 25}e^{-x}(( 10x+4)\sin(2x)+ (3-5x)\cos(2x)) \right) \\={1\over 10}e^{-x}(10x+4) \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos(2x) +c_2\sin(2x)+{1\over 10}e^{-x}(10x+4)}$$

解答：$$(x^3+17){dy\over dx} =x^2 y \Rightarrow {1\over y}dy={x^2\over x^3+17}dx \Rightarrow \ln y={1\over 3}\ln(x^3+17)+c_1 \\ \Rightarrow \bbox[red, 2pt]{y=c_2(x^3+17)^{1/3}}$$

解答：$$f(x)為\text{Gaussian distribution} \Rightarrow \int_{-\infty}^\infty f(x)\,dx =1 \Rightarrow  \int_{-\infty}^\infty e^{-x^2/2a}\,dx =1  \\\Rightarrow a=1/2\pi \Rightarrow f(x)=e^{-\pi x^2 }\\假設\mathcal F(f(x))= F(\omega),現在f(x)=e^{-x^2/2a } \Rightarrow f'(x)=-{x\over a}f(x) \\\Rightarrow \mathcal F(f'(x))=\mathcal F(-{x\over a}f(x)) \Rightarrow i\omega F(\omega)=-{1\over a} iF'(\omega) \Rightarrow F'(\omega)=-a\omega F(\omega)\\ \Rightarrow\bbox[red, 2pt]{ F(\omega)= e^{-a\omega^2/2}= e^{-\omega^2/4\pi}}\\ 註：f(x)為\text{Gaussian distribution},初始值F(0)=0$$

解答：$$\cases{x=r\cos\theta \\ y=r\sin \theta} \Rightarrow \int_{-4}^4 \int_0^{\sqrt{16-x^2}} exp(-(x^2+y^2))\,dydx = \int_0^\pi \int_0^4 r e^{-r^2}\,drd\theta \\ =\int_0^\pi \left. \left[ -{1\over 2} e^{-r^2}\right] \right|_0^4 d\theta=\int_0^\pi -{1\over 2}e^{-16}+{1\over 2}\,d\theta = \bbox[red, 2pt]{{\pi\over 2}\left( 1-e^{-16}\right)}$$

解答：$$\vec F=xy^2\vec i+yz^2\vec j+zx^2 \vec k \Rightarrow div(\vec F)=y^2+z^2+x^2 \\ 依散度定理：封閉曲面的流量(Flux)= \iiint_V div(\vec F)\,dV =\iiint_v (x^2+y^2+z^2)\,dxdydz\\ 球座標\cases{x=\rho \sin \phi \cos\theta \\ y=\rho \sin \phi \sin \theta\\ z=\rho \cos \phi} \Rightarrow Flux=\int_0^{2\pi} \int_0^\pi \int_{\sqrt 2}^{\sqrt 6} \rho^2 \rho^2 \sin \phi  \,d\rho d\phi d\theta \\=\int_0^{2\pi} \int_0^\pi {1\over 5}(36\sqrt 6-4\sqrt 2)\sin \phi d\phi d\theta =\int_0^{2\pi}{2\over 5}(36\sqrt 6-4\sqrt 2) d\theta =\bbox[red, 2pt]{{4\pi\over 5}(36\sqrt 6-4\sqrt 2)}$$

=========================== END ==============================

解題僅供參考,其他歷年試題及詳解