112年台科大自動控制碩士班-工程數學詳解

國立臺灣科技大學112學年度碩士班招生試題

系所組別:自動化及控制研究所碩士班
科目:工程數學

解答:$$y''-y=0 \Rightarrow \lambda^2-1=0  \Rightarrow \lambda=\pm 1 \Rightarrow y_h=c_1e^x +c_2e^{-x} \\ y''-y=5\sin^2(x) ={5\over 2}(1-\cos(2x)) \Rightarrow y_p=A+ B\cos(2x)+C\sin(2x) \\ \Rightarrow y_p'=-2B\sin(2x) +2C\cos(2x) \Rightarrow y_p''=-4B\cos(2x)-4C\sin(2x) \\ \Rightarrow y_p''-y_p=-A-5B\cos(2x)-5C\sin(2x) ={5\over 2}-{5\over 2}\cos(2x) \Rightarrow \cases{A=-5/2\\ B=1/2\\ C=0} \\ \Rightarrow y_p=-{5\over 2}+{1\over 2}\cos(2x) \Rightarrow y=y_h+y_p \Rightarrow y=c_1e^x +c_2e^{-x}-{5\over 2}+{1\over 2}\cos(2x) \\ \Rightarrow y'=c_1e^x -c_2e^{-x}-\sin(2x) \Rightarrow \cases{y(0)=c_1+c_2-5/2+1/2=2\\ y'(0)=c_1-c_2= -4} \Rightarrow \cases{c_1=0\\ c_2=4} \\ \Rightarrow \bbox[red, 2pt]{y=4e^{-x}-{5\over 2}+{1\over 2}\cos(2x)}$$

解答:$$f(t+2) =f(t) \Rightarrow L\{f(t)\} ={\int_0^2 f(t)e^{-st}\,dt \over 1-e^{-2s}} ={1\over 1-e^{-2s}} \int_0^2 te^{-st}\,dt \\={1\over 1-e^{-2s}} \cdot {1\over s^2}\left( 1-e^{-2s}(2s+1)\right) = \bbox[red, 2pt] {{1\over s^2(1-e^{-2s})}(1-(2s+1)e^{-2s})}$$

解答:$$y'={dy\over dx} ={y\over x+y} \Rightarrow ydx-(x+y)dy=0 \Rightarrow \cases{P(x,y)=y\\ Q(x,y)=-(x+y)} \Rightarrow \cases{P_y=1\\ Q_x=-1} \\ \Rightarrow u'=-{P_y-Q_x\over P}u=-{2\over y}u \Rightarrow y^2u'+2yu=0 \Rightarrow (y^2u)'=0 \Rightarrow y^2u=c_1 \\ \Rightarrow \text{integration factor }u={1\over y^2} \Rightarrow \cases{uP=1/y\\ uQ=-(x+y)/y^2} \Rightarrow (uP)_y=-{1\over y^2} =(uQ)_x \\ \Rightarrow \Phi(x,y)= \int{1\over y}dx =\int -{x+y\over y^2}\,dy \Rightarrow \Phi(x,y)={x\over y} +\phi(y) ={x\over y}-\ln y +\rho(x) \\ \Rightarrow \bbox[red, 2pt]{{x\over y}-\ln y=c_1}$$

解答:$$a_0= {1\over \pi}\int_{-\pi}^\pi (x-x^2)\,dx = -{2\over 3}\pi^2 \\a_n= {1\over \pi} \int_{-\pi}^\pi (x-x^2) \cos(nx)\,dx =-{4\over n^2}(-1)^n \\ b_n={1\over \pi} \int_{-\pi}^\pi (x-x^2) \sin(nx)\,dx =-{2 \over n}(-1)^n \\ \Rightarrow f(x)\sim\bbox[red, 2pt]{ -{1\over 3}\pi^2 - \sum _{n=1}^\infty{2 \over n}(-1)^n \left[ {2\over n} \cos (nx)+\sin(nx)\right] }$$

解答:$$\mathcal F\left\{ {d\over dt} te^{-2t} \cos(t)u(t) \right\} =i\omega\mathcal F\left\{  te^{-2t} \cos(t)u(t) \right\} =i\omega \cdot i {d\over d\omega}\mathcal F\left\{   e^{-2t} \cos(t)u(t) \right\}\\ =-\omega  {d\over d\omega}  \mathcal F\left\{   e^{-2t} \cos(t)u(t) \right\} =-\omega \cdot  {d\over d\omega} \left({2 +i\omega\over 1+(2+i\omega)^2} \right) \\=-\omega\cdot \left({i\over 1+(2+i \omega)^2} -{(2+i\omega) (2(2+i\omega)i) \over (1+(2+ i\omega)^2)^2} \right) =-\omega \cdot {i\omega^2+4\omega -3i\over (1+(2+i\omega)^2)^2} =-{i\omega^3+4\omega^2 -3i \omega\over (1+(2+i\omega)^2)^2} \\ \Rightarrow \mathcal F\{x(t)\} = i{d\over d\omega} \left(-{i\omega^3+4\omega^2 -3i \omega\over (1+(2+i \omega)^2)^2} \right) \\\qquad =i\cdot \left(-{3i\omega^2+ 8\omega -3i \over (1+(2+i\omega)^2)^2} +{ -4i\omega^4 -24\omega^3 +44i \omega^2 + 24\omega\over (1+(2+i\omega)^2)^3}\right)\\= {3\omega^2- 8i\omega -3 \over (1+(2+i\omega)^2)^2} +{ 4\omega^4 -24i\omega^3 -44\omega^2 +24i\omega\over (1+(2+i\omega)^2)^3} = \bbox[red, 2pt]{\omega^4-4i\omega^3 +6\omega^2-28i\omega -15\over (1+(2+i\omega)^2)^3} $$

解答:$$\textbf{(1) }A=\begin{bmatrix}\sin \theta &-\cos\theta & 0\\ \cos\theta & \sin \theta & 0\\ 0 & 0 & 1 \end{bmatrix} \Rightarrow AA^T= \begin{bmatrix}\sin \theta &-\cos\theta & 0\\ \cos\theta & \sin \theta & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}\sin \theta &\cos\theta & 0\\ -\cos\theta & \sin \theta & 0\\ 0 & 0 & 1 \end{bmatrix} \\ =\begin{bmatrix} \cos^2 \theta+\sin^2\theta & \sin \theta\cos \theta -\sin \theta \cos \theta & 0\\ \cos \theta\sin \theta-\cos \theta \sin \theta& \cos^2 \theta+ \sin^2 \theta & 0\\ 0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 1& 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1\end{bmatrix} =I\\ \Rightarrow A \text{ is orthogonal} \quad \bbox[red, 2pt]{QED.}\\ \textbf{(2) }\det(A-\lambda I) = -\lambda^3+(2 \sin x+1)\lambda^2-(2 \sin x+1)\lambda+1=-(\lambda-1)(\lambda^2- (2\sin x) \lambda +1) =0 \\ \qquad \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{1, \sin x \pm i\cos x   }$$

解答:$$\text{Let }\cases{x=(a_1,a_2)\\ y=(b_1,b_2)} \Rightarrow \langle T(x),y \rangle =\langle x,T^*(y) \rangle \Rightarrow \langle (-2ia_1+3a_2,a_1-a_2), (b_1,b_2)\rangle =\langle(a_1,a_2), T^*(y)) \rangle \\ \Rightarrow \langle (a_1,a_2),T^*(y) \rangle= -2ia_1b_1+3a_2b_1+a_1b_2-a_2b_2 =a_1(-2ib_1+b_2)+ a_2(3b_1-b_2) \\ \Rightarrow \bbox[red, 2pt]{T^*(b_1,b_2) =(-2ib_1+b_2, 3b_1-b_2)}$$

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解題僅供參考，碩士班歷年試題及詳解