110年中山大學資工碩士班-工程數學詳解

 國立中山大學110學年度碩士班招生考試

科目名稱:工程數學【資工系碩士班乙組】

解答: $$A= \begin{bmatrix}1 & 5 & 0 & 2 & -2 & 4\\0 & 1 & 0 & 0 & 4 & 8\\0 & 0 & 0 & 1 & 7 & -2 \end{bmatrix} \xrightarrow{R_1-5R_2 \to R_1} \begin{bmatrix}1 & 0 & 0 & 2 & -22 & -36\\0 & 1 & 0 & 0 & 4 & 8\\0 & 0 & 0 & 1 & 7 & -2 \end{bmatrix}  \xrightarrow{R_1-2R_3\to R_1} \\ \begin{bmatrix}1 & 0 & 0 & 0 & -36 & -32\\0 & 1 & 0 & 0 & 4 & 8\\0 & 0 & 0 & 1 & 7 & -2 \end{bmatrix} \Rightarrow \cases{v-36z=-32\\ w+4z=8\\ y+7z=-2} \\ \Rightarrow (v,w,x,y,z)= \bbox[red, 2pt]{(36z-32,-4z+8,x,-2-7z,z),x,z \in \mathbb R}$$

解答: $$\textbf{2.1 } A=\begin{bmatrix} 3 & 0 & 0\\ 0 & 2& 1\\ 0& 1& 2 \end{bmatrix} \Rightarrow \det(A-\lambda I) =(3-\lambda)(2-\lambda)^2-(3-\lambda)= \bbox[red, 2pt]{-(\lambda-3)^2 (\lambda-1)} \\\textbf{2.2 }\det(A-\lambda I)=0 \Rightarrow \lambda=\bbox[red, 2pt]{1,3}$$

解答: $$H(s)={2s^2-10s+1 \over s^2+3s+2} =2+{13\over s+1}-{29 \over s+2} \\ \Rightarrow h(t)= L^{-1}\{H(s)\} =L^{-1}\{ 2\} +L^{-1}\{{13\over s+1}\} -L^{-1}\{{29\over s+2}\}= \bbox[red, 2pt]{2\delta(t)+13e^{-t}-29e^{-2t}}$$

解答: $$\textbf{4.1 }a_k={1\over 8} \int_{-4}^0 (4+t)e^{-j(2\pi/8)kt}\, dt \Rightarrow \bbox[red, 2pt]{x(t)=\begin{cases} 0, & 0\lt t\le 4\\ 4+t, & -4\le t\lt 0\end{cases}} \\\textbf{4.2 }$$

$$\textbf{4.3 }a_0={1\over 8} \int_{-4}^0 (4+t) \, dt  ={1\over 8} \left. \left[ 4t+{1\over 2}t^2 \right] \right|_{-4}^0 ={1\over 8} \cdot 8= \bbox[red, 2pt] 1$$

解答: $$\text{Let }\cases{x_1=y \\ x_2=y' \\x_3=y''} \Rightarrow y'''-4y'=0 \Rightarrow x_3'=4x_2 \Rightarrow \begin{bmatrix} x_1'\\ x_2'\\ x_3'\end{bmatrix} =\begin{bmatrix} 0& 1& 0\\ 0& 0& 1\\ 0& 4& 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} +\begin{bmatrix} 0\\ 0\\ t+ 3\cos t+e^{-2t}\end{bmatrix}\\ C=\begin{bmatrix} 0& 1& 0\\ 0& 0& 1\\ 0& 4& 0\end{bmatrix} \Rightarrow \det(C- \lambda I)=-(\lambda^3-4\lambda)=0 \Rightarrow \lambda=-2,0,2 \\ \Rightarrow V=\begin{bmatrix} 1 & 1 & 1 \\\lambda_1 & \lambda_2 &  \lambda_3\\\lambda_1^2  & \lambda_2^2 & \lambda_3^2 \end{bmatrix} =\begin{bmatrix}1 & 1 & 1 \\-2 & 0 & 2 \\4 & 0 & 4 \end{bmatrix} \Rightarrow V^{-1} =\begin{bmatrix}0 & \frac{-1}{4} & \frac{1}{8} \\1 & 0 & \frac{-1}{4} \\0 & \frac{1}{4} & \frac{1}{8} \end{bmatrix} \\ \Rightarrow \begin{bmatrix} z_1'\\ z_2'\\ z_3'\end{bmatrix} = \begin{bmatrix}-2 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix} z_1\\ z_2\\ z_3\end{bmatrix} +\begin{bmatrix} {1\over 8}\\ -{1\over 4}\\ {1\over 8}\end{bmatrix} (t+ 3\cos t+e^{-2t}) \Rightarrow \cases{z_1'= -2z_1 +{1\over 8} (t+ 3\cos t+e^{-2t}) \\z_2'=-{1\over 4}(t+ 3\cos t+e^{-2t}) \\z_3'=2z_3+{1\over 8}(t+ 3\cos t+e^{-2t})} \\ \Rightarrow \cases{z_1=c_1e^{-2t}+{1\over 8}te^{-2t}+{1\over 16}t+ {3\over 40}\sin t+{3\over 20}\cos t-{1\over 32} \\z_2= c_2-{1\over 8}t^2 -{3\over 4}\sin t +{1\over 8}e^{-2t} \\z_3= c_3e^{2t} -{t\over 16}-{1\over 32}e^{-2t}+ {3\over 40}\sin t-{3\over 20} \cos t-{1\over 32}} \\\Rightarrow \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 \\-2 & 0 & 2 \\4 & 0 & 4 \end{bmatrix} \begin{bmatrix} z_1\\ z_2\\ z_3\end{bmatrix} \Rightarrow y(t)=x_1=z_1+z_2+z_3 \\ \Rightarrow y= \left( c_2-{1\over 16}\right)-{1\over 8}t^2+ c_3e^{2t}+ \left(c_1+{1\over 8}-{1\over 32} \right) e^{-2t} +{1\over 8}te^{-2t}-{3\over 5}\sin t\\ \Rightarrow \bbox[red, 2pt]{y=c_4-{1\over 8}t^2 +c_3e^{2t} +c_5 e^{-2t} +{1\over 8}te^{-2t}-{3\over 5}\sin t}$$

解答: $$y''+4y=3\sin(2t) \Rightarrow L\{y'' \}+4L\{y\} =3L\{\sin(2t)\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+4Y(s) =3\cdot {2\over s^2 +2^2} \Rightarrow (s^2+4)Y(s) ={6\over s^2+4}+2s-1 \\ \Rightarrow Y(s)={6\over (s^2+4)^2} +{2s-1\over s^2+4} ={6\over s^2+4}+ {2s\over (s^2+4)^2}-{1\over (s^2+4)^2} \\ \Rightarrow y(t)= L^{-1}\{Y(s)\}=L^{-1}\{{6\over s^2+4}\}+ L^{-1}\{{2s\over (s^2+4)^2} \}- L^{-1} \{{1\over (s^2+4)^2}\} \\\qquad =  {6\over 16}(\sin(2t)-2t\cos(2t))+2\cos(2t)-{1\over 2}\sin(2t)\\ \Rightarrow \bbox[red, 2pt]{y(t)=-{1\over 8}\sin(2t) + 2\cos(2t)-{3\over 4}t\cos(2t)}$$

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解題僅供參考，碩士班歷年試題及詳解