114年中山大學電機碩士班-工程數學甲詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學甲【電機系碩士班戊組選考、庚組、通訊所碩士班乙組選考、電波聯合碩士班選考】

第1-6題為複選題,每題5分,總分30分。每題有5個選項,其中至少有1個是正確答案,答錯1個選項者,得3分;答錯2個選項者,得1分;答錯多於2個選項或未作答者,該題以零分計算。

解答:$$(E)\bigcirc: \mathbf a_1+2\mathbf a_2=3\mathbf a_3 \Rightarrow \mathbf a_1+2\mathbf a_2-3\mathbf a_3=0, \text{the coefficients (1,2,-3) are not all zero.} \\ \quad \text{By definition, the vectors }\mathbf a_1, \mathbf a_2,\mathbf a_3 \text{ are linearly dependent.} \\(C) \bigcirc: \text{ Matrix }A\text{ has 3 columns, and }\mathbf a_1, \mathbf a_2,\mathbf a_3 \text{ are linearly dependent.} \Rightarrow rank(A)\le 2 \\(A)\times:  \mathbf a_1+ \mathbf a_2+\mathbf a_3 =\mathbf b \Rightarrow A \begin{bmatrix}1\\1\\1 \end{bmatrix}=\mathbf b \Rightarrow \text{ at least one solution }[1,1,1]^T\\\quad \text{And }rank(A)\le 2 \Rightarrow \text{ there are infinitely many solutions.} \\(B) \bigcirc: \text{ the linear system is consistent }\Rightarrow rank(A)= rank([A,b]) \\(D)\times: [1,2,3]^T \text{ is a solution} \Rightarrow \mathbf a_1+2 \mathbf a_2+3\mathbf a_3 =\mathbf b =\mathbf a_1+\mathbf a_2+\mathbf a_3 \Rightarrow \mathbf a_2+2\mathbf a_3=0 \\ \quad \text{This contradicts the given condition }\mathbf a_2+2\mathbf a_3\ne 0 \Rightarrow [1,2,3]^T \text{ is NOT a solution} \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(BCE)}$$

解答:$$\textbf{(A) }\bigcirc: A= \begin{bmatrix} 2 & 0 & 0 & 0 & 2\\0 & 1 & 1 & 1 & 0\\0 & 1 & 1 & 1 & 0\\0 & 1 & 1 & 1 & 0\\2 & 0 & 0 & 0 & 2\end{bmatrix} \xrightarrow{R_5-R_1\to R_5, R_3-R_2\to R_3, R_4-R_2\to R_4} \begin{bmatrix}2 & 0 & 0 & 0 & 2\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix} \\\qquad \xrightarrow{R_1/2\to R_1} \begin{bmatrix}1 & 0 & 0 & 0 & 1\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow rref(A)= \begin{bmatrix}1 & 0 & 0 & 0 & 1\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \\\textbf{(B)}\bigcirc: trace(A)=2+1+1+1+2= 7 \\\textbf{(C)} \times: rank(A)=2 \ne 5 \Rightarrow \det(A)=0 \\\textbf{(D)} \bigcirc: \det(A-\lambda I) =-\lambda^3(\lambda-3)(\lambda-4) =0 \Rightarrow 3 \text{ is an eigenvalue of }A \\\textbf{(E)}\bigcirc: \text{All eigenvalues (0,3,4) are non-negative} \\ \Rightarrow \text{The true statements are }\bbox[red, 2pt]{(ABDE)}$$

解答:$$Q \text{ is an orthogonal matrix} \Rightarrow Q^TQ=I \Rightarrow A^TA=(QR)^T(QR) =R^T(Q^TQ)R =R^TIR \\ \Rightarrow A^TA=R^TR = \begin{bmatrix}1& 0& 0\\ 2&4& 0\\ 3& 5& 6 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 0&4& 5\\ 0& 0& 6 \end{bmatrix} = \begin{bmatrix}1& 2& 3\\ 2& 20& 26\\ 3& 26& 70 \end{bmatrix} = (d_{i ,j})_{3\times 3} \\ (A) \bigcirc: \mathbf a_3^T \mathbf a_1 = d_{3,1} =3 \\(B)\times :\mathbf a_3^T \mathbf a_2 = d_{3,2}=26 \ne 24\\ (C) \bigcirc: \mathbf a_3^T \mathbf a_3 =  d_{3,3}=70\\ (D)\times \mathbf a_2^T \mathbf a_1 = d_{2,1}=2 \ne 3\\ (E) \times: \mathbf a_2^T \mathbf a_2 = d_{2,2}=20\ne 30 \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(AC)}$$

解答:$$C=[\mathbf a_1\; \mathbf a_2] = \begin{bmatrix}1& 2\\ 0& 1\\ 1& 0\\ 2& 2 \end{bmatrix} \Rightarrow P=C(C^TC)^{-1}C^T = \left[\begin{matrix}\frac{1}{2} & \frac{1}{3} & - \frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\- \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} & \frac{1}{3}\\\frac{1}{3} &0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right] \\ \Rightarrow P \mathbf a_3 = \left[\begin{matrix}\frac{1}{2} & \frac{1}{3} & - \frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\- \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} & \frac{1}{3}\\\frac{1}{3} &0 & \frac{1}{3} & \frac{2}{3}\end{matrix} \right] \begin{bmatrix} 3\\4\\5\\6 \end{bmatrix} = \begin{bmatrix}4\\{2\over 3} \\{8\over 3} \\{20\over 3} \end{bmatrix} =\mathbf q_3 \Rightarrow \cases{q_{31} =4\\ q_{32} =2/3 \ne 4/3\\ q_{33} =8/3 \ne 10/3\\ q_{34} =20/3\ne 22/3} \\ \Rightarrow \mathbf a_1^T \mathbf q_3 = \begin{bmatrix} 1& 0& 1& 2 \end{bmatrix} \begin{bmatrix} 4 \\{2\over 3} \\{8\over 3} \\{20\over 3} \end{bmatrix} =20 \Rightarrow \text{ True statements: }\bbox[red, 2pt]{(AE)}$$

解答:$$(A)\times: A= \begin{bmatrix}0&-1\\ 1&0 \end{bmatrix} \Rightarrow AA^T= I, \text{ but trace}(A)=0\\ (B)\times: A= \begin{bmatrix} 1&0\\0&-1\end{bmatrix} \Rightarrow AA^T=I, \text{ but det}(A)=-1\ne 0 \\(C)\bigcirc: AA^T=I \Rightarrow A^{-1} =A^T \text{ exists  } \Rightarrow \text{rank}(A)=n \\(D)\bigcirc: A \text{ is an orthogonal matrix} \Rightarrow |\lambda_i| =1, \lambda_i \text{ is an eigenvalue of }A \Rightarrow \text{trace}(A) = \sum \lambda_i \le 3 \\(E)\times: A= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2+1=0 \Rightarrow \lambda=\pm i \not \in \mathbb R \\ \Rightarrow \text{True statements: }\bbox[red, 2pt]{(CD)}$$

解答:$$(A)\times: A= \begin{bmatrix}1& 0\\0&1\\ 0& 0 \end{bmatrix} \Rightarrow \begin{bmatrix}1\\0\\0 \end{bmatrix} \text{ and }\begin{bmatrix}1\\0\\0 \end{bmatrix} \text{ are linearly independent, but }\\\qquad AA^T= \begin{bmatrix}1& 0& 0\\ 0& 1&0\\ 0&0&0 \end{bmatrix} \text{ is singular} \\(B) \bigcirc: A\text{ is }m\times n \Rightarrow A^T\text{ is }n\times m \Rightarrow \text{rank}(A^T)+ \dim(N(A^T)) =m \\(C)\times: \cases{AA^T\text{ is }m\times m\\ A^TA\text{ is }n\times n} \Rightarrow R(AA^T) \ne R(A^TA), \text{if }m\ne n \\(D)\times: A= \begin{bmatrix}1& 2 \end{bmatrix} \Rightarrow \cases{AA^T= [5] \\ A^TA = \begin{bmatrix}1&2\\3& 4 \end{bmatrix} } \Rightarrow \cases{\det(AA^T)=5\ne 0\\ \det(A^TA)=0} \\(E)\times: [3,4,5] \cdot [1,-1,2]^T=9 \ne 0 \\ \Rightarrow \text{True statement: }\bbox[red, 2pt]{(B)}$$

第7-13題為單選題,總分35分。每題答對5分,答錯或未作答者以0分計。

解答:$$L^{-1} \left\{{-5\over s+16} \right\}= -5L^{-1} \left\{{1\over s+16} \right\} =-5e^{-16t} \Rightarrow \bbox[red, 2pt]{(C)}$$

解答:$$\mathcal F\{u(t)\} =\pi \delta(\omega)+{1\over j\omega} \ne \pi\delta(\omega) \Rightarrow \text{The wrong statement is }\bbox[red, 2pt]{(B)}$$

解答:$$(B)\times: \text{Taking the limit to an infinite number of terms ($N \to \infty$) in the Fourier series,}\\\text{ the oscillatory overshoot at the discontinuity does not disappear.}\\\text{The amplitude of the overshoot approaches a constant finite limit, not zero.} \\ \Rightarrow \text{The wrong statement: }\bbox[red, 2pt]{(B)}$$

解答:$$N=10 \Rightarrow a_k=a_{k+10} \Rightarrow a_1=a_{11}=5 \\x[n] \text{ is real and even }\Rightarrow a_k=a_{-k} \Rightarrow a_{-1}=a_1=5 \\ \text{By Parseval's realtion, we have }{1\over N}\sum_{n=0}^{N-1} |x[n]|^2 = \sum_{k=0}^{N-1}|a_k|^2\\ \text{Now, }|a_1|^2+|a_{-1}|^2=5^2+5^2=50 \Rightarrow a_k=0 \text{ for }k\ne 1,9 (\text{mod }10) \\ x[n] =\sum_{k=0}^{N-1}a_ke^{jk\omega_0 n}, \text{ where }\omega_0 ={2\pi\over N=10} ={\pi\over 5} \Rightarrow x[n]=a_1e^{j\pi n/5}+a_{-1}e^{-j\pi n/5} \\= 5e^{j\pi n/5}+ 5e^{-j\pi n/5} =5\cdot 2\cos{\pi n\over 5} =10\cos {n\pi\over 5} \Rightarrow \cases{A=10\\ B=\pi/5\\ C=0} \Rightarrow \bbox[red, 2pt]{(A)}$$

解答:$$\mathcal F\{\delta(t+1)\}+\mathcal F\{ \delta(t-1)\} = \int_{-\infty}^\infty \delta(t+1)e^{-j\omega  t}\,dt + \int_{-\infty}^\infty \delta(t-1)e^{-j\omega  t}\,dt  =e^{-j\omega(-1)} +e^{-j\omega(1)} \\=e^{j\omega } +e^{-j\omega } \Rightarrow \bbox[red, 2pt]{(A)}$$

解答:$$x[n]= \cos {3\pi n\over 5} ={1\over 2} \left( e^{j3\pi n/5}+e^{-j 3\pi n/5} \right)\\ \Rightarrow X(e^{j\omega}) ={1\over 2} \sum_{\ell=-\infty}^\infty \left( 2\pi\delta(\omega-{3\pi\over 5}-2\pi \ell)+2\pi \delta(\omega+{3\pi\over 5}-2\pi \ell) \right) \\= \sum_{\ell=-\infty}^\infty \pi \left(   \delta(\omega-{3\pi\over 5}-2\pi \ell)+  \delta(\omega+{3\pi\over 5}-2\pi \ell) \right) \Rightarrow \bbox[red, 2pt]{(C)}$$

解答:$$L\{y''\}-2L\{y'\}-8L\{y\}=L\{f(t)\} \Rightarrow s^2Y(s)-s-2sY(s)+2-8Y(s) =F(s) \\ \Rightarrow Y(s) ={F(s)+s-2\over s^2-2s-8} = \left( {1/6\over s-4}-{1/6\over s+2} \right)F(s)+{1/3\over s-4}+{2/3\over s+2} \\ \Rightarrow y(t) =L^{-1}\{Y(s)\} ={1\over 6}e^{4t}*f(t)-{1\over 6}e^{-2t}*f(t) +{1\over 3}e^{4t}+{2\over 3}e^{-2t} \\ \Rightarrow \cases{A=1/6\\ B=-1/6\\ C=1/3\\ D=2/3} \Rightarrow \text{The correct option is }\bbox[red, 2pt]{(A)}$$

第14題到第17題需要詳明推導計算過程。如推導計算過程錯誤,將酌扣分數或不給分。

解答:$${dM_z(t) \over dt}={M_0-M_z(t) \over T_1} \Rightarrow \int {dM_z\over M_0-M_z} =\int{1\over T_1}dt \Rightarrow -\ln(M_0-M_z)={t\over T_1}+c_1 \\ \Rightarrow M_0-M_z=c_2e^{-t/T_1} \Rightarrow M_z(t)=M_0-c_2e^{-t/T_1} \Rightarrow M_z(t)=0=M_0-c_2 \Rightarrow c_2=M_0 \\ \Rightarrow \bbox[red, 2pt]{M_z(t)=M_0 \left( 1-e^{-t/T_1} \right)}$$

解答:$$u(x,y) =X(x)Y(y) \Rightarrow {1\over y}X'Y={1\over 3x}XY' \Rightarrow {3xX'\over X} ={yY'\over Y} =k \\ \text{Solving for }X(x): {X'\over X}={k\over 3x} \Rightarrow \ln|X|={k\over 3}\ln |x|+c_1 \Rightarrow X=c_2x^{k/3} \\\text{Solving for }Y(y): {Y'\over Y} ={k\over y} \Rightarrow \ln |Y|= k\ln |y|+c_3 \Rightarrow Y=c_4y^k \\ \Rightarrow u(x,y)= (c_2x^{k/3}) (c_4y^k) \Rightarrow \bbox[red, 2pt]{u(x,y)=c_5 \left( x^{1/3}y \right)^k, c_5\text{ and }k \text{ are constants}}$$

解答:$$\textbf{(1) }\text{Top Surface }S_1: f(x,y,z)=x^2+y^2+z-2=0 \Rightarrow \nabla f=(f_x,f_y,f_z) =(2x,2y,1) \\ \qquad \Rightarrow \text{unit normal vector }\vec n_1 ={\nabla f\over |\nabla f|} \Rightarrow \bbox[red, 2pt]{\vec n_1= {(2x,2y,1)\over \sqrt{4x^2+4y^2+1}}} \\ \text{Bottom Surface }S_2:z=0 \Rightarrow \text{unit normal vector } \bbox[red, 2pt]{\vec n_2=(0,0,-1)} \\ \textbf{(2) }\text{Area of }S_1= \iint_D \sqrt{1+z_x^2+z_y^2}\,dA= \iint_D \sqrt{1+4x^2+ 4y^2}\,dA =\int_0^{2\pi} \int_0^{\sqrt 2} \sqrt{1+4r^2}\, r\,drd\theta \\\qquad =2\pi \cdot {1\over 8} \int_1^9 \sqrt u\,du ={13\pi\over 3} \quad (u=1+4r^2) \\\text{Area of }S_2=(\sqrt 2)^2 \pi=2\pi \\ \Rightarrow \text{ total area: }{13\pi\over 3}+2\pi = \bbox[red, 2pt]{19\pi\over 3}$$

解答:$$\textbf{(1) }z=x+iy \Rightarrow z\bar z=2 \Rightarrow (x+iy)(x-iy)=x^2+y^2=2 \\ \qquad \Rightarrow \bbox[red, 2pt]{\text{It is a circle centered at the origin with a radius of }R=\sqrt 2} \\\textbf{(2) } f(z)={e^{-z} \over z^2+1} \Rightarrow \cases{\text{Res}(f,i) =e^{-i}/2i \\ \text{Res}(f,-i)= e^i/(-2i)} \Rightarrow \oint_C f(z)dz = 2\pi i(\text{Res}(f,i)+ \text{Res}(f,-i)) \\\qquad =2\pi i \left( {e^{-i}\over 2i} -{e^i\over 2i} \right) =2\pi (-\sin 1) =\bbox[red, 2pt]{-2\pi i \sin 1}$$

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解題僅供參考，碩士班歷年試題及詳解