115年中央大學電機碩士班-工程數學詳解

國立中央大學115學年度碩士班考試入學

系所:電機工程碩士班
科目:工程數學(不含複變)

解答:$$\textbf{A. } C = \frac{K - P_0}{P_0}={100-10\over 10}= \bbox[red, 2pt]9 \\ \textbf{B. }P(t) = \frac{K}{1 + C e^{-rt}} ={100\over 1+9e^{-0.3t}} \Rightarrow P(5) =  {100\over 1+9e^{-0.3\times 5}} \approx  \frac{100}{1 + 9(0.2231)} =  \frac{100}{3.0079} \approx \bbox[red, 2pt]{33.25 } \\ \textbf{C. } 50 = \frac{100}{1 + 9 e^{-0.3t}} \Rightarrow e^{0.3t}=9 \Rightarrow 0.3t= \ln 9 \Rightarrow t={\ln 9\over 0.3}  \approx \frac{2.1972}{0.3} \approx \bbox[red, 2pt]{7.324} \text{ years} \\ \textbf{D. } \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \Rightarrow P'(0)= 0.3\times 10 \times \left( 1-{10\over 100} \right)= \bbox[red, 2pt]{2.7} \\ \textbf{E. }P(t)=20={100\over 1+9e^{-0.3t}} \Rightarrow e^{0.3t}={9\over 4} \Rightarrow t={\ln(9/4) \over 0.3} \approx{0.8109\over 0.3}= \bbox[red, 2pt]{2.703} \text{ years}$$

解答:$$\textbf{A. }  \text{characteristic roots: }\pm 2i \Rightarrow \omega = \bbox[red, 2pt]2 \text{ rad/s} \\ \textbf{B. }Q_p=A \Rightarrow Q_p''+4Q_p=4A= 20 \Rightarrow A= \bbox[red, 2pt]5 \\ \textbf{C. }Q(t)=Q_h+Q_p = c_1\cos (2t)+c_2 \sin(2t)+5 \Rightarrow Q'(t)=-2c_1\sin(2t)+2c_2\cos(2t) \\ \Rightarrow Q(0)=c_1+5=0 \Rightarrow c_1=-5 \Rightarrow Q'(0) =2c_2=0 \Rightarrow c_2=0 \Rightarrow Q(t)= -5\cos(2t)+5 \\ \Rightarrow Q(\pi/6)=  5 - 5\cos\left(2 \cdot \frac{\pi}{6}\right)  = \bbox[red, 2pt]{5\over 2} \\ \textbf{D. }Q(\pi/4)  = 5 - 5\cos\left(2 \cdot \frac{\pi}{4}\right) = 5 - 5\cos\left(\frac{\pi}{2}\right) =  \bbox[red, 2pt] 5  \\ \textbf{E. }Q(t)=0 \Rightarrow -5\cos(2t)+5=0 \Rightarrow \cos(2t)=1 \Rightarrow t= \bbox[red, 2pt]\pi$$

解答:$$\textbf{A. }\cases{dx/dt =4x-xy=0\\  dy/dt=-2y+xy=0} \Rightarrow \cases{x(4-y)=0 \Rightarrow x=0,y=4\\ y(-2+x)=0 \Rightarrow y=0,x=2} \Rightarrow (x,y)= \bbox[red, 2pt]{(0,0),(2,4)} \\\textbf{B. }{dy\over dx}={ -2y+xy \over 4x-xy} \Rightarrow \left.{dy\over dx}\right|_{(1,3)} ={-6+3\over 4-3}= \bbox[red, 2pt]{-3} \\\textbf{C. }  H(x, y) = x + y - 2 \ln x - 4 \ln y \Rightarrow H(4,1)=5-2\ln 4-0  = 5 - 2\times 1.3863 = \bbox[red, 2pt]{2.2274} \\ \textbf{D. } \left. {dy\over dx}\right|_{(3,2)} ={-4+6\over 12-6}= \bbox[red, 2pt]{1\over 3}$$

解答:$$[A\mid I]= \left[ \begin{array}{cccc\|cccc} 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 2 & 1 & -3 & 2 & 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 2 & 0 & 0 & 1 & 0 \\ -3 & 0 & 0 & -1 & 0 & 0 & 0 & 1 \end{array} \right]  \xrightarrow{\matrix{R_2 \leftarrow R_2 - 2R_1\\ R_3 \leftarrow R_3 - R_1\\R_4 \leftarrow R_4 + 3R_1}}  \left[ \begin{array}{cccc\|cccc} 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & -5 & 0 & -2 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & 0 & 1 & 0 \\ 0 & 0 & 3 & 2 & 3 & 0 & 0 & 1 \end{array} \right] \\ \xrightarrow{ \matrix{R_1 \leftarrow R_1 - R_3 \\R_2 \leftarrow R_2 + 5R_3 \\R_4 \leftarrow R_4 - 3R_3} }  \left[\begin{array}{cccc\|cccc} 1 & 0 & 0 & 0 & 2 & 0 & -1 & 0 \\ 0 & 1 & 0 & 5 & -7 & 1 & 5 & 0 \\ 0 & 0 & 1 & 1 & -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 & 6 & 0 & -3 & 1 \end{array}\right] \xrightarrow{ \matrix{R_2 \leftarrow R_2 - 5R_4 \\R_3 \leftarrow R_3 - R_4}} \left[ \begin{array}{cccc\|cccc} 1 & 0 & 0 & 0 & 2 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 & 23 & 1 & -10 & 5 \\ 0 & 0 & 1 & 0 & 5 & 0 & -2 & 1 \\ 0 & 0 & 0 & 1 & -6 & 0 & 3 & -1 \end{array} \right]\\ \Rightarrow  A^{-1} = \bbox[red, 2pt]{\begin{bmatrix} 2 & 0 & -1 & 0 \\  23 & 1 & -10 & 5 \\  5 & 0 & -2 & 1 \\  -6 & 0 & 3 & -1 \end{bmatrix} }$$

解答:$$ B = \begin{bmatrix} 3 & 3 \\ 2 & 0 \\ 0 & 2 \end{bmatrix} \Rightarrow B^TB = \begin{bmatrix} 3 & 2 & 0 \\ 3 & 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 3 \\ 2 & 0 \\ 0 & 2 \end{bmatrix}  = \begin{bmatrix} 13 & 9 \\ 9 & 13 \end{bmatrix}  \Rightarrow \det(B^TB-\lambda I)= \lambda^2-26\lambda+88=0 \\ \Rightarrow (\lambda-22)(\lambda-4)=0 \Rightarrow \cases{\lambda_1= 22\\ \lambda_2=4} \Rightarrow \cases{\sigma_1= \sqrt{\lambda_1} = \sqrt{22} \\ \sigma_2= \sqrt{\lambda_2}=2} \Rightarrow  \Sigma = \begin{bmatrix} \sqrt{22} & 0 \\ 0 & 2 \\ 0 & 0 \end{bmatrix} \\ \cases{\lambda_1 \Rightarrow \text{unit eigenvector }v_1= \begin{bmatrix}1/\sqrt 2\\ 1/\sqrt 2 \end{bmatrix} \\ \lambda_2 \Rightarrow \text{unit eigenvector }v_2 = \begin{bmatrix}1/\sqrt 2\\ -1/\sqrt 2 \end{bmatrix}} \Rightarrow  V = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix}  \\ \Rightarrow \cases{ u_1 = \frac{1}{\sqrt{22}} \begin{bmatrix} 3 & 3 \\ 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix} = \frac{1}{\sqrt{44}} \begin{bmatrix} 6 \\ 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 6/\sqrt{44} \\ 2/\sqrt{44} \\ 2/\sqrt{44} \end{bmatrix} = \begin{bmatrix} 3/\sqrt{11} \\ 1/\sqrt{11} \\ 1/\sqrt{11} \end{bmatrix} \\  u_2 = \frac{1}{2} \begin{bmatrix} 3 & 3 \\ 2 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 0 \\ 2/\sqrt{2} \\ -2/\sqrt{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 1/\sqrt{2}  \\ -1/\sqrt{2} \end{bmatrix} } \Rightarrow u_3= {u_1\times u_2 \over |u_1\times u_2|} \\ \Rightarrow u_3=   \begin{bmatrix} -2/\sqrt{22} \\ 3/\sqrt{22} \\ 3/\sqrt{22} \end{bmatrix} \Rightarrow  U = \begin{bmatrix} 3/\sqrt{11} & 0 & -2/\sqrt{22} \\ 1/\sqrt{11} & 1/\sqrt{2} & 3/\sqrt{22} \\ 1/\sqrt{11} & -1/\sqrt{2} & 3/\sqrt{22} \end{bmatrix}  \\ \Rightarrow B=\bbox[red, 2pt]{ \begin{bmatrix} 3/\sqrt{11} & 0 & -2/\sqrt{22} \\ 1/\sqrt{11} & 1/\sqrt{2} & 3/\sqrt{22} \\ 1/\sqrt{11} & -1/\sqrt{2} & 3/\sqrt{22} \end{bmatrix} \begin{bmatrix} \sqrt{22} & 0 \\ 0 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix}}$$

解答:$$\cases{ F(t) = [\sin(3(t - 1)) + \cos(3(t - 1))] U(t - 1) \\  g(t) = [\sin(3t) + \cos(3t)] U(t) } \Rightarrow F(t)=g(t-1) \Rightarrow \hat F(\omega) = \mathcal F\{F(t)\} = e^{-j\omega}\mathcal F\{g(t)\} \\ \cases{\mathcal F\{\sin(3t)U(t)\} =  \frac{3}{9 - \omega^2} + \frac{\pi}{2j} [\delta(\omega - 3) - \delta(\omega + 3)] \\ \mathcal{F}\{\cos(3t) U(t)\} = \frac{j\omega}{9 - \omega^2} + \frac{\pi}{2} [\delta(\omega - 3) + \delta(\omega + 3)] } \\\Rightarrow \mathcal F\{g(t)\} = \mathcal F\{\sin(3t)U(t)\}+ \mathcal F\{\cos(3t)U(t)\} =  \frac{3 + j\omega}{9 - \omega^2} + \frac{\pi}{2} \left[ (1 - j)\delta(\omega - 3) + (1 + j)\delta(\omega + 3) \right] \\ \Rightarrow  \hat F(\omega) =  \bbox[red, 2pt]{e^{-j\omega} \left( \frac{3 + j\omega}{9 - \omega^2} + \frac{\pi}{2} \left[ (1 - j)\delta(\omega - 3) + (1 + j)\delta(\omega + 3) \right] \right) }$$

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解題僅供參考，碩士班歷年試題及詳解