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:δ=[01/201/200001]=[01−1011100][10001/2000−1/2][0011/21/20−1/21/20]≡PDP−1⇒cosδ=P[cos1000cos(1/2)000cos(1/2)]P−1=[cos(1/2)000cos(1/2)000cos1]⇒sinδ=P[sin1000sin(1/2)000−sin(1/2)]P−1=[0sin(1/2)0sin(1/2)0000sin1]因此eiδ=cosδ+isinδ=[cos(1/2)000cos(1/2)000cos1]+[0isin(1/2)0isin(1/2)0000isin1]=[cos(1/2)isin(1/2)0isin(1/2)cos(1/2)000isin1+cos1]解答:L{y″−3y′+2y}=L{4t−8}⇒(s2−3s+2)Y(s)−2s−1=4s2−8s⇒Y(s)=4s2(s−2)(s−1)−8s(s−2)(s−1)+2s+1(s−2)(s−1)=(3s+2s2+1s−2−4s−1)−(4s+4s−2−8s−1)+(5s−2−3s−1)=−1s+2s2+2s−2+1s−1⇒y(t)=L−1{Y(s)}⇒y(t)=−1+2t+2e2t+et
解答:y″+4y=0⇒yh=c1cos(2x)+c2sin(2x)令{y1=cos(2x)y2=sin(2x)⇒W=|y1y2y′1y′2|=|cos(2x)sin(2x)−2sin(2x)2cos(2x)|=2⇒yp=−cos(2x)∫sin(2x)5xe−x2dx+sin(2x)∫cos(2x)5xe−x2dx=−52cos(2x)(125e−x((3−5x)sin(2x)−2(5x+2)cos(2x)))+52sin(2x)(125e−x((10x+4)sin(2x)+(3−5x)cos(2x)))=110e−x(10x+4)⇒y=yh+yp⇒y=c1cos(2x)+c2sin(2x)+110e−x(10x+4)

解答:(x3+17)dydx=x2y⇒1ydy=x2x3+17dx⇒lny=13ln(x3+17)+c1⇒y=c2(x3+17)1/3

解答:f(x)為Gaussian distribution⇒∫∞−∞f(x)dx=1⇒∫∞−∞e−x2/2adx=1⇒a=1/2π⇒f(x)=e−πx2假設F(f(x))=F(ω),現在f(x)=e−x2/2a⇒f′(x)=−xaf(x)⇒F(f′(x))=F(−xaf(x))⇒iωF(ω)=−1aiF′(ω)⇒F′(ω)=−aωF(ω)⇒F(ω)=e−aω2/2=e−ω2/4π註:f(x)為Gaussian distribution,初始值F(0)=0
解答:{x=rcosθy=rsinθ⇒∫4−4∫√16−x20exp(−(x2+y2))dydx=∫π0∫40re−r2drdθ=∫π0[−12e−r2]|40dθ=∫π0−12e−16+12dθ=π2(1−e−16)
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解題僅供參考,其他歷年試題及詳解
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