Processing math: 100%

2024年4月24日 星期三

113年中山光電碩士班-工程數學詳解

國立中山大學113學年度碩士班招生考試

科目名稱:工程數學【光電系碩士班】

 

解答:y+2y=4e2xe2xy+2e2xy=4(e2xy)=4e2xy=4x+c1y=4xe2x+c1e2x
解答:cos(x+y)dx+(3y2+2y+cos(x+y))dy=0{P(x,y)=cos(x+y)Q(x,y)=3y2+2y+cos(x+y)Py=sin(x+y)=QxexactΦ(x,y)=Pdx=QdyΦ=cos(x+y)dx=(3y2+2y+cos(x+y))dyΦ=sin(x+y)+ϕ(y)=y3+y2+sin(x+y)+ρ(x)Φ=sin(x+y)+y2+y2+c1=0


解答:y+3y+2=0λ2+3λ+2=0(λ+2)(λ+1)=0λ=1,2y=c1ex+c2e2x
解答:dydx=(x+1)exy21y2dy=(x+1)exdx1y=xex2ex+c1y=1xex+2ex+c2

解答:cos(wt)=12(eiwt+eiwt)L{f(t)}=0eatcos(wt)estdt=120(e(iw+as)t+e(iw+as)t)dt=12[1iw+ase(iw+as)t+1iw+ase(iw+as)t]|0=12[1iw+as1iw+as]=sa(as)2+w2

解答:L{y}L{y}=L{t}s2Y(s)s1Y(s)=1s2Y(s)=1s2(s21)+s+1s21Y(s)=1s2(s21)+1s1=1s212(s+1)+32(s1)y(t)=L1{Y(s)}y(t)=t12et+32et

解答:{x1x2+x3=0x1+x2x3=010x2+25x3=9020x1+10x2=80=[1111110102520100][x1x2x3]=[009080]augmented matrix:[1110111001025902010080]R1+R2R2,R420R1R4[1110000001025900302080]R3/10R3[11100000015290302080]R1+R3R1,R430R3R4[107290000015290095190]R4/(95)R4[107290000015290012]R1(7/2)R4R2,R3(5/2)R4R3[1002000001040012]{x1=2x2=4x3=2

解答:[AI]=[112100311010134001]R1R1[112100311010134001]R23R1R2,R1+R3R3[112100027310022101]R2/2R2[112100017232120022101]R1+R2R1,R32R2R2[103212120017232120005411]R3/(5)R3[103212120017232120001451515]R11.5R3R1,R23.5R3R2[10071015310010131015710001451515]A1=[71015310131015710451515]
解答:F(f(x))=f(x)eiωtdt=11eiωtdt=[1iωeiωt]|11=1iω(eiωeiω)=1iω(2isinω)=2sinωω

解答:ex.digitalsignalprocessing...

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解題僅供參考,其他歷年試題及詳解

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