關於一些工程數學的題目

1.F(s)=1/[(s^2-2)^2+1]=1/(s^4-4s^2+4+1)=1/(s^4-4s^2+5)判別式: D=4^2-4*5=-4<0分母=虛根不能轉換
2.F(s)=1/(s^2-4s+13)=1/[(s^2-4s+4)+9]=1/[(s-2)^2+9]=(1/3)*3/[(s-2)^2+9]ans: f(t)=(1/3)*e^2t*sin(3t)公式: L{e^at*f(t)}=F(s-a)
3.F(s)=(s-2)/(s^2-4s+13)=(s-2)/[(s-2)^2+9]ans: f(t)=e^2t*cos(3t)公式: L[sin(at)]=a/(s^2+a^2)L[cos(at)]=s/(s^2+a^2)
4.F(s)=s/(s^2-2s+5)=[(s-1)+1]/[(s-1)^2+4]=(s-1)/[(s-1)^2+4]+0.5*2/[(s-1)^2+4]=e^t*cos(2t)+0.5e^t*sin(2t)=e^t{cos(2t)+0.5sin(2t)}
5.F(s)=(4s-23)/(s^2-4s+13)=4(s-23/4)/[(s-2)^2+9]=4(s-2-15/4)/[(s-2)^2+9]=4(s-2)/[(s-2)^2+9]-5*3/[(s-2)^2+9]=4e^2t*cos(3t)-5e^2t*sin(3t)=e^2t*{4cos(3t)-5sin(3t)}

6.F(s)=s/(s^2-4s+13)=(s-2+2)/[(s-2)^2+9]=(s-2)/[(s-2)^2+9]+(2/3)*3/[(s-2)^2+9]=e^2t*cos(3t)+(2/3)*sin(3t)=e^2t*{cos(3t)+(2/3)*sin(3t)}

>這家不錯 lｖ3３3。ｃC買幾次啦真的一樣
匤俲匢

1. 請檢查題目
2. (1/3) e^(2t) sin(3t)
3. e^(2t) cos(3t)
4. (1/2) e^t [sin(2t) + 2 cos(2t)]
5. e^(2t) [4cos(3t) - 5sin(3t)]
6. (1/3) e^(2t) [2sin(3t) + 3cos(3t)]