作业帮拉普拉斯变换 F(s)=(s+3)/[(s+1)(s+2)] 求z变换有两题,F(s)=(s+3)/[(s+1)(s+2)] F(s)=1/[s(s^2+1)]求z变换
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 17:02:00
![拉普拉斯变换 F(s)=(s+3)/[(s+1)(s+2)] 求z变换有两题,F(s)=(s+3)/[(s+1)(s+2)] F(s)=1/[s(s^2+1)]求z变换](/uploads/image/z/2431610-26-0.jpg?t=%E6%8B%89%E6%99%AE%E6%8B%89%E6%96%AF%E5%8F%98%E6%8D%A2+F%28s%29%3D%28s%2B3%29%2F%5B%28s%2B1%29%28s%2B2%29%5D+%E6%B1%82z%E5%8F%98%E6%8D%A2%E6%9C%89%E4%B8%A4%E9%A2%98%2CF%28s%29%3D%28s%2B3%29%2F%5B%28s%2B1%29%28s%2B2%29%5D+F%28s%29%3D1%2F%5Bs%28s%5E2%2B1%29%5D%E6%B1%82z%E5%8F%98%E6%8D%A2)
拉普拉斯变换 F(s)=(s+3)/[(s+1)(s+2)] 求z变换有两题,F(s)=(s+3)/[(s+1)(s+2)] F(s)=1/[s(s^2+1)]求z变换
拉普拉斯变换 F(s)=(s+3)/[(s+1)(s+2)] 求z变换
有两题,F(s)=(s+3)/[(s+1)(s+2)]
F(s)=1/[s(s^2+1)]求z变换
拉普拉斯变换 F(s)=(s+3)/[(s+1)(s+2)] 求z变换有两题,F(s)=(s+3)/[(s+1)(s+2)] F(s)=1/[s(s^2+1)]求z变换
先化成a/(s+1)+b/(s+2)
a(s+2)+b(s+1)=s+3
a+b=1
2a+b=3
a=2
b=-1
=2/(s+1)-1/(s+2)
f(t)=2e^(-t)-e^(-2t)
G(z)=(f(1)z^(-1)+f(2)z^(-2)+...
=(2[1+(ez)^(-1)+(ez)^(-2)+...]-[1+e^(-2)z^(-1)+e^(-4)z^(-2)+...])
={2[1/(1-(ez)^(-1))]-[1/(1-e^(-2)z^(-1)]}
=2ez/(ez-1)-e²z/(e²z-1)
查表也可知(e^at) 转换为 z/(z-e^(a))
2z/(z-e^(-1))-z/(z-e^(-2)) 一样
----------------------------
1/[s(s^2+1)]
a/s+(bs+c)/(s²+1)=1/[s(s^2+1)]
as²+a+bs²+cs=1
(a+b)=0
c=0
a=1
b=-1
=1/s-s/(s²+1)
f(t)=1-cos(t)
G(z)= z/(z-1)-z(z-cos(1))/(z²+1-2zcos(1))
=z[z²+1-2zcos(1)-(z-1)(z-cos(1)]/[(z²+1-2zcos(1))(z-1)]
=z[1-zcos(1)+z-cos(1)]/[(z²+1-2zcos(1))(z-1)]
=z(1+z)(1-cos(1))/[(z²+1-2zcos(1))(z-1)]