习题
2-1、试列写图2-40中各电路的微分方程,图中u1、u2为输入、输出电压。
i1 u1 R1 C1 i2 R2 i3 C2 u2 u1 i1 R1 i2 L (b)
图2-40 习题2-1用图
R2 C i3 u2 (a)
答:(a)
设中间变量i1,i2,i3如图所示,则根据基尔霍夫电压回路、节点电流定律,得 u1?i1R1?i2?C1dt -------------------(1)
i2?C1dt?i3R2?u2 --------------------(2) u2??i3dt -------------------(3) C2i1?i2?i3 --------------------(4)
将(3)整理得: i3?C2u'2 ---------------------(5) 将(5)带入到(2),得
i2?C1dt?R2C2u'2?u2 ---------------------(6)
即 i2?C1C2R2u\2?C1u'2 ---------------------(7) 将(5)、(7)带入到(4)及(6)带入到(1),整理得
u1?i1R1??i2dt?(i2?i3)R1?R2C2u'2?u2 C1?(C1C2R2u\2?C1u'2?C2u'2)R1?C2R2u'2?u2
即
R1C1R2C2u\2?R1C1u'2?R1C2u'2?R2C2u'2?u2?u1
设T1?R1C1,T2?R2C2,T3?R1C2,则
T1T2u\2?(T1?T2?T3)u'2?u2?u1
(b)
设中间变量i1,i2,i3如图所示,则根据基尔霍夫电压回路、节点电流定律,得 u1?i1R1?Li'2 -------------------(1) Li'2?i3R2?u2 --------------------(2)
iu2??3dt -------------------(3)
Ci1?i2?i3 --------------------(4)
将(3)变形得: i3?Cu'2 ---------------------(5) 将(5)带入到(2),得
Li'2?R2Cu'2?u2 ---------------------(6)
即 i2?R2C1u2??u2dt ---------------------(7) LL将(5)、(7)带入到(4)及(6)带入到(1),整理得
u1?i1R1?Li'2?(i2?i3)R1?R2Cu'2?u2
?(R2C1u2??u2dt?Cu'2)R1?R2Cu'2?u2 ------(8) LL对(8)两边求导,整理得
(R1C?R2C)u\2?(R1RR2C?1)u'2?1u2?u'1 LL设T1?R1C,T2?R2C,T3?L,则有 R1(T1?T2)u\2?(或者为
T21?1)u'2?u2?u'1 T3T3(T1?T2)T3u\2?(T2?T3)u'2?u2?T3u'1
2-2 、设机械系统如图2-41所示,图中?r、?c为输入、输出角位移(角度),试列写该机械
系统的微分方程。
[提示]扭力矩M与角位移?、弹性系数k成正比。风阻M与角速度?'、阻尼系数f成正比。
J1 J2 k1 k2 f ?r ?a 图2-41 习题2-2用图 ?c \解:根据牛顿第二定律:(对直线运动
?F?ma,而旋转运动?M?J?'?J?)
(?r??a)k1?J1(?r??a)\ -------------------(1) (?a??c)k2?f?'c?J2(?a??c)\ -------------------(2)
在零初始条件下,对式(1)、(2)两试进行拉氏变换,得:
?r(s)k1??a(s)k1?J1s2?r(s)?J1s2?a(s) -------(3)
?a(s)k2??c(s)k2?fs?c(s)?J2s2?a(s)?J2s2?c(s) -------(4)
由式(3)、(4)得:
k1?J1s2?a(s)??(s) -------------------(5) 2rk1?J1s
自动控制系统的数学模型习题答案
![](/skin/haowen/images/icon_star.png)
![](/skin/haowen/images/icon_star.png)
![](/skin/haowen/images/icon_star.png)
![](/skin/haowen/images/icon_star.png)