南航胡寿松自动控制原理课后题烟台大学文经学院

解：根据复数阻抗 得：
C1 + ui R1 R2 C2 + u0 -
uo ui
1 R2 C2 s
1 R1 C1s 1 R2 C2 s R 1 1 C1s
R1 R2C1C2 s 2 s ( R1C1 R2C2 ) 1 R1 R2C1C2 s 2 s ( R1C1 R2C2 R1C2 ) 1
C(s) 1 (s) R1 (s) 2 (s) R2 (s)
所以系统传递函数为：
C ( s) P 11 P 22 R( s) G1 ( s )G2 ( s )G3 ( s ) G4 ( s) 1 G2 ( s) H1 ( s ) G1 ( s )G2 ( s ) H1 (s ) G2 (s )G3 (s ) H 2 (s )
U o ( s) U i (s)
1 R2 C2 s 1 R1 C1s 1 R2 C2 s R 1 1 C1s
R1 R2C1C2 s 2 s ( R1C1 R2C2 ) 1 R1 R2C1C2 s 2 s ( R1C1 R2C2 R1C2 ) 1
所以得系统时域数学模型为：
d 2uo (t ) duo (t ) R1 R2C1C2 ( R1C1 R2C2 R1C2 ) uo (t ) 2 dt dt d 2ui (t ) dui (t ) R1 R2C1C2 ( R1C1 R2C2 ) ui (t ) 2 dt dt
1 1 1 拉斯变换得：R2i (s) [ i (s) i`(0)] U o ( s) C2 s s
C2 sU o ( s) i (s) 1 R2C2 s
1
C2 s i (t) L [ U o ( s)] 1 R2C2 s R1 C2 s 1 则： 2 2 L [ U o ( s)] U o (t ) U i (t ) - C1 R1 jC1 1 R2C2 s
C ( s) i 1 i i R( s ) G1G2G3G4G5G6 G7G3G4G5G6 (G1G8G6 G7 H1G8G6 )(1 G4 H 2 )
P
4
1 [G2 H1 G4 H 2 G6 H 3 G3G4G5 H 4 G8 H 4 H1 G1G2G3G4G5G6 H 5 G1G8G6 H 5 G7G3G4G5G6 H 5 G7 H1G8G6 H 5 ] [G2 H1G4 H 2 G2 H1G6 H 3 G4 H 2G6 H 3 G4 H 2G1G8G6 H 5 G4 H 2G8 H 4 H1 G4 H 2G7 H1G8G6 H 5 ] G2 H1G4 H 2G6 H 3
C (s) le(1 cf ) (leha ) (ehbc ) 2 ( s) R2 ( s) 1 (cf eg bcdeh adeh) cfeg
( s) 1 ( s) 2 ( s) C ( s) C ( s) 错！！ R1 ( s) R2 ( s)
2-3 (a)解：设回路电流为i(t),则有：
+ ui R1
C1 + R2 C2 u0 -
1 j C1 1 i (t ) R2i (t ) i (t )dt ui (t ) 1 C 2 j // R1 C1 1 R2i (t ) i (t )dt U o (t ) C2
2-3 (a)绘制系统结构图 列写每个元器件 + 的输入输出方程
I(s) Ui(s) -
I1(s) 1/sC1 R1 + U1(s) R2 + U0(s) -
方法一：
1/sC2
U i ( s ) U1 ( s ) U o ( s ) I R1 ( s ) U1 ( s ) / R1 U R 2 ( s ) I ( s ) R2 U o (s) U R 2 (s) U C 2 (s)
Ui(s)
U1(s)
Uo(s)
U1(s)
U1(s)
1/R1
sC1
I1(s)
I2(s)
I1(s) I2(s)
I(s)
I(s)
I(s)
UC2(s)
R2
UR2(s)
1/sC2
UR2(s)
UO(s)
U o ( s) U R 2 ( s) U C 2 ( s)
UC2(s)
Ui(s)
U1(s)
U1(s)
2-15(e)：
R1 1 b
a c -f -h d
R2 e -g 1 1 C
1 (cf eg bcdeh adeh) cfeg
C ( s) bcde ade a(1 eg ) bc(1 eg ) 1 ( s) R1 ( s) 1 (cf eg bcdeh adeh) cfeg
U1(s)
UR2(s)
UC2(s)
Uo(s)
1/R1
sC1
I1(s)
I2(s)
I1(s) I2(s)
I(s)
I(s)
I(s)
R2
1/sC2
UR2(s)
UO(s)
UC2(s)
UFra Baidu bibliotek(s) U1(s)
Uo(s)
1/R1 sC1
I1(s) I2(s)
I(s)
R2 1/sC2
UR2(s) UC2(s)
UO(s)
方法二：
+ I(s) Ui(s) U1(s) I1(s) R1
I1(s) 1/sC1 R1 + U1(s) R2 + U0(s) UO(s) R2 1/sC2
1/sC2
Ui(s) Uo(s)
I2(s) sC1
1/R 1
I(s)
2-13(e)：
R(s) 1 1 G1(s) 1
-H2(s) G2(s) G3(s) -1 H1(s) 1 C(s) 1
G7 1 R(s) G1 G2 -H1 -H5 G3
G8 G4 -H4 -H2 G5 G6 -H3 1 C(s)
单个回路9条，
两两互不接触回路6组，三三互不接触回路1组
1 [G2 H1 G4 H 2 G6 H 3 G3G4G5 H 4 G8 H 4 H1 G1G2G3G4G5G6 H 5 G1G8G6 H 5 G7 G3G4G5G6 H 5 G7 H1G8G6 H 5 ] [G2 H1G4 H 2 G2 H1G6 H 3 G4 H 2G6 H 3 G4 H 2G1G8G6 H 5 G4 H 2G8 H 4 H1 G4 H 2G7 H1G8G6 H 5 ] [G2 H1G4 H 2G6 H 3 ]
d uo duo R1 R2C1C2 2 ( R1C1 R2C2 R1C2 ) uo dt dt 2 d ui dui R1 R2C1C2 2 ( R1C1 R2C2 ) ui dt dt
2
解：建立s域电路模型得：
1/sC1 + Ui(s) R1 1/sC2 R2 + U0(s) -
1 I C1 ( s ) U 1 ( s ) / sC1 I ( s ) I C 1 ( s ) I R1 ( s ) 1 U C 2 (s) I (s) sC2
U i ( s ) U1 ( s ) U o ( s ) 1 I C1 ( s ) U1 ( s ) / sC1 I R1 ( s ) U1 ( s) / R1 I ( s ) I C1 ( s ) I R1 ( s ) U R 2 ( s ) I ( s ) R2 1 U C 2 (s) I (s) sC2
2-15(b)：
1 R(s) G1
G7 G2 -H1 -H5 G3
G8 G4 -H2 -H4 G5 G6 -H3 1 C(s)
系统前向通道4条 P1=G1G2G3G4G5G6 P2=G7G3G4G5G6 P3=－G7H1G8G6 P4=G1G8G6 Δ1=1 ; Δ2=1 ; Δ3=1+G4H2 ; Δ4=1+G4H2
-H2(s) R(s) 系统前向通道2条： 1 G1(s) 1 G2(s) G3(s) -1 H1(s) 1 C(s) 1 1
P 1 G1 ( s )G2 ( s )G3 ( s ); 1 1 P2 G4 ( s );
G4(s)
2 1 [G2 ( s) H1 ( s) G1 ( s)G2 ( s) H1 ( s) G2 ( s)G3 ( s) H 2 ( s)] 1 G2 ( s ) H1 ( s ) G1 ( s )G2 ( s ) H1 ( s ) G2 ( s )G3 ( s ) H 2 ( s )
G4(s)
解：系统信号流图如图所示， 系统单个回路3条，没有两两互不接触回路，所以
1 [G2 (s) H1 (s) G1 (s)G2 (s) H1 (s) G2 (s)G3 (s) H 2 (s)] 1 G2 (s) H1 (s) G1 (s)G2 (s) H1 (s) G2 (s)G3 (s) H 2 (s)