2-1试建立图所示电路的动态微分方程
-
u o
+
u o
解:
u 1=u i -u o
i 2=C du 1dt i 1=i-i 2u o i=R 2u 1i 1=R 1=u i -u o
R
1
dt
d (u i -u o )=C
(a)u C d (u i -u o )
dt
u o -R 2=i -u o R 1
i=i 1+i 2i 2=C du 1
dt u o i 1=R 2
u 1-u o =L R
2du o
dt
R 1
i=(u i -u 1)
(b)解:
)
-R 2(u i -u o )=R 1u 0-CR 1R 2(du i
dt dt du
o CR 1R 2du o dt du i
dt +R 1u o +R 2u 0=CR 1R 2+R 2u i
u o
+C R 2
du 1dt o +L R 2
du o
dt
u du o dt R 1R 2L du o dt +
CL R 2d 2u o dt 2=--i R 1u o R 1u o
R 2+C )u o R 1R 2L du o dt ) CL R 2d 2u o dt 2
=++(u i R 11R 11R 2+(C+
2-2 求下列函数的拉氏变换。(1) f(t)=sin4t+cos4t
L [sin ωt ]= ωω2+s 2
=s+4s 2+16
L [sin4t+cos4t ]= 4s 2+16s s 2+16+s ω2+s 2
L [cos ωt ]=解:
(2) f(t)=t 3+e 4t
解:L [t 3+e 4t ]= 3!s 41s-4+
6s+24+s 4s 4(s+4)=
(3) f(t)=t n e at
L [t n e at ]=n!(s-a)n+1
解:
(4) f(t)=(t-1)2e 2t
L [(t-1)2e 2t ]=e -(s-2)2(s-2)3
解:
2-3求下列函数的拉氏反变换。A 1=(s+2)s+1(s+2)(s+3)s=-2=-1=2
f(t)=2e -3t -e -2t
(1) F(s)=s+1(s+2)(s+3)解:A 2=(s+3)s+1(s+2)(s+3)s=-3F(s)= 2s+31s+2-= A 1s+2s+3+ A 2
(2) F(s)=s (s+1)2(s+2)f(t)=-2e -2t -te -t +2e -t
解:= A 2s+1s+2+
A 3+ A 1(s+1)2A 1=(s+1)2s (s+1)2(s+2)s=-1A 3=(s+2)s (s+1)2(s+2)s=-2d ds s
s+2][A 2= s=-1
=-1
=2
=-2
(3) F(s)=2s 2-5s+1s(s 2+1)F(s)(s 2+1)s=+j =A 1s+A 2s=+j
A 2=-5A 3=F(s)s s=0f(t)=1+cost-5sint
解:= s + A 3s 2+1A 1s+A 2=1
2s s 2-5s+1=A 1s+A
2 s=j s=j
j -2-5j+1=jA 1+A 2
-5j-1=-A 1+jA 2
A 1=1F(s)= 1s s 2+1s -5s 2+1++
(4) F(s)=s+2s(s+1)2(s+3)解:=+s+1
A 1s+3A 2(s+1)2+s A 3+A 4
-12
A 1= 23A 3= 112A 4= A 2= d [s=-1
ds ](s+2)s(s+3) -34
= -34A 2= +-43+f(t)=e -t 32e -3t 2-t e -t 12
1= s=-1 [s(s+3)]2[s(s+3)-(s+2)(2s+3)]
(2-4)求解下列微分方程。
y(0)=y(0)=2 ·+6y (t )=6+5d 2y (t )dt 2dy (t )dt
(1)解:s 2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)= 6s
A 1=1
y(t)=1+5e -2t -4e -3t
A 2=5 A 3=-4
Y(s)=6+2s 2+12s s(s 2+5s+6)= A 1s+2s+3+ A 3s + A 2
2-5试画题图所示电路的动态结构图,
并求传递函数。
c
+
( U r (s)U c (s)=1R 11+(+sC)R 21
R 1+sC)R 2
=R 2+R 1R 2sC R 1+R 2+R 1R 2
sC
(2)c
L 1
=-R 2 /Ls L 2=-/LCs 2L 3=-1/sCR 1Δ1=1L 1L 3=R 2/LCR 1s 2
P 1=R 2/LCR 1s 2
=
R 1CLs 2+(R 1R 2C+L)s+R 1+R 2
U r (s)U c (s)R 2
2-8 设有一个初始条件为零的系统,系统的输入、输出曲线如图,求G(s)。
δ(t)
δ解:
K t-T c(t)=T (t-T)K C(s)=Ts K (1-e )2-TS C(s)=G(s)
2-9 若系统在单位阶跃输入作用时,已知初始
条件为零的条件下系统的输出响应,求系统的传递函数和脉冲响应。 r(t)=I(t)c(t)=1-e +e
-2t -t R (s )=1s 解:G(s)=C(s)/R(s)s+21-1s C(s)=1s +1+=s(s+1)(s+2)
(s 2+4s+2)=(s+1)(s+2)
(s 2+4s+2)
C(s)=(s+1)(s+2)
(s 2+4s+2)c(t)=δ(t)+2e -2t -e -t
2-10 已知系统的拉氏变换式,试画出系统的动态结构图并求传递函数。解:X 1(s)=R(s)G 1(s)-G 1(s)[G 7(s)-G 8(s)]C(s)
X 2(s)=G 2(s)[X 1(s)-G 6(s)X 3(s)]
X 3(s)=G 3(s)[X 2(s)-C(s)G 5(s)]C(s)=G 4(s)X 3(s)
={R(s)-C(s)[G 7(s)-G 8(s)]}G 1(s)
C 1+G 3G 2G 6 +G 3G 4G 5+G 1G 2G 3G 4(G 7 -G 8
)
G 1G 2G 3G 4R (s )(s )
=
2-11求系统的传递函数
解:L 1=-G 2H 1L 2=-G 1G 2H 2P 1=G 1G 2P 2=G 3G 2Δ1 =1
Δ2 =1R (s )C (s )=Σn
k=1P k Δk Δ
Δ=1+G 2H 1+G 1G 2H 2
1+G 2H 1+G 1G 2H 2G 2G 1+G 2G 3=
解:R (s )C (s )=1+G 1G 2H +G 1G 4H G 1G 2+G 2G 3+G 1G 2G 3G 4 H
L 1=-G 1G 2H L 2=-G 1G 4H P 1=G 1
G 2Δ1 =1P 2=G 3G 2Δ=1+G 1G 4H+G 1G 2H Δ2=1+G 1G 4H
(c)
C(s)R(s)1+G 1G 2+G 1H 1–G 3H 1G 1G 2(1–G 3H 1)
=
(d)
解: L 1=-G 2H
P 1=G 1Δ1 =1P 2=G 2Δ2 =1
1+G 2
H
1(G 1+G 2 )R(s)C(s)=
(e)
解: L 2=G 1G 4
L 3=-G 2G 3L 4=G 2G 4L 1=-G 1G 3P 1=G 1Δ1=1P 2=G 2Δ2=1
1+G 1G 3+G 2G 3–G 1G 4-G 2G 4
=(G 1+G 2)C(s)
R(s)
(f)
解: L 1=-G 1G 2L 2=G 2P 1=G 1Δ1=1-G 2
Δ=1+G 1G 2-G 2C(s)R(s)1+G 1G 2–G 2G 1
(1–G 2)
=
L 1=G 2H 2L 2=-G 1G 2H 3Δ1=1
P 1=G 1G 21-G 2H 2+G 1G 2H 3
G 2G 1=R (s )C (s )P 1=G 2Δ1=1P 2=-G 1G 2H 1Δ2=1
1-G 2H 2+G 1G 2H 3
G 2(1-G
1H 1 )=D (s )C (s )
(b)
解:L 1=-G 1G 2L 2=-G 1G 2H Δ1=1
P 1=G 1G 21+G 1G 2H+G 1G 2
G 1G 2=R (s )C (s )P 1=G n G 2Δ1=1P 2=1Δ2=1+G 1G 2H
D (s )C (s )1+G 1G 2+G
1G 2H =1+G n G 2+G 1G 2H
解:L 1=-G 2L 2=-G 1G 2G 3P 1=G 2G 3P 2=G 1G 2G 3R (s )C (s )=1+G 2+G 1G 2G 3
G 2G 3+G 1G 2G
3
Δ1=1
Δ2=1P 1=-G 2G 3Δ2=1+G 2
Δ1=1P 2=1R (s )C (s )=1+G 2+G 1G 2G 3-G 2G 3+1+G 2
解:L 1=-G 3G 4L 2=-G 2G 3G 5Δ1=1P 1=G 1G 5P 1=G 1G 5Δ1=1P 2=1Δ2=1+G 3G 4
P 2=G 2G 3G 5Δ2=11+G 2G 3G 5+G 3G 4
=R (s )C (s )G 1G 2G 5+G 1G 5
1+G 2G 3G 5+G 3G 4
=R (s )E (s )G 1G 5+(1+G 1G 5 )
C(s)R(s)=1+G 3(G 1+G 2)(G 1+G 2)(G 3+G 4)
解:L 1=-G 1G 3L 2=-G 2G 3Δ1=1P 1=G 1G 3P 2=G 2G 3Δ2=1
P 3=G 1G 4Δ3=1P 4=G 2G 4Δ4=1
E (s )R (s )=1+G 3(G 1+G 2)
1(C D s )(s )
=1=G 2(s)E (s )X (s )
解:L 1=G 1G 2L 3=-G 4
L 2=-G 1G 4G 5H 1H 2P 1=G 1G 2G 3Δ=1-G 1G 2+G 1G 4G 35H 1H 2+G 4 -G 1G 2G 4Δ1=1+G 4
1+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4
G 1G 2G 3(1+G 4 )=
C 1(s)R 1(s)
1+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4G 4G 5G 6(1-G 1G 2)=C 2(s)
R 2(s) 1+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4-G 1G 2G 3G 4G 5H 1=C 1(s)R 2(s)
1+G 4+G 1G 4G 5H 1H 2-G 1G 2-G 1G 2G 4G 1G 4G 5G 6H 2
=C 2(s)R 1(s)
解:c(t)=c(∞)98%t=4T=1 min r(t)=10t
e(t)=r(t)-c(t)c(t)=10(t-T+e )-t/T =10(T-e )-t/T e ss =lim t →∞
e(t)=10T =2.5
T=0.25
解:R 1Cs+1R 1/R 0G (s )= u c
(t)=K(1–e
t T -)K Ts +1
=T=R 1C=0.5 K=R 1/R 0=10 =10(1–e -2t )8=10(1–e -2t )0.8=1–e -2t
e -2t =0.2
t=0.8
3-1 设温度计需要在一分钟内指示出响应值的98%,并且假设温度计为一阶系统,求时间常数T 。如果将温度计放在澡盆内,澡盆的温度以10o C/min 的速度线性变化,3-2 电路如图,设系统初始状态为零。
(1)求系统的单位阶跃响应,及u c (t 1)=8时的1 R 0=20 k Ω R 1=200 k Ω C=2.5μF
(2) 求系统的单位脉冲响应,单位斜坡响应,及单位抛物响应在t 1时刻的值.
g(t)=e -t/T T K
t 1=0.8=4解:
u c (t)=K(t-T+T e -t/T )=4R(s)=1s 2
R(s)=1R(s)=1s 3s s+1/T +T 2=K(T s 2-1s 3-T 2)
=1.2Ts 1s 3K +1U c (s)= -0.5t+0.25-0.25e -2t )12
t 2u c (t)=10(
4s(s+5)G(s)=
解:R(s)=s 2+5s+4C(s)4s(s+1)(s+4)C(s)=4R(s)=s 1s+41+1/3s =4/3s +1
-13c(t)=1+ -4t -t 43-e e
G(s)=1s(s+1)
解:C(s)
=s 2+s+1
R(s)1
2= 1ωn
2ωn ζ=1ζ=0.5ω=1n ω= =0.866
d ωn 2 ζ1-=60o -1ζ
=tg β1-2ζt r =d ωπβ-= 3.14-3.14/30.866
=2.42t p =d ωπ 3.140.866= =3.63σ%=100%e -ζζπ1-2
=16%-1.8
e
t s = ζ3ωn
=6t s = ζ4ωn =8
3-6 已知系统的单位阶跃响应:c(t)=1+0.2e -60t
-10t
-1.2e
(1) 求系统的闭环传递函数。
(2) 求系统的阻尼比和无阻尼振荡频率。
解:s+60+0.21s C(s)= 1.2s +10-s(s+60)(s+10)
=600=s 2+70s+600C(s)R(s)600R(s)=s 12=600ωn 2ωn ζ=70ζ=1.43ω=24.5n 1.3
t
c(t)
01
0.1
解:
t p ==0.121-ζπωn =0.3e -ζζπ1-2e ζζπ1-2=3.3ω1-n 2 ζ 3.140.1==31.4ζ21-ζπ/=ln3.3=1.19π21-ζ)2/ζ(=1.429.862ζ=1.42-1.422ζζ=0.35=33.4ωn ωn ωn
ζs(s+2 )G(s)=21115.6s(s+22.7)=
3-3 已知单位负反馈系统的开环传递函数,求系统
3-4 已知单位负反馈系统的开环传递函数,求系统的上
升时间t 、峰值时间t 、超调量σ% 和调整时间t 。
3-7 设二阶系统的单位阶跃响应
曲线如图,系统的为单位反
(1)s 3+20s 2+9s+100=0
解:劳斯表如下:s 1s 0 s 3 s 2 1 9 20 100 4100
系统稳定。
(3) s 4+8s 3+18s 2+16s+5=0118 5 s 4 s 3 8 16 劳斯表如下:s 2 16 5 s 121616s 0 5系统稳定。
G(s)=s(s+1)(0.5s 2+s+1)K(0.5s+1)
解:s 4+3s 3+4s 2+2s+Ks+2K=0
1 4 2K s 4 s 3
3 2+K s 2 b 31b 31= 10-2K 3 (K-1.7)(K+11.7)>0
K<10s 1
2K b 41
b 41= K 2+10K-2010-K
解:10(1+G(s)=s 2+s+10
s τ)s 1=s(s 2+s+1010(s+1)s τ)Φ(s)=s 3 +s 2+1010(s+1)s 2+10s+10τ110
s 3 s 2 (1+10τ)10 s 1 b 31s 0 10
b 31= 10(1+10τ)-10 1+10τ
>0τ>0
解:G(s)=s 2(s+1)
10(τs+1)Φ(s)=s 3 +s 2+10s+10τ10(τs+1)s 3 s 2 1 10
110
τs 1 b 31
s 0 10b 31= 10τ-10 1>0τ>1 r(t)=I(t)+2t+t 2
s 2R(s)=1s
2+s 3
2+
3-12 已知单位负反馈系统的开环传递函3-13 已知系统结构如图,试确定系
统稳定时τ值范围。
3-14 已知系统结构如图,试确定系
3-16 已知单位反馈系统的开环传递函数,
解:(0.1s+1)(0.2s+)
(1) G(s)=20K p =20
υ=0e ss1=R 01+K =21
1K υ=0 e ss2=∞
K a =0
e ss3=∞
e ss =∞
υ=1K υ
=10
K a =0
e ss3=∞
e ss =∞
s(s+2)(s+10)(2) G(s)=200s(0.5s+1)(0.1s+1)=10K p =∞
e ss1=0
e ss2=K 2=102
10(2s+1)
s 2(s 2+4s+10)(3) G(s)=υ=2
K p =∞e ss1=0s 2(0.1s 2+0.4s+1)=(2s+1)K υ=∞e ss2=0K a =1
e ss3=2
e ss =2
3-17 已知系统结构如图。
(1)单位阶跃输入:
确定K 1 和τ值σ%=20%t s =1.8(5%)解:s K G(s)=s 2+K 1τ1Φ(s)=s 2+K 1s+K 1τK 12ωn ζ=K 1
τ2=K 1
ωn =0.2e -ζζπ1-2t s = ζ3ωn
=1.8ζ=0.45
ωn 31.8*0.45==3.72ωn K 1==13.7τ=0.24
(2) 求系统的稳态误差:1t r(t)=I(t), t ,2
2
解:G(s)=s 2+K 1s τK 1=s(s+1)τK 11τ1υ=1K p =∞
e ss1=0
R(s)=1s R(s)=s 2
1=K K υe ss2=τ=0.24R(s)=s 31K a =0
e ss3=∞
解:G(s)=s 2+2s+K s τK K
=s(s+1)τ
2+K 1τ2+K Φ(s)=s 2+(2+K )s+K τK 2ωn ζ=2+K τ2=K ωn =2*0.7 K e ss = 2+K τK
=0.25= 0.25K-2 τK K =31.6τ=0.186
3-18 已知系统结构如图。为使ζ=0.7时,单位斜坡输入的稳态误差e ss =0.25确定K
和τ值 。
3-19 系统结构如图。
r(t)=d 1(t)=d 2(t)=I(t)
(1)求r(t)解:e ssr =lim s·1+G(s)F(s)s →0s 1=1+G(0)F(0)
1(2) 求d 1(t)和d 2(t)同时作用下的稳态误差.
e ssd = lim s -F(s)1+G(s)F(s)s →0
-11+G(s)F(s)+s 1[]E d (s)= -G 2(s)H(s)
1+G 1(s)G 2(s)H(s)·
D(s)=1+G(0)F(0)-[1+F(s)] (3) 求d 1(t)作用下的稳态误差.
G(s)=K p +K s Js 1F(s)=e ssd = lim s -F(s)1+G(s)F(s)s →0s 1-s →0
s 1= lim s 1+(K p +K s Js
1)Js 1=0
解:
K r
(s+1)
G(s)=
解:K r
Φ(s)=s+1+K
r
K r =0s=-1-K r
s=-1
K r =→∞s=-∞
s=-2+j0s=0+j1s=-3+j2
4-3 已知系统的开环传递函数,
试绘制出根轨迹图。
解:s(s+1)(s+5)
(1) G(s)=K r (s+1.5)(s+5.5)
4-1 已知系统的零、极点分布如图,大致绘制出系统的根轨迹。
4-2 已知开环传递函数,试用解析法绘制出系统的根轨
1)开环零、极点p 1=0p 2=-1p 3=-52)实轴上根轨迹段p 1~p 2z 1=-1.5z 2=-5.5z 1~p 3z 2~-∞3)根轨迹的渐近线n-m= 1θ= +180o
4)分离点和会合点A (s )B'(s )=A'(s )B (s )A(s)=s 3+6s 2+5s B(s)=s 2+7s+8.25A(s)'=3s 2+12s+5B(s)'=2s+7s 1=-0.63s 2=-2.5s 3=-3.6s 4=-7.28
K s(s+1)(s+4)(2) G(s)=r (s+1.5)
1)开环零、极点p 1=0p 2=-1p 3=-4
2)实轴上根轨迹段
p 1~p 2z 1=-1.5
p 3~z 1
3)根轨迹的渐近线n-m= 2θ= +90o 2
σ=-1-4+1.5=-1.75
4)分离点和会合点
A(s)=s 3+5s 2+4s B(s)=s+1.52+10s+4B(s)'=1
s=-0.62
K s(s+1)2
(3) G(s)=r 1)开环零、极点
p 1=0p
2=-1p 3=-1
2)实轴上根轨迹段p 1~p 2
p 3~-∞3)根轨迹的渐近线n-m=33σ=-1-1=-0.674)根轨迹与虚轴的交点
θ= +180o +60o ,
s 3+2s 2+s+K r =0
K r
=0
K r =2 ω2,3=±1ω1=0 5)分离点和会合点
A(s)=s 3+2s 2+s B(s)=1
A(s)'=3s 2+4s+1B(s)'=0s=-0.33
1)开环零、极点
p 1=0p 2=-3p 3=-72)实轴上根轨迹段p 1~p 2p 4=-15z 1=-8p 3~z 1p 4~-∞3)根轨迹的渐近线
n-m=3s(s+3)(s+7)(s+15)(4) G(s)=K r (s+8)
3σ=-3-7-15+8=-5.67
θ= +180
o +60o , 4)根轨迹与虚轴的交点s 4+25s 3+171s 2+323s+8K r =0
K r =0 ω1=0K r =638
ω2,3=±6.2
5)分离点和会合点A(s)=s 4+25s 3+171s 2+315s B(s)=s+8
A(s)'=4s 3+75s 2+342s+315
B(s)'=2s+7s=-1.4
解:s(s+1),
K r (s+2)
G (s )=p 1=0p 2=-1z 1=-2p 1~p 2z 1~-∞分离点和会合点
s 2+4s+2=0s 1=-3.41s 2=-0.59s 2+s+K r s+2K r =0闭环特征方程式s=-2+j ω
(-2+j ωω)2+(-2+j )(1+K r )+2K r =0 ω-4+(1+K r )ω=04-ω2-2(1+K r )+2K r =0
K r =3ω=±1.41
解:p 1=0p 2=-1p 3=-3p 1~p 28
p 3~ -+60o θ
= +180o , 3
σ= -1-3=-1.3根轨迹的分离点:
A (s )B'(s )=A'(s )
B (s )3s 2+8s+3=0s 1=-0.45s 2=-2.2舍去s(s+1)(s+3)
K r
G (s )H (s )=与虚轴交点
s 3+4s 2+3s+K r =02=0ωK r -4ω3+3
ω=0-K r
=0 K r =12
ω2,3=±1.7ω1=0(1)
ζ=0.5得s 1=-0.37+j0.8K r =|s 3||s 3+1||s 3+3|
s 3=-4+0.37×2=-3.26=3.26×2.26×0.26=1.9
4-5 已知系统的开环传递函数。(1)试绘制出根轨迹图。(2)增益K r 为何值时,
复数特征根的实部为-2。
4-6 已知系统的开环传递函数,试确定闭环极点ζ=0.5时的K r 值。
s(s+3)(s G(s)H(s)=
2+2s+2)
K r
(2)解:p 3.4=-1±j p 2=-3p 1=0p 1~p 2
4
σ=-3-1-1=-1.25θ= +135o +45o , 根轨迹的出射角
θ3=+πθ
1-θ2-θ4-=+π-135o-90o-26.6o=-71.6o
与虚轴的交点
s(s+3)(s 2+2s+2)+K r =0s 4+5s 3+8s 2+6s+K r =0
(j ωω)4+5(j ω)3+8(j ω)2+j6
+K r =0ω4-8
ω2+K r =0-5ω3+6ω=0
K r =0 ω1=0K r =8.16 ω2,3=±1.1
135
分离点和会合点4s 3+15s 2+16s+6=0
解得s=-2.3ζ=0.5得s 1=-0.36+j0.75
K r =|s 1||s 1+3||s 1+1+j||s 1+1-j|=2.92
4-7 已知系统的开环传递函数,
(1) 试绘制出根轨迹图。
解:p 1=0p 2=-2p 3=-4p 1~p 28
p 3~ -+60o θ= +180o , 3
σ= -2-4=-2根轨迹的分离点:A (s )B'(s )=
A'(s )B (s )
s 1=-0.85s 2=-3.15舍去(3)与虚轴交点
s 3+6s 2+8s+K r =0s(s+2)(s+4)
K r
G (s )H (
s )=(2) 阻尼振荡响应的K r 值范围
s=-0.85K r =0.85×1.15×3.15=3.1s=±j2.8K r =48
2=0ωK r -6-ω3+8ω=0K r =0
K r =48
ω2,3=±2.8ω1=0(4)ζ=0.5
s 1=-0.7+j1.2s 3=-6+0.7×2=-4.6K r =4.6×2.6×0.6=7.2
(s+1)G(s)=10(1) 解:(s+11)
(s)=10φA(10)=
ω112+()2ω√=0.905=112+110√=12210√=-5.2o φ(ω)ω11
=-tg -1111
=-tg -1c s (t)= 0.9sin(t+24.8o )
5-1 已知单位负反馈系统开环传递函数,当输入信号r(t)=sin(t+30o ),试求系
统的稳态输出。 5-2 已知单位负反馈系统开环传递函数,试绘制系统开环幅相频率特性曲线 。
s(s+5)(s+15)(1) G(s)=750解:
I 型系统ω=0ω=∞A()=∞ω-90)=-90o φ(
ω)=-270)=-270o φ(ω)=0)=
A(ω
(2s+1)(8s+1)
(3) G(s)=10
解:n-m=2
0型系统ω=0ω=∞A()=10 ωφ(ω)=0)=
A(ω0)=0o φ(
ω)=解:s(s-1)(5) G(s)=
10解:n-m=2
I 型系统ω=0ω=∞)=∞A(ω-270)=-270o φ(
ω)=-180)=-180o φ(ω)=0)=
A(ω
s 2(s+0.1)(s+15)(7) G(s)=10(s+0.2)
解:II 型系统ω=0ω=∞
A()=∞ω-180)=-180o φ(ω)=-270)=-270o φ(ω)=
0A()=
ωω
s(s+5)(s+15)(1) G(s)=750解:
s(G(s)=1051s+1)s+1)(15
1ω1=5ω2=1520lgK=20dB ω=0ω=∞
-90()=-90o φω)=-270)=-270o φ(ω)=
(2s+1)(8s+1)
(3) G(s)=10解:20lgK=20dB
ω1=0.125ω2=0.5
ω=0ω=∞0)=0o
φ(ω)=-180)=-180o φ(
ω)=
s(s-1)(5) G(s)=10解:20lgK=20dB
ω1=1
ω=0ω=∞-270)=-270o φ(ω)=-180)=-180o
φ(
ω)=5-2 试绘制系统开环对数频率特性曲线。
s 2(s+0.1)(s+15)(7) G(s)=10(s+0.2)解:
s 2(10s+1)(0.67s+1)= 1.33(5s+1)20lgK=2.5
dB
ω1=0.1ω2=0.2
ω3=15
ω=0ω=∞-180φ)=-180o (ω)=
-270()=-270o φω)=
(a)
20lg K =20
K =10
10G(s)=(0.1s+1)
(b)
20lg K =-20K =0.10.1s G(s)=(0.05s+1)
(d)20lg K =48
K =251
251G(s)=(s+1)(0.1s+1)(0.01s+1)
(c)
s 100G(s)=(100s+1)K =100
(0.01s+1)
(c)s 100G(s)=(100s+1)K =100
(0.01s+1)
(e)由图可得:20lgM r =4.58dB M r =1.7得:1
=1-ζ2 2ζ1=±0.94ζ2=±0.32ζ=0.3
ζωr ω=1-2ζ2 n ω=50
n s 100G(s)=[(0.02s)2+0.01s+1)]=1000ωK=2T ζ=0.011)2T 2=(=0.022n ω
(a)
系统不稳定
(b)
5-4
已知系统的开环幅频率特性曲线,写出传递函数并画出对数相频特性曲线。
5-7 已知奈氏曲线,p 为不稳定极点个数,υ为积分环节个数,试判别系统稳定性。
(c)
ω=0+
(d)
(e)
ω
系统稳定
(f)
(g)
(h)
解:s 10G(s)=(10s+1)K =10
(0.05s+1)
c
ω1010≈12=1c ω=180o -90o -tg -110-tg -10.05γ=180o +
)(ωφc =90o -84.3o -2.9o = 2.8o
5-17 已知系统开环幅频率特性曲线(1)写出传递函数。(2)利用相位裕量判断稳定性(3)
5-18 已知系统结构,试绘制系统的开环对数幅频特性曲线,并计算相角稳定裕量。
s(0.5s+1)(0.02s+1)G(s)=10解:
c
ω0.510≈12=4.5c ω-tg -10.02×4.5
γ=180o -90o -tg -1 0.5×4.5
=90o -66o -2.6o = 21.4o
G 0(s)=s(s+5)
500K o
解:K =K v =100
G 0(s)=s(0.2s+1)10020lgK=40dB γ=12.6°
取Δ=5.6o =45o –12.6o +5.6o =38°
1+sin φm
1–sin φm
a=φm =γ-′γ+Δ
=4.2
c
ω0.2100≈12=22.4c ω
=6.2dB
aT=0.04G c (s)=1+0.04s 1+0.01s G(s)=G 0(s)G c (s)
)=10lga L 0( ωm = ωm =40ωc
'a =82= ω2ωm aT
1=23.8= ω182
1T==0.01
G 0(s)=s(0.5s+1)(0.2s+1)
K
解:K=K v =10
10G 0(s)=s(0.5s+1)(0.2s+1)
20lg10=20dB γ=-18°
φ=-180o +γ'+Δ= -180o +50o +10o =-120o
c
ω0.510≈1
2=4.5c ω
6-1 已知单位负反馈系统开环传递函数,采用超前校正满足
K υ =100, γ≥45o 。 6-5 已知单位负反馈系统开环传递函数,试 设计串联校正装置以满足K υ =10, o
β=0.05
=0.1β1ω==
15ωT c
'21=ωβω2=0.005
取G c (s)=1+Ts β1+Ts =200s+110s+1G(s)=G 0(s)G c (s)
)=26dB L 0( ωc '=-20lg
β取=0.5ωc '
解:
(a)G 0(s)=s(0.1s+1)20
G c (s)=10s+1
s+120(s+1)G(s)=s(0.1s+1)(10s+1)
(b)G 0(s)=s(0.1s+1)20G c (s)=0.01s+10.1s+1
G(s)=s(0.01s+1)
20
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6-12 已知系统G 0(s) 和校正装置G c (s)的对数频率特性曲线,要求绘制校正