1 1 1 = = =0.4 K v K 2.5
当 r (t ) = t 时：
ess =
6-10
解（1）当 T D = 0 时 φ s = ( )
25 s 2 + 3 s + 25 3 3 = = 0.3 ⇒ ξ= 2ξ wn = 3 2wn 2 × 5
wn2 = 25
tr =
⇒
wn = 5
π −β
当 r (t ) = t 时， 1
e ss = KV
1 1+ K
= 0
p
=
1 K
当 r (t ) = t 2 时，
e ss = 2 =∞ Ka
6-4（3），（4）
（3）解：G(s) =
K s(s 2 + 4s + 200) K p = lim G ( s ) = ∞ s →o K Kv = lim sG(s) = s→o 200 2 K a = lim s G ( s ) = 0
⇒
α=
4
α2
1 = 0.84 2
6-4（1），（2）
(1)解:
G ( s) =
s→o
K p = lim G ( s ) = 50 K v = lim s G ( s ) = 0
s→o
50 (0.1s + 1)(2 s + 1)
(2)解:G(s) =
K s(0.1s + 1)(0.5s + 1)
K p = lim G(s) = ∞ s →o K v = lim s G(s) = K
s →o
（4）解： (s) = K (s 2+ 1)(4s + 1) G 2
s →o
s (s + 2s + 10) K p = lim G ( s ) = ∞
K v = lim s G ( s ) = ∞
s →o
s→ o
K a = lim s 2 G ( s ) =
当 r (t ) = 1(t ) 时，
1 =0 KV
e ss =
2 = ∞ Ka
2 20 = Ka K
6-5
解： G ( s ) =
100 s (0.1s + 1)
K p = lim G ( s ) = ∞
s →o
K v = lim s G ( s ) = K = 100
K a = lim s 2 G ( s ) = 0
s→ o
s →o
故该系统不稳定.
5-2
4 3 2 解:(1) s + 2 s + s + 2 s + 1 = 0
(2) s5 + 2s4 + s3 + 3s2 + 4s + 5 = 0
s5 s4
⇒
s4 s3
⇒
1 2
1 1 2 0
1 2 − 1 2 9
1 3 3 2 5 0 0
4 5 0 0 0 0
s2 s1 s0
⇒ ess =
1 1 2a 1 + + = 0+ +∞ = ∞ 1 + K p Kv Ka 100
6-6
解：
G(s) =
10 s (0.1s + 1)
K p = lim G ( s ) = ∞
s→o
Kv = limsG(s) = K =10
s→o
K a = lim s 2 G ( s ) = 0
s→o
ess =
1 1 6 = = = 0.24 Kv K 25
s
（1） ts = （2） t s =
3 = 2( s ) 0.3 × 5
3 = 1( s ) 0.6 × 5
M p = 37.23% M p = 9.478%
ess = 0.12
ess = 0.24
令 Im G ( jw) = 0 ,得: 10 cos wτ − w sin wτ = 0 −2 w( w cos wτ + 10sin wτ ) 令 ,得: Re G ( jw ) = = −1 w4 + 100 w2
2( w cos wτ + 10sin wτ ) = w( w2 + 100)
(1)
(2)
续5-5
由(1)式得: 故:
tan τ w =
sinτ w =
10 w
1+ tan2 τ w =
1 cos2 τ w
10 w2 +2 + 100
代入(2)中,得:
2( w 2 + 100) w2 + 100
= w( w 2 + 100)
w4 +100w2 −4=0
0 < K < 16
5-5
解:
2e−τ s G(s) = s(s +10) 2e− jwτ 2(cos w − j sin w ) τ τ G( jw) = = jw( jw+10) −w2 + j10w −2w(wcos w +10sin w ) − j2w(10cos w − wsin w ) τ τ τ τ = w4 +100w2
1 3
当 x (t ) = t 时， e ss = K = 25 = 0.12 （2）“小”闭环：) = G(s 闭环传递函数： 25 25 = ； s(s + 3) + 25TD s s(s + 3 + 25TD ) 25 25 = 2 s(s + 3 + 25TD ) + 25 s + (3 + 25TD )s + 25
与K无关.
闭环特征方程为:
⇒ (Ts +1)(T s +1)(T +1) + K = 0 ⇒ TTTs3 +(TT +TT +TT ) s +(T +T +T ) s + K +1= 0 1 2 3 1 2 3 1 2 1 3 2 3 1 2 3
⇒ (T +T +T )(TT +TT +TT ) > TTT (K +1) 1 2 3 1 2 1 3 2 3 1 2 3
5-1
解:(1)特征方程为:
⇒
1 + G ( s) = 0
⇒
1+
3s + 1 =0 s 2 + 2 s + 50
s 2 + 5s + 51 = 0
(2)特征方程为:
⇒
故该系统稳定. 1 + G(s) = 0
⇒
1+
1050 =0 s ( s + 1)(0.1s + 1)
0.1s 3 + 1.1s 2 + s + 1050 = 0
s →o
当 r (t ) = 1(t ) 时,
K a = lim s 2 G ( s ) = 0
s→o
K a = lim s 2 G ( s ) = 0
s→ o
当 r (t ) = 1(t ) 时，
e ss =
1 1 ε ss = ess = = 1 + K p 51 r (t ) = t 时, 当 1 ε ss = ess = =∞ Kv 当 r ( t ) = t 2 时, 2 ε ss = ess = =∞ Ka
K = wn 2 = 476.4
⇒
= 1.04
ξ = 0.6
⇒
wr = wn 1− 2ξ 2 = 11.55
−πξ
wn = 21.827
ts =
a = 2ξ wn = 26.19
Mp = e
1−ξ 2
= 0.09478 = 9.478%
3 3 = = 0.229 ξ wn 0.6 × 21.827
w =w 1−2ξ2 + 2−4ξ2 +4ξ4 =25.06s−1 b n
2
=
π − 0.927
5 1− 0.6
2
= 0.5536(s)
wn 1−ξ
=
π
5 1− 0.6
2
= 0.785(s)
Mp =e
= 0.09478 = 9.478%
开环传递函数为：
当 x (t ) = t 时 ， 又有： t
= 3 ξ wn
25 25 25 6 G(s) = = = s(s + 3 + 25TD ) s(s + 6) s( 1 s +1) 6
⇒ e ss =
a0 a a + 1 + 2 =∞ Kv Ka 1+ K p
6-8
解：G ( s ) =
K s(s + a)
⇒
⇒ φ ( s) = 1+ G(s) = s(s + a) + K = s2 + sa + K
wn = K
⇒
G(s)
K
K
wn 2 = K
Mr = 1 2ξ 1 − ξ
2
2ξ w n = a
e ss
当 r (t ) = 1(t ) 时，
e ss =
K 10
当 r ( t ) = t 时，
2 当 r (t ) = t 时，
1 = =∞ 1+ K p
当 r ( t ) = t 时， 当 r (t ) = t 2 时，
e ss =
1 1+ K
= 0
p
e ss =
1 200 = KV K