第三周作业_参考答案

x(t ) = x ss + x t = 3 + 1.5e −3t − 4.5e −t
Solution: 3.8(d): Also there are two methods to solve the problem. Using first method.
∵ X (s) =
1 100 100 ⋅ 2 = 2 s + 3s + 4s + 2 s + 100 ( s + 1)( s + 2s + 2)( s 2 + 100) A Bs + C Ds + E = + + 2 2 s + 1 ( s + 1) + 1 s + 100
3
Using Laplace transformation method: .
I 2 ( s) =
10 100 s 1000 ⋅ 3 = 3 2 2 s s + 115s + 3500 s + 100000 s + 115s + 3500 s + 100000 1000 = 2 ( s + 88.2)( s + 26.8s + 1136)
x(t ) = 0.99e − t − e −t cos t + 0.01 cos 10t − 0.001 sin 10t = 0.99e −t − e −t cos t + 0.01 cos(10t + arctg 0.1)
4．P659. 3.9
Solution: 3.9 The loop equations of Prob.2.2: E = ( R1 +
We can use the table of Laplace transform pairs in P636 to get the solution:
i 2 (t ) = L−1 [ I 2 ( s )] == L−1{
1000 } ( s + 88.2)[( s + 13.4) 2 + 312 ]
Method 2: Solving differential equation.
∵ x = x ss + xt .
The characteristic equation is: λ2 + 4λ + 3 = 0 , λ 1 = −1 , λ 2 = −3 Then the transient response is: x(t ) t = a1e −t + a 2 e −3t The steady state response is: x(t ) s = b0 =
第三周作业
English version book P.651.Problems Chapter
1．(P656) 2.20
Solution: Let x1 = h1 , x2 = h2 , u1 = q A ,
R2 + R3 ⎡ ⎢− A ( R R + R R + R R ) 1 1 2 1 3 2 3 x=⎢ ⎢ R3 ⎢ ⎣ A2 ( R1 R2 + R1 R3 + R2 R3 )
可见： wn = 600 = 24.5 ； ξ = 70
2 ⋅ 24.5
= 1.43
6．已知控制系统的传递函数为 G ( s ) =
10(2 s + 1) ，试求系统的单位脉冲 k(t)和单位 ( s + 1)( s 2 + 4 s + 8)
阶跃响应 h(t)。 解： （1）因为单位脉冲输入为： u (t ) = δ(t ) ；其拉氏变换为： U ( s ) = 1 故单位脉冲响应
u2 = qB , then
⎤ 0⎥ qA ⎤ ⎥⎡ ⎢ 1 ⎥ ⎣ qB ⎥ ⎦ A2 ⎥ ⎦
R3 ⎤ ⎡1 ⎥ ⎢ A1 ( R1 R2 + R1 R3 + R2 R3 ) A ⎥x+⎢ 1 ⎥ ⎢ R1 + R3 − ⎥ ⎢0 A2 ( R1 R2 + R1 R3 + R2 R3 ) ⎦ ⎣
2．(胡寿松 P.71 习题 2-6）在液压系统管道中，设通过阀门的流量 Q 满足如下流量方程：
Q=K P
式中，K 为比例常数；P 为阀门前后的压差，若流量 Q 与压差 P 在其平衡点(Q0,P0)附近作微 小变化，试导出线性化流量方程。 解：将非线性的流量方程在平衡点附近根据泰勒级数展开
y = f ( x0 ) + f ′( x0 )( x − x0 ) + ........ Q = K P0 +
R0 9 = =3 A0 3
Considering zero initial conditions, we have:
x(0) = a1 + a 2 + 3 = 0
and
x ′(0) = −a1 − 3a 2 = 0 , solving then got: a1 = −4.5, a 2 = 1.5
Therefore the solution of the problem is as same as method 1:
= 0.153e −88.2t + 0.4e −88.2t sin(31t − 22.4 )
（胡寿松 P.134 习题 3-6）已知控制系统的单位阶跃响应为 5．
h(t ) = 1 + 0.2e −60t − 1.2e −10t ；
试确定系统的阻尼比 ξ 和自然频率 wn 。
解：因为：系统的单位脉冲响应 k(t)的象函数为系统闭环传递函数φ(s)，故可以通过对 k(t) 求拉氏变换得到系统闭环传递函数φ(s)，而 k(t)与单位阶跃响应成 D 关系：
∵ D 2 x + 4 Dx + 3x = 9 ; ∴ X ( s ) =
9 3 4 .5 1.5 = − + s ( s + 4 s + 3) s s + 1 s + 3
2
3 4.5 1.5 x(t ) = L−1 [ X ( s )] = L−1 { − + } = 3 + 1.5e −3t − 4.5e −t s s +1 s + 3
写成增量式：
K 2 P0
( P − P0 ) + ........ = Q0 + K P 0 ΔP
ΔQ = K P 0 ΔP
其中， K P 0 =
K 2 P0
3．P659.
3.8 (b), (d)
Solution: 3.8(b) There are two methods to solve the problem Method 1: Using Laplace transformation.
k (t ) = L−1 [G ( s )] = L−1{−
s+2 11 2 + 2[ + ]} 2 2 s +1 ( s + 2) + 2 ( s + 2) 2 + 2 2
= −2e −t + 2e − 2t cos 2t + 11e − 2t sin 2t
或： k (t ) = L−1 [G ( s )] = L−1 {−
k (t ) = h ′(t ) = −12e −60t + 12e −10t = 12(e −10t − e −60t )
故
2 wn 1 1 600 G ( s ) = L[k (t )] = 12[ − = ]= 2 s + 10 s + 60 ss 2 + 70 s + 600 s 2 + 2ξwn s + wn
I 2 ( D) C2 D = 3 E ( D ) R1C1C 2 LD + ( R1 R2 C1C 2 + LC 2 ) D 2 + ( R1C 2 + R1C1 + R2 C 2 ) D + 1 = D 100 D = 3 2 2 0.01D + 1.15 D + 35 D + 1000 D + 115 D + 3500 D + 100000
ϕ = tg −1 (
w 2 ) = tg −1 = 10.3 α0 −α 11
故： k (t ) = −2e −2t + 11.18e −2t sin( 2t + 10.3 ) （2）因为单位阶跃输入为： u (t ) = 1(t ) ；其拉氏变换为： U ( s ) =
1 s
H (s) =
10(2 s + 111 ⋅ = + − 2 2 s ( s + 1)( s + 4 s + 8) s s + 1 s + 4s + 8
3 2
By residue’s theorem: A=0.99, B=C= -1, D=0.01 and E= -0.01
X ( s) =
− ( s + 1) 0.99 0.01s − 0.01 + + s + 1 ( s + 1) 2 + 1 s 2 + 100
The inverse transform of X(s), the solution is
1 1 )i1 − ( )i 2 C1 D C1 D
0 = −(
1 1 1 )i1 + ( R2 + LD + + )i 2 C1 D C1 D C 2 D
Due to the loop equations of Prob.2.2, the transfer function of i2(t) to e(t) can be obtained