离散时间控制系统作业答案

Judge the stability of the system:
A( z) 1 Q( z) 0 z 2 (1 e T ) z e T k (1 e T ) z 0 A(1) k (1 e T ) 0 k 0(0 e T 1)
X ( z) 3z 2 5 z ( z 1) 2 ( z 2)
Solution Note that the X(z) has two poles z1=1 (two order) and z2=2. Hence, from residue computation equation, we obtain
A( z ) 1 KG( z ) 0 z 2 (0.368K 1.368) z 0.368 0.264K 0 A(1) 0.632K 0 K 0 A(1) 2.736 6.104K 0 K 26.382 0 K 2.394 A(0) 1 5.182 K 2.394
x( k )
d 3z 2 5 z ( z 1) 2 z k 1 2 dz ( z 1) ( z 2)
3z 2 5 z ( z 2) z k 1 2k 1 2 k 2 ( z 1) ( z 2) z 2 z 1
k 0
③Find the initial value and final value for each of the following functions:
X ( z)
10 z 1 1 ， X ( z) 1 1 z (1 z 1 ) 2
Solution x(0)=1，x()=1 x(0)=0，x()= ④有关 z 传递函数比较综合的题目： 由下述差分方程描述的系统
Y ( z) 5 X ( z ) ( z 1) 2 ( z 5)
Solution:
Y ( z) 5 5 0 3 2 2 X ( z ) ( z 1) ( z 5) z 7 z ＋11z 5
Biblioteka Baidu
1 0 0 0 0 0 1 , 0 , C 5 0 0 5 11 7 1
y(k ) y(k 1) 0.24 y(k 2) u(k ) u(k 1)
式中 u(k)、 y(k)为系统的输入输出信号， 假设 y(k)=0 k<0， 当输入为单位阶跃序列时， 确定 y(k)的解析解。 解：
G( z ) Y ( z) 1 z 1 z2 z z z 2 ( z 1) Y ( z ) G ( z ) U ( z ) U ( z ) 1 z 1 0.24 z 2 z 2 z 0.24 z 1 ( z 1)( z 0.4)( z 0.6) Y ( z) z ( z 1) 25 / 3 14 / 3 12 z ( z 1)( z 0.4)( z 0.6) z 1 z 0.4 z 0.6 25 14 y(k ) 1(k ) (0.4) k 12(0.6) k 3 3
A(1) 0 k 2(1 e T ) 1 e T
A(0) 1
a. when T = 0.5, k should satisfy 0 < k < 8.15, so we should only need to compute ess for k = 6.
K p lim Q( z ) , K v lim
X (s) 1 1 1 1 Tz 1 1 1 X ( z ) 2 2 1 2 1 s s 1 s ( s 1) s (1 z ) 1 z 1 e T z 1
X ( z ) z 1Z{ 2 1 2 1 } z 1{ } s 1 s 2 1 e T z 1 1 e 2T z 1
*图讲完后的作业： ①Ppt 表中倒数第一、第二行的脉冲传递函数的求解。 C(s) R(s) G1(s) G2(s) + H(s)
C(z)
C ( z)
G2 ( z )G1 R( z ) 1 G1G2 H ( z )
R(s) +

G(s) H(s)
C(s)
C(z)
-
C( z)
GR( z ) 1 GH ( z )
0.3720 - 0.3652i >> abs(ans) ans = 3.1894 0.5213 0.5213 ③ G(z) R(s) T _
D(z) ZOH
G(s)
Y(s)
G( s)
1 , D( z ) k , T 0.1 s(0.1s 1)
Determine steady-state error coefficients and steady-state error when the input is t and t2 respectively, where k =1.
③When time delay is larger than the sampling period, try to obtain the state-space model of zero-order-hold sampling of the continuous-time system. ④2.12(a) 外部模型讲完后的作业： ①Let us find the z-transform of the（部分分式法或留数法均可） 1 s3 X (s) 2 ， X ( s) e Ts ( s 1)( s 2) s ( s 1) Solution
②
(a) Find the x(kT) k 0, for（long－division）
X ( z) 10 z 5 ( z 1)( z 0.2)
Solution
X ( z) 10z 1 17 z 2 18.4 z 3 ...
(b) Find the x(kT) k 0 for（部分分式法）
1 r (t ) 1(t ) t , ess=essp+essv=0+(1/3k) 0.05. K6.67 23.94 > k 6.67 3
④ Ppt 例子 4.6 系统框图如③中所示： where
G( s ) 1 kz , D( z ) , r (t ) 1(t ) 0.5t , s 1 z 1
X ( z) 10 z ( z 1)( z 0.2)
Solution X ( z) 10 12.5 12.5 12.5 z 12.5 z ， X ( z) ， x(kT) 12.5(1 0.2k ), k 0 z ( z 1)( z 0.2) z 1 z 0.2 z 1 z 0.2 (c) Find the x(kT) k 0 for （留数法）
②use Jury stability criterion and 修正 Routh criterion to analyze the stability of the system :
A( z) 45z 3 177z 2 119z 39 0
roots([45 -177 119 -39]) ans = 3.1894 0.3720 + 0.3652i
脉冲传递函数变成状态空间表示： ①脉冲传递函数具有 m 重根时，从脉冲传递函数推导 Jordan 形式的状态空间表示！ ②Obtain the state-space equation for the following pulse-transfer function, and then draw the block diagram for the state-space form.
a. T = 0.5, k = 6 and 10, compute the ess respectively; b. T = 0.5, determine the range of the k for satisfying ess 0.05. 解：
G( z ) (1 z 1 )Z{ G( s) 1 e T k (1 e T ) z } , Q ( z ) D ( z ) G ( z ) s z e T ( z 1)( z e T )
1 0 1 , 1, C 1 0 0 . 16 1
Solution: Y ( z) z2 2 U ( z ) z z 0.16 第四章作业 ①Ppt24 页，例子 4.5
G(z) T _
D(z) ZOH
R(s)
G(s)
Y(s)
1 When r (t ) 1(t ) t , determine the interval of the k which should make ess 0.05. 3
Solution: First we will analyze the stability of the system: G(s) G(z) 判断 G(z)的稳定性 0 < k < 23.94 r(t) = t, Kv=1. ess=1 r(t) = t2, Ka= 0, ess=inf
第一、 二章没有作业 第三章的作业 内部模型讲完后的作业： ①2.1、2.2(a)(c)、2.3(a)(b)、2.3(b)draw the block-diagram of the state-space model. ②Try to compute the transform matrix T when the original system and the changed system are described as follows:
Determine the range of K when system is stable where 1 G( s) , T 1, D( z ) k s( s 1) Solution：
G ( z ) (1 z 1 ) Z { G( s) 0.368z 0.264 } s ( z 1)( z 0.368)
③ Obtain the state-space equation for the following difference equation, and then draw the block diagram for the state-space form. y(k 2) y(k 1) 0.16 y(k ) u(k 1) 2u(k ) y(0)=y(1)=u(0)=0 Solution: Y ( z) z2 2 U ( z ) z z 0.16 1 0 0 , , C 2 1 0.16 1 1 ④compute the z-pulse transfer function.
z 1
1 z 1 k Q( z ) , z 1 T T
ess
1 0.5 T 0.042 1 K p K v 2k
T 0.05 k 5 2k
b.
ess
with the stability condition 0 < k < 8.15, we conclude that