第八章习题参考答案

自动控制原理的课后习题答案,讲解非常详细!

第八章习题参考答案8-7 解： Q L 1[ 1 1 ] = L 1[ 1 1 + 1 ] = t 1(t ) + e t s s ( s + 1) s2 s s +1 1 1 ∴ Z[ ] = Z [t 1(t ) + e t ] s s ( s + 1) Tz z z = + 2 ( z 1) z 1 z e tTz ( z e T ) z ( z 1)( z e T ) + z ( z 1) 2 = ( z 1) 2 ( z e T ) z ( z 0.368) z ( z 1)( z 0.368) + z ( z 1) 2 = ( z 1) 2 ( z 0.368)

自动控制原理的课后习题答案,讲解非常详细!

1 1 ] G0 ( z ) = (1 z ) Z [ s s ( s + 1) 1

z z ( z 0.368) z ( z 1)( z 0.368) + z ( z 1) 2 = × z 1 ( z 1) 2 ( z 0.368) z 0.368 ( z 1)( z 0.368) + ( z 1) 2 = ( z 1)( z 0.368) 0.368 z + 0.264 = 2 z 1.368 z + 0.368 0.368 z + 0.264 G0 ( z ) G0 ( z ) z 2 1.368 z + 0.368 Gc ( z ) = = = 1 + GH ( z ) 1 + G0 ( z ) 1 + 0.368 z + 0.264 z 2 1.368 z + 0.368 0.368 z + 0.264 = 2 z z + 0.632

自动控制原理的课后习题答案,讲解非常详细!

0.368 z + 0.264 z C ( z ) = Gc ( z ) R ( z ) = 2 × z z + 0.632 z 1 0.368 z 2 + 0.264 z = 3 z 2 z 2 + 1.632 z 0.632

z R( z ) = z 1

用长除法将C(z)展开成幂级数：C ( z ) = 0.368 z 1 + z 2 + 1.4 z 3 + 1.4 z 4 + 1.147 z 5 + 0.895 z 6 + L

取Z的反变换得：

c * (t ) = 0.368δ (t T ) + δ (t 2T ) + 1.4δ (t 3T ) + 1.4δ (t 4T ) + 1.147δ (t 5T ) + 0.895δ (t 6T ) + L

自动控制原理的课后习题答案,讲解非常详细!

8-11 解： 系统的开环脉冲传递函数为：K K (1 e T ) z G( z) = Z[ ]= 2 s ( s + 1) z (1 + e T ) z + e T

闭环脉冲传递函数为：C ( z) G( z) G( z) = = R( z ) 1 + GH ( z ) 1 + G ( z ) K (1 e T ) z z 2 (1 + e T ) z + e T = K (1 e T ) z 1+ 2 z (1 + e T ) z + e T K (1 e T ) z = 2 z + [ K (1 e T ) (1 + e T )]z + e T

自动控制原理的课后习题答案,讲解非常详细!

闭环特征方程为：

z + [ K (1 e ) (1 + e )]z + e2

T

T

T

=0

作W变换得：

K (1 e T ) w2 + 2(1 e T ) w + [ K (1 e T ) + 2(1 + e T )] = 0由劳斯判据令各系数均大于零，即：

K (1 e ) > 0 2(1 e T ) > 0 K (1 e ) + 2(1 + e ) > 0解出系统的稳定域为： T T

T T

1+ e 0< K <2 T 1 e T >0

T

自动控制原理的课后习题答案,讲解非常详细!

8-17 解：

1 G ( z ) = Z [Gh ( z ) G0 ( z )] = (1 z ) Z [ G0 ( z )] s 2 1 1/ 2 1/ 2 1 1 = (1 z ) Z [ 2 ] = (1 z ) Z [ 2 + ] s ( s + 2) s s s+2 1 1 2t 1 = (1 z ) Z [t 1(t ) + e ] 2 2 Tz 1 z 1 z 1 ] = (1 z ) [ + 2 2T ( z 1) 2 z 1 2 z e 1

0.184 z + 0.132 z (0.184 + 0.132 z ) = = 1 1 ( z 1)( z 0.368) (1 z )(1 0.368 z )

1

1

自动控制原理的课后习题答案,讲解非常详细!

单位阶跃输入时的最小拍控制器为：

z 1 D( z ) = 1 1 z G( z) z (1 z )(1 0.368 z ) = 1 1 1 1 z z (0.184 + 0.132 z ) 1 0.368 z 5.435(1 0.368 z ) = = 1 1 0.184 + 0.132 z 1 + 0.717 z 1 1 1 1 1

1

自动控制原理的课后习题答案,讲解非常详细!

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