第三章答案
??Ax的状态转移矩阵?(t)如下,求其逆矩阵?(t)和系统1. (6分)已知齐次状态方程x矩阵A。
?1?3e?t?2e?2t2e?t?2e?2t?。 ?(t)???t?2t?t?2t??2e?3e???3e?3e?3et?2e2t2et?2e2t??1解: ?(t)??(?t)?? (3分) t2tt2t??2e?3e???3e?3e? A??(t)|t?0???12? (3分) ??-3-4?2. (8分)求定常控制系统的状态响应。
?01??0??1?&x?t????x?t????u?t?,t?0,x?0????,u?t??1?t?
?1?2???1??0??e?t?te?t解:e???t??teAtt0t?te?t??t?1?t??e?? (4分)
?t1?te?t?te?t???x(t)??(t)x(0)???(?)Bu(t??)d??e?t?te?t?t??e????1? (4分)
??d???????0????t????tee??e?0???????Ax的状态转移矩阵?(t)如下,求其系统矩阵A。 3.(3分) 已知齐次状态方程x?3e?t?2e?2t?(t)???t?2t??3e?3e2e?t?2e?2t?。 ?t?2t??2e?3e??解:A??(t)|t?0???12? (3分) ??-3-4??10??1??x??x?u, ????11??1?4.(8分)已知系统的状态方程为:
初始条件为x1(0)?1,x2(0)?0。求系统在单位阶跃输入作用下的响应。
?1???ets?10???1???解:解法1:?(t)?L??t???1s?1???te???0?; (4分) t?e?
?etx??t?te解法2:
t?0??1?et?????t???0e???0?(t??)et???et??et?1??2et?1?0??1?d???t???t???。 (4分) t?????t?1e????te??te??2te??s2?1??s?1?11?s?1?1x(s)?(sI?A){Bu(s)?x(0)}????; ??2?2?2?s1s(s?1)??(s?1)??s(s?1)?2s??11
?2et?1?。 x?L[x(s)]??t??2te??15.(8分) 已知系统的状态空间描述为
?12??0?&x???x??1?u?3?4???? y??10?x1)求系统的状态转移矩阵。
2)初始条件为x(0)??01?,求系统在u?t??1?t?作用下的系统输出y(t)。 解:1)
T?s?1?2?sI?A??s?4??3?s?4??s?42??(s?1)(s?2)1?1(sI?A)???????3(s?1)(s?2)??3s?1???(s?1)(s?2)?2?3??s?2?1?1?1s?1??(t)?L?(sI?A)?L?????3?3??s?1s?2(5分)
2)
2?(s?1)(s?2)??s?1?(s?1)(s?2)??2e?t?2e?2t???2e?t?3e?2t?22???t?2ts?1s?2??3e?2e????23???3e?t?3e?2t?s?1s?2???2e?t?2e?2t?t?2e???2e?2??x(t)??(t)x(0)???(?)Bu(t??)d???????d??t?2t?00?2e???3e?2??2e?3e????t??2e?t?e?2t?1???e?2t?1??2e?2e?????????3131?t?2t??t?2t?3t2e?e???e????2e?3e????22????22??y(t)??10?x(t)??e?2t?1?t?2t
(3分)
或
??e?2t?1?x(t)?L?1[(sI?A)?1x(0)?(sI?A)?1BU(s)]??3?2t1??e?? ?2??2?y(t)??10?x(t)??e?2t?16.(8分) 已知系统的状态空间描述为
2
?01??0?&x???x??1?u?2?3???? y??10?x1)求系统的状态转移矩阵。
2)初始条件为x(0)??01?,求系统在u?t??1?t?作用下的系统输出y(t)。 解:sI?A??T?s?1? (1分) ??2s?3?
s?3??s?31??(s?1)(s?2)1?1(sI?A)???????2(s?1)(s?2)??2s???(s?1)(s?2)?1?2??s?2?1?1?1s?1??(t)?L?(sI?A)?L?????2?2??s?1s?2
1?(s?1)(s?2)?? (2分)
s?(s?1)(s?2)??e?t?e?2t???e?t?2e?2t?(2分)
11???t?2ts?1s?2??2e?e????12???2e?t?2e?2t?s?1s?2???e?t?e?2t?t?e???e?2??x(t)??(t)x(0)???(?)Bu(t??)d????t?????d??2t??2??00?e?e?2e?2e????11??11???e?t?e?2t???e?t?e?2t????e?2t?????t?22?22?2t??????t?2t?2t??e?2e?e?ee????t
(2分)
11y??10?x??e?2t?22(1分)
?1?2t1??e???1?1?1或 x(t)?L[(sI?A)x(0)?(sI?A)BU(s)]??22 (2分)
???2te??11y??10?x??e?2t? (1分)
22?1?t3t?2(e?e)7.(3分)设状态转移矩阵?(t)????e?t?e3t??1?(?e?t?e3t)?4?,试确定其矩阵A。
1?t3t?(e?e)?2?3
&解: A??(t)t?01?t?1?t3t3t?(?e?3e)(e?3e)??2?11?4 ??????1?41??e?t?3e3t(?e?t?3e3t)????2?t?08.(10分) 已知系统的状态空间描述为
?01??0?&??x?x??2?u?2?3???? y??31?x若初始条件为x(0)??01?,求系统在u?t??1?t?作用下的系统输出y(t)。 解:1)
T?s?1?sI?A????2s?3?s?31???s?31??(s?1)(s?2)(s?1)(s?2)?1?1?(sI?A)???????2s(s?1)(s?2)??2s????(s?1)(s?2)(s?1)(s?2)???111??2?????2e?t?e?2ts?2s?1s?2?1?1?1s?1?(t)?L?????(sI?A)???L??22?12?2e?t?2e?2t??????s?1s?2s?1s?2???(6分)
2)
e?t?e?2t???e?t?2e?2t??e?t?e?2t?t?2e???2e?2??x(t)??(t)x(0)???(?)Bu(t??)d????t???d??2t????2??00??e?2e???2e?4e??e?t?e?2t???2e?t?e?2t?1???e?t?1? (4???t?????t??2t??t?2t???e?2e??2e?2e??e?y(t)??31?x(t)??2e?t?3t分)
或用另一种方法
?t??e?1??1?1?1x(t)?L[(sI?A)x(0)?(sI?A)BU(s)]???t??e? (4分)
y(t)??31?x(t)??2e?t?3?3e?t?2e?2t9.(3分)设状态转移矩阵?(t)???t?2t??3e?3e2e?t?2e?2t?,试确定其矩阵A。 ?t?2t??2e?3e?4
&解: A??(t)t?0?12? (3分) ?????3?4?10.(10分) 已知系统的状态空间描述为
?01??0?&??xx???1?u?2?3???? y??10?x初始条件为x(0)??01?,求系统在u?t??1?t?作用下的系统输出y(t)。 解:(1)
T?s?1?sI?A???2s?3??s?31???s?31??(s?1)(s?2)(s?1)(s?2)?1?1?(sI?A)???????2s(s?1)(s?2)??2s????(s?1)(s?2)(s?1)(s?2)???111??2?????2e?t?e?2ts?2s?1s?2?1?1?1s?1?(t)?L?????(sI?A)???L??22?12???2e?t?2e?2t?????s?1s?2s?1s?2??e?t?e?2t???e?t?2e?2t? (6分)
(2)
t?e?t?e?2t?t?e???e?2??x(t)??(t)x(0)???(?)Bu(t??)d????t?????d??2t??2??00?e?e?2e?2e????11??11???e?t?e?2t???e?t?e?2t????e?2t?????t? (4分) 22?22?2t??????t?2t??e?2e?e?2t?e?e???11y??10?x??e?2t?22或用另一种方法
?1?2t1??e??x(t)?L?1[(sI?A)?1x(0)?(sI?A)?1BU(s)]??22???2t e?? (4分)
11y??10?x??e?2t?2211. (10分)已知系统的状态空间描述及初始条件如下:
?01??0?&??xx?u,y??1?????6?5??1?
0?x x1?0??0,x2?0??1
5
数学物理方法第三章答案完整版
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