机械控制工程基础答案提示
第二章 系统的数学模型
2-1 试求如图2-35所示机械系统的作用力F(t)与位移y(t)之间微分方程和传递函数。
F(t)y(t)f
图2-35 题2-1图
解：依题意： m?dy?t?2dt22?ab?F?t??f?dy?t?dt?ky?t?
故 mdy?t?dt2?f?dy?t?dta?ky?t??ab?F?t?
传递函数: G?s??Y?s?F?s??ms2b
?fs?k2-2 对于如图2-36所示系统，试求出作用力F1(t)到位移x2(t)的传递函数。其中，f为粘性阻尼系数。F2(t)到位移x1(t)的传递函数又是什么？
k1 m1 x1(t) F1(t) k2 f F2(t) m2 x2(t) 图2-36 题2-2图
解：依题意：
dx1?t??dx1?t?dx2?t????F?kxt?f??mm 对1： 1 1112?dtdt?dt??2 1
对两边拉氏变换：F1?s??k1x1?f?sX1?s??sX2?s???2m1sX1?s? ①
dx2?t??dx?t?dx2?t?? 对m2： F2?t??f?1 ????kxt?m2222?dtdtdt??2对两边拉氏变换：F2?s??f?sx1?s??sx2?s???k2x2?s??m2s2X2?s? ② ?m1s2?fs?k1x1?s??fsx2?s??F1?s?故： ? 2??fsx1?s??m2s?fs?k2x2?s??F2?S?????2?F1?s??m2s?fs?k2?fsF2?s??x1?s??222?m1s?fs?k1m2s?fs?k2??fs?故得：? 2fsF1?s??F2?s?m1s?fs?k2?x?s??2222???ms?fs?kms?fs?k?fs1122?????????????故求F1?t?到x2?t?的传递函数 令：F2?s??0
G1?s??x2?s?F1?s?4?fs
??ms12?fs?k1??m2s?fs?k2???fs?22
fsm1m2s?f?m1?m2?s??m1k2?m2k1?s?f?k1?k2?s?k1k232求F2?t?到x1?t?的传递函数 令：F1?s??0
G1?s??x1?s?F2?s?4?fs?ms12?fs?k1??m2s?fs?k2???fs?22
?fsm1m2s?f?m1?m2?s??m1k2?m2k1?s?f?k1?k2?s?k1k232
2-3 试求图2-37所示无源网络传递函数。
i1(t)R1ui R1 uo L1 uii2(t)i(t)R2uoC1 C2 L2 R2 a)图2-37 题2-3图
解 （a）系统微分方程为
b） 2
1C?i?t?dt?i?t?R121 ui?i2?t?R1?i?t?R2 u0?i?t?R2i?t??i1?t??i2?t?拉氏变换得 1CsI1?s??R1I2?s?Ui?s??I2?s?R1?I1?s?R2 U0?s??I1?s?R2I?s??I1?s??I2?s?R2消去中间变量I1?s?,I2?s?,I?s?得：G?s??U0?s?Ui?s??R2?CsR1?1?R1?R2?CsR1?1??R1?R2?R1Cs?1?
Cs?1R1R2R1?R2（b）设各支路电流如图所示。
ui i1 i3 R1 i2 L1 i4 i5 C1 C2 L2 i6 R2 uo b）
系统微分方程为
ui?t??R1i3?t??uR1i3?t??Lu0?t??1C2di2?t?1 0?t??1??2??3??4??5??6?
dt4
?i?t?dtdi5?t?dtu0?t??L2u0?t??R2i6?t?i2?t??i3?t??i4?t??i5?t??i6?t?由（1）得：Ui?s??R1I3?s??Uo?s? 由（2）得：R1I3?s??L1sI2?s? 由（3）得：Uo?s??1C2si4?s?
3
由（4）得：Uo?s??L2sI5?s? 由（5）得：Uo?s??R2I6?s?
由（6）得：I2?s??I3?s??I4?s??I5?s??I6?s?
故消去中间变量I1?s?,I2?s?,I3?s?,I4?s?,I5?s?,I6?s?得： L2?Us??L1s?1?U?L?L?o?12?R1? i?s?L1L2LL?R?R?LC2?12122ss?11?L2?L1?L2?R1R22-4 证明L?cos?t??ss2??2
证明：设f?t??cos?t
由微分定理有L?d2f??t????s2F?s??sf?0??(1)?0? ?dt2f?2由于f?0??cos0?1，f?1??0????sin0?0，
df?t?2dt2???cos?t将式（2）各项带入式（1）中得
L?2???cos?t???s2F?s??s 即 ??2F?s??s2F?s??s
整理得F?s??ss2??2
2-5 求f(t)?12t2的拉氏变换。
解：F?s??L?12??2t??????12?st02t2e?stdt?1?12?0s3?st?ed?st?
令st?x，得
F?s??12s3??2?x0xedx
由于伽马函数??n?1????xne?x0dx?n!，在此n?2
所以F?s??12s32!?1s3
2-6 求下列象函数的拉氏反变换。 （１）X(s)?5s?3(s?1)(s?2)(s?3)
4
1）2） （
（
（２）X(s)?s?2s?3(s?1)32
（３）X(s)?解：（1） X(s)?1s(s?1)(s?2)3
5s?3(s?1)(s?2)(s?3)?A1s?1?A2s?2?A3s?3
A1??X(s)(s?1)?s??1?5??1??3(?1?2)(?1?3)??1
同理 A2?7，A3??6
X(s)??1s?1?7s?2?6s?3
拉式反变换得
x(t)??e?t?7e?2t?6e?3t
2（2）X(s)?s?2s?3(s?1)32??s?1??23(s?1)?2(s?1)3?1s?1
拉式反变换得
x(t)?te2?t?e?t
A1(s?1)3（3）X(s)?1s(s?1)(s?2)3??A2(s?1)2?A3(s?1)?A4s?A5s?2
A1?1s(s?2)s??1?1?1(?1?2)??1
A2??d?1??ds?s(s?2)?1d2?s??1??2s?2?s(s?2)22s??1?0
??1A3??2?2ds?s(s?2)?s??11d???2s?2????22?2ds?s(s?2)??s??1s(s?2)?(s?1)(4s?12s?8s)s(s?2)44s??12232??1
A4?1(s?1)(s?2)1s(s?1)3s??23s?0?12
A5??12
所以X(s)??1(s?1)3?1s?1?12s?12(s?2)
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