第六章 习 题
6.1 图题6.1所示电路,求u(t)对i(t)的系统函数H(s)=U(s)
/I(s)。
1?1F1HLCU(t)_R2(a)R1I(s)+U(s)_LsL1CsR2
(b)
图题 6.1
答案
有
ll thin解:图解6.1(a)电路的s域电路模型图解6.1(b)所示。故
gse and A代入数据得
hing at a tim in th?1???LsR2????Cs?????U(s)I(s)?R1?1??Ls?R2??Cs???U(s)2s2?2s?1H(s)??2I(s)s?s?1eir being are gi(t)+ood for somethin1F3+U1C1?+LR21HU2(t)-
(a)-R1LsU1(s)U2R2-(b)- 图题 6.2
6.2 图解6.2(a)所示电路,求u2?t?对u1?t?的系统函数
答案
示。故有
ll thin解:图解6.2(a)所示电路的s域电路模型如图解6.2(b)所
gs inH(s)?U2(s)U1(s)。
the and AH(s)?代入数据得
hing at a timeirU2?s?s2?2sH(s)??2U1?s?s?5s?3 beU2(s)Ls?R2?U1(s)Ls?R2?1?R2Csing ar++e g1Csood f2?or somethin?5t 6.3 已知系统的单位冲激响应h(t)?5teU(t),零状态响应
y(t)?U(t)?2e?5tU(t)?5te?5tU(t)。求系统的激励f(t)。
答案
5s?5解:
故得
in th
y/0??-2。
e a?? timt ahing and All thins2?5H(s)?20???0,s?5s?5,初始状态为y?6.4 已知系统函数
(1).求系统的单位冲激响应h(t);
(2). 当激励f(t)=δ(t)时,求系统的全响应y(t);(3)当激励f(t)=U(t)时,求系统的全响应y(t)。
gs33f(t)?δ(t)?U(t)?e-5tU(t)?δ(t)??1?e-5t?U(t)55eir125??Y(s)ss?5?s?5?2311F(s)?????5H(s)5ss?5s?5 be
ing故得激励f(t)的像函数为
ar125Y(s)???ss?5?s?5?2e gooh?s??d for somethin答案
解: (1)
故
h(t)?δ(t)-2e-2tcostU(t)?e-2tsinU(t)?δ(t)-2e-2t?cos2t?2sin2t?U(t)(2)系统的微分方程为
y//?t??2y/?t??5y?t??f//?t??5f/?t? be对上式等号两边同时求拉普拉斯变换,并考虑到拉普拉斯变换
的微分性质,有
ings2Y(s)?sy0??y/0??2sY(s)?2y0??5Y(s)?s2F(s)?5F(s ①
?/0????2, 代入上式得今 F(s)?1,y(0)?0,y? inll thin故得全响应为
gsnd Ae a1F(s)?,y(0?)?0,y/0???2s(3)将 代入上式①,有
th timeir???? ar?e g????s2?32?s?1?Y(s)?2?1?s?2s?5?s?1?2?4y(t)?δ(t)-2e-tcostU(t)s2?2s?512Y(s)?2??2?ss?5s?5s?s?1?2?4?hing at a故得全响应为
ood fors2?5?2s?2s?s?1?2H(s)?2?1?2?1??s?5s?5s?2s?5?s?2?2?1?s?1?2?4 somethiny(t)?1?2e?tsin2tU(t)??6.5 图题6.5所示电路。(1)求电路的单位冲激响应h(t);
i0?和u0?;
????????(3)今欲使电路的单位阶跃响应g(t)=U(t),求电路的初始状态
i0?和u0?。
答案
根据该图得
in th解(1)零状态条件下的s域电路模型如图解6.5(b)所示。故
eir bell thin故得单位冲激响应为
h(t)?te?tU(t)?V?e a6.5(c)所示。故
nd A(2)非零状态条件下求零输入响应ux?t?的s域电路模型如图
gsH(s)?U?s?11??2?F?s?2?s?1s?2s?1?s?1?2shing at a tim1i0??u0?11sux?s????u0?1ss2?s?s??ing ar1s??e g(2)今欲使电路的零输入响应ux(t)=h(t),求电路的初始状态
oo??d for somethin