in2?1HRL++f(t)1FCU(s)--(a)
2?sI(s)gR ++F(s)1esU(s)r--a g(b)n
ieb2?rs i-i(0-)+Re+h1+f(t)tU(s)- 1snsu(0?)-i sg(c)
2?s-i(0-n)+iht+ 1+1llsU(s)s-A1 u(0?)+-ds-na(d) e依题意要求，应使Ux?s??H(s),即应有
htemos rof do6.5
timhing at a图解o1i(0?)?u(0?)111s??u(0?)?21ss??s?12?s?s从而有
(s?2)u(0?)?i(0?)?1u(0?)?0,i(0?)?1A（3）非零状态条件下求电路单位阶跃响应g（t）的s域电路模
型，如图解6.5（d）所示。故
gs in1G(s)?,s即应有依题意要求，应使
e and All thin0???i?0???s?2从而有 ?s?2?u?0???1V,i?0???0故得 u?U2?s?U1?s?；
their11?i(0?)?u(0?)11sG(s)?s??u(0?)1ss2?s?s11?i(0?)?u(0?)111ss??u(0?)?1sss2?s?s tim6.6 图题6.6所示电路。(1)求
hing at a(2)若u1?t??cos2tU(t)?V?,C?1F,求零状态响应u2?t?；
beingH(s)? are good for somethin(3)在u1?t?不变的条件下，为使响应u2?t?中不存在正弦稳态响
应，求C的值及此时的响应u2?t?。
somethin答案
示。故
na e r1Hof? C?du1u2o??g?oa? era sg?ni?us?e1?b1u2?s?? Csr?ie?b?ht解（1）图解6.6 n（a）电路的s域电路模型如图题6.6（b）所
i sg1nH2s2?1?i?s??h1s?1s2?2C1tCsCs?C 2?ls?1lCs（2）
U1?s??ss2?4,C?1F, 则
H?s??s2?1s2?1s2?2s?1??s?1?2故
Ad timhing at a
K3s2?1sK11K12K2U2?s??H?s?U1?s????????s?1?2s2?4?s?1?2?s?1?s?j2s?j2K11?（s??1）
??d?s2?1s162???K12??s?1??ds??s?1?2s2?4?s??125???eirK3?K2? be?53.1。故得
inll things A th。。21633u2?t???te?t?e?t?e?j2te?j53.1?ej2tej53.1?5251010???3?8??t3???te?cos2t?53.1?????u?t??V??2?5?5??????????????正弦稳态响应?瞬态响应?s2?s2?1Css2?4ndU2?s??H?s?U1?s??ingK2??1s3??s?J2?10?s?1?2?s?j2??s?j2?s??j2?s2310 ar??53.1。e a（3）
hing at a tim由此式可见，欲使u2?t?中不存在正弦稳态响应，就必须有
s2?1?s2?4,C故得C?0.25F。代入上式有
e g?21s?CCoo?d f?1s22??s?1??5 ?s?1?2s2?4?s2?or somethinU2?s??s1.077?0.077??s2?8s?4s?4?23s?4?23
6.7 图题6.7所示电路。
U2?s?U1?s?；
H(s)?(3)求K=2时，系统的单位冲激响应h(t)。
eir be(2)求K满足什么条件时系统稳定；
ing(1)求
ar1?1?+U(s)1s1s the g+U1(s)oo-????4?23?t??4?23?t?V?u2?t???1.077e?0.077e??????????????U?t???瞬态响应??d f+or+U2(s)故得
inll things-图解 6.7
答案
A解（1）对节点①列写KCL方程为
time a????11??s??δ(s)-sKU?s??U1?s?1?s??s??ndδ(s)?又
hing at aU?s??1??1??1?s?s somethin--
西工大,西电 孙肖子版 模电第六章 复频域系统函数与系统模拟--答案
