文档介绍：信号与系统练****及答案一、单项选择题)f(t)的表达式为(()的波形如题1图所示,则2(t-1)u(t-1).tu(t-1)DB.(t-1)u(t-1)C..Atu(t)4?22)]dt-?(t)?2?(t(t?3t?2)[)的积分结果是(4?()(t)的波形如题3(a)图所示,则f(5-2t).周期矩形脉冲的谱线间隔与()),则它的频谱带宽(.,得到相同的响应,)(())(teu((t)=))=,且收敛域为(的拉氏变换为F1s?Re[s]<[s]>1Re[s]<[s]>??dxf(t)?)?(x))等于(的单边拉氏变换F(??11-2s-..ess)?2?(se=的原函数f(t)等于()(s)?2-2t-2(t-1)-2t-2(t-2)u(t-2).eu(t-2)Deu(t-1)(t-1).BDCDA:(t),则该系统的阶跃响应g(t)为_________。(t)的傅里叶变换为X(jω),那么x(t-t)的傅里叶变换为_________________。(t),零状态响应为y(t)=2f(t-t),则该系统的单位冲激0f。_________________为h(t)(t)=u(t),则当该系统的输入信号f(t)=tu(t)时,其零状态响应为_________________。-t)u(t),则其系统函数H(s)=__________。(t)=(1-et?????d1()?6.。??。)?u(t)f(tr(t)?e(2t)是否是线性、时不变、因果的系统?8、判断系统。。)(t)(ft)?(t?1u答案:t????d((t)?)hgtj??eX(j)02、1、??12tu(t)?)t(t2?3、4、021u(t?1)6、5、s(s?1)1???)(?、8、线性、时变、非因果7?js?1、?jt?2t?u(t?1)??(t2)f(t)f(t?e)e(2)(1)?2j???(t?2)?e1?t)(,再由频移性质得,由时移性质得解:(1)已知???1()??1)j2?j?jt2(??e)?eF(je?2(t?),即?j?2e?????t)tj??t2?(?2j?jwt????dtf(?(Fj)dtu?e?dt)1?(e)te?te)2(?j?21?????-t,求系统的零状u(t),系统的冲激响应h(t)=(t)=sintu(t)态响应y(t)的象函数Y(s)。ff)F(s)?H(t)?f()?h(t)Y(s)?sY(t解:ff111