信号与系统(应自炉)习题答案第2章习题解

#y(t)=£{(—4。一2i)+6)[w((r-2i)—1)— —2i)-2)]+(4(—2i)—10)R((—2i)—2)— —2i)-3)]}SO+ —u(t—1)]当输入为]任)时：6yQ)=2(1—cos~u(t—1)]+2[cos(，—1)—cos["Q—1)—u(t—2p，)] 2[1—cos(f 1)][〃Q—2pi)—n(t—2pi—1)]2-28题图2-28所示系统是由几个“子系统”组成，各子系统的冲激响应分别为:h(t)=u(t)(积分器)1h)=8Q—1)(单位延时器)2h(t)=-5(t)(倒相器)3题图2-28求系统总的冲激响应。TOC\o"1-5"\h\z解：y(t),x(t)*[h(t)+h(t(t)*h(t)]=x(t)*[u(t)-u(t)*5(t-1)*„(t)]112 3=x(t)*[u(t)-u(t-1)]，二h(t),u(t)-u(t-1)2-29已知LTI系统框图如题图2-29所示，各子系统的冲激响应分别为：h(t),u(t)-u(1-1)，h(t),u(t)，h(t),5(t)，1 2 3求总系统的冲激响应h(t)。题图2-29解：y(t)，x(t)*[h(t)+h(t)]*h(t))=x(t)*[u(t)-u(t-1)]+x(t)*[u(t)-u(t-1)]*u(t)2 3 1将x(t),5(t)带入得到：h(t)，[u(t)—u(t—1)]+[u(t)—u(t—1)]*u(t)，u(t)+1[u(t)—u(t—1)]2-30已知系统的冲激响应h(t)，e-2tu(t)(1)若激励信号为x(t),e-t[u(t)-u(t-2)]+ (t—2)，式中P为常数，试确定零状态响应y(t)，要求系统在t>2的零状态响应为零，试确定P值应等于多少？若激励信号表示为x(t),f(t)[u(t)-u(t-2)]+ (t-2)，式中f(t)为任意t函数，要求系统在t>2的零状态响应为零，试确定P值应等于多少？并验证(1)解：易求系统的零状态响应：(I)y(t)=h(t)*x(t)=e-2tu(t)*[e-t(u(t)-u(t-2))+P8(t-2)]=Pe-2(t-2)+e-2(e2-1) (在t>2时)=0,易求：P=e-4一e-2(2)y(t)=h(t)*x(t)=e-2tu(t)*{f(t)[u(t)-u(t-2)]+P,(t-2)}=e-2tu(t)*{f(t)[u(t)-u(t-2)]}+Pe-23-2)u(t-2)因系统在t>2的零状态响应为零有邛=-J02f(T炉dTe4e-21J2f(T)e2T邛=-J02f(T炉dTe40带入验证P=e-4-e-22-20求下列各函数")与負的卷积f1(t)*f2(t)(1) f(t),u(t),f(t),e-3tu(t)1 2(2) f(t),tu(t),f(t),u(t)1 2(3) f(t),tu(t),f(t),u(t)一uG-2„12(4) f(t),tu(t-1),f(t),u(t+2)12(5) f(t),8(t-2),f(t),cos(…t+45。)12(6) f(t),(1+t)["(t)-u(t-1)],f(t),u(t-1)-u(t—2)12(7) f(t),cos(®t),f(t),8(t+1)-8(t-1)1 2(8) f(t)，e-2tu(t),f(t)，sintu(t)12解：⑴f(t)，f(t)*f(t)，顼-e-3,)u(t)(2)f(t)，f(t)*f(t)，*112u(t)1 2 3 1 2 2(3)f(t),212u(t)+2(t-2)2u(t-2) (4)f(t),2(t+1)2u(t+D⑸f(t),cos(…(t—2)+4