-6/(s+2)/(s-1)=2/(s+2)-2/(s-1),==== 2e^(-2t)u(t)+2e^(t)u(-t)
6/(s-2)/(s-1)=6/(s-2)-6/(s-1),===== -6e^(2t)u(-t) +6e^(t)u(-t)
简单吧,
δ(t)是单位脉冲信号,如果某个信号f(t)与δ(t+a)卷积,就是将f(t)移a个单位,变成f(t+a)。因此u(t+1)与f2(t)卷积后,得到u(t+1+5)+u(t+1-5),而u(t-1)与f2(t)卷积后,得到u(t-1+5)+u(t-1-5),将u(t+1+5)+u(t+1-5)与u(t-1+5)+u(t-1-5)相减就得到答案u(t+6)+u(t-4)-u(t+4)-u(t-6),也就是s(t)=f1(t)*f2(t)=[u(t+6)-u(t+4)]+[u(t-4)-u(t-6)]欢迎分享,转载请注明来源:内存溢出
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