令f1(t)=2ε(t)–2ε(t –1),f2(t)=ε(t+1)–ε(t–1)f1(t), f1(t)*f2(t) 的结果是:
A、2(t+1)ε(t+1)- 2(t–1)ε(t–1)
B、2tε(t)–2(t–2)ε(t–2)
C、2(t+1)ε(t+1)- 2(t–1)ε(t–1)–2tε(t)–2(t–2)ε(t–2)
D、2(t)ε(t)- 2(t)ε(t)–2tε(t)–2(t–2)ε(t–2)
发布时间:2025-08-20 13:43:28
A、2(t+1)ε(t+1)- 2(t–1)ε(t–1)
B、2tε(t)–2(t–2)ε(t–2)
C、2(t+1)ε(t+1)- 2(t–1)ε(t–1)–2tε(t)–2(t–2)ε(t–2)
D、2(t)ε(t)- 2(t)ε(t)–2tε(t)–2(t–2)ε(t–2)