2. 如果JK触发器的J=0,K=1,则当时钟脉冲出现时,Qn+1为 ( )

A.1 B.0 C.Q D.-Q
2025-03-24 17:33:48
推荐回答(1个)
回答1:

J = K = 1 时,Qn+1 = Qn‘,即触发器翻转。
J = 1,K = 0 时,Qn+1 = 1 ;
J = 0,K = 1 时,Qn+1 = 0 ;
J = K = 0 时,Qn+1 = Qn ;
J = K = 1 时,Qn+1 = Qn' ;