已知某JK触发器的输入信号J、K及CP波形如图所示,设触发器的初始状态为0,试画出Q的波形图?

2025-03-15 21:15:16
推荐回答(4个)
回答1:

先画出来A和B的与、与非的信号波自形,对准画,再对着CP看,在CP上升沿JK触发器触发,根据JK触发器的特征方程: 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;就可以画出输出波形。

J=Q’,K=Q,Z=Q异或CP,

故,Q(n+1)=Q’(n),

Q波形:在CP的每一个下降沿,Q翻转一次,

Z波形:将CP波形与Q波形异或。

动态数据类型

波形图还接收动态数据类型,用于Express VI。动态数据类型除包括对应于信号的数据外,还包括信号信息的各种属性,如信号名称、数据采集日期和时间等。属性指定了信号在波形图中的显示方式。当动态数据类型包含多个通道时,波形图可显示每个通道的曲线并自动格式化图例以及图形x标尺的时间标识。

以上内容参考:百度百科-波形图

回答2:

JK触发器有cp上升沿触发,和下降沿触发两种,都给你画出来了;

JK触发器的特征方程为 Qn = J*Q' + K' *Q;

初态:Q=0,那么

CP1:Qn = J*Q' + 0 = 1;

CP2:Qn = 0 + K' *Q = 1;

cp3、cp4,自己想想;

回答3:

回答4:

第一张是答案 再看第二章你就明白了。

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