Question
Download Solution PDFIf the present state is 0 and the next state is 1, then:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFTruth Table of JK flip flop
|
J |
K |
Qn+1 |
|
0 |
0 |
Qn |
|
0 |
1 |
0 |
|
1 |
0 |
1 |
|
1 |
1 |
\(\overline{Q_n}\) |
When both inputs are low, the next state output will be the present state output.
For J=0 & K=1, the next state output will always be 0 irrespective of the present state.
For J=1 & K=0, the next state output will always be 1 irrespective of the present state.
When both inputs are high, the next state output will be the complement of the present state output.
Characteristics table of JK flip flop
|
J |
K |
Qn |
Qn+1 |
|
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
1 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
0 |
1 |
|
1 |
0 |
1 |
1 |
|
1 |
1 |
0 |
1 |
|
1 |
1 |
1 |
0 |
Excitation table of JK flip flop
|
Qn |
Qn+1 |
J |
K |
|
0 |
0 |
0 |
X |
|
0 |
1 |
1 |
X |
|
1 |
0 |
X |
1 |
|
1 |
1 |
X |
0 |
From the excitation table, it is observed that:
If the present state is 0 and the next state is 1, then J = 1 & k = don't care.
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