Question
Download Solution PDFAn X-Y flip-flop whose characteristic table is given below is to be implemented using a J-K flip-flop
|
X |
Y |
Qn+1 |
|
0 |
0 |
1 |
|
0 |
1 |
Qn |
|
1 |
0 |
Q̅n |
|
1 |
1 |
0 |
This can be done using-
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The characteristic equation of a J-K flip flop is given by-
⇒ Qn+1 = JQ̅n + K̅Qn
Calculation:
Let Qn is the present state and Qn+1 is the next state of the given X-Y flip-flop.
|
X |
Y |
Qn |
Qn+1 |
|
0 |
0 |
0 |
1 |
|
0 |
0 |
1 |
1 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
|
1 |
0 |
0 |
1 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
0 |
Solving the above using K-map, we get the characteristic equation of X-Y flip-flop is-
Qn+1 = Y̅Q̅n + X̅Qn
Characteristic equation of a J-K flip flop is given by
Qn+1 = JQ̅n + K̅Qn
Comparing, J = Y̅ and K = XLast updated on Jun 23, 2025
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.