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Synchronous Logic Circuit Revision Q

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The document discusses a synchronous sequential circuit consisting of three J-K flip-flops, requiring the demonstration of state changes over six clock cycles from an initial state of '0'. It details the logic input descriptions and transition states for each flip-flop, as derived from the J-K flip-flop transition table. The process includes populating a table with initial conditions and tracking each clock cycle's output based on the current state.

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Synchronous Sequential Circuits Class Revision Question A synchronous sequential circuit is shown below which is comprised of three J-K flip-flops. (Let the flip-flops be numbered 1, 2 and 3 from the left hand side). 1 2 3
The J-K Transition Table is: Synchronous Sequential Circuits Class Revision Question Assuming that the outputs Q1, Q2 and Q3 of the flip-flops are all initially in the '0' state, demonstrate the sequence that the circuit will clock through over the next 6 clock cycles by obtaining logic descriptions for each of the J-K inputs, and deriving each 'next-state' output from each 'present-state' output using the J-K flip-flop transition table. (Ignore the set and clr of each flip-flop). J K Q(n+1) 0 0 Qn No Change 0 1 0 Clear 1 0 1 Set 1 1 Qn Toggles
The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Looking at Flip Flop 1:- J1 = Q2 Similarly at Flip Flop 2:- J2 = Q3 And also:- K1 = Q2 And also:- K2 = Q3
The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Looking at Flip Flop 3:- J1 = 1 And also:- K3 = Q1.Q2 = (Q1 AND’ed with Q2)
The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Now let’s tabulate the synchronous conditions J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2)
The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions And add the initial conditions J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition
The J-K Transition Table is: Synchronous Sequential Circuits Starting to populate the table As Q2 = 0, J1 and K1 = 0. As Q3 = 0, J2 and K2 = 0. As Q1 = 0 & Q2 =0; Q1.Q2 = 0 so K3 = 0. (J3 is ALWAYS 1) J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial Condition 0 0 0 0 0 0 0 1 0
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 0:- Q1 (stays same) = 0. 0 1 0 J2 and K2 = 0:- Q2 (stays same) = 0. 1 0 1 J3 = 1 & K3 = 0:- Q3 = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial Condition 0 0 0 0 0 0 0 1 0 Clock 1 0 0 1
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table As Q2 = 0, J1 and K1 = 0. J K Q(n+1) As Q3 = 1, J2 and K2 = 1. 0 0 Qn As Q1.Q2 = 0 so K3 = 0. 0 1 0 (J3 is ALWAYS 1) 1 0 1 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q 2) Initial Condition 0 0 0 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 0:- Q1 (stays same) = 0. 0 1 0 J2 and K2 = 1:- Q2 (Toggles) = 1. 1 0 1 J3 = 1 & K3 = 0:- Q3 (Set) = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table As Q2 = 1, J1 and K1 = 1. J K Q(n+1) As Q3 = 1, J2 and K2 = 1. 0 0 Qn As Q1.Q2 = 0 so K3 = 0. 0 1 0 (J3 is ALWAYS 1) 1 0 1 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q 2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 1:- Q1 (Toggles) = 1. 0 1 0 J2 and K2 = 1:- Q2 (Toggles) = 0. 1 0 1 J3 = 1 & K3 = 0:- Q3 (Set) = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1
The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J K Q(n+1) As Q2 = 0, J1 and K1 = 0. 0 0 Qn As Q3 = 1, J2 and K2 = 1. 0 1 0 As Q1.Q2 = 0 so K3 = 0. 1 0 1 (J3 is ALWAYS 1) 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1 0 0 1 1 1 0
The J-K Transition Table is: Can you complete Populating the table? J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1 0 0 1 1 1 0 Clock 4 Clock 5 Clock 6

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Synchronous Logic Circuit Revision Q

  • 1. Synchronous Sequential Circuits Class Revision Question A synchronous sequential circuit is shown below which is comprised of three J-K flip-flops. (Let the flip-flops be numbered 1, 2 and 3 from the left hand side). 1 2 3
  • 2. The J-K Transition Table is: Synchronous Sequential Circuits Class Revision Question Assuming that the outputs Q1, Q2 and Q3 of the flip-flops are all initially in the '0' state, demonstrate the sequence that the circuit will clock through over the next 6 clock cycles by obtaining logic descriptions for each of the J-K inputs, and deriving each 'next-state' output from each 'present-state' output using the J-K flip-flop transition table. (Ignore the set and clr of each flip-flop). J K Q(n+1) 0 0 Qn No Change 0 1 0 Clear 1 0 1 Set 1 1 Qn Toggles
  • 3. The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Looking at Flip Flop 1:- J1 = Q2 Similarly at Flip Flop 2:- J2 = Q3 And also:- K1 = Q2 And also:- K2 = Q3
  • 4. The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Looking at Flip Flop 3:- J1 = 1 And also:- K3 = Q1.Q2 = (Q1 AND’ed with Q2)
  • 5. The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions Now let’s tabulate the synchronous conditions J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2)
  • 6. The J-K Transition Table is: Synchronous Sequential Circuits Identifying Initial Conditions And add the initial conditions J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition
  • 7. The J-K Transition Table is: Synchronous Sequential Circuits Starting to populate the table As Q2 = 0, J1 and K1 = 0. As Q3 = 0, J2 and K2 = 0. As Q1 = 0 & Q2 =0; Q1.Q2 = 0 so K3 = 0. (J3 is ALWAYS 1) J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial Condition 0 0 0 0 0 0 0 1 0
  • 8. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 0:- Q1 (stays same) = 0. 0 1 0 J2 and K2 = 0:- Q2 (stays same) = 0. 1 0 1 J3 = 1 & K3 = 0:- Q3 = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial Condition 0 0 0 0 0 0 0 1 0 Clock 1 0 0 1
  • 9. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table As Q2 = 0, J1 and K1 = 0. J K Q(n+1) As Q3 = 1, J2 and K2 = 1. 0 0 Qn As Q1.Q2 = 0 so K3 = 0. 0 1 0 (J3 is ALWAYS 1) 1 0 1 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q 2) Initial Condition 0 0 0 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0
  • 10. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 0:- Q1 (stays same) = 0. 0 1 0 J2 and K2 = 1:- Q2 (Toggles) = 1. 1 0 1 J3 = 1 & K3 = 0:- Q3 (Set) = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1
  • 11. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table As Q2 = 1, J1 and K1 = 1. J K Q(n+1) As Q3 = 1, J2 and K2 = 1. 0 0 Qn As Q1.Q2 = 0 so K3 = 0. 0 1 0 (J3 is ALWAYS 1) 1 0 1 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q 2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0
  • 12. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J-K flip-flop transition table J K Q(n+1) (changes on the CLOCK pulse) 0 0 Qn J1 and K1 = 1:- Q1 (Toggles) = 1. 0 1 0 J2 and K2 = 1:- Q2 (Toggles) = 0. 1 0 1 J3 = 1 & K3 = 0:- Q3 (Set) = 1. 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1
  • 13. The J-K Transition Table is: Synchronous Sequential Circuits Populating the table J K Q(n+1) As Q2 = 0, J1 and K1 = 0. 0 0 Qn As Q3 = 1, J2 and K2 = 1. 0 1 0 As Q1.Q2 = 0 so K3 = 0. 1 0 1 (J3 is ALWAYS 1) 1 1 Qn J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1 0 0 1 1 1 0
  • 14. The J-K Transition Table is: Can you complete Populating the table? J1 K1 Q1 J2 K2 Q2 J3 K3 Q3 (Q2) (Q2) (Q3) (Q3) 1 (Q1.Q2) Initial 0 0 0 Condition 0 0 0 0 1 0 Clock 1 0 0 1 0 0 1 1 1 0 Clock 2 0 1 1 1 1 1 1 1 0 Clock 3 1 0 1 0 0 1 1 1 0 Clock 4 Clock 5 Clock 6