This design uses just 3 JK flip-flops. The usual documents are required - the documents below,...

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This design uses just 3 JK flip-flops. The usual documents are required - the documents below, Karnaugh maps etc (looped), with a screenshot of your counter where M=1 and the Gray code = 101. Design a synchronous 3-bit counter with outputs Qa,Qb & Qc made from JK flip-flops; Qc is the m.s.b. The counter has two modes, controlled by an input M. 1) Pure Binary Mode, when M-0. The counter produces a modulo-5 pure-binary up-sequence thus: [QcQbQa-] 001,010,011,100,101 (repeat). States zero, six, and seven are not used. 2) Gray Code Mode, when M-1.The counter produces a modulo-8 Gray code sequence, thus:[QcQbQa=] 000,100,110,010,011,111,101,001 (repeat.). There are no unused states. Neighboring codes in a Gray code differ in just one bit -a Gray code marks the co-ordinates of a Karnaugh map, as stated in the box above. The usual documents are needed – this paperwork stapled to a screenshot of Logisim while it is in mode M-1 and in state 101. STAPLE PLEASE. Name and ID on everything. THIS IS PART 1 of a 2 PART ASSIGNMENT. I will post part 2 a.s.a.p. Sorry for the delay. A Gray code is what makes the Karnaugh map work! Look at any code in the provided sequence and observe that the code before it, and also the code after it differ from the code you've chosen only in ONE (1) bit! This means that, on a Karnaugh map, the minterms between two adjacent 1's will differ in such a way that a variable can be eliminated For example A.B.C.D + A.B.C.D' = A.B.C.(D + D') = A.B.C. Correct reading of the map will yield a minimized Boolean expression. You need to understand this %3! Current state Controls Required Next State Je Kc Jb Kb Ja Ka M Qc ob Qa State M Qb Qa State 1 1 3 1 1 4 1 5 1 1 0. 6 1 1 1 1 1 1 1 1 1 10 1 1 11 1 1 1. 12 1 1 1 13 1 1 1 14 1 1 1 15 1 1 Minimize each map individually, and mark your looping!!!!!!!! DaQb> 00 01 11 10 00 01 11 10 QeM QcM 00 00 01 01 11 11 10 10 Ke = Je = O O - - DaQb→ QaQb> 00 01 11 10 00 01 11 10 QcM QcM 00 M 00 01 01 11 11 10 10 Jb = Kb = QaQb→ QaQb→ 00 01 11 10 00 01 11 10 QcM QcM 00 00 01 01 11 11 10 10 Ja = Ка- Did you minimize each map individually, and mark your looping???? What are the equations? Jc= Kc= Jb= Kb= Ja= Ка- What is the successor state to zero (0), when M=0? What is the successor state to six (6), when M=0? What is the successor state to seven (7), when M=0? CERTIFICATION: I certify that this represents my own work exclusively. Sign: My NYIT I.D. is Last Name: First Name: Thank you! This design uses just 3 JK flip-flops. The usual documents are required - the documents below, Karnaugh maps etc (looped), with a screenshot of your counter where M=1 and the Gray code = 101. Design a synchronous 3-bit counter with outputs Qa,Qb & Qc made from JK flip-flops; Qc is the m.s.b. The counter has two modes, controlled by an input M. 1) Pure Binary Mode, when M-0. The counter produces a modulo-5 pure-binary up-sequence thus: [QcQbQa-] 001,010,011,100,101 (repeat). States zero, six, and seven are not used. 2) Gray Code Mode, when M-1.The counter produces a modulo-8 Gray code sequence, thus:[QcQbQa=] 000,100,110,010,011,111,101,001 (repeat.). There are no unused states. Neighboring codes in a Gray code differ in just one bit -a Gray code marks the co-ordinates of a Karnaugh map, as stated in the box above. The usual documents are needed – this paperwork stapled to a screenshot of Logisim while it is in mode M-1 and in state 101. STAPLE PLEASE. Name and ID on everything. THIS IS PART 1 of a 2 PART ASSIGNMENT. I will post part 2 a.s.a.p. Sorry for the delay. A Gray code is what makes the Karnaugh map work! Look at any code in the provided sequence and observe that the code before it, and also the code after it differ from the code you've chosen only in ONE (1) bit! This means that, on a Karnaugh map, the minterms between two adjacent 1's will differ in such a way that a variable can be eliminated For example A.B.C.D + A.B.C.D' = A.B.C.(D + D') = A.B.C. Correct reading of the map will yield a minimized Boolean expression. You need to understand this %3! Current state Controls Required Next State Je Kc Jb Kb Ja Ka M Qc ob Qa State M Qb Qa State 1 1 3 1 1 4 1 5 1 1 0. 6 1 1 1 1 1 1 1 1 1 10 1 1 11 1 1 1. 12 1 1 1 13 1 1 1 14 1 1 1 15 1 1 Minimize each map individually, and mark your looping!!!!!!!! DaQb> 00 01 11 10 00 01 11 10 QeM QcM 00 00 01 01 11 11 10 10 Ke = Je = O O - - DaQb→ QaQb> 00 01 11 10 00 01 11 10 QcM QcM 00 M 00 01 01 11 11 10 10 Jb = Kb = QaQb→ QaQb→ 00 01 11 10 00 01 11 10 QcM QcM 00 00 01 01 11 11 10 10 Ja = Ка- Did you minimize each map individually, and mark your looping???? What are the equations? Jc= Kc= Jb= Kb= Ja= Ка- What is the successor state to zero (0), when M=0? What is the successor state to six (6), when M=0? What is the successor state to seven (7), when M=0? CERTIFICATION: I certify that this represents my own work exclusively. Sign: My NYIT I.D. is Last Name: First Name: Thank you!