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Wednesday, March 31. Review Homework Step-by-step Design. J = 0 K=0 J = 0 K=1. J = 1 K=0 J = 1 K=1. J = 0 K=1 J = 1 K=1. J = 0 K=0 J = 1 K=0. 0 0. 0 1. 1 0. 1 1. J=0 K = -. J=1 K = -. J = - K=1. J = - K=0. 1.) Excitation table for a J-K flip-flop. possible

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## Wednesday, March 31

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**Wednesday, March 31**• Review Homework • Step-by-step Design**J = 0 K=0**J = 0 K=1 J = 1 K=0 J = 1 K=1 J = 0 K=1 J = 1 K=1 J = 0 K=0 J = 1 K=0 0 0 0 1 1 0 1 1 J=0 K = - J=1 K = - J = - K=1 J = - K=0 1.) Excitation table for a J-K flip-flop possible inputs transition q Q minimum inputs**q2q1**q2q1 q2q1 1 00 1 00 00 01 01 1 1 1 01 1 11 - - - - - 11 11 - 1 1 10 10 0 10 1 0 1 q0 q0 q0 0 0 0 1 1 1 T2= q2 + q1 T1= T0= q2 Using T flip-flops, design a counter 000 010 100 001 011 101 (000 and start over). q2 q1q0 Q2 Q1 Q0 T2 T1 T0 0 0 0 0 1 0 0 1 0 0 0 1 0 1 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 0 1 1 0 1 1 0 1 0 0 0 1 0 1 1 1 0 - - - - - - 1 1 1 - - - - - - q2’**Counter Circuit**000 010 100 001 011 101 (000 and start over). T2 = q2 + q1 T1 = q2’ T0 = q2**Design step-by-step**0) Obtain a state transition diagram or state transition table 1) Transfer the information in the state transition diagram or state transition table into truth table form. The number of rows is determined by the number of flip-flops and the number of external inputs. 2) For each flip-flop, compare the present state to the next state and use the appropriate excitation table to complete the truth table for that flip-flop. 3) Find a minimum expression for each flip-flop control and for the output 4) Draw the circuit**Design example**Build a synchronous state machine that will observe a single input and output a “1” whenever the sequence 1, 1, 0 has been observed. Mealy Moore**00**10 01 Design example x q1 q0 Q1 Q0 Z 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 - - - 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 - - - Assign values to the states “no part” = 00 “seen 1” = 01 “seen 11” = 10**Design example**Assume that both flip-flops will be J-K flip-flops. The excitation table is: 00: J=0 K=-, 01: J=1 K=-, 10: J=- K=1, 11: J=- K=0 x q1 q0 Q1 Q0 Z 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 - - - 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 - - - J1 K1 0 - 0 - - 1 - - 0 - 1 - - 0 - -**Design example**Assume that both flip-flops will be J-K flip-flops. The excitation table is: 00: J=0 K=-, 01: J=1 K=-, 10: J=- K=1, 11: J=- K=0 x q1 q0 Q1 Q0 Z 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 - - - 1 0 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 - - - J1 K1 0 - 0 - - 1 - - 0 - 1 - - 0 - - J0 K0 0 - - 1 0 - - - 1 - - 1 0 - - -**x q1**- 1 00 00 00 - - 00 - - 01 - - - 01 - 1 - 01 01 11 11 - - 11 - - - - 11 - 1 1 10 10 10 - - - 10 1 q0 0 1 q0 0 1 q0 0 1 q0 0 1 x q1 q0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 J1 K1 0 - 0 - - 1 - - 0 - 1 - - 0 - - J0 K0 0 - - 1 0 - - - 1 - - 1 0 - - - J0 = xq1’ J1 = xq0 K0 = 1 K1 = x’ xq1 xq1 xq1 Design example**x q1**00 01 1 - 11 - 10 q0 0 1 Z = x’q1 Final circuit J1 = xq0 K1 = x’ J0 = xq1’ K0 = 1**Homework for Monday, April 5**1. (20 points)Draw the circuit for the Moore machine of the example using D flip-flops. 2. (30 points)Find the State Transition Table and State Transition Diagram of the circuit shown below. Show your work.**Problem 3 for Monday, April 5**Draw the circuit that realizes the state transition diagram shown below. Use a J-K flip-flop for Q0 and use an S-R flip-flop for Q1.

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