Wednesday, March 31. Review Homework Stepbystep 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 JK flipflop. possible
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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 JK flipfloppossible
inputs
transition
q Q
minimum
inputs
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 flipflops, design a counter000 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’
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 flipflops and the number of external inputs.
2) For each flipflop, compare the present state to the next state and use the appropriate excitation table to complete the truth table for that flipflop.
3) Find a minimum expression for each flipflop control and for the output
4) Draw the circuit
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
10
01
Design examplex 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
Assume that both flipflops will be JK flipflops. 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
 
Assume that both flipflops will be JK flipflops. 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 
 

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 example1. (20 points)Draw the circuit for the Moore machine of the example using D flipflops.
2. (30 points)Find the State Transition Table and State Transition Diagram of the circuit shown below. Show your work.
Draw the circuit that realizes the state transition diagram shown below. Use a JK flipflop for Q0 and use an SR flipflop for Q1.