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

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Presentation Transcript
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:

00: J=0 K=-, 01: J=1 K=-, 10: J=- K=1, 11: 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:

00: J=0 K=-, 01: J=1 K=-, 10: J=- K=1, 11: 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.