# Exercise 5  Sequential Circuits: I

## Question 1. Flip-Flops and state tables [ 2 marks ]

A sequential circuit has two D flip-flops A and B, two inputs X and Y, and one output Z. The flip-flop input equations and the circuit output are as follows.

DA = X' Y + X A

DB = X' B + X A

Z = B

Note that this question was originally entitled "Count-down Counter" in error. This question is simply about flip-flop circuits and state tables.

a. Draw the logic diagram of the circuit. b. Tabulate the state table. ## Question 2. Count-down Counter [ 3 marks ]

Design a 2-bit count-down counter. This is a sequential circuit with two flip-flops and one input X. When X=0, the state of the flip-flops does not change. However, when X=1, the state sequence is 11, 10, 01, 00, then back to 11 and the sequence repeats. (Under normal use, X=1 is applied for a short time during which the circuit counts down by one to obtain the final stored result. Additional pulses of X=1 continue the counting down operation.)

The state table is shown below: This leads to the Boolean logic expressions:

A+ = XA + X(AB + AB)        B+ = XB + XB

These expressions can be drawn as block circuit diagrams using AND-OR gates, or including XOR and XNOR gates. The A and B inputs on the right hand side are the original stored values in the flip-flops and the resulting expression values are fed back to the inputs of the flip-flops to produce the final outputs A+ and B+. For this type of circuit the D type flip-flop is quite suitable.

The block diagram is left as a straightforward exercise for students.

## Question 3. J-K Up-Down Counter [ 5 marks ]

Design a sequential circuit with two J-K flip-flops A and B and two inputs E and X. If E=0, the circuit remains in the same state regardless of the value of X. When E=1 and X=1, the circuit goes "up" through the state transitions from 00 to 01 to 10 to 11 and back to 00, then repeating. When E=1 and X=0, the circuit goes "down" through the state transitions from 00 to 11 to 10 to 01 back to 00, then repeating. (Note that if the E=1 and X inputs are applied only long enough to produce a single state transition, then a new set of inputs E=1 and the complement value of the previous X input is applied, the counter can be used to count up/down first, then count down/up).

The state table is shown below: This leads to the Boolean logic expressions:

A+ = EX(AB+AB) + EX(AB+AB) + EA        B+ = EB + EB

These expressions can be drawn as block circuit diagrams using AND-OR gates, or including XOR and XNOR gates. The A and B inputs on the right hand side are the original stored values in the flip-flops and the resulting expression values are fed back to the inputs of the flip-flops to produce the final outputs A+ and B+.

It is required that you use J-K flip-flops for this question. By connecting the expression outputs to the J-input, and the complement of the expression to the K-input, you basically transform the JK to a D type flip-flop.

The block diagram is left as a straightforward exercise for students.