Ch5 HW solution v1 - N/A PDF

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ENS 220 / CSC 346 Chapter 5 exercise Problem 1 – Explain the differences among a truth table, a state table, a characteristic table, and an excitation table. Also explain the difference among a Boolean equation, a state equation, a characteristic equation, and a flip-flop input equation. The truth table describes a combinational circuit. The state table describes a sequential circuit. The characteristic table describes the operation of a flip-flop. The excitation table gives the values of flip-flop inputs for a given state transition. The four equations correspond to the algebraic expression of the four tables. Problem 2 - Construct a JK flip-flop, using a D flip-flop, a two-to-one-line multiplexer, and an inverter. D = JQ’+K’Q -> use Q as a select control for 2x1 Mux

Problem 3 – Show that the characteristic equation for the complement output of a JK flip-flop is

Q' (t  1)  J ' Q' KQ Q(t+1)=JQ’+K’Q -> Q’(t+1) = (JQ’+K’Q)’=(J’+Q)(K+Q’)=J’K+J’Q’+KQ+QQ’ =J’K(Q+Q’)+J’Q’+KQ=J’KQ+J’KQ’+J’Q’+KQ=(J’+1)KQ+J’Q’(K+1)=KQ+J’Q’ Problem 4 – A sequential circuit with two D flip-flops A and B, two inputs x and y, and one output z is specialized by the following next-the state and output equations:

a) b) c)

A(t  1)  x ' y  xB B(t  1)  x ' A  xB zA Draw the logic diagram of the circuit. Tabulate the state table. Draw the corresponding state diagram.

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A

Present states B 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1

inputs y

x 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

A 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Next states B 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1

output z 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

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xy/z –00/0,10/0,11/0

00/0

00

01

10/1,11/1 01/00 01/0 00/1 10/0,11/0 01/1,10/1,11/1

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00/1 01/1

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Problem 5 – A sequential circuit has one flip-flop Q, two inputs x and y, and one output S. It consists of a full adder circuit connected to a D flip-flop, as shown in Figure 1. Derive the state table and state diagram of the sequential circuit.

Figure 1 Circuit diagram

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Problem 6 - Starting from state 00 in the state diagram of Figure 2 , determine the state transitions and output sequence that will be generated when an input sequence of 010110111011110 is applied.

Figure 2 State diagram

Present state: Input: Output: Next state:

00 0 0 00

00 1 0 01

01 0 1 00

00 01 11 00 1 1 0 1 0 0 1 0 01 11 00 01

01 1 0 11

11 1 0 10

10 0 1 00

00 1 0 01

01 11 10 10 00 1 1 1 0 0 0 0 1 11 10 10 00

Problem 7 – Reduce the number of states in the following state table, and tabulate the reduced state table: Table 1 State table

Next State Present State a b c

x=0 f d f

x=1 b c e

Output x=0 0 0 0

x=1 0 0 0 4

d e f g h

g d f g g

a c b h a

1 0 1 0 1

0 0 1 1 0

Problem 8 – Starting from state a and the input sequence 01110010011, determine the output sequence for a) The state table of Table 1 and b) The reduced state table from the previous problem. Show that the same output sequence is obtained for both (a) State: a f b c e d g h g g h a Input: 0 1 1 1 0 0 1 0 0 1 1 Output: 0 1 0 0 0 1 1 1 0 1 0 (b) State: a f b a b d g d g g d a Input: 0 1 1 1 0 0 1 0 0 1 1 Output: 0 1 0 0 0 1 1 1 0 1 0

Problem 9 – Derive the state table and the state diagram of the sequential circuit shown in Figure 3 . Explain the function that the circuit performs.

Figure 3 Circuit diagram

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Problem 10 – A sequential circuit has two JK flip-flops A and B and one input x. The circuit is described by the following flip-flop input equations:

JA=x KA=B’ JB=x KB=A a)

Derive the state equations A(t+1) and B(t+1) by substituting the input equations for the J and K variables. A(t+1)=JAA’+KA’A=xA’+BA B (t+1)=JBB’+KB’B=xB’+A’B

b)

Draw the state diagram of the circuit

Problem 11 – A sequential circuit has two JK flip-flops A and B, two inputs x and y, and one output z. The flip-flop input equations and circuit output equation are

JA=Bx+B’y’ KA=B’xy’ JB=A’x KB=A+xy’ z=Ax’y’+Bx’y’ a) Draw the logic diagram of the circuit. b) Tabulate the state table. c) Derive the state equation for A and B. 6

Problem 12 – Design a sequential circuit with two D flip-flops A and B and one input x_in. a) When x_in=0, the state of the circuit remains the same. When x_in=1, the circuit goes through the state transitions from 00 and 01, to 11, to 10, back to 00, and repeat.

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b)

When x_in=0, the state of the circuit remains the same. When x_in=1, the circuit goes through the state transitions from 00 and 11, to 01, to 10, back to 00, and repeat.

Problem 13 –Design a one-input, one-output serial 2’s complementer. The circuit accepts a string of bits from the input and generates the 2’s complement at the output. The circuit can be reset asynchronously to start and end the operation (hint: using the direct control inputs).

Problem 14 –Design a sequential circuit with two JK flip-flops A and B and two inputs E and F. If E=0, the circuit remains in the same state regardless of the value of F. When E=1 and F=1, the circuit goes through the sate transitions from 00 to 01, to 10, to 11, back to 00, and repeats. When E =1 and F=0, the circuit goes through the state transitions from 00 to 11, to 10, to 01, back to 00, and repeats.

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Problem 15 – A sequential circuit has three flip-flops A, B, and C; one input x_in; and one output y_out. The state diagram is shown in Figure 4. The circuit is to be designed by treating the unused states as don’t care conditions. Analyze the circuit obtained from the design to determine the effect of the unused states. a) Use D flip-flop in the design. b) Use JK flip-flop in the design.

A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

Present states B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

Input x_in 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

A 0 1 0 1 0 0 0 0 0 0 x x x x x x

next states B 1 0 0 0 1 0 0 1 1 1 x x x x x x

C 1 0 1 0 0 0 1 0 0 1 x x x x x x

output y_out 0 1 0 1 0 1 0 1 0 0 x x x x x x

a) Use D flip-flops

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b) Use JK flip-flops

Problem 16 – Design the sequential circuit specified by the state diagram of Figure 4, using T flip-flop.

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Figure 4 State diagram

A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

Present states B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

Input x_in 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

A 0 1 0 1 0 0 0 0 0 0 x x x x x x

next states B 1 0 0 0 1 0 0 1 1 1 x x x x x x

C 1 0 1 0 0 0 1 0 0 1 x x x x x x

output y_out 0 1 0 1 0 1 0 1 0 0 x x x x x x

TA 0 1 0 1 0 0 0 0 1 1 x x x x x x

TB 1 0 0 0 0 1 1 0 1 1 x x x x x x

TC 1 0 0 1 0 0 0 1 0 1 x x x x x x

TA = A+B’x TB=A+B’C’x’+BC’x+BCx’ TC=A’B’C’x’+Ax+Cx Y_out = A’x

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