Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

Chapter 17 Combinational Circuits: Implementation and Design

1. Simplify using Boolean algebra techniques: (a) A (A + B) + (B + AA)(A + B

)

(b) ))(( BBBABA Solution: (a) Let f = A (A + B) + (B + AA) (A + B

). This can be reduced as:

f = ( A A + A B) + (B + A)(A + B

) f = A B+AB+A+A B

f = B (A+ A ) + A (1+ B

) f = B+A

(b) Let f = ))(( BBBABA . This can be reduced as: f = )1)(( BABA

f = 0BA f = BA

2. Simplify the following Boolean functions using K-maps and realize the gate circuits

using only NAND gates.

(a) f (A,B,C) = ABC + AB + ABC + AC (b) f (A,B,C) = AB + BC + BC + ABC (c) f (A,B,C, D) = (0,1,2,3,8,9,10,11)

Solution:

(a) f (A,B,C) = ABC + AB + ABC + AC

f (A,B,C) = ABC + AB (C+ C ) + ABC + AC(B+ B )

f (A,B,C) = ABC + ABC + ABC + ABC + ACB+ ABC

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

f = B + AC+ AC

(b) f (A,B,C) = AB + BC + BC + ABC

f (A,B,C) = AB (C+C )+ BC (A+ A )+ BC(A+ A ) + A BC

f (A,B,C) = ABC + AB C + BC A+ BA C + BCA+ BC A +A BC

f(A,B,C) = ABC + AB C + BC A + BCA+ A BC

f = B + AC

(c ) f(A,B,C,D) = (0,1,2,3,8,9,10,11)

f = B

3. Simplify and implement using (i) only NAND gates; and (ii) only NOR gates:

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

f(A,B,C,D)= (0,1,2,3,8,9,10,11) d(A,B,C,D)= (12,13) Solution:

f = B = ( )( )B A A C C = ABC ABC ABC ABC = ( )( )( )( )ABC ABC ABC ABC

Fig.3.2 Realization of B using NAND gates

)()()()(

))()()((

))((

)A)(B(

CC AA

CBACBACBACBA

CBACBACBACBA

CCABCCBA

CCAB

BBf

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

Fig. 3.2 Realization of B using NOR gates

4. What is the output Q of the circuit shown in Fig.17p.4? Implement the circuit using only NAND gates.

Fig.17p.4 The given gate circuit

Solution:

f = )( BAAB

ABBABAAB

BABAABAB

BABAABBABAAB

BABAABBABAAB

ABBABABABAAB

ABBABABABAAB

BABAABBABAAB

)1(

)()((

))((

))((

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

5. Draw and use a K-map to produce a minimized SOP and POS expression for:

f (w, x, y, z) = (0, 2, 6, 7, 8, 10, 11, 15) Solution:

f = yxwxyzzywwyzzxxyw

6. Simplify the following Boolean function using the K-map. f = RPQPQRRQPQRPRQPRQP Solution: Mapping the product terms, as shown in Table 6.1.

Table 6.1 K-map for RPQPQRRQPQRPRQPRQP

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

There are two groups. We have two product terms in the simplified Boolean function. For the sub-cube covered by minterms 0, 1, 2 and 3, P is unchanged and for the sub-cube covered by minterms 2, 3, 7 and 6, Q is unchanged. Hence the minimal solution is f = QP .

7. Simplify f = RPQRQPRQPRQP using a K-map. Solution: Mapping the product terms, as shown in Table. 7.1.

Table 7.1 K-map for RPQRQPRQPRQP

If the K-map is rolled backwards as a cylinder, the two extreme pairs of cells are also adjacent. Grouping the product terms, the simplified Boolean function is f = R .

8. Simplify the Boolean function f = RPQRQPRQPQRPRQPRQP . Solution: Mapping the product terms, as shown in Table 8.1.

Table 8.1 K-map for RPQRQPRQPQRPRQPRQP

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

The simplified function obviously has two terms. For the sub-cube covered by minterms 0,1,2,3, P is unchanged and for the sub-cube covered by minterms 0,2,4,6, R is unchanged. Hence the minimal solution is f = RP .

9. Simplify f = PQRRPQRQPQRP : (a) using Boolean simplification; (b) using K-map. Solution: (a) f = PQRRPQRQPQRP Factoring terms 1 and 4 results in

F = RPQRQPP)PQR( = RPQRQPQR (since P + 1P ) Factoring 1 and 3 results in

f = RQP)RPQ(R Since RPPRR , f simplifies to

f = RQPP)Q(R = RQPPQQR Factoring terms 2 and 3 gives

F = )( RQQPQR which simplifies to

f = )( RQPQR , which is the same as

f = PRPQQR (b) Now use the K-map Mapping the product terms, as shown in Table 9.1.

Table 9.1 K-map for PQRRPQRQPQRP

For the sub-cube covered by minterms 3, 7, Q and R are unchanged, for the sub-cube covered by minterms 5, 7, P and R are unchanged and for the sub-cube covered by minterms 6, 7, P and Q are unchanged. Hence the minimal solution is f = . QRPQPR

10. A combinational circuit has four inputs and one output. The output is 1 when (a) all the inputs are equal to 1; (b) none of the inputs is equal to 1; and (c) when the odd number of inputs is equal to 1. Design a logic circuit to implement the above.

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Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

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Solution: Step 1: The number of inputs is 4 and output is 1. Let the inputs be P,Q,R,S and output f. Step2: Truth table is presented in Table 10.1, satisfying the conditions.

Table 10.1 Truth table

P Q R S f 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1

The contents of the truth table are mapped in Table. 10.2.

Table 10.2 K-map

The minimal solution is f = QRSPRSPQRPQSSRQRQPSRPSQP From the above Boolean expression, the logic diagram can be easily drawn.