SOLUTION OF COMBINATIONAL CIRCUIT MORRIS MANO(2ND EDITION)

 4.1: Design a combinational circuit with three input and one output. The output is equal to logic 1 when the binary value of the input is less than 3. The output is logic 0 otherwise.

Solution:           

A	B	C	Y
0	0	0	1
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	0
1	0	1	0
1	1	0	0
1	1	1	0

                       

             Y=A'C'+A'B'

4.2: Design a combinational circuit with three inputs x, y and z and three outputs A, B, C. When the binary input is 0,1,2 or 3, the binary output is one greater than the input. When the binary input is 4,5,6,7 the binary output is one less than the input.

Solution:

x	y	z	A	B	C
0	0	0	0	0	1
0	0	1	0	1	0
0	1	0	0	1	1
0	1	1	1	0	0
1	0	0	0	1	1
1	0	1	1	0	0
1	1	0	1	0	1
1	1	1	1	1	0

K map for A

A=xz + yz + xy

K map for B

B=xy'z'+x'y'z+x'yz'+xyz

K map for C

C=z

4.3: A majority function is generated in a combinational circuit when the output is equal to one if the input variable has more 1's than 0's. The output is 0 otherwise. Design a 3 input majority function.

Solution:

x	y	z	O
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	1

                O=xy+yz+zx

4.4: Design a combinational circuit that adds one to a 4 bit binary number A3 A2 A1 A0. For example if the input of the circuit is A3 A2 A1 A0=1101, THE OUTPUT IS 1110. The circuit can be designed using four half adders.

Solution:

A3	A2	A1	A0	S4	S3	S2	S1	S0
0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	0	1	0
0	0	1	0	0	0	0	1	1
0	0	1	1	0	0	1	0	0
0	1	0	0	0	0	1	0	1
0	1	0	1	0	0	1	1	0
0	1	1	0	0	0	1	1	1
0	1	1	1	0	1	0	0	0
1	0	0	0	0	1	0	0	1
1	0	0	1	0	1	0	1	0
1	0	1	0	0	1	0	1	1
1	0	1	1	0	1	1	0	0
1	1	0	0	0	1	1	0	1
1	1	0	1	0	1	1	1	0
1	1	1	0	0	1	1	1	1
1	1	1	1	1	0	0	0	1

4.5: A combinational circuit produce the binary sum of two 2 bit numbers x1 x0 and y1 y0. The outputs are C, S1 and S0. Provide a truth table of the combinational circuit.

Solution:

X1	X0	Y1	Y0	C	S1	S0
0	0	0	0	0	0	0
0	0	0	1	0	0	1
0	0	1	0	0	1	0
0	0	1	1	0	1	1
0	1	0	0	0	0	1
0	1	0	1	0	1	0
0	1	1	0	0	1	1
0	1	1	1	1	0	0
1	0	0	0	0	1	0
1	0	0	1	0	1	1
1	0	1	0	1	0	0
1	0	1	1	1	0	1
1	1	0	0	0	1	1
1	1	0	1	1	0	0
1	1	1	0	1	0	1
1	1	1	1	1	1	0

 

4.6: Design the circuit of 4.5 using two full adders.

Solution: 

4.7: Design a combinational circuit that multiplies two 2 bit numbers, a1 a0 and b1 b0, to produce a 4 bit product c3 c2 c1 c0. Use AND gates and half adders.

Solution: 

4.8: Show that a full subtractor can be constructed with two half subtractors and an OR gate.

Solution: 

In Full Subtractor: D= x'y'z + x'yz' + xy'z'+ xyz

                                  =z XOR x XOR y

                                B=x'y+x'z+yz

Half Subtractor:     D=x'y+xy'

                                   =x XOR y

                                B=x'y

4.9: Design a combinational circuit with three inputs and six outputs. The output binary number should be the square of the input binary numbers.

Solution: 

x	y	z	A	B	C	D	E	F
0	0	0	0	0	0	0	0	0
0	0	1	0	0	0	0	0	1
0	1	0	0	0	0	1	0	0
0	1	1	0	0	1	0	0	1
1	0	0	0	1	0	0	0	0
1	0	1	0	1	1	0	0	1
1	1	0	1	0	0	1	0	0
1	1	1	1	1	0	0	0	1

 

K MAP FOR A: 

        A= xy

K MAP FOR B

            B= xy'+ xz

K MAP FOR C

            C=xy'z+ x'yz

K MAP FOR D

                D=yz'

K MAP FOR E

        E=0

K MAP FOR F

            F=z

4.10: Design a combinational circuit with 4 inputs and that represent a decimal digit in BCD and 4 outputs that produce the 9's complement of the input digit. The six unused combinations can be treated as don't care conditions.

Solution:  

A	B	C	D	w	x	y	z
0	0	0	0	1	0	0	1
0	0	0	1	1	0	0	0
0	0	1	0	0	1	1	1
0	0	1	1	0	1	1	0
0	1	0	0	0	1	0	1
0	1	0	1	0	1	0	0
0	1	1	0	0	0	1	1
0	1	1	1	0	0	1	0
1	0	0	0	0	0	0	1
1	0	0	1	0	0	0	0

 K MAP FOR w

            w=A'B'C'

K MAP FOR x

        x=BC'+B'C

K MAP FOR y

        y=C

K MAP FOR z

            z=D'

4.11: Design a combinational circuit with four inputs and four outputs. The output generates the 2's complement of the input binary number.

Solution:

A	B	C	D	w	x	y	z
0	0	0	0	0	0	0	0
0	0	0	1	1	1	1	1
0	0	1	0	1	1	1	0
0	0	1	1	1	1	0	1
0	1	0	0	1	1	0	0
0	1	0	1	1	0	1	1
0	1	1	0	1	0	1	0
0	1	1	1	1	0	0	1
1	0	0	0	1	0	0	0
1	0	0	1	0	1	1	1
1	0	1	0	0	1	1	0
1	0	1	1	0	1	0	1
1	1	0	0	0	1	0	0
1	1	0	1	0	0	1	1
1	1	1	0	0	0	1	0
1	1	1	1	0	0	0	1

K MAP FOR w

        w=A'B+A'D+A'C+AB'C'D'

k map for x

        x=B'D+B'C+BC'D'

K MAP FOY y

        y=C'D+CD'

K MAP FOR z

        z=D

4.12: Design a combinational circuit that detects an error in the representation of a decimal digit in BCD. The output of the circuit must be equal to logic 1 when the input contains any one of the unused six bit combination in the BCD code.

Solution: 

A	B	C	D	Y
0	0	0	0	0
0	0	0	1	0
0	0	1	0	0
0	0	1	1	0
0	1	0	0	0
0	1	0	1	0
0	1	1	0	0
0	1	1	1	0
1	0	0	0	0
1	0	0	1	0
1	0	1	0	1
1	0	1	1	1
1	1	0	0	1
1	1	0	1	1
1	1	1	0	1
1	1	1	1	1

        Y=AC+AB

4.13: Design a code converter that converts a decimal digit 84-2-1 to BCD code

Solution: 

A	B	C	D	w	x	y	z
0	0	0	0	0	0	0	0
0	1	1	1	0	0	0	1
0	1	1	0	0	0	1	0
0	1	0	1	0	0	1	1
0	1	0	0	0	1	0	0
1	0	1	1	0	1	0	1
1	0	1	0	0	1	1	0
1	0	0	1	0	1	1	1
1	0	0	0	1	0	0	0
1	1	1	1	1	0	0	1

 K MAP FOR w

        w=AB+AC'D'

K MAP FOR x

        x=B'D+B'C+BC'D'

K MAP FOR y

            y=C'D+CD'

K MAP FOR z

            z=D

4.14:  Design a combinational circuit that converts decimal digit from the 2421 code to the 84-2-1 code.

Solution: 

w	x	y	z	A	B	C	D
0	0	0	0	0	0	0	0
0	0	0	1	0	1	1	1
0	0	1	0	0	1	1	0
0	0	1	1	0	1	0	1
0	1	0	0	0	1	0	0
0	1	0	1	X	X	X	X
0	1	1	0	X	X	X	X
0	1	1	1	X	X	X	X
1	0	0	0	X	X	X	X
1	0	0	1	X	X	X	X
1	0	1	0	X	X	X	X
1	0	1	1	1	0	1	1
1	1	0	0	1	0	1	0
1	1	0	1	1	0	0	1
1	1	1	0	1	0	0	0
1	1	1	1	1	1	1	1

        K MAP FOR A

        A=  w

        K MAP FOR B

        B=    w'x + xyz + w'z + w'y

    K MAP FOR C

    C= w'y'z + w'yz' + wy'z' + wyz

    K MAP FOR D 

    D=  z

4.15: Design a combinational circuit that converts binary number of 4 bits to a decimal number in BCD. Note that the BCD number is the same as binary number as  long as the input is less than or equal to 9. The binary number from 1010 to 1111 converts to BCD numbers from  10000  10101.

Solution:

w	x	y	z	A	B	C	D	E
0	0	0	0	0	0	0	0	0
0	0	0	1	0	0	0	0	1
0	0	1	0	0	0	0	1	0
0	0	1	1	0	0	0	1	1
0	1	0	0	0	0	1	0	0
0	1	0	1	0	0	1	0	1
0	1	1	0	0	0	1	1	0
0	1	1	1	0	0	1	1	1
1	0	0	0	0	1	0	0	0
1	0	0	1	0	1	0	0	1
1	0	1	0	1	0	0	0	0
1	0	1	1	1	0	0	0	1
1	1	0	0	1	0	0	1	0
1	1	0	1	1	0	0	1	1
1	1	1	0	1	0	1	0	0
1	1	1	1	1	0	1	0	1

K MAP FOR A

    A=wy + wx

    K MAP FOR B

       B=wx'y'

    K MAP FOR C

        C=w'x + xy

    K MAP FOR D

        D= w'y + wxy'

    K MAP FOR E

        E=z

4.17: Analyze the two output combinational circuit shown in figure. Find the boolean function for the two outputs as a function of the three inputs and explain the circuit operation.

Solution: 

T1=AB

T2=A+B

T3=T2C

    =(A+B)C

    =AC+BC

T4=T2+C

    =A+B+C

T5=T1C

    =ABC

T6=T1+T3

    =AB+BC+AC

T7=T6' T4

    =(AB+BC+AC)'(A+B+C)

F2=T6

    =AB+BC+CA

F1=T7+T5

    =(AB)'(BC)'(AC)'(A+B+C)+ABC

    =(A'+B')(B'+C')(A'+C')(A+B+C)+ABC

    =(A'B'+A'C'+B'B'+B'C')(A'+C')(A+B+C)+ABC

    =(B'(A'+1)+A'C'+B'C')(A'+C')(A+B+C)+ABC

    =(B'(1+C')+A'C')(A'+C')(A+B+C)+ABC

    =(B'+A'C')(A'+C')(A+B+C)+ABC

    =(A'B'+B'C'+A'A'C'+A'C'C')(A+B+C)+ABC

    =(A'B'+B'C'+A'C')(A+B+C)+ABC

    =A'B'A+A'B'B+A'B'C+AB'C'+BB'C'+B'C'C+A'C'A+A'C'B+A'C'C+ABC

    =A'B'C+AB'C'+A'C'B+ABC

F1 is equivalent to the sum of full adder and F2 is equivalent to the carry of the full adder. So, this circuit will act as full adder.

4.18: Derive the truth table of the following circuit:

Solution:

As we have found that that the circuit is equivalent to full adder. So, the truth table will be same as full adder.

A	B	C	F2	F1
0	0	0	0	0
0	0	1	0	1
0	1	0	0	1
0	1	1	1	0
1	0	0	0	1
1	0	1	1	0
1	1	0	1	0
1	1	1	1	1

4.19: Draw the NAND logic diagram for each of the following expression using multilevel NAND  gate circuits:

a) (AB'+CD')E+BC(A+B)

Solution:

First, we will replace all the AND gates with AND INVERT and all the OR gates with INVERT OR. And then will check if there is any input which is connected to only one small circle and will replace it with it's complement.

b) w(x+y+z)+xyz

Solution: 

First, we will replace all the AND gates with AND INVERT and all the OR gates with INVERT OR. And then will check if there is any input which is connected to only one small circle and will replace it with it's complement.

4:20 is same as 4:19

4.21: Determine the boolean function for outputs F and G as a function of four inputs A,B,C and D.

Solution: 

T1=(A'(A'D)')'

     =A+A'D

T2=A'+BC

F=T2T1

  =(A'+BC)(A+A'D)

  =AA'+A'A'D+ABC+A'BCD

  =A'D+ABC+A'BCD

G=T2(A'D)'

   =(A'+BC)(A+D')

   =A'A+A'D'+ABC+BCD'

   =A'D'+ABC+BCD'

4.22: Verify the circuit below generates the exclusive NOR function:

Solution:

Output of exclusive NOR function,

G = (x'y+xy')'

  =(x'+y)(x+y')

  =xx'+x'y'+xy+yy'                                                  

  =xy+x'y'                                                                          

Now,

F=((x+ (x+y)')'+ (y+(x+y)')')'

  =(x+(x+y)')(y+(x+y)')

  =(x+x'y')(y+x'y')

  =xy+xx'y'+x'y'x'y'+x'y'y

  =xy+x'y'

  =G

Additional questions:

1. How can I design a combinational circuit with three inputs x,y and z, and three outputs A,B and C. When the binary input is 0,1,2 or 3, the binary output is four greater than the input?

Solution: 

x	y	z	A	B	C
0	0	0	1	0	0
0	0	1	1	0	1
0	1	0	1	1	0
0	1	1	1	1	1
1	0	0	X	X	X
1	0	1	X	X	X
1	1	0	X	X	X
1	1	1	X	X	X

    K MAP FOR A

         

      A=1

    K MAP FOR B

     B=y

    K MAP FOR C

        C=z