Chapter 4, Problem 72E

Explanation of Solution

Circuit diagram for combining the two full adders and its equivalent truth table

Full adder:

• The full adder is an electric circuit which is used to perform the addition of three binary digits.
• It takes the inputs of two binary digits and another input is the carry-in value.
• It produces the output as sum and carry-out.
• Logic diagram for the full adder takes three inputs A, B, and Carry-in and produces the output Sum and Carry-out.

Circuit diagram for combining the two full adders:

• The first full adder takes the input as B, D, Carry-in (assume 0) and produces the output Z and Carry.
• The second full adder takes the input as A, C, and output of the first full adder “Carry” and produces the output as Y and X.
• Diagrammatic representation after combining two full adders is shown below:

• Circuit diagram:
• Combine the logic diagram of the two full adders that adds two two-bit binary numbers. The circuit diagram is shown below:

Truth table explanation:

• From the above circuit diagram:
  • Boolean expression for Z:
    • First, the inputs B and D are passed to XOR gate to produce the output as $\boxed{\text{B}\oplus \text{D}}$.
    • Next, pass the output of first XOR gate “$\text{B}\oplus \text{D}$” and “Carry-in” as the input of second XOR gate to produce the output as $\boxed{\text{Z = ((B}\oplus \text{D)}\oplus \text{Carry-in)}}$.
  • Boolean expression for Carry:
    • Next, pass the same inputs B and D from the first XOR gate to AND gate to produce the output as $\boxed{\text{B}\cdot \text{D}}$.
    • Next, pass again the output of first XOR gate “$\text{B}\oplus \text{D}$” and “Carry-in” as the input of second AND gate to produce the output as $\boxed{\text{((B}\oplus \text{D)}\cdot \text{Carry-in)}}$.
    • Finally, pass the output of first AND gate “$\text{B}\cdot \text{D}$” and the output of second AND gate “$\text{((B}\oplus \text{D)}\cdot \text{Carry-in)}$” as the input of OR gate to produce the output as $\boxed{\text{Carry = (B}\cdot \text{D) + ((B}\oplus \text{D)}\cdot \text{Carry-in)}}$.
  • Boolean expression for Y:
    • First, the inputs A and C are passed to XOR gate to produce the output as $\boxed{\text{A}\oplus \text{C}}$.
    • Next, pass the output of first XOR gate “$\text{A}\oplus \text{C}$” and “Carry” as the input of second XOR gate to produce the output as $\boxed{\text{Y = ((A}\oplus \text{C)}\oplus \text{Carry)}}$.
  • Boolean expression for X:
    • Next, pass the same inputs A and C from the first XOR gate to AND gate to produce the output as $\boxed{\text{A}\cdot \text{C}}$.
    • Next, pass again the output of first XOR gate “$\text{A}\oplus \text{C}$” and “Carry” as the input of second AND gate to produce the output as $\boxed{\text{((A}\oplus \text{C)}\cdot \text{Carry)}}$.
    • Finally, pass the output of first AND gate “$\text{A}\cdot \text{C}$” and the output of second AND gate “$\text{((A}\oplus \text{C)}\cdot \text{Carry)}$” as the input of OR gate to produce the output as $\boxed{\text{X = (A}\cdot \text{C) + ((A}\oplus \text{C)}\cdot \text{Carry)}}$.
• Therefore, the Boolean expression for combining two full adders produce X, Y, and Z.

$\boxed{\begin{array}{l}\text{X = (A}\cdot \text{C) + ((A}\oplus \text{C)}\cdot \text{Carry)}\\ \text{Y = ((A}\oplus \text{C)}\oplus \text{Carry)}\\ \text{Z = ((B}\oplus \text{D)}\oplus \text{Carry-in)}\end{array}}$

Truth table for combining the two full adders:

A	B	C	D	Carry-in	$B\oplus D$	$A\oplus C$	Z	Carry	Y	X
0	0	0	0	0	0	0	0	0	0	0
0	0	0	1	0	1	0	1	0	0	0
0	0	1	0	0	0	1	0	0	1	0
0	0	1	1	0	1	1	1	0	1	0
0	1	0	0	0	1	0	1	0	0	0
0	1	0	1	0	0	0	0	1	1	0
0	1	1	0	0	1	1	1	0	1	0
0	1	1	1	0	0	1	0	1	0	1
1	0	0	0	0	0	1	0	0	1	0
1	0	0	1	0	1	1	1	0	1	0
1	0	1	0	0	0	0	0	0	0	1
1	0	1	1	0	1	0	1	0	0	1
1	1	0	0	0	1	1	1	0	1	0
1	1	0	1	0	0	1	0	1	0	1
1	1	1	0	0	1	0	1	0	0	1
1	1	1	1	0	0	0	0	1	1	1

For example:

• If the inputs are A as 0, B as 0, C as 1, D as 0, and Carry-in as 0, then the full adder adds the two two-bit binary numbers as shown below:

• Then, it produces the output of X, Y, Z, and Carry. Truth table is shown below:

A	B	C	D	Carry-in	$B\oplus D$	$A\oplus C$	Z	Carry	Y	X
0	0	0	0	0	0	0	0	0	0	0
0	0	0	1	0	1	0	1	0	0	0
0	0	1	0	0	0	1	0	0	1	0
0	0	1	1	0	1	1	1	0	1	0
0	1	0	0	0	1	0	1	0	0	0
0	1	0	1	0	0	0	0	1	1	0
0	1	1	0	0	1	1	1	0	1	0
0	1	1	1	0	0	1	0	1	0	1
1	0	0	0	0	0	1	0	0	1	0
1	0	0	1	0	1	1	1	0	1	0
1	0	1	0	0	0	0	0	0	0	1
1	0	1	1	0	1	0	1	0	0	1
1	1	0	0	0	1	1	1	0	1	0
1	1	0	1	0	0	1	0	1	0	1
1	1	1	0	0	1	0	1	0	0	1
1	1	1	1	0	0	0	0	1	1	1