Answered

1.3 The XOR function is usually expressed as:
[tex]\[ A \oplus B = A \cdot \bar{B} + \bar{A} \cdot B \][/tex]

Use Boolean algebra to show that this expression is also equivalent to:
(a) [tex]\[ \overline{A \cdot B + \bar{A} \cdot \bar{B}} \][/tex]
(b) [tex]\[ (A + B) \cdot (\bar{A} + \bar{B}) \][/tex]
(c) [tex]\[ (A + B) \cdot \overline{(A \cdot \bar{B})} \][/tex]

1. Draw the logic required to directly implement (a).
2. Derive the dual of this circuit and state which of the two remaining expressions it directly represents.
3. Finally, dual one of the gates in this second circuit to obtain the implementation of (c).



Answer :

GinnyAnswer
Let's tackle each part of the given problem step by step, using Boolean algebra to show that [tex]\( A \oplus B = A \cdot \overline{B} + \overline{A} \cdot B \)[/tex] is equivalent to the other expressions.

### Part (a)
Show that [tex]\( A \oplus B \)[/tex] is equivalent to [tex]\(\overline{A \cdot B + \overline{A} \cdot \overline{B}}\)[/tex].

Step-by-Step Solution:
1. Using the properties of Boolean algebra, we know that the XOR function is expressed as:
[tex]\[ A \oplus B = A \cdot \overline{B} + \overline{A} \cdot B \][/tex]

2. By De Morgan's laws, we can transform the expression [tex]\(\overline{A \cdot B + \overline{A} \cdot \overline{B}}\)[/tex]:
[tex]\[ \overline{A \cdot B + \overline{A} \cdot \overline{B}} = \overline{A \cdot B} \cdot \overline{\overline{A} \cdot \overline{B}} \][/tex]

3. Applying De Morgan's laws again, we get:
[tex]\[ \overline{A \cdot B} = \overline{A} + \overline{B} \][/tex]
[tex]\[ \overline{\overline{A} \cdot \overline{B}} = A + B \][/tex]

4. Substituting these into the expression, we get:
[tex]\[ \overline{A \cdot B + \overline{A} \cdot \overline{B}} = (\overline{A} + \overline{B}) \cdot (A + B) \][/tex]

Thus, we have shown that:
[tex]\[ \overline{A \cdot B + \overline{A} \cdot \overline{B}} = (\overline{A} + \overline{B}) \cdot (A + B) \][/tex]

### Part (b)
Show that [tex]\( A \oplus B \)[/tex] is equivalent to [tex]\( (A + B) \cdot (\overline{A} + \overline{B}) \)[/tex].

Step-by-Step Solution:
1. From Part (a), we have:
[tex]\[ \overline{A \cdot B + \overline{A} \cdot \overline{B}} = (\overline{A} + \overline{B}) \cdot (A + B) \][/tex]

2. Observing the right-hand side, we see that it directly matches the expression:
[tex]\[ (A + B) \cdot (\overline{A} + \overline{B}) \][/tex]

Thus, we have shown that:
[tex]\[ A \oplus B = (A + B) \cdot (\overline{A} + \overline{B}) \][/tex]

### Part (c)
Show that [tex]\( A \oplus B \)[/tex] is equivalent to [tex]\( (A + B) \cdot \overline{(A \cdot \overline{B})} \)[/tex].

Step-by-Step Solution:
1. We start with the XOR function:
[tex]\[ A \oplus B = A \cdot \overline{B} + \overline{A} \cdot B \][/tex]

2. Let's consider the expression [tex]\( \overline{(A \cdot \overline{B})} \)[/tex]:
[tex]\[ \overline{(A \cdot \overline{B})} \][/tex]

3. Using De Morgan's law, we have:
[tex]\[ \overline{(A \cdot \overline{B})} = \overline{A} + B \][/tex]

4. Now the expression:
[tex]\[ (A + B) \cdot \overline{(A \cdot \overline{B})} = (A + B) \cdot (\overline{A} + B) \][/tex]

So, we have shown that:
[tex]\[ A \oplus B = (A + B) \cdot (\overline{A} + B) \][/tex]

This completes the Boolean algebra derivations proving the equivalence of the given expressions.

### Logic Circuit Implementation
To implement the circuit for part (a):
1. Represent [tex]\( A \cdot B + \overline{A} \cdot \overline{B} \)[/tex] using AND and OR gates.
2. Use a NOT gate to invert the output.

#### Circuit Diagram (a):
```
A ---AND---+ +-- NOT -- OUTPUT
AND --- OR ---
B ---AND---+ | |
OR---------|
\bar{A}--AND---\bar{B} |
| \bar{} __
\bar{} -------------|
```

#### Dual Circuit
To derive the dual circuit:
1. Replace every AND gate with an OR gate and vice versa.
2. Replace every variable with its complement.

The dual of this circuit [tex]\( \overline{(A \cdot B + \overline{A} \cdot \overline{B})} \)[/tex] is [tex]\( (A + B) \cdot (\overline{A} + \overline{B}) \)[/tex].

This directly represents expression (b).

#### Modified Circuit for (c)
Finally, to obtain the implementation of (c) [tex]\( (A + B) \cdot \overline{(A \cdot \overline{B})} \)[/tex], dual one of the gates in this second circuit.

1. The expression [tex]\( \overline{(A \cdot \overline{B})} \)[/tex] can be obtained by taking NOT of [tex]\( A \cdot \overline{B} \)[/tex].
2. Implement [tex]\( (A + B) \cdot (\overline{A} + B) \)[/tex].

Thus, the final logic circuit for expression (c) requires one NOT, one AND, and two OR gates:

#### Circuit Diagram (c):
```
A -- OR ---+ +-- AND -- OUTPUT
OR ------- B |
B -- OR ---+ \bar{A}---+
|
B --+
```

Other Questions