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Sagot :
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 --+
```
### 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 --+
```
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