How Can DeMorgan's Law Be Applied to Implement (A+B).(A+C) Using NAND Gates?

  • Thread starter aurao2003
  • Start date
  • Tags
    Law
In summary, the conversation is about implementing the expression (A+B).(A+C) using NAND gates and DeMorgans law. The original poster is struggling with drawing the diagram and asks for help. Another user suggests expanding and simplifying the expression before drawing the diagram and provides a resource for constructing the circuit. The original poster thanks them and asks for further guidance.
  • #1
aurao2003
126
0

Homework Statement


Hi
I am struggling with the diagram of this question. Can anyone help? I would have loaded my diagram but unsure how to do it. This is the question:

Starting with the following statement (A+B).(A+C), show how this can be implemented using NAND gates using DeMorgans law.



Cheers


Homework Equations





The Attempt at a Solution


After doubly negating, I obtained (A.B) + (A.C). And this is where I get stuck.
Please help. I have a deadline looming this week.
 
Physics news on Phys.org
  • #2
Try to expand and then simplify the expression first.

ehild
 
  • #3
Thanks for a response. I am okay with the maths but unsure about the actual diagram. Any pointers? Thanks.
 
  • #4
The double-negated (A+B).(A+C) is not A.B+A.C.

If you do not know how to draw the diagram, see http://www.kpsec.freeuk.com/gates.htm. Copy the figures and construct the circuit from them.
If you do not know how to upload it into your post, click on "Go Advanced".

ehild
 
  • #5


DeMorgans law states that the negation of a conjunction (AND) is the disjunction (OR) of the negations of the individual statements. In this case, (A+B).(A+C) can be written as ~(~(A+B) V ~(A+C)). We can use this to implement the expression using NAND gates as follows:

1. Start by creating two NAND gates, one for each term (A+B) and (A+C).
2. For the first NAND gate, connect A and B as inputs. The output of this gate will be ~(A+B).
3. For the second NAND gate, connect A and C as inputs. The output of this gate will be ~(A+C).
4. Use a third NAND gate to connect the outputs of the first two NAND gates. This will give us ~(~(A+B) V ~(A+C)), which is equivalent to (A+B).(A+C).
5. Finally, use a fourth NAND gate to negate the output of the third NAND gate, giving us the final expression (A+B).(A+C) implemented using NAND gates.

I hope this helps. Let me know if you have any further questions. Good luck with your deadline!
 

Related to How Can DeMorgan's Law Be Applied to Implement (A+B).(A+C) Using NAND Gates?

1. What is DeMorgans law in terms of NAND gates?

DeMorgans law states that the NAND operation of two inputs is equivalent to the inverted NOR operation of the same inputs. In other words, the output of a NAND gate is the inverse of the output of a NOR gate with the same inputs.

2. How is DeMorgans law applied to simplify logic circuits using NAND gates?

By applying DeMorgans law, complex logic circuits can be simplified by replacing AND, OR, and NOT gates with NAND gates. This reduces the number of gates required and can lead to more efficient and cost-effective circuits.

3. Can DeMorgans law be applied to any logic circuit?

Yes, DeMorgans law can be applied to any logic circuit, as it is a fundamental principle that holds true for all Boolean operations.

4. What is the truth table for DeMorgans law using NAND gates?

The truth table for DeMorgans law using NAND gates is as follows:

Input A Input B NAND Output NOR Output
0 0 1 1
0 1 1 0
1 0 1 0
1 1 0 0

5. How does DeMorgans law using NAND gates affect the output of a logic circuit?

DeMorgans law using NAND gates does not change the output of a logic circuit, but it can simplify the circuit and reduce the number of gates required. This can lead to cost savings and improved efficiency in digital systems.

Back
Top