Simplifying Boolean Circuits: W NAND X & YNORZ

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SUMMARY

The discussion focuses on simplifying a Boolean circuit represented by the equation f = (W NAND X) NOR (Y NOR Z). The participants apply De Morgan's Theorem to transform the expression into a simplified Sum of Products (SoP) form. The final simplified equation is f = wx(y + z), demonstrating the effective use of Boolean algebra techniques to achieve simplification.

PREREQUISITES
  • Understanding of Boolean algebra concepts
  • Familiarity with De Morgan's Theorem
  • Knowledge of NAND and NOR gate operations
  • Ability to manipulate logical expressions in SoP form
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  • Study advanced Boolean algebra techniques for circuit simplification
  • Learn about the practical applications of NAND and NOR gates in digital circuits
  • Explore further examples of De Morgan's Theorem in circuit design
  • Investigate tools for simulating Boolean circuits and verifying simplifications
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Students and professionals in electrical engineering, computer science, and anyone involved in digital circuit design and optimization.

shamieh
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Find the simplified equation (SOP FORM) corresponding to the following circuit (i.e. apply Demorgan's theorem to bring the negations inside the parenthesis).

View attachment 1466So is it saying W NAND X which means (w!x!) + YNORZ so (w!x!) + (y!z!) = wxyz?
 

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shamieh said:
Find the simplified equation (SOP FORM) corresponding to the following circuit (i.e. apply Demorgan's theorem to bring the negations inside the parenthesis).

View attachment 1466So is it saying W NAND X which means (w!x!) + YNORZ so (w!x!) + (y!z!) = wxyz?

Your notation is a bit confusing to me. So let me write it as I'm used to.

You have f = (w NAND x) NOR (y NOR z).

Written in boolean form this is:
$$f = \overline{\overline{wx} + \overline{y+z}}$$
Applying De Morgan's Theorem ($\overline{p+q} = \overline p \cdot \overline q$) yields:
$$f = \overline{\overline{wx}} \cdot \overline{\overline{y+z}} = wx(y+z)$$

Perhaps you can simplify it further to SoP form?
 

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