How To Implement This Function With 2 Input Nand

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SUMMARY

This discussion focuses on implementing a logic function using two-input NAND gates. The function F(w,x,y,z) is derived from the expression F(w,x,y,z)=(w'+x'+y')(+)xyz, leading to a simplified form of F = \overline{x} + \overline{y} + (\overline{w})(\overline{z}) + wxyz. The conversation emphasizes the importance of transforming the function into a Sum of Products (SOP) format to facilitate implementation with NAND gates. Additionally, it highlights the application of DeMorgan's theorem in simplifying the expression further.

PREREQUISITES
  • Understanding of logic gates, specifically NAND gates.
  • Familiarity with Boolean algebra and simplification techniques.
  • Knowledge of DeMorgan's theorem and its applications.
  • Experience with Sum of Products (SOP) forms in digital logic design.
NEXT STEPS
  • Research how to implement logic functions using two-input NAND gates.
  • Study Boolean algebra simplification techniques for complex expressions.
  • Learn about DeMorgan's theorem and its implications in digital circuits.
  • Explore practical applications of Sum of Products (SOP) in circuit design.
USEFUL FOR

Digital logic designers, electrical engineers, and students studying digital circuit design who are looking to implement functions using NAND gates.

transgalactic
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Default How To Implement This Function With 2 Input Nand..
(+) =XOR

F(w,x,y,z)=(w'+x'+y')(+)xyz

i tried to :

F(w,x,y,z)=(w'+x'+y')(+)xyz=(w'+x'+y')' *(xyz) + (w'+x'+y')(xyz)'=
=(wxy)(xyz)+(w'+x'+y')(x'+y'+z')
..
..
in the end i get
=wxyz+w'x'+w'y'+w'z'+x'+x'y'+x'y'+y'+y'z'+y'z'

what to do know?
how to transform implement it with a two input line NANDs
?
 
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Well, the word NAND should scream Sum of Products (SOP), so try to get your function to that point, then you can modify it further. After playing around with your function, you should get something like:

F = \overline{x} + \overline{y} + (\overline{w})(\overline{z}) + wxyz

(Keep in mind that A+AB = A; your final equation doesn't take that into account).

I'm going to show you another useful proof:

\overline{ABCD} = \overline{A}+\overline{B}+\overline{C}+\overline{D} = (\overline{A}+\overline{B})+(\overline{C}+\overline{D}) = \overline{AB}+\overline{CD} = \overline{(\overline{AB})(\overline{CD})}

The function is also equal to (through DeMorgan):
F = \overline{xy ( \overline{ \bar{w} \bar{z} })( \overline{wxyz}) }

Use the proof above on that form of F. Use the proof above to get \overline{(wxyz)}, too. Use NANDs as inverters to get w' and z'.
 
I think There is a mistake it should be \overline{AB}+\overline{CD} = \overline{({AB})({CD})}
 

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