Simple boolean simplification - can i simplify it further?

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Discussion Overview

The discussion revolves around the simplification of a Boolean function represented in sum-of-products form, specifically F(A,B,C,D) = (sum of) m (2,3,5,7,11,13). Participants explore methods to design a two-level network for implementing this function using logic gates.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant presents an initial attempt at simplifying the Boolean function using a Karnaugh map and proposes a potential expression for F.
  • Another participant questions the definition of a two-level network, indicating a need for clarification.
  • A later reply suggests a corrected expression for F, highlighting the common factor of C in the simplification process.
  • Another participant provides a definition of a two-level circuit and presents an alternative sum-of-products expression for F, indicating a method for drawing the corresponding circuit using NAND gates.
  • One participant expresses gratitude for assistance and mentions successfully figuring out the problem independently.

Areas of Agreement / Disagreement

There is no clear consensus on the simplification of the Boolean function, as participants present different expressions and approaches without resolving the discrepancies.

Contextual Notes

Participants' discussions include various interpretations of the two-level network concept and the implications of using different forms of the Boolean function. Some assumptions about the simplification process and the definitions of terms remain unaddressed.

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Homework Statement



F(A,B,C,D) = (sum of) m (2,3,5,7,11,13)
Design a two level network to implement the above sum of products:

Homework Equations



F(A,B,C,D) = (sum of) m (2,3,5,7,11,13)

The Attempt at a Solution


Code:
K-map 
___Ab
CD|00..01..11..10 
00|
01|1...1...1
11|1...1...1
10|

My attempted solution:
F = CA' + CDB + CD'B'
F = CA' + C (DB + D'B')
F = CA' + C(D xor B)'
F = CA' + CD xor CB'

now can i factor out the C? This has to be two level so I'm thinking it'd take up two different blocks of logic gates.

Any help?
 
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What is a two level network may I ask?
 
the last line of it should be: F = C (A' + D xor B')

Since all have C in common. I'm probably wrong though.
 
OK. I found what the definition of a two-level circuit is: the implementation of a Boolean function with NAND gates is simplest if the function is in sum-of-products form. This form corresponds to a two-level circuit.

So, as a sum of products, I get that F = BC'D + A'BD + B'CD + A'B'C. You should be able to quickly draw the two-level circuit from it using NAND gates without any problems.
 
thanks a tonne i managed to figure it out a while back.
thanks
 

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