How can I simplify this circuit using boolean algebra for XOR and XNOR gates?

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SUMMARY

The discussion focuses on simplifying the boolean expression M'(A'B'C + ABC') + M(AB'C' + A'BC) using XOR and XNOR gates. Participants highlight the limitations of traditional methods like K-maps and suggest alternative approaches, including the use of multiplexers (specifically the 4067 model) for circuit synthesis. The conversation emphasizes the complexity of achieving a minimal circuit design while considering the number of packages and transistors required. Ultimately, the consensus is that while simplification is possible, it may not yield the least number of components in practical applications.

PREREQUISITES
  • Understanding of boolean algebra and simplification techniques
  • Familiarity with K-map methodology for boolean expressions
  • Knowledge of XOR and XNOR gate functionalities
  • Experience with multiplexers, particularly the 4067 model
NEXT STEPS
  • Study advanced boolean algebra techniques for circuit simplification
  • Learn how to effectively use K-maps for complex expressions
  • Explore the design and implementation of multiplexers in digital circuits
  • Investigate the transistor requirements for XOR and XNOR gates
USEFUL FOR

Electrical engineers, digital circuit designers, and students studying digital logic who are looking to optimize circuit designs using boolean algebra and gate-level simplifications.

speck
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I want to simplfy M'(A'B'C+ABC')+M(AB'C'+A'BC) to as simple a circuit as possible.

I don't know the boolean algebra to simplfy the ABC terms. Help please, Speck
 
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Tried K-maps?

From quick Venn inspection of (A'B'C+ABC') and (AB'C'+A'BC), I don't think you can simplify them further using AND, OR, NOT only
 
K-map is how I initial got the Eq. , right, it won't simplify with AND, OR, NOT. I want to use XOR with XNOR gates. I would really like it to simplify to something like (AB Oplus C) using XOR, but it does not. Thks, Speck
 
Does anyone think that the (A'B'C+ABC') part of the Eq. will reduce to (A Oplus B Oplus C)?
 
By simple, do you mean the least number of packages? It's trivial with a single PLA, but you'd need a burner...
 
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I tried to put it into XOR/XNOR but I really couldn't find any way.

P.S. (I learned this stuff few weeks ago, so all I know is that there should be checkboard pattern)

Now that I said that I realized that there is infact a pattern and it is easier to isolate it when you look at it. You got to approach it differently.
See K-Map When A = 0 and C = 1
A = 1 C = 0

I get something like

A!C!(B XOR M) + A!C (M XOR B)

So far, I look at K-Map and try to isolate 2 literal K-Maps that look like XOR and "and" it with conditions like A = 1 and C = 0 .. It works so far
 
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Another PLA-type cheater's answer: use a multiplexer. Input ABCM as the addresses, hardwire the 16 inputs to 1 or 0 to synthetise the desired logic function. The 4067 is such a 16-to-1 mux-demux and seems to be still relatively common (hey, I just feel younger!). One single package, no programming needed.

For a non-cheater answer, you'll have to wait a bit more. M and B have similar roles, as do A and C, so combining these pairs first could bring something.
 
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(A xnor C) nor (B xor M)
please check!

It's not "the simplest" form for the number of packages on a breadboard.

In a chip, I guess it's not the minimum number of Mos neither, as an xor or an xnor needs two inverters and 8 transistors, and a 16-to-1 mux also needs one inverter per input and this function consumes 16 N-channels and 16 P-channels.

Well, this form must be the simplest in the mind of some teacher at least.
 
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