Converting to just NANDs with DeMorgan's

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SUMMARY

The discussion focuses on converting the Boolean expression a3'a1 + a3'a2 + a3'a0 + a3a2'a1'a0' into a circuit using only NAND and NOR gates, applying DeMorgan's Law. The initial transformation using DeMorgan's resulted in the expression ( (a3'a1)' (a3'a2)' (a3'a0)' (a3a2'a1'a0')')', which is valid but not aesthetically pleasing. Participants suggest drawing the circuit first and then converting AND and OR gates to NAND and NOR gates, while eliminating redundant inverters for a cleaner realization of the circuit.

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  • Understanding of Boolean algebra and expressions
  • Familiarity with DeMorgan's Theorems
  • Knowledge of digital logic gates, specifically NAND and NOR
  • Experience with circuit simplification techniques
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  • Learn about DeMorgan's Theorems in depth
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Melawrghk
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Homework Statement


Convert a3'a1 + a3'a2 + a3'a0 + a3a2'a1'a0'

to an expression (well, it's a circuit eventually) that would only employ NAND and NOR gates. Use DeMorgan's Law.


Homework Equations


a'+b'=(ab)'

The Attempt at a Solution


I applied DeMorgan's as the problem suggested...

(a3'a1)''+(a3'a2)''+(a3'a0)'' + (a3a2'a1'a0')''

Which yielded:
( (a3'a1)' (a3'a2)' (a3'a0)' (a3a2'a1'a0')')'

This is in just NAND and NOR gates, but the circuit isn't really aesthetically appealing... Is there a simplifying trick I'm missing? Or is this just what it is supposed to be?

Thanks in advance!
 
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I would draw out the circuit first before doing any Boolean algebra using regular AND OR and INVERTER gates. Then when its all drawn, convert the AND and NOR gates to NAND and NOR. Put inverter circles on the inputs of the and OR gates to turn them into NANDs, and inverter circles on the output of the AND gates to turn them into NAND gates. Cancel out and redundant inverters. Should results in a full NAND-NAND realization of you circuit.
 

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