Solving Logic Function w/ NOR Gates

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The discussion focuses on solving a logic function using NOR gates based on a provided truth table. The initial expression derived from the truth table is Q = B~C, which simplifies the logic function. Participants discuss how to redesign the circuit using only NOR gates, applying De Morgan's laws to convert expressions. A key point is that a NOR gate can be used as an inverter, and the transformation of BC' into (B' + C)' is highlighted. The conversation concludes with confirmation of the proposed circuit design using NOR gates.
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Homework Statement


Consider the truth table below.
(a) Write an expression for the logic function that it represents, and simplify that expression as much as possible, showing all steps.
(b) Draw a logic circuit implementing this.
(c) Redesign your circuit to use only NOR gates.

Homework Equations


A B C Q(output)
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0

The Attempt at a Solution


a) I chose to concentrate on when Q has values of 1. so I came up with the following equation: (where ~ means not, ie. ~A means that A is false)
Q = ~AB~C + AB~C
= B~C(~A+A) [~A+A=1]
= B~C
b) See attached.
c) that's where I could use some help... I'm not really sure how to write an expression or change my circuit diagram using what I already have.

Thanks!
 

Attachments

  • circuit.jpg
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Do you know De Morgan's laws? They should help you work this out.

(A + B)' = A'B'

(AB)' = A' + B'

Try drawing logic gates for each of those and they'll show you how else your function could be represented.
 
I drew them, and I see how something can go from AND to NOR for examples but I still don't understand how I can break BC' into something that I can draw using only NORs.
If I put the whole expression like so: (BC')' would I be able to do anything with it?
 
theuniverse said:
I drew them, and I see how something can go from AND to NOR for examples but I still don't understand how I can break BC' into something that I can draw using only NORs.
If I put the whole expression like so: (BC')' would I be able to do anything with it?

Remember that you can make an inverter out of a NOR. Put one of its inputs to 0 and it will invert the other input.
 
I think I get it. I will need to use 1 NOR for my inverter, and 3 NORs for my AND, am I right?
 
theuniverse said:
I think I get it. I will need to use 1 NOR for my inverter, and 3 NORs for my AND, am I right?

You can do better. Think about applying DeMorgans law to your BC'.
 
Using two NOR gates: BC'=B''C'=(B'+C)'.
So one of my NOR gates is used to invert the B, and then the other one connects the B' and a C which makes it equal to (B'+C)'.
Would that be correct?
 
theuniverse said:
Using two NOR gates: BC'=B''C'=(B'+C)'.
So one of my NOR gates is used to invert the B, and then the other one connects the B' and a C which makes it equal to (B'+C)'.
Would that be correct?

Looks good to me. You can always check what the output your circuit gives.
 
Last edited:
Thanks for the help LCKurtz!
 

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