XOR Gate Circuits: Draw Z with Lowest # of 2 Input Gates

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Discussion Overview

The discussion revolves around designing a circuit for the function Z(A,B,C,D) using the lowest number of 2-input XOR gates. Participants explore methods to simplify the expression using Karnaugh maps and discuss the construction of the circuit.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the function Z(A,B,C,D) = SUM-M(1,2,4,7,8,11,13,14) and requests assistance in drawing the circuit.
  • Another participant notes that their attempt at a Karnaugh map resulted in too many terms and seeks clarification on the approach taken by others.
  • A participant observes that the first row of the K map resembles an XOR operation.
  • One participant suggests considering the identical nature of the third row of the K map and how to construct a NOT gate from an XOR gate.
  • Another participant proposes using XOR gates to derive outputs P and Q from inputs A, B, C, and D, and questions how to combine these outputs effectively.
  • There is a suggestion to explore using fixed inputs (TRUE or FALSE) with the XOR gate to manipulate outputs creatively.
  • Participants discuss the potential of using different signals in conjunction with the XOR gate to achieve the desired outputs.
  • Ultimately, one participant summarizes their final answer as Q = XOR(A, B), P = XOR(C, D), and ANS = XOR(Q, P), indicating that this configuration seems to work.

Areas of Agreement / Disagreement

Participants express various approaches and ideas, but there is no clear consensus on the optimal method for constructing the circuit. Multiple competing views and techniques are presented throughout the discussion.

Contextual Notes

Participants reference the use of Karnaugh maps and the construction of logical gates, but there are unresolved details regarding the simplification process and the specific configurations of the gates.

kukumaluboy
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Z(A,B,C,D) = SUM-M(1,2,4,7,8,11,13,14)

Draw circuit of Z with lowest number of 2 inputs xor-gates.

Can i have a clue? I tried drawing k map and had too many terms.
 
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kukumaluboy said:
I tried drawing k map and had too many terms.
Can you show us what you've done?
 
dqoopu.jpg

So from the K map, in the 1st row, AB looks like a xor.
 
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bump.!
 
Your observation about the first row of the map is good. Note how the third row is identical. Note also how the other rows are the negation of them. Contemplate how you might build a NOT gate from an XOR. Does it give you any ideas?
 
Thanks for tip!

Ok so With a xor gate i can get a'b+ ab' as an output. Let's call this output as P.

Next step is for me to get c'd' + cd which is simply xnor? So i will use a xor(c,d)=Q. Th3n ill xor(Q,1) to get the inverse?

Now to combine both P and Q do i need to construct AND gate with xor?
 
kukumaluboy said:
Thanks for tip!

Ok so With a xor gate i can get a'b+ ab' as an output. Let's call this output as P.
Okay, fine. That's a XOR b. It'll come in handy.
Next step is for me to get c'd' + cd which is simply xnor? So i will use a xor(c,d)=Q. Th3n ill xor(Q,1) to get the inverse?

Now to combine both P and Q do i need to construct AND gate with xor?
It's getting complicated again. If I may suggest a short digression:

Did you spend any time playing with the XOR gate to try to make a NOT gate? Hint: you can assume access to constants TRUE and FALSE (or 1 and 0) to use as inputs anywhere you like. If you feed one input of an XOR with a fixed 0, what does it do to the other input? How about if you use a fixed 1? Can you think of any interesting uses for such a beast? How about if the "fixed" input were actually another signal?
 
Hi! In my reply i used a fixed "1". Xor(Q, 1)
 
kukumaluboy said:
Hi! In my reply i used a fixed "1". Xor(Q, 1)
Okay, but maybe you can think of another signal to use for the "1" that would select the appropriate action for that XOR to match the other rows in the table? What values of cd are associated with the "straight" a XOR b, and which with its complement?
 
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  • #10
Challenge accepted :cool:
 
  • #11
Hi my finam answeer
answer.

Q = XOR(A , B)
P = XOR(C , D)
ANS = XOR(Q , P)
 
  • #12
kukumaluboy said:
Hi my finam answeer
answer.

Q = XOR(A , B)
P = XOR(C , D)
ANS = XOR(Q , P)
That seems to work.
 
  • #13
kukumaluboy said:
Hi my finam answeer
answer.

Q = XOR(A , B)
P = XOR(C , D)
ANS = XOR(Q , P)
Yes, well done.
 

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