MHB Truth Tables/Minimizing/SOP/POS Part 2

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The discussion focuses on verifying and minimizing Boolean expressions related to truth tables for functions of three variables, X, Y, and Z. Participants confirm the correctness of the truth tables and canonical sum of products (SOP) expressions for the function defined by the sum of minterms. There is a challenge in minimizing the SOP expression, with suggestions for alternative approaches to achieve a minimal product of sums (POS) expression. The final consensus indicates that the minimized POS expression is likely (x + y!)(x + z!). The conversation emphasizes the importance of careful factorization and transformations in Boolean algebra.
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Need someone to check my answers once more. (Promise this is the last time lol)(Wasntme)

Draw the truth table corresponding to $f$(X,Y,Z) = $$\sum$$m(0,1,2,6,7)

Answer:
  1. x y z | f
  2. 0 0 0 |1
  3. 0 0 1 |1
  4. 0 1 0 |1
  5. 0 1 1 |0
  6. 1 0 0 |0
  7. 1 0 1 |0
  8. 1 1 0 |1
  9. 1 1 1 |1


Write out the canonical sum of products SOP expression for $f$(X,Y,Z) of problem above.

ANSWER:
x!y!z! + x!y!z + x!yz! + xyz! + xyz


Minimize the expression above.

x!y!z!+x!y!z+x!yz!+xyz!+xyz = x!y!(z! + z) + y(x!z! + xz! + xz) --->
= y[z!(x! + x) + xz] = (x! + x) + xz = 1 + xz = x!y! + xz?

Draw the truth table corresponding to $f$(X,Y,Z)= POSM(1,2,3) (product of sums symbol M)

ANSWER:
  • x y z | f
  • 0 0 0 |1
  • 0 0 1 |0
  • 0 1 0 |0
  • 0 1 1 |0
  • 1 0 0 |1
  • 1 0 1 |1
  • 1 1 0 |1
  • 1 1 1 |1


write out the canonical sums POS expression for $f(x,y,z) of the prob above.

ANSWER:
(x + y + z!)(x + y! + z)(x + y! + z!)


minimize the expression...

Just need someone to check my answers and help me solve the last problem.

Don't know how to minimize it when I can't factor it out. I also have 9 terms.. So I need to distribute it?

HELP

thank you.
Sham
 
Last edited:
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shamieh said:
Draw the truth table corresponding to $f$(X,Y,Z) = $$\sum$$m(0,1,2,6,7)

Answer:
  1. x y z | f
  2. 0 0 0 |1
  3. 0 0 1 |1
  4. 0 1 0 |1
  5. 0 1 1 |0
  6. 1 0 0 |0
  7. 1 0 1 |0
  8. 1 1 0 |1
  9. 1 1 1 |1
Correct.

shamieh said:
Write out the canonical sum of products SOP expression for $f$(X,Y,Z) of problem above.

ANSWER:
x!y!z! + x!y!z + x!yz! + xyz! + xyz
Correct.

shamieh said:
Minimize the expression above.

x!y!z!+x!y!z+x!yz!+xyz!+xyz = x!y!(z! + z) + y(x!z! + xz! + xz) --->
= y[z!(x! + x) + xz] = (x! + x) + xz = 1 + xz = x!y! + xz?
The first equality is correct. But what does ---> mean? Does it mean that you are dropping the first term x!y!(z! + z)? Next I don't understand where y from y[z!(x! + x) + xz] and z! from z!(x! + x) disappeared. If you do it correctly, you'll get x!y! + yz! + xyz. It's possible to turn it into x!y! + yz! + xy using some additional transformations, but I believe minimal POS will have three minterms.

shamieh said:
Draw the truth table corresponding to $f$(X,Y,Z)= POSM(1,2,3) (product of sums symbol M)

ANSWER:
  • x y z | f
  • 0 0 0 |1
  • 0 0 1 |0
  • 0 1 0 |0
  • 0 1 1 |0
  • 1 0 0 |1
  • 1 0 1 |1
  • 1 1 0 |1
  • 1 1 1 |1
Correct.

shamieh said:
write out the canonical sums POS expression for $f(x,y,z) of the prob above.

ANSWER:
(x + y + z!)(x + y! + z)(x + y! + z!)
Correct.

shamieh said:
minimize the expression...
I believe the answer is (x + y!)(x + z!). Instead of minimizing POS, you can take the dual expression x!y!z + x!yz! + x!yz, minimize it and take the dual again.
 
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