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Truth Tables/Minimizing/SOP/POS Part 2

shamieh

Active member
Sep 13, 2013
539
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) = \(\displaystyle \sum\)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:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Draw the truth table corresponding to $f$(X,Y,Z) = \(\displaystyle \sum\)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.

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.

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.

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.

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.

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.