1
\$\begingroup\$

I got this question from a university site and thought of attempting it to test my knowledge in Logic circuits, Boolean alg and Karnaugh graph. Problem is, I have been scratching my what logic circuit is coming out; I am not able to translate this into a known Boolean Algebra.

The Question:

A vehicle seat belt circuit is such that the car should only start if the driver’s seat belt is fastened and either the front passenger seat is unoccupied or the front passenger seat is occupied and the passenger seat belt is fastened. Obtain a truth table, boolean equation and Karnaugh graph.

I need help on the truth table.

Suppose:

A: Driver seat belt

B:Passenger seat occupied

C:Passenger Seat Belt

Considering A, B and C, I get this truth table, but see my comments below:

   A    B     C   B??C    A AND(B??C)
  ---  ---  ---   ----    --------
    0    0    0   1       0
    0    0    1   1       0
    0    1    0   0       0
    0    1    1   1       0
    1    0    0   1       1
    1    0    1   1       1
    1    1    0   0       0
    1    1    1   1       1

When I consider B and C (Passenger seat occupancy and Passenger seat belt: When there is no passenger, Passenger seat belt does not matter as long as driver seat belt on. these are represented by B??C. The problem is, I am not able to figure out what Boolean condition this is: AND, XOR, NOT, etc. This is what is making it difficult for me to produce the resulting Boolean Algebra for subsequent questions. Help.

\$\endgroup\$

2 Answers 2

Reset to default
2
\$\begingroup\$ 
 B   !B    C   B??C
---  ---  ---  ----
 0    1    0    1
 0    1    1    1
 1    0    0    0
 1    0    1    1
 0    1    0    1
 0    1    1    1
 1    0    0    0
 1    0    1    1

Do you see it now?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ it seems the resulting Boolean equation is A.(B'+C) which should expand to $$ A.(B'+C) $$ $$ AB'+AC $$ This does not seem plottable on a 3-var Karnaugh Graph using either A.B|C or A|B.C map. \$\endgroup\$
    – Sylvester
    Commented Oct 19, 2014 at 6:19
  • \$\begingroup\$ If you can generate a truth table for it, you can generate a K-map for it. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Oct 19, 2014 at 6:23
  • \$\begingroup\$ Picture this: <pre> C| C' C AB--- --- --- A'B' | AB' | AB | A'B | </pre> I do not find an intersection of AB and C or either to plot to K-map. \$\endgroup\$
    – Sylvester
    Commented Oct 19, 2014 at 6:37
2
\$\begingroup\$

Here's the logic diagram, the raw truth table, and the minimized truth table:

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ looking at the truth table, you have added D. But, I figure out that: $$ D = A.(B'+C) $$ So I was thinking the K-map ought to be a 3-varible one with A, B, C conditions. Is this incorrect? \$\endgroup\$
    – Sylvester
    Commented Oct 19, 2014 at 8:26
  • \$\begingroup\$ I think it's D = AB'+ ABC. As for the Kmap, try it. Go here for a lot of examples. \$\endgroup\$
    – EM Fields
    Commented Oct 19, 2014 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.