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I got a digital design problem. My final implementation of the circuit program didn't achieve the required criteria, so I brought my problem and my works here to be suggested by you if I mis-manipulated any details.

Logic System: Binary--1,0.

Issue after Simulation and Test on Circuit Board: LED lighted up at the begining without any switches were turned ON. (Would it make sense?)

Problem Statement:

Use 4 switches A, B, C, D to control a LED. LED is ON if the following conditions are satisfied,

D and B turn ON,

or A turns OFF while C turns ON,

or A and B turn ON and the rest turn OFF,

or C and D turn OFF,

or all switches turn ON.

End of Problem Statement.

Notations and Circuit Combinations we use:

AND gate: · (dot)

OR gate: +

"Not/Invert something": ~

SOP: Sum of Products

POS: Product of Sums

SOP and POS are equivilant to each other.

My works:

True Table based on the 5 criteria

ON:1, OFF:0.

Collection of LED results is {1011111110001101}.

I can handle operating the software part, but I really want to check with you guys if I use POS method and based on the 5 criteria given in the problem, whether or not my true table is correct.

Please help me to point out my blind spots. Thanks

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    \$\begingroup\$ how about instead of a delete you removed your answer from the question and post your answer with details of how you accomplished it to help future users. \$\endgroup\$ – Kortuk Sep 19 '12 at 2:51
  • \$\begingroup\$ @Kortuk Then I'd spare this question's life, if you think it deserves mercy. Thanks anyway. \$\endgroup\$ – Joyful Sylph Sep 19 '12 at 21:16
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Going by the problem definition directly:

  1. D and B turn ON: B•D

  2. A turns OFF while C turns ON: ~A•C

  3. A and B turn ON and the rest turn OFF: This condition is already covered by condition 4.

  4. C and D turn OFF: ~C•~D

  5. all switches turn ON: This condition is already covered by 1.

So your final function is: B•D + ~A•C + ~C•~D

3 ands, 2 ors.

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  • \$\begingroup\$ I tested my true table again without interchanging between 0&1 for the program and it was correct. It also verified your statement that the 3rd condition is irrelevant because of condition 4, and I finally understand why you concluded that. I think what messed me up previously was that I changed 1 for 0 and vice versus in my program so that absurdity occurred. Thanks. \$\endgroup\$ – Joyful Sylph Sep 18 '12 at 0:59

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