Combinational Circuit with LED Lighting

Question

Combinational Circuit design question.

    A
   ____
  |    |
F |    | B
  |    |
   ____ 
  |  G |
E |    | C
  |    |
   ____
    D

Suppose this is a LED display. It would take input of 4 bit
(0000)-(1111) and display the Hex of it. For example
if (1100) come in it would display C by turning on AFED and turning off BCG.

If (1010) comes in it would display A by turning on ABCEFG 
and turn off D.

These display will all be Capital letters so there is no visual
difference between 0 and D and 8 and B.

Develop a truth table and an optimized expression using Karnaugh Maps.

I'm not exactly sure how to begin. For the truth table would I be using (w,x,y,z) as input variable or just the ABCDEFG variable since it's the one turning on and off?

input (1010)-->A--> ABCEFG~D (~ stand for NOT)
input (1011)-->B--> ABCDEFG
input (1100)-->C--> ADEF~B~C~G

So would I do for all hex 0-F then that would give me the min. term canonical then use Karnaugh Map to optimize it? Any help would be grateful!

Answer 1

You need to do 7 of these: Each for one segment in the 7-segment display. This figure is for illustration only. It doesn't necessarily map to any segment in your problem.

    cd=00 01 11 10  <-- where abcd = 0000 for 0  : put '1' if the light is on
ab= 00  1  1  1  1                 = 0001 for 1  : put '0' if it's off for
ab= 01  1  1  1  0                 = 0010 for 2 ...      the given segment
ab= 11  0  1  1  1        
ab= 10  1  1  1  0                 = 1111 for f
           ^^^^ = d=1 region
              ^^^^ = c==1 region

The two middle rows represent "b==1" region and the two last rows are a==1 region.

From that map find maximum size rectangles (that are of size [1,2 or 4] x [1, 2 or 4]); that can be overlapping. The middle 2x4 region is coded as 'd'. The top row is '~a~b'. The top left 2x2 square is '~a~c'. A bottom left square that wraps from row 4 to row 1 is '~b~c'. Finally the small 2x1 region that covers position x=4, y=3 is 'abc'.

This function would thus be 'd + ~a~b + ~a~c + ~b~c + abc'. If there are no redundant squares (that are completely covered by other squares), then this formula should be optimal canonical form. (not counting XOR operation). Repeat for 7 times for the real data!

Any selection/permutation of the variables should give the same logical circuit, whether you use abcd or dcba or acbd etc.

Answer 2

1) Map your lights to bits:

ABCDEFG, so truth table will be:

                   ABCDEFG
input (1010)-->A-->1110110

and so on.

You will have big table (with 16 rows).

2) Then follow sample on wikipedia for every output light.