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Boolean Algebra, Logic Diagram, K-Map, Nor gates HELP

  • Thread starter l46kok
  • Start date
  • #1
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Boolean Algebra, Logic Diagram, K-Map, Nor gates... HELP!!!!

Homework Statement


F(A,B,C,D) = Sigma(2,4,6,10,12)
d(A,B,C,D) = Sigma(0,8,9,13) [Dont Care Functions]

Implement the function using no more than 2 NOR gates.

Homework Equations


K-map


The Attempt at a Solution


First of all, I am wondering if this is PHYSICALLY POSSIBLE.

Anyways, if you draw the K-map for it, you get
AB\CD
d 0 0 1
1 0 0 1
1 0 0 0
d d 1 1

As for the grouping, I grouped the Top 4 0s (0100, 1100, 0101, 1101), mid 4 0s (0101, 1101, 0111, 1111) and 2 0s on the right and middle (1011, 1111)

If you're doing Nor implementation.. it's usually easier to group the 0s so
F' = BD + A'D + ABC

(As for the notation, ' are inversion, + is or and * is And EG : AB is A and B)

But we want F.. so we apply the demorgan's theorem

F = (B'D')(AD')(A'B'C')

and that is our function.

I looked at this function and I thought there is no way to solve this problem only by using NOR gates.

Please help!
 
Last edited:

Answers and Replies

  • #2
161
0
F(A,B,C,D) = Sigma(2,4,6,10,12)
d(A,B,C,D) = Sigma(0,8,9,13) [Dont Care Functions]

AB\CD
d 0 0 1
1 0 0 1
1 0 0 0
d d 1 1
I think you got the table wrongly. There are 5 minterms in the problem statement but you have 6 in your table; and there should be 4 dont cares while you have only 3 in the table.
 
  • #3
161
0
F' = BD + A'D + ABC

But we want F.. so we apply the demorgan's theorem
F = (B'D')(AD')(A'B'C')

and that is our function.
On a related note, De Morgan does not give you the transformation above. Read up on De Morgan again.
 
  • #4
296
0

Homework Statement


F(A,B,C,D) = Sigma(2,4,6,10,12)
d(A,B,C,D) = Sigma(0,8,9,13) [Dont Care Functions]

Implement the function using no more than 2 NOR gates.

Homework Equations


K-map


The Attempt at a Solution


First of all, I am wondering if this is PHYSICALLY POSSIBLE.

Anyways, if you draw the K-map for it, you get
AB\CD
d 0 0 1
1 0 0 1
1 0 0 0
d d d 1

As for the grouping, I grouped the Top 4 0s (0100, 1100, 0101, 1101), mid 4 0s (0101, 1101, 0111, 1111) and 2 0s on the right and middle (1011, 1111)

If you're doing Nor implementation.. it's usually easier to group the 0s so
F' = BD + A'D + ABC

(As for the notation, ' are inversion, + is or and * is And EG : AB is A and B)

But we want F.. so we apply the demorgan's theorem

F = (B'D')(AD')(A'B'C')

and that is our function.

I looked at this function and I thought there is no way to solve this problem only by using NOR gates.

Please help!
Oops, I meant to put a d instead of that 1. It's fixed now.

And yeah, F is supposed to be

F = (B'+D')(A+D')(A'+B'+C')

But this still does not answer my question. It is utterly impossible to represent this just using two nor gates.
 
  • #5
161
0
AB\CD
d 0 0 1
1 0 0 1
1 0 0 0
d d d 1

Oops, I meant to put a d instead of that 1. It's fixed now.
With this K-map, shouldn't there be only two groups of 0's?
 
  • #6
296
0
?

With this K-map, shouldn't there be only two groups of 0's?
? I can't see how you can have only two groupings of 0's and cover all 0's here.

Can you please explain how that might be done?
 
  • #7
296
0
?

With this K-map, shouldn't there be only two groups of 0's?
AH CRAP YOU'RE RIGHT

Now I can do it from here.

Thanks!
 

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