# Digital logic design

1. Nov 16, 2007

### hazim

[SOLVED] digital logic design

i want to simplify this function using four-variable map: F(w,x,y,z) = $$\sum$$(0,1,2,4,5,7,11,15) , the map contains one "1" at the upper right corner. my question is how can i take this 1 in a group? or what can i do with it???

2. Nov 16, 2007

### hazim

another question is if i have a 3-variable map ( i mean the k-map) full of 1s and Xs(representing the don't care conditions), can i take all as one group??? if yes then the function will be zero!??

3. Nov 17, 2007

### Ian_Brooks

Hazim,

for your first question can you give us your truth table -

what I understand is that you gave us a sum of minterms and stated that there's only a 1 in the logic table "at the upper right corner" - however this is for your k-map am I right?

If that's the only case that gives you a logic of 1 - then what does that say about your other logic combinations? Which one should you be concerned with then?

4. Nov 17, 2007

### Ian_Brooks

If your k-map is only filled with 1's an x's
then the function will be true for all cases so to speak

why would the function be 0?

Last edited: Nov 17, 2007
5. Nov 17, 2007

### hazim

you don't need the truth table, and you can get it from the sum of minterms if you want....i ment that there is one 1 at the corner alone, and i have to simplify the function by taking 1's in a group (2 ones or 4 or 8...) but now i solved this, i took it alone and it worked....

6. Nov 17, 2007

### hazim

the function is with variables x, y, z; when taking all in a group and even in more groups, but all th literals are for sure taken in groups, then x and x' will cancel, and y will cancel with y', and z with z'

7. Nov 17, 2007

### Ian_Brooks

hmm i see what you are trying to do

but then for what cases will your system give you a logic of 1?

if all cases are true then it should be just a logic 1 for your function.

You may try PM'ing berkman for help - he helped many people with Digital design questions.

8. Nov 17, 2007

### hazim

sorry but what is PM'ing berkman?

9. Nov 17, 2007

### Ian_Brooks

you send a personal message to a member called berkman - he's a member on these forums that helped me back when i was taking digital design a couple years back.

I'm still certain that for a function if all cases are true or don't cares - then you get a f() = 1
- double check with your professor if you have to.

10. Nov 19, 2007

### Staff: Mentor

I've read the thread, and sorry, I don't understand the question. Could you please re-state it? And if you could post a copy of your K-map along with your restatement of the question, that would help.

11. Nov 29, 2007

### mohammed1

Answer is << complete solution deleted by berkeman >>

Last edited by a moderator: Nov 30, 2007