Digital Logic Design: Simplifying 4-Var Map | F(w,x,y,z)

[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?

 

[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!??

 

[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?

 

[Ian_Brooks]

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!??

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?

 

[hazim]

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?

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...

 

[hazim]

why would the function be 0?

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'

 

[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.

 

[hazim]

sorry but what is PM'ing berkman?

 

[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.

 

[berkeman]

sorry but what is PM'ing berkman?

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.

 

[mohammed1]

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?

Answer is << complete solution deleted by berkeman >>