How To Implement This Function With 2 Input Nand

  • #1
1,395
0
Default How To Implement This Function With 2 Input Nand..
(+) =XOR

F(w,x,y,z)=(w'+x'+y')(+)xyz

i tried to :

F(w,x,y,z)=(w'+x'+y')(+)xyz=(w'+x'+y')' *(xyz) + (w'+x'+y')(xyz)'=
=(wxy)(xyz)+(w'+x'+y')(x'+y'+z')
..
..
in the end i get
=wxyz+w'x'+w'y'+w'z'+x'+x'y'+x'y'+y'+y'z'+y'z'

what to do know???
how to transform implement it with a two input line NANDs
???
 

Answers and Replies

  • #2
19
0
Well, the word NAND should scream Sum of Products (SOP), so try to get your function to that point, then you can modify it further. After playing around with your function, you should get something like:

[tex]F = \overline{x} + \overline{y} + (\overline{w})(\overline{z}) + wxyz[/tex]

(Keep in mind that [tex]A+AB = A[/tex]; your final equation doesn't take that into account).

I'm going to show you another useful proof:

[tex]\overline{ABCD} = \overline{A}+\overline{B}+\overline{C}+\overline{D} = (\overline{A}+\overline{B})+(\overline{C}+\overline{D}) = \overline{AB}+\overline{CD} = \overline{(\overline{AB})(\overline{CD})} [/tex]

The function is also equal to (through DeMorgan):
[tex]F = \overline{xy ( \overline{ \bar{w} \bar{z} })( \overline{wxyz}) }[/tex]

Use the proof above on that form of F. Use the proof above to get [tex]\overline{(wxyz)}[/tex], too. Use NANDs as inverters to get w' and z'.
 
  • #3
21
0
I think There is a mistake it should be [tex] \overline{AB}+\overline{CD} = \overline{({AB})({CD})} [/tex]
 

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