Hi, Bean. Straight homework help here. Cool.
Beginning digital logic classes are given three tools -- truth table, Karnaugh
mapping, and Demorgan, right? First thing I did was just draw a truth table.
Inspection shows there are only five cases where F is not true, so I decided to
do this inverted -- that is, solve for what's not true, rather than what's
true.
Let Q = (A! AND B! AND C!)
Let R = (A! AND B! AND D!)
Let S = (A! AND C! AND D!)
Let T = (B! AND C! AND D!)
It should be obvious looking at the table that F! = Q + R + S + T (with A! AND
B! AND C! AND D! being the redundant case which is true for all). Now, you can
Demorganize that, and come up with what you have below, with the last NAND
being used to invert again from F! to F. View in fixed font (like M$
Notepad)...
.---.
A!----| |
B!----| |o----------
C!----|& | |
| | |
'---' |
.---. |
A!----| | | .-----.
B!----| |o------- | | |
D!----|& | | '-- | | __
| | | | | |-----| |
'---' '----- | |o------| |& |o-
.---. | & | ------|__|
A!----| | .------| |
C!----| |o------- | |
D!----|& | .---| |
| | | '-----'
'---' |
.---. |
B!----| | |
C!----| |o----------
D!----|& |
| |
'---'
created by Andy´s ASCII-Circuit v1.22.310103 Beta
www.tech-chat.de
This is pretty similar to factoring quadratics back in high school -- your
ability improves with practice. Try solving problems in different ways, and
just do what works.
Good luck.
Chris