Back
to top
Boolean algebra and logic gate help
Watch this threadPage 1 of 1
Go to first unread
Skip to page:
Shiinobii
Badges:
0
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
0
reply
Algebraholic
Badges:
0
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report
#2
So the first logic gate at the top is a NAND gate- for inputs it has A and B- so this is 'A NAND B'
The gate underneath it is a NOR gate- for inputs it has B and C- so this is 'B NOR C'
The output from these two gates are the inputs to a second NAND gate- so this is (A NAND B) NAND (B NOR C)
The output from this gate becomes one of the the inputs to a second NOR gate- the other input being C
So the system is...
[ (A NAND B) NAND (B NOR C) ] NOR C
This can be simplified by first of all getting rid of the last NOR I mentioned.
Recall that 'A NOR B' means ¬(A OR B) and by DeMorgans law is the same as ¬A AND ¬B
So we have ¬[ (A NAND B) NAND (B NOR C) ] AND ¬C
So what does 'A NAND B' mean? Apply DeMorgans Law to this- what is it the same as? Now apply this to the line of Boolean Algebra above.
The gate underneath it is a NOR gate- for inputs it has B and C- so this is 'B NOR C'
The output from these two gates are the inputs to a second NAND gate- so this is (A NAND B) NAND (B NOR C)
The output from this gate becomes one of the the inputs to a second NOR gate- the other input being C
So the system is...
[ (A NAND B) NAND (B NOR C) ] NOR C
This can be simplified by first of all getting rid of the last NOR I mentioned.
Recall that 'A NOR B' means ¬(A OR B) and by DeMorgans law is the same as ¬A AND ¬B
So we have ¬[ (A NAND B) NAND (B NOR C) ] AND ¬C
So what does 'A NAND B' mean? Apply DeMorgans Law to this- what is it the same as? Now apply this to the line of Boolean Algebra above.
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
