Boolean algebra and logic gate help

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Shiinobii
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#1
Report Thread starter 7 years ago
#1
Name:  Screenshot_2014-12-29-13-36-02-1.png
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Size:  35.2 KBHey guys I'm new to this boolean algebra and logic gates module and was wondering if someone could help me with this question? Any help will be appreciated.
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Algebraholic
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#2
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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.
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