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Is This Logic Diagram Correct?

  • Thread starter ibcoding
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  • #1
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The boolean expression I am working with is:

((ab)'(b'c)' + a'bpc') where ' is the NOT symbol.

I have the following logic circuit diagram. Is it correct?








 

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  • #2
cepheid
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Welcome to PF!

The diagram is confusing a bit, because one of the a lines is very close to one of the b lines. However it looks correct to me.
 
  • #3
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Ok, I know exactly which line you are talking about; I thought the same thing. I'll change that around. Thanks!
 
  • #4
cepheid
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I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

= a'b + a'c' + b'b + b'c'

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.
 
Last edited:
  • #5
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Alright, variable p is missing from that equation, but I get the idea. On that note, is this the correct truth table outputs:

1
1
0
0
1
1
1
1
1
1
0
0
0
0
0
0
 
  • #6
11
0
I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

(ab)' * (b'c)'

Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

= (a' + b') * (b + c') [eq. 1]

and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

= a'b + a'c' + b'b + b'c'

and using the associative property:

a'(b + c') + 0 + b'c'

Anything OR'd with 0 is just itself, so this becomes:

a'(b + c') + b'c' [eq. 2]

I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.
Ahh, the a' a' are adjacent so it is absorbed, got ya
 
  • #7
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So the equation should be a'(b + c') + b'pc'
 
  • #8
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which gives me three and gates and two or gates
 
  • #9
cepheid
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Alright, variable p is missing from that equation, but I get the idea.

Well, I was ONLY considering the (ab)'(b'c)' in the original equation, as an example. I wasn't trying to simplify the whole thing.

On that note, is this the correct truth table outputs:
1
1
0
0
1
1
1
1
1
1
0
0
0
0
0
0
Here's what I get:

Code:
a	|	b	|	c	|	p	|	f
0 	|	0 	|	0 	|	0 	|	1
0 	|	0 	|	0 	|	1 	|	1
0 	|	0 	|	1 	|	0 	|	0
0 	|	0 	|	1 	|	1 	|	0
0 	|	1 	|	0 	|	0 	|	1
0 	|	1 	|	0 	|	1 	|	1
0 	|	1 	|	1 	|	0 	|	1
0 	|	1 	|	1 	|	1 	|	1
1 	|	0 	|	0 	|	0 	|	1
1 	|	0 	|	0 	|	1 	|	1
1 	|	0 	|	1 	|	0 	|	0
1 	|	0 	|	1 	|	1 	|	0
1 	|	1 	|	0 	|	0 	|	0
1 	|	1 	|	0 	|	1 	|	0
1 	|	1 	|	1 	|	0 	|	0
1 	|	1 	|	1 	|	1 	|	0
 
  • #10
11
0
Sorry, I wasn't complaining. :) Alright, that's the same as what I have.
 
  • #11
11
0
The equation breaks down to a'b + b'c' + a'bp, I believe.
 

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