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Combinational Logic + Boolean equation help

Blogs > Raithed
Post a Reply
1 2 Next All
Raithed
Profile Blog Joined May 2007
China7078 Posts
Last Edited: 2008-04-21 02:27:56
April 21 2008 02:27 GMT
#1
1. When people write B' does it mean B(not)? Like B with a bar on top of it?
......._
B' = B

Also...

2.
X = B(A+C) + C
X = BA + BC + C
X = BA + C(B+1) <-- this part confuses me, can someone explain how the BC + C turn into C(B+1)?
X = BA + C*1 (B + 1 = 1 this is from rules)
x = BA + C (C * 1 = C this is from rules)

And 3. This NAND gate, with the inverter bubble there, would it be A' B' into A+B? Since the inverters cancel?

[image loading]


****
NathanSC
Profile Blog Joined February 2008
United States620 Posts
April 21 2008 02:33 GMT
#2
As far as BA + BC + C = BA + C(B + 1), that's really simple factoring.

Look at it as 1BC + 1C. Factor out the C.

C(1B + 1).

And clearly the coefficient 1 before the B is extraneous, so it's not written.
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
Last Edited: 2008-04-21 02:39:12
April 21 2008 02:38 GMT
#3
1) Yes

2) See above post, just factoring.

3) So this is an AND gate with inversions on all inputs and outputs?

Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'

Then you have a NOR with an inverted output which is the same as OR.
Do you really want chat rooms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 02:38 GMT
#4
I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..

x^2 + 4x + 4 would be (x+2)(x+2) ...

Can you explain further, I really don't see it.
Muirhead
Profile Blog Joined October 2007
United States556 Posts
April 21 2008 02:40 GMT
#5
In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.

Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB
starleague.mit.edu
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 02:41 GMT
#6
Forget the BA for a moment.

BC + C = C(B + 1)
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Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 02:43 GMT
#7
On April 21 2008 11:38 fight_or_flight wrote:
1) Yes

2) See above post, just factoring.

3) So this is an AND gate with inversions on all inputs and outputs?

Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'

Then you have a NOR with an inverted output which is the same as OR.

For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:

A' B' = AB? It doesn't work that way?
Myxomatosis
Profile Blog Joined July 2006
United States2392 Posts
April 21 2008 02:44 GMT
#8
On April 21 2008 11:38 Raithed wrote:
I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..

x^2 + 4x + 4 would be (x+2)(x+2) ...

Can you explain further, I really don't see it.

x^2 + 4x + 4 = (x+2)(x+2)
x^3 + x^2 + 4x + 4 = x^3+(x+2)(x+2)

BC + C = C(B+1)
BA + BC + C = BA + C(B+1)
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 02:46 GMT
#9
On April 21 2008 11:40 Muirhead wrote:
In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.

Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB

Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1?
NathanSC
Profile Blog Joined February 2008
United States620 Posts
April 21 2008 02:53 GMT
#10
On April 21 2008 11:46 Raithed wrote:
Show nested quote +
On April 21 2008 11:40 Muirhead wrote:
In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.

Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB

Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1?

Lol, you kinda missed the point.

He was using an example to demonstrate the factorization of BC + C = C(B + 1).

He pretended that the 1 in the above example was an A, which would have been written as:

BC + CA = C(B + A).

Exact same problem, just with another variable.
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 02:53 GMT
#11
On April 21 2008 11:43 Raithed wrote:
Show nested quote +
On April 21 2008 11:38 fight_or_flight wrote:
1) Yes

2) See above post, just factoring.

3) So this is an AND gate with inversions on all inputs and outputs?

Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'

Then you have a NOR with an inverted output which is the same as OR.

For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:

A' B' = AB? It doesn't work that way?

Does not work that way. Look at the truth tables:

out = A'B'

A B out
0 0 1
0 1 0
1 0 0
1 1 0

----------------------------

out = AB

A B out
0 0 0
0 1 0
1 0 0
1 1 1

------------------------------

Therefore you conclude that A'B' = NOR. (look up demorgan's theorem)
Do you really want chat rooms?
NathanSC
Profile Blog Joined February 2008
United States620 Posts
Last Edited: 2008-04-21 02:57:55
April 21 2008 02:56 GMT
#12
On April 21 2008 11:53 fight_or_flight wrote:
Show nested quote +
On April 21 2008 11:43 Raithed wrote:
On April 21 2008 11:38 fight_or_flight wrote:
1) Yes

2) See above post, just factoring.

3) So this is an AND gate with inversions on all inputs and outputs?

Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'

Then you have a NOR with an inverted output which is the same as OR.

For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:

A' B' = AB? It doesn't work that way?

Does not work that way. Look at the truth tables:

out = A'B'

A B out
0 0 1
0 1 0
1 0 0
1 1 0

----------------------------

out = AB

A B out
0 0 0
0 1 0
1 0 0
1 1 1

------------------------------

Therefore you conclude that A'B' = NOR. (look up demorgan's theorem)

Yeah, this is exactly accurate. Not A AND Not B evaluates as true in very different circumstances than A AND B evaluates as true.
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 03:01 GMT
#13
Without looking at the truth table, I thought not from input, with a not in output would cancel, this just confirms that it's NOR. I didn't mean exactly as (AB)' == AB, I just meant that the ' cancels.
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 03:06 GMT
#14
No, they don't cancel. If you have 2 bubbles on the input of an AND, you can make it into an OR with a bubble on the output.

Same thing if the inputs on an OR have bubbles...it becomes an AND with a bubble on the output.
Do you really want chat rooms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 03:08 GMT
#15
So basically, the bubbles DON'T cancel, whatsoever?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 03:12 GMT
#16
But wait, the gate is a NAND gate...
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 03:12 GMT
#17
Except with inverters.
Do you really want chat rooms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 03:14 GMT
#18
So if it's

Bubbled input NAND --- inverter --- bubbles cancel? Only this way?

[image loading]
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 03:15 GMT
#19
Which is basically

NAND + inverter = comes out as AB? Instead of (AB)' ?
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 03:33 GMT
#20
[image loading]
Do you really want chat rooms?
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