A little electronics question about logical gates

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The discussion centers on understanding logical gates, specifically OR and AND gates, and their simplification using Boolean algebra. Participants clarify the definitions of logical addition, multiplication, and negation, emphasizing the importance of truth tables for beginners. Errors in truth tables are identified, particularly regarding the outputs of AND and OR operations, with corrections provided through examples. The conversation also touches on De Morgan's Laws and the simplification of expressions, leading to a final output that is dependent on the inputs. The thread concludes with a collaborative effort to ensure accurate understanding and application of Boolean algebra principles.
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Oh, I mean an AND gate, then. sorry.
 


In that case, erase that "OR".
And then the output is correct, but can be simplified.

ehild
 
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:frown: Let's start with the basic things. How are the logical addition, multiplication and negation defined?

ehild
 


Hi Fp! :smile:

Perhaps you would like to try to fill out the following table?

$$
\begin{array}{|c|c|c|c|c|c|c|c|}
a & b & c & ab & \overline{ab} & \overline{ab} c & cc & \overline{ab} c + cc \\
\hline \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & & & \\
0 & 1 & 0 & 0 & 1 & & & \\
0 & 1 & 1 & 0 & & & & \\
1 & 0 & 0 & 0 & & & & \\
1 & 0 & 1 & 0 & & & & \\
1 & 1 & 0 & 1 & & & & \\
1 & 1 & 1 & & & & & \\
\hline \end{array}
$$
 
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  • #10


Good point, edited.
I don't like how Latex shows it though.
 
  • #11


Just found it myself too ;) Fixed it.
 
  • #12


ehild said:
:frown: Let's start with the basic things. How are the logical addition, multiplication and negation defined?

ehild

Well, it depends on the scenario. What bridge are we talking about? What function do we have?

Perhaps you would like to try to fill out the following table?

I did my own, similar one, a while ago... except I didn't include AC(capped)
http://img684.imageshack.us/img684/4424/chart3u.jpg

And this is how I got F

http://img402.imageshack.us/img402/7498/chart1ar.jpg
I presume 1+1 reboots the thing and makes it 0?
 
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  • #13


Femme_physics said:
Basically I want to see if I got it right?

http://img853.imageshack.us/img853/7681/laster.jpg

A truth table is the most basic way to do it, but it is tedious. This is very easy to simplify using the basic laws of Boolean algebra (BA).

Let's review some basic rules of BA. Let x be a binary variable.

How would you simplify xx?

What is (x + 1)? (1 + x)?

What is x.1? 1.x?
 
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  • #14
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  • #15


A truth table is the most basic way to do it, but it is tedious. This is very easy to simplify using the basic laws of Boolean algebra (BA).
Well, we were told to always do truth tables since we're just beginners.

Let's review some basic rules of BA. Let x be a binary variable.

How would you simplify xx?

X^2

What is (x + 1)? (1 + x)?

2X + X^2 +1

What is x.1? 1.x?

That would be 1 times x? that's X.

Some entries are wrong. For instance 1 + 1 is not 0, but 1. Remember that '+' here signifies the OR function.
Did not know that! So,

http://img225.imageshack.us/img225/5489/fscan.jpg
 
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  • #16


Femme_physics said:
X^2

"xx" represents "x AND x".
Do you know what that is?


Femme_physics said:
2X + X^2 +1

"x+1" represents "x OR 1", which is?


Femme_physics said:
That would be 1 times x? that's X.

"1x" represents "1 AND x".
But yes, that would be the same as x.


Femme_physics said:
Did not know that! So,

http://img225.imageshack.us/img225/5489/fscan.jpg

Yes! :smile:

Can you see which columns are identical to F?
 
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  • #17


"xx" represents "x AND x".
Do you know what that is?

Multiplication.

"x+1" represents "x OR 1", which is?

Addition.

"1x" represents "1 AND x".
But yes, that would be the same as x.

Also multiplication!

Can you see which columns are identical to F?
Yes, CC. But, we only know how to simplify using Karno Map (translating the name from Hebrew).

So,

http://img576.imageshack.us/img576/92/image201201010014.jpg
 
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  • #18


Femme_physics said:
http://img225.imageshack.us/img225/5489/fscan.jpg

OK, so the above Truth Table is now correct. Do you see that the output is exactly c? The output is 0 when c=0 and 1 when c=1. So the simplified final answer is just c.

Here's how to see that by doing the Boolean Algebra. I'm going to answer my own post here:

xx (can also be written x.x) = (x AND x) = x. This is because when x = 0, 0 AND 0 = 0, and when x = 1, 1 AND 1 = 1.

(x+1) = (x OR 1) = (1+x) = (1 OR x) = 1. A '1' OR anything (or vice versa) is still '1'. This line also demonstrates commutativity of addition (the OR function).

(1.x) = (1 AND x) = (x.1) = (x AND 1) = x. A '1' AND anything (or vice versa) depends wholly on x, as you can see from the truth table for AND. This line also demonstrates commutativity of multiplication (the AND function).

OK, so back to the problem. The first line uses the distributive law over multiplication (AND), which you already seem to know. It is used again in regrouping terms.

(\bar{ab}+c).c = \bar{ab}c + c.c = \bar{ab}c + c = (\bar{ab}+1).c = 1.c = c

See how easy that was? :wink:

As a final note, please don't get confused between binary number addition and the Boolean OR, and binary number multplication and the Boolean AND. As an exercise you may want to work out the other simple rules in Boolean Algebra, (the ones for 0.x, 0 + x, and maybe inspect the slight complexities of the XOR function, and culminating in doing your own proof of De Morgan's Laws).
 
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  • #19


Thanks Curious! We actually only studied Boolean Algebra yesterday! I know how to tackle it all now :) if I'll run into any problems I'll post back. Thanks everyone :)
 
  • #21


This might help:

OR is like a parallel connection in circuit (where 1 means key closed and 0 means key open)... Thus for x+1 whatever be x, current always flow and thus its 1

and for AND ... its like series, thus 1ANDx means only x because even if 1 is always closed, current flow depends on x
 
  • #23


Femme_physics said:
This is the original problem I was trying to minimize its function...did I get it right?

http://img221.imageshack.us/img221/4978/solutionmpa.jpg

Your final answer is \overline{c}? Afraid that's not right.

I get the final answer as \overline{c}(\overline{a} + \overline{b}) with no further reduction possible, since the output is equally dependent on the three inputs. The reduction involved one application of De Morgan's Law.

In the output of the inverter (NOT gate) Z, the double-negation of (ab) simply gives ab. So just write ab as one of the inputs to the NOR gate P rather than \overline{\overline{ab}}, which is unnecessarily cumbersome.

Also, check the way you wrote down the output of the NOR gate P. The unsimplified output (I'm leaving your double-negation notation intact here) will look like:

\overline{\overline{\overline{ab}}+(\overline{ab}c+cc)}

which is *not* the same as what you wrote. Remember that, in general, \overline{a + b} \neq \overline{a} + \overline{b}. Negation does not distribute that way over OR or AND. In fact, this is why you need De Morgan's Law to "break up" this sort of "whole negation".

Please *carefully* recheck your truth table as well. It's wrong in quite a few rows.
 
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  • #24


Curious3141 said:
I get the final answer as \overline{c}(\overline{a} + \overline{b})

You got this as equivalent for Fout
But I'm getting \overline{abc}
 
  • #25


cupid.callin said:
You got this as equivalent for Fout
But I'm getting \overline{abc}

It is wrong, remember de Morgan's Laws. Curious' result is right. ehild
 
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  • #26


Check 6) line

I still don't get what is wrong about my truthtable.

The way I understand,

0 + 0 + 0 = 0
0 + 0 + 1 = 1
0 + 1 + 1 = 1
1 + 1 + 1 = 1

Where is my mistake exactly?
 
  • #27


Femme_physics said:
I still don't get what is wrong about my truthtable.

The way I understand,

0 + 0 + 0 = 0
0 + 0 + 1 = 1
0 + 1 + 1 = 1
1 + 1 + 1 = 1

Where is my mistake exactly?

What you write here are correct equations.

If we zoom in on your line 6, you have for instance:

a=1, b=1, c=0

In your table you should have:
$$\overline {\overline {ab}c} = \overline {\overline {11}0} = \overline {00} = 1$$
but this is not the result you have in your table.

Anyway, the actual complete expression should be:

$$\overline{\overline{\overline{ab}}+(\overline{ab}c + cc)}
= \overline{\overline{\overline{11}}+(\overline{11}0 + 00)}
= \overline{1+(00+0)}
= \overline{1}
= 0$$


Btw, best wishes for 2012!
 
  • #28
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  • #29


How did you get ##\overline { \overline{ab}c } = \overline{ab} \cdot \overline{abc}##?
 
  • #30


I like Serena said:
How did you get ##\overline { \overline{ab}c } = \overline{ab} \cdot \overline{abc}##?

Isn't that the definition of ##\overline { \overline{ab}c }## ?
 
  • #31


Femme_physics said:
Isn't that the definition of ##\overline { \overline{ab}c }## ?

Errr... no?
Why would you think so?
 
  • #32


Femme_physics said:
Isn't that the definition of ##\overline { \overline{ab}c }## ?

No you can't write it like that but you can use de morgan law to solve that:

\overline{\overline{ab}c}

\overline{\overline{ab}} + \overline{c}

ab + \overline{c}

But using: \overline{ab} = \overline{a} + \overline{b}
 
  • #33
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  • #34


Looks like you got the hang of boolean algebra! :approve:

However, you dropped something when you had ##\overline{cc}## and went to ##cc##.EDIT: I missed the abc part which is not right. Fixed it in post #38.
 
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  • #35


I like Serena said:
Looks like you got the hang of boolean algebra! :approve:

However, you dropped something when you had ##\overline{cc}## and went to ##cc##.

True, but at the end of the day the result is still 0 cause we got a and a(capped) in the same multiplication line so everything gets nullified, right?
 
  • #36


Femme_physics said:
True, but at the end of the day the result is still 0 cause we got a and a(capped) in the same multiplication line so everything gets nullified, right?

Actually, you don't have ##\overline{a}##, but you have ##\overline{ab}##.
But yes, in combination with ##ab##, everything gets nullified.
 
  • #37


I like Serena said:
Actually, you don't have ##\overline{a}##, but you have ##\overline{ab}##.
But yes, in combination with ##ab##, everything gets nullified.

Ah, gotcha! Now I am asked to write the logical value of the function Out, when the values of all the ports
C = '0'
A = B = '1'

But still, everything gets nullified right?
 
  • #38


Hold on, I missed something.

It does not get nullified.

You got ##\overline{\overline{ab}c}##, which you took to ##abc##, but that is not right.
That changes everything.
 
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  • #41


Could you add parentheses in the 3rd line?
 

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