A little electronics question about logical gates

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  • #5
Femme_physics
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Oh, I mean an AND gate, then. sorry.
 
  • #6
ehild
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In that case, erase that "OR".
And then the output is correct, but can be simplified.

ehild
 
  • #8
ehild
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:frown: Let's start with the basic things. How are the logical addition, multiplication and negation defined?

ehild
 
  • #9
I like Serena
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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
I like Serena
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Good point, edited.
I don't like how Latex shows it though.
 
  • #11
I like Serena
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Just found it myself too ;) Fixed it.
 
  • #12
Femme_physics
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: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 [Broken]

And this is how I got F

http://img402.imageshack.us/img402/7498/chart1ar.jpg [Broken]
I presume 1+1 reboots the thing and makes it 0?
 
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  • #13
Curious3141
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Basically I want to see if I got it right?

http://img853.imageshack.us/img853/7681/laster.jpg [Broken]

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
Curious3141
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Well, it depends on the scenario. What bridge are we talking about? What function do we have?



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

And this is how I got F

http://img402.imageshack.us/img402/7498/chart1ar.jpg [Broken]
I presume 1+1 reboots the thing and makes it 0?

Some entries are wrong. For instance 1 + 1 is not 0, but 1. Remember that '+' here signifies the OR function.
 
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  • #15
Femme_physics
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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 [Broken]
 
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  • #16
I like Serena
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  • #17
Femme_physics
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"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 [Broken]
 
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  • #18
Curious3141
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OK, so the above Truth Table is now correct. Do you see that the output is exactly [itex]c[/itex]? The output is 0 when [itex]c=0[/itex] and 1 when [itex]c=1[/itex]. So the simplified final answer is just [itex]c[/itex].

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

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

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

[itex](1.x) = (1 AND x) = (x.1) = (x AND 1) = x[/itex]. 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.

[tex](\bar{ab}+c).c = \bar{ab}c + c.c = \bar{ab}c + c = (\bar{ab}+1).c = 1.c = c[/tex]

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
Femme_physics
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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
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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
Curious3141
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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 [Broken]

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

I get the final answer as [itex]\overline{c}(\overline{a} + \overline{b})[/itex] 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 [itex]ab[/itex] as one of the inputs to the NOR gate P rather than [itex]\overline{\overline{ab}}[/itex], 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:

[tex]\overline{\overline{\overline{ab}}+(\overline{ab}c+cc)}[/tex]

which is *not* the same as what you wrote. Remember that, in general, [itex]\overline{a + b} \neq \overline{a} + \overline{b}[/itex]. 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
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I get the final answer as [itex]\overline{c}(\overline{a} + \overline{b})[/itex]

You got this as equivalent for Fout
But i'm getting [itex]\overline{abc}[/itex]
 
  • #25
ehild
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You got this as equivalent for Fout
But i'm getting [itex]\overline{abc}[/itex]

It is wrong, remember de Morgan's Laws. Curious' result is right.


ehild
 
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