# Homework Help: A little electronics question about logical gates

1. Dec 31, 2011

### Femme_physics

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2. Dec 31, 2011

### ehild

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3. Dec 31, 2011

### Femme_physics

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4. Dec 31, 2011

### ehild

Re: A little electronics question about logical gates....

Well, OR adds the inputs. But the symbol for an OR gate is not what you drew. It is spiky. Do you mean an OR or an AND gate?
ehild

Last edited: Dec 31, 2011
5. Dec 31, 2011

### Femme_physics

Re: A little electronics question about logical gates....

Oh, I mean an AND gate, then. sorry.

6. Dec 31, 2011

### ehild

Re: A little electronics question about logical gates....

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

ehild

7. Dec 31, 2011

### Femme_physics

Last edited by a moderator: May 5, 2017
8. Dec 31, 2011

### ehild

Re: A little electronics question about logical gates....

ehild

9. Dec 31, 2011

### I like Serena

Re: A little electronics question about logical gates....

Hi Fp!

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

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

Last edited: Dec 31, 2011
10. Dec 31, 2011

### I like Serena

Re: A little electronics question about logical gates....

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

11. Dec 31, 2011

### I like Serena

Re: A little electronics question about logical gates....

Just found it myself too ;) Fixed it.

12. Jan 1, 2012

### Femme_physics

Re: A little electronics question about logical gates....

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?

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13. Jan 1, 2012

### Curious3141

Re: A little electronics question about logical gates....

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. Jan 1, 2012

### Curious3141

Re: A little electronics question about logical gates....

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. Jan 1, 2012

### Femme_physics

Re: A little electronics question about logical gates....

Well, we were told to always do truth tables since we're just beginners.

X^2

2X + X^2 +1

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

Did not know that! So,

http://img225.imageshack.us/img225/5489/fscan.jpg [Broken]

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16. Jan 1, 2012

### I like Serena

Re: A little electronics question about logical gates....

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

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

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

Yes!

Can you see which columns are identical to F?

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17. Jan 1, 2012

### Femme_physics

Re: A little electronics question about logical gates....

Multiplication.

Also multiplication!

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. Jan 1, 2012

### Curious3141

Re: A little electronics question about logical gates....

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?

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. Jan 2, 2012

### Femme_physics

Re: A little electronics question about logical gates....

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 :)

20. Jan 2, 2012

### Femme_physics

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