# A little electronics question about logical gates

Gold Member

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?

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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!

Gold Member

Well, fine, I redid 6 and also got "1". However, now everything in my Karnaugh Map equals 1 which means the result is 0 because they all cancel each other out!

http://img831.imageshack.us/img831/9793/abceh.jpg [Broken]

Best wishes for 2012, Klass

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

Gold Member

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

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

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

Errr... no?
Why would you think so?

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

No you cant 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: $\overline{ab} = \overline{a} + \overline{b}$

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Looks like you got the hang of boolean algebra!

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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Gold Member

Looks like you got the hang of boolean algebra!

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?

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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.

Gold Member

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?

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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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How did you get the forth line??
check it again !!

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Could you add parentheses in the 3rd line?