# Need Boolean Equation for Simple Circuit

Discussion in 'Homework Help' started by AARRGGHHH, Feb 20, 2013.

1. ### AARRGGHHH Thread Starter New Member

Feb 20, 2013
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Okay, this is from a test, I got it wrong. My answer was (AB + CD). I thought I was right. Can someone explain where I went wrong?

Thank you

Last edited: Feb 20, 2013

Apr 5, 2008
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Hello,

Have a look at this drawing with some hints made by Bill_Marsden:

Bertus

3. ### AARRGGHHH Thread Starter New Member

Feb 20, 2013
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Sorry bertus, your drawing just confused me further, but thanks for trying.

Given the resolution to another question I posted, I obviously missed the small circles at the ends of the gates. Apparently these invert the values.

The image is called ((AB)'+(CD)'), something we couldn't see when taking the test. One thing I'm still confused about: Since there are three small inversion circles, shouldn't there be a third ' sign in the equation?

In other words, instead of ((AB)'+(CD)'), would the answer be ((AB)'+(CD)')' ?

Many thanks

4. ### tshuck Well-Known Member

Oct 18, 2012
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Yes, that would be the case.

Sometimes, it is good to draw the function implemented at each point of the diagram...

• ###### func.png
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5. ### AARRGGHHH Thread Starter New Member

Feb 20, 2013
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Visuals are always helpful, thank you.

6. ### tshuck Well-Known Member

Oct 18, 2012
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Well, I think it is important to understand what bertus was saying...

An inverted input to a NOR gate has the same functionality of an AND gate.

So, you could do this:

Which, looking at it from a Boolean viewpoint is:

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7. ### t06afre AAC Fanatic!

May 11, 2009
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As a little aside. In the early days of the TTL logic. The 7400 gate costed less than others gates. So I am told is was cost effective(to some limit) to at least try to just use as many NAND gates as possible. But still to day. It is a quite common question on test and exams. To ask for a specific boolean algebra function. To be made just with NAND gates

8. ### tshuck Well-Known Member

Oct 18, 2012
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This is because it is easier to create a NAND gate than most others...
It is still a relevant question.

9. ### WBahn Moderator

Mar 31, 2012
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Yes - the name of the file doesn't match the diagram contained therein.

The answer should just be ABCD -- this is a four input AND gate.

To some of the other comments:

Having students implement logic in just NAND gates is not only a good academic exercise on a number of levels, but the NAND gate is the preferred gate for CMOS IC logic design as it takes up the smallest room and has the smallest prop delay of any logic gate with comparable number of inputs. This includes when compared to the NOR due to differences in majority carrier mobilities in NFETs and PFETs.

May 28, 2009
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The dot in Boolean means to AND, not to multiply as in normal algebra. Example: A . B = Y means input A AND input B equals output Y.

A macron (overscore) will NOT (or invert) the output denoted by the bubble at the gate output. Example: A NOT . B NOT = Y (or AB (with macron) = Y) This turns the AND gate into a NOT AND gate or NAND gate.

The output of each NAND is fed into the inputs of the NOR gate.

The plus (+) symbol in Boolean terms means OR, not to add. Example: A + B = Y means input A OR input B equals Y (the output). A circled plus symbol will denote an exclusive (OR or NOR).

A macron (overscore) will NOT (or invert) the output as denoted by the bubble at the output of the gate, making it a NOR gate. Since the gate is a NOR, you must include a macron above the inputs. AB NOT + CD NOT equals Y.

In my opinion, your instructor marked your answer wrong because you forgot to include the macrons over AB and CD in your answer for the input to the NOR gate. AB NOT + CD NOT = Y

11. ### tshuck Well-Known Member

Oct 18, 2012
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Actually, this is multiplication:
1 * 1 = 1
and
x * 0 = 0.

This is addition, however, it isn't decimal addition. This is binary addition, where only 1 or 0 is representable, so, 1+1 equals the highest possible output, 1.

..the instructor marked the answer wrong because the answer is wrong. Even if the macrons were present on (AB) and (CD), because the whole this is NORed, not ORed...

Your NOT representation is a bit sloppy, and the reason parenthesis are a must with describing Boolean operations.
is this (AB) NOT or is it A(B) NOT ?

Either way, the use of letters in a Boolean expression is asking for trouble. The most common method is using the apostrophe to denote negation, so the above should read:
(AB)' + (CD)'
...actually should be ((AB)' + (CD)')', but I was making another point...

What would happen if you tried to write (A)'B ?
A NOT B?

May 28, 2009
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I understand the attention to detail. Notwithstanding how I tried to explain the problem, the point is the student forgot to include the macron in the answer.

13. ### tshuck Well-Known Member

Oct 18, 2012
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Sorry, if that came across a bit harsh (that wasn't the intention), I was trying to clarify some points I felt could trip up a newcomer....

14. ### WBahn Moderator

Mar 31, 2012
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But, as you say, it is a NOR gate in the schematic, so the answer would be (using your notation):

(AB NOT + CD NOT) NOT = Y

which reduces to simply

ABCD = Y

May 28, 2009
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Agreed. That would be the logical way to look at it.

16. ### t06afre AAC Fanatic!

May 11, 2009
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I remember from school. I did a lot of boolean simplifying pre work. by using Shannons rule(had to look that up), a pencil and eraser. Every time you split a an inverterline. You change a dot to a plus sign, and vica verca.

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17. ### WBahn Moderator

Mar 31, 2012
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I'm going to have to disagree on this one. Binary addition in an N-bit world is normal addition reduced modulo 2^N. So 1-bit addition is a mod-2 world making 1+1=0. Binary addition is an XOR gate (and, it turns out, is indistinguishable from subtraction, which is a very useful property in cryptography and coding theory).

18. ### tshuck Well-Known Member

Oct 18, 2012
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True. I should have specified that I meant a binary magnitude addition(the signal can only take two magnitudes). Adding 5V and 5V can't exceed the 5V supply, so the maximum it can output is 5V. 10V extends past the supply, and 0V wouldn't make sense. A carry in a magnitude simply doesn't exist, so 5V must be the output.