Need help badlyabout full adder circuit

[marlboro73]

hey guys.. can u pls help me solve this problem. i have some answers but I am just not really sure if they're right. pls help me. my exam starts in a few hours

here's the problem:


The truth table of a full adder circuit is given below, where A and B are two inputs and Cin is the carry from the previous bits.

http://img513.imageshack.us/img513/8765/truthtableeu5.th.jpg http://g.imageshack.us/thpix.php

i) Get the output expression for sum and carry from the truth table.

ii) use Boolean Algebra to simplify Sum only to an XOR gate

iii) Derive the expression for the carry and simplify it using Boolean algebra.
iv) Draw the full circuit showing the inputs, sum and the carry.

e) Using Boolean algebra and De- Morgan’s or otherwise prove the property
f) State and prove De- Morgan’s law for three input variables

 

[Ouabache]

Welcome to PF forums! You will find there are many fascinating useful discussions here on all kinds of topics (most of them science related).

That's an interesting question. It is good, that you have worked out some soln's to your problem. Before we can help you, you need to post your thoughts, a sequence of how you attempted a solution. Then ask questions regarding your attempt.. You probably want to reread this https://www.physicsforums.com/showthread.php?t=94388"to help guide you on posts. It is also listed at the top of this homework area under "why hasn't anybody answered my question".

 

[piercas]

I am happy to assist you with your problem. It seems like you are working on a full adder circuit and need help with some of the steps. Let's go through each part of the problem and see if we can find the correct answers together.

i) To get the output expression for sum and carry, we can look at the truth table and see what the output is for each combination of inputs. For the sum, we can see that it is equal to A XOR B XOR Cin. For the carry, it is equal to A AND B OR (A XOR B) AND Cin.

ii) To simplify the expression for sum to an XOR gate, we can use Boolean algebra. We can rewrite the expression as A XOR B XOR Cin, and then use the properties of XOR gates to simplify it to just A XOR B.

iii) To derive the expression for the carry, we can use the same method as in part i. We can see that the expression is equal to A AND B OR (A XOR B) AND Cin. We can then simplify this using Boolean algebra to just A AND B OR (A XOR B).

iv) To draw the full circuit, we can use the simplified expressions from parts ii and iii. The inputs will be A, B, and Cin. The sum output will be A XOR B, and the carry output will be A AND B OR (A XOR B).

e) To prove the property using Boolean algebra and De Morgan's law, we can rewrite the expression as (A AND B) OR (A XOR B) using the properties from part iii. We can then use De Morgan's law to rewrite this as (A OR (A XOR B)) AND (B OR (A XOR B)). This can be further simplified to (A OR B) AND (A XOR B). Finally, using the properties of XOR gates, we can simplify this to just A OR B, which is the desired result.

f) De Morgan's law for three input variables states that the complement of the OR function of three variables is equal to the AND function of the complement of each variable. Mathematically, it can be written as:

(A OR B OR C)' = A' AND B' AND C'

To prove this, we can use a similar approach as in part e. We can rewrite the expression as (A OR B OR C)' = (A' AND B') AND