MHB Logic Circuits and Truth Tables

[gEOdude]

Hey guys, I'm in a pickle with this question. Its really not my cup of tea, its a struggle factor really but I don't really have a choice whether I like it or not. A worked example such as this allows me to see what it is I need to do. If anyone can provide a helping hand it would assist me tons.View attachment 4301The questions is as follows: For each circuit write the equivalent logic expression, truth table showing the final output (O) for inputs (A,B and C) and find if these expressions give the same output (comparing the final output columns)?

 

[Sudharaka]

Hey guys, I'm in a pickle with this question. Its really not my cup of tea, its a struggle factor really but I don't really have a choice whether I like it or not. A worked example such as this allows me to see what it is I need to do. If anyone can provide a helping hand it would assist me tons.The questions is as follows: For each circuit write the equivalent logic expression, truth table showing the final output (O) for inputs (A,B and C) and find if these expressions give the same output (comparing the final output columns)?

Hi, :)

Welcome to MHB!

So first you have to find a logic expression for each circuit. For this you need to know what each component (gate) of the circuit outputs when given an input. I suggest trying to write the expression for the second circuit first since it might be a bit easier than the first.

In the second circuit you have three gates. A "NOT" gate, a "OR" gate and an "AND" gate. All you got to know is that,

1) "NOT" gates inverts the input. An input of '1' for a NOT gate would result in a '0' and an input of '0' would result in a '1'. So if the input is say $x$ we denote the output by $x'$ where $x'$ denote the opposite of $x$.

2) "OR" gates take two inputs and add then together. Basically this means if the two inputs are $x$ and $y$ the output will be $x+y$.

3) "AND" gates take two inputs and multiplies then together. Hence if the two inputs are $x$ and $y$ the output will be $xy$.

Using these you'll be able to write down the expression for the second circuit. We can do this step by step. What is inserted to gate number 1 (which is a NOT gate) is the input "B". Then what will be it's output? :)

 

[gEOdude]

Hi, :)

Welcome to MHB!

So first you have to find a logic expression for each circuit. For this you need to know what each component (gate) of the circuit outputs when given an input. I suggest trying to write the expression for the second circuit first since it might be a bit easier than the first.

In the second circuit you have three gates. A "NOT" gate, a "OR" gate and an "AND" gate. All you got to know is that,

1) "NOT" gates inverts the input. An input of '1' for a NOT gate would result in a '0' and an input of '0' would result in a '1'. So if the input is say $x$ we denote the output by $x'$ where $x'$ denote the opposite of $x$.

2) "OR" gates take two inputs and add then together. Basically this means if the two inputs are $x$ and $y$ the output will be $x+y$.

3) "AND" gates take two inputs and multiplies then together. Hence if the two inputs are $x$ and $y$ the output will be $xy$.

Using these you'll be able to write down the expression for the second circuit. We can do this step by step. What is inserted to gate number 1 (which is a NOT gate) is the input "B". Then what will be it's output? :)

Is this right?: (B' + C) * A Circuit 2
(A* (A' * C') * B') + (A * C * B) Circuit 1

Truth table is a tricky one for me but I attempted them are they right is this what I am supposed to do.

View attachment 4307 (CIRCUIT 2)
View attachment 4308 (CIRCUIT 1)

 

[Sudharaka]

Is this right?: (B' + C) * A Circuit 2
(A* (A' * C') * B') + (A * C * B) Circuit 1

Truth table is a tricky one for me but I attempted them are they right is this what I am supposed to do.

(CIRCUIT 2)
(CIRCUIT 1)

Great work. You have done everything almost correctly. :)

The only mistake I see is in the expression for circuit 1. Note that gate 4 in that circuit is a little different from the ones I mentioned. It's called a NAND gate. A "NAND" gate is basically a AND gate followed by a NOT. That is if the inputs to a NAND gate is say $x$ and $y$ it will output, $(xy)'$. That is it multiplies the inputs and then take the compliment.

So there's a little thing that should be corrected in your expression for circuit 1. And accordingly the truth table will also have to be changed. :)

 

[gEOdude]

Great work. You have done everything almost correctly. :)

The only mistake I see is in the expression for circuit 1. Note that gate 4 in that circuit is a little different from the ones I mentioned. It's called a NAND gate. A "NAND" gate is basically a AND gate followed by a NOT. That is if the inputs to a NAND gate is say $x$ and $y$ it will output, $(xy)'$. That is it multiplies the inputs and then take the compliment.

So there's a little thing that should be corrected in your expression for circuit 1. And accordingly the truth table will also have to be changed. :)

Is this right: View attachment 4312 AND how do I compare do I just say that they are the same.

View attachment 4313

 

[Sudharaka]

Is this right: AND how do I compare do I just say that they are the same.

Please do not duplicate posts, just one reply would be more than enough as I get a notification for each of the threads I have subscribed. :)

What you have written in post #3 is almost correct. All you got to do is to take the compliment after you take the product for gate #4. That is the expression should be,

\[(A* (A' * C')' * B') + (A * C * B)\]

Note that since the inputs of gate 4 is $A'$ and $C'$ we have $(A'C')'$ as the output for that gate. Can you write the truth table for this?

 

[gEOdude]

Please do not duplicate posts, just one reply would be more than enough as I get a notification for each of the threads I have subscribed. :)

What you have written in post #3 is almost correct. All you got to do is to take the compliment after you take the product for gate #4. That is the expression should be,

\[(A* (A' * C')' * B') + (A * C * B)\]

Note that since the inputs of gate 4 is $A'$ and $C'$ we have $(A'C')'$ as the output for that gate. Can you write the truth table for this?

This is what I got, is this right: View attachment 4317

 

[Sudharaka]

This is what I got, is this right:

Almost. Check the fourth row again. Everything else is correct. :)

 

[gEOdude]

I suppose the two circuits are supposed to be equal. Am I right? Is the equation that I have correct or do I have to change it. Is my table right now.

 

[Sudharaka]

I suppose the two circuits are supposed to be equal. Am I right? Is the equation that I have correct or do I have to change it. Is my table right now.

Great. Your table is correct now. Using Boolean logic it can be shown that the two circuits are in fact equivalent. Let us try to simplify the expression for the 1st circuit.

\[(A * (A' * C')' * B') + (A * C * B)\]

There's a law called the "De Morgan's law" which basically says,

\[(A * B)'=A'+B'\]

Using this we have,

\begin{eqnarray}
(A * (A' * C')' * B') + (A * C * B)&=&(A * (A+C) * B') + (A * C * B)\\
&=&(A * A * B')+(A * C * B')+(A * C * B)
\end{eqnarray}

Since \(A * A = A\) we have,

\[(A * (A' * C')' * B') + (A * C * B)= (A * B') + A * C * (B' + B)\]

Since \(B + B' = 1\) we have,

\[(A * (A' * C')' * B') + (A * C * B)= (A * B') + (A * C) = A * (B' + C)\]

So the two circuits are indeed equivalent. Even using the truth tables you can conclude that the two circuits are equivalent. :)