Hello my friends i have 3 boolean expressions and i wonder if there is a way to simplify them:

1. F1= C'B + CA' + CB' + BF

2. F2= ACB' + C'B + C'A'D

3. F3 = CB' + CA' + B'A'E + C'BA

after the simplification i will create the circuit but i dont know if i can first simplify the expressions above.

thanks

 

There are quite a way of doing this. For example, have you tried to draw the gates version of this expressions?

 

Not yet. I thought it needs to be simplify first and then to draw the gates.

 

It is a worthy goal, I do it both ways, generally because I am trying to use specific gates. Pick a problem, work on it and present where you get stuck. We can help you at these points, you will find it is very algebraic in nature.

 

For signal F1 i implement the circuit :

(F signal is asserted low)

and then i use inverted gates for the above circuit:

is this correct? and something like this you mean by drawing the gates?

 

That's what I meant. I'm going to bed to go to work tonight, but I will be back.

There is another convention, a simple circle in front of the gate input indicates inversion.

From my library here at AAC

 

The simplification of those expressions are literally not possible unless you do what our MOD just said but the order of the evalutaion of the gates can be changed in all the expression s

 

Thanks about the reply Jeffrey Samuel , but i dont understand what do you mean by the "evaluation of the gates".

 

Hello my friends i have 3 boolean expressions and i wonder if there is a way to simplify them:

1. F1= C'B + CA' + CB' + BF

2. F2= ACB' + C'B + C'A'D

3. F3 = CB' + CA' + B'A'E + C'BA

after the simplification i will create the circuit but i dont know if i can first simplify the expressions above.

thanks

It depends on how you define "simple". Fewest terms? Fewest gates and, if so, which gates are fair game? Fewest propogation delays? Fewest transistors and, if so, in what technology?

Depending on your metric, F1 is already about as "simple" as it is going to get.