Drawing Logic Gates From Boolean Expressions | Important Questions 4 | Digital Electronics

So here we are with another question about designing

logic

gates

to get a

logic

output, a

boolean

expression output, so here there are three signal values ​​present a b c but the number of terms that are present in this

boolean

expression is 3, so We have increased the number of terms and there are three signal values ​​until now the

questions

we have discussed was a simple expression involving two, three or four signal values, but it was a simple expression involving two terms or one term, but here we have three signs. values ​​a b and c but there are also three terms okay ac b bar c abc bar so this is something new so let's design it so first we will assign the different signal values ​​to b c then looking at the expression we will try to calculate determine which logic

gates

are will require, so here is an operation between a and c, so gates will be required and, the plus sign indicates or gates will be required, the inversion in b and c, bar c, bar and gates, will not be gate or inverter. required, so the first term we will point to each term separately, we will try to get each term separately and then we will combine them well, first a and c and the operation, so we will give a and c a two inputs and gate to get a c, then it is b bar c so for b to pass through the inverter or not the gate to get bar b then finish with c to get bar b c now this bar abc now here uh the question is it depends on what is given in the question now we have logic gates that can have three inputs, multi input logic gates are available, you can have two inputs, three inputs, four inputs, five inputs, whatever, if it has been given in the question that you can use one gate and three inputs or one gate or three inputs or three input gate, then you can do this directly just in a, you give a, you give b and you give c with an inverter to one gate and three inputs and you will directly get the expression a b c bar, but nothing has been mentioned here. like this, so we can't use three entrances and one door, so we have to do it in steps, so how will we do it?

Let's see, first we'll have to get a b, okay, first we'll have to get a b, okay, for that you take one here. It might get a little confusing here, so follow me carefully. We'll take a first okay, so this is a, we'll give it to two inputs and door a and then we'll get b, so I'm taking b from here a and b, so this. gave us a b so we got a b then c for bar c we will have to take c here from here we will pass it through an inverter we have bar c now we have to do it and these two bars a b and c we give them to our two inputs and gate to get a bar b c, okay, look to get a bar b c, we can't use three entrances and one door here because nothing like that has been mentioned, we can only use two entrance doors, so first we have to divide it into two parts . we get a b, so we take a b, we gave it to two inputs and a gate, we got a b, then we take the signal c, we pass it through an inverter or a gate no to get the c bar, then we add a bar b and c to get a bar b c, it's fine this way. we have a bar b c now we have all the terms we have ac we have a bar b c we have a bar b c again the question here, if we are allowed to use three input gates then we can directly pass it through three inputs or gate ac then b slash c then a b c slash we can pass it directly through a three entrance door and we will get the exit directly, but here we are not, there is none of that mentioned in the question, so we have to go through the two entrance doors.

Okay, so for that first we have to pass these two terms through two inputs or gates that will get ac plus b slash c then we will pass this output with two inputs or gates that will get ac plus b slash c plus a b c slash plus here The middle or the operation is okay so this may be a little confusing for you so follow the steps carefully, pause the video at any time you don't understand again, rewind it and watch it because a lot of connections have been made here so that I did it deliberately.

I took this question because many concepts are involved here, it involves three signal values ​​and three terms. Here we have a term that involves three signal values ​​and I deliberately told you that we cannot use a three input get directly because it has not been mentioned in the question so we have to follow the two gate concept. Well, we finally got the result. It was a little long and a little complex, but we finally got the result. There are much more complicated

expressions

than this, so we'll do it. Discuss it in the next few videos, but here I wanted you to understand this concept.

Well, again, the same steps first identify the signal values, identify which logic gates are required, then target the individual terms that are present in the expression, and then combine all the terms. finally to get the result this is the result, okay?