Practice Problems on Combinational Circuits (Part 4)

welcome to the fourth part of practice problems on combinational circuits after this we are going to start a new series in which we will talk about the sequential circuits in this part I'm having two problems the first problem is for me I'm going to solve it and the second problem is for you you can consider it as your homework I will just give a small hint and tell you how to approach for the solution and you will solve it post answer in the comment box so that everyone can take help of it and um that's all for the second problem now we will move to the first problem which will be solved by me in this it says implement the following Boolean expression using an 8 is to one multiplexor a very simple problem we have already done this when we were studying the multiplexors and it's a kind of revision for you the function f is there having the four variables a b c d the Min terms are given as you can see and also the don't care the don't care is 02 and the mean terms are 1 3 5 10 11 13 and 14 now what is the step number one the step number one is to make the K map depending upon the number of variables I'm having a 16 cell K map I have made it already because I want to save the time and the step number two is definitely filling this map with this minum and the don't care after doing this we will make a 8 is to one multiplexor we will assign randomly the variables as the selector variable in them and then the one variable the left variable will act as the input these are all the steps that we have to follow to get to our answer now let's fill this map with the minan terms the mean term is 1 3 5 10 so 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 and this is 15 which is definitely not in our main terms and the don't cares are 0er and two so Zer is my don't care and two is my don't care now we have to make a 8 is to one multiplexor I will make a 8 is to one multiplexor here and the selector variables are three in this case so these are my three select variables s0 S1 S2 the output is y and there are eight inputs 3 4 5 6 7 8 the first input is i0 i1 I2 I3 I4 I5 I6 and i7 now the important part comes we have to assign these four variables as the input and also as the selector variable so I will take for the convenience I will take s0 = to C S1 = to B and S2 equal to a so this is what I have decided and depending upon this I will make um truth table in which the input is s0 S1 S2 three variabl so definitely eight combinations 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 and finally 1 one 1 so these are the eight combinations and the output is y when all all these selector variables are zero it means the output is going to be i0 similarly i1 I2 I3 I4 I5 I6 and i7 this is what we have learned from the multiplexers now I will use this K map and try to find out the value of i0 i1 up to i7 depending upon the values of S2 S1 and s0 and as I have already made a assumption that s0 is C S1 is B and S2 is a so we can use this K map easily now let's see in this K map when we are having a b and c as zero a is zero for this entire row but C is zero for the first two cells so I will take these first two cells and in this I'm having a don't care and one the don't care I will take as one for the combination purpose and you can see overall this cell is giving me one that's why i0 is equal to 1 very simple similarly we will see for the case number two in which a 0 0 it means this entire row whereas C is one C is one for this two last cells so I will take these two last cells and these two last cells are also giving me one so i1 is also one when AB is 01 it means this row and C is zero so again the first two cells of this row it gives me D because here I'm having zero and one is there in this cell and for this cell the value of D is 1 it means I2 is equal to D now we will see when C is 1 whereas AB is 01 it means this two cells in this two cells the values are 0 0 so I'm having I3 as zero it's a very simple simple process to do the implementation using the marks you will find a very different process in the books which is very very inconvenient I really don't like that method you can follow this method it's very easy and uh let's see for this case when AB is 1 Z it means this row C is0 then again the first two columns of this row and it gives us zero now we will check for this two cells because C is one in this case and I'm having I5 as 1 now when a is 1 one and C is zero it means this column I'm again having input as D so I6 is D we will check finally for the last case when all these three are one It Means A B C are on and for this we will see the last two cells and D is zero in this case so I'm having D complement so it's a very simple way to implement and you can see we are having the implementation of this Boolean expression using a 8 cos one marks this is C this is B this is a and the first input is one the second input is one again the third input is D and the fourth is zero fifth is zero sixth is 1 seventh is D and eth is D complement in this way we have implemented this function and Y is F this F so this is all that we have to do in this problem now let's move to the problem number two that you have to solve let's first read what it says design a combinational circuit with three inputs there are three inputs so x y z are my three inputs and three outputs are also there a b c are the three outputs so a b c are the three outputs now let's see what it says when the binary input is 0 1 2 or three it means 0 0 0 0 0 1 0 1 0 or 0 1 1 this is the binary input of 0 1 2 or 3 the binary output is two greater than the input it means when the input is 0 0 0 it is two greater and the two greater of 0 0 0 is 0 1 0 similarly you have to see for these three cases also and there is one more statement it says when the binary input is 4 5 6 or 7 the binary output is to less than the input so if I say I'm having four 4 is 1 0 0 it is 2o less than the four so it should be two again so 0 1 0 in this way you have to fill all the values of your output and then you have to use the kmap and minimize the function and implement it as your circuit you can write down the minimized function in the comment section and we will see whether it is correct or not so this is all for this presentation I hope you enjoyed it and more than enjoyment I hope you learned something and if you have any doubt regarding any part of this presentation you can ask in the comment section this is all I will end this here see you in the next one