 Hello everyone, Myself Prithviraj Pitambir working as an assistant professor in department of Electronics Engineering at Vulture National Technology, SolarPool. Welcome to this video session on multiplexers. Let us truly learn how to implement Boolean functions using multiplexers. At the end of this session, students will be able to examine working of multiplexers and they can implement simple logic functions using multiplexers. So, multiplexer is a many input device and only one output device. So, the relation is defined by multiplexer having n select lines, can have two ratio and data input lines and it can be used to implement a Boolean function of n variables. So, let us learn method one, a multiplexer with the help of selection inputs, generate some of mean terms or product of maximum expressions and individual mean term or maximum can be defined by data inputs. So, here you will see that you have this 4 is to 1 multiplexer having two select lines S1 is 0 and four data lines D0 D1 D2 D3 can be used to implement SOP like this S1 bar S0 bar D0 plus S1 bar S0 D1 plus S1 S0 bar D2 plus S1 S0 D3. In POS form S1 plus S0 plus D0 plus S1 plus S0 bar plus D1 S1 bar plus S0 plus D2 and S1 bar plus S0 bar plus S3 bar. So, here maximums are used and in the first SOP form mean terms are used and each term is defined by data inputs. So, let us see one example implementation of function AB summation of mean terms 1 2 so SOP. So, the mean terms are given and the mean terms are 1 and 2. So, the logic is connect the data inputs corresponding to the mean terms to logic 1 and data inputs corresponding to the max terms to logic 0. So, here D0 D3 are the max terms so connect them to logic 0 and D1 and D2 are the data inputs corresponding to mean term that is why connect them to VCC. So, you if you place the actual values of S1 S0 and D0 D1 D2 D3 in the previous expression you will have SOP is equal to A bar B bar dot 0 plus A bar B dot 1 plus AB bar dot 1 plus AB dot 0. So, the final expression will reduce to A bar B plus AB bar or if you consider the max terms the expression will become A plus B plus 0 A plus B bar plus 1 A bar plus B plus 1 A bar plus B bar plus 0 which reduces to A plus B dot A bar plus B bar. So, we have SOP form as well as POS form and each term individually defined by corresponding data inputs. So, let us solve this problem quickly write down your answer implementation of f of ABC is equal to summation of 1 2 4 7 mean terms. So, here the solution is here you we have 1 2 4 and 7 mean terms and obviously 0 3 5 6 max terms. So, data lines corresponding to mean terms which are D1 D2 D4 D7 should be connected to 1 and data lines corresponding to these max terms 0 3 5 6 goes to logic 0 or ground. So, the expression implemented here is A bar B bar C plus A bar B dot C bar plus AB bar C bar plus ABC or you can have this POS form also. We have one more method we can use a multiplexer having n select lines means 2 H 2 n data lines can be used efficiently to implement a Boolean function of n plus 1 variables. So, for example, a 4 variable Boolean function can be implemented using please calculate the multiplexer yes the multiplexer required is 8 is to 1 multiplexer. So, if you put here the values so, 3 select lines so, 8 data lines you can use 8 is to 1 multiplexer to implement 4 variable function. So, there are some steps you need to follow first a tabulated truth table for the given Boolean function use first n variables as a select lines then evaluate the output as a function of the last unused variable or variables for each combination of selection inputs. So, for example, the function can be 0 1 or the variable or the complements of variable or the combination of variable etc. So, you can have multiple combinations now apply these new values in proper order to the corresponding data inputs. So, you may require some extra logic circuit you can use if required for example, some gates let us implement this previous function f of ABC is equal to summation of minimums 1 2 4 7. So, the first step is the tabulation of this table so, we have this table so, f is f column so, these are the main terms 1 2 4 and 7 and the maximum 0 3 5 and 6. Now, second step is to use AB as a select lines. So, here A and B are used as a select line so, here C is still unused so, let us compute f in terms of C. So, we can divide this table into 4 parts in the first part AB is 0 0 and f is C in the second part AB is 0 1 and the relation of f and C is C bar. In the third part AB is 1 0 and the relation is again C bar and the last part AB is 1 1 and the relation is C. Now, let us do the final connections. So, here this you can reduce this table with AB as a two input columns and f and here you will see that f is represented in terms of the last unused variable. So, for 0 0 f is C 0 1 f is C bar 1 0 f is C bar 1 1 f is C. So, here the definitions for D 0 to D 3 are like this C C bar C bar C and AB. So, you can again still write f of S O P is equal to A bar B bar C A bar B C bar A B bar C bar and A B C. So, this C is nothing but D 0 this C bar is nothing but D 1 this C bar is nothing but D 2 and this C is nothing but D 3. So, here you have used a small multiplexer for the implementation of the same function. Instead of 80s to 1, we have implemented this function using 4s to 1mux with some extra hardware in this case one inverter. You can use these textbooks for your further readings and to solve some more problems based on this content. Thank you.