Hey all, Welcome to Geeksforjobs
A gate is a digital circuit which either allows a signal to pass through it or stop it.
A logic gate is a digital circuit which allows a signal to pass through it only if certain logical conditions are satisfied.
The basic logic gates are AND,OR,NAND,NOR,XOR,INV and BUF.The last two logic gates are not standard terms.They stand for “inverter” and “buffer” respectively.The symbols and their corresponding expressions given in below table.
Let A,B are two inputs for AND logic gate.
2.”A+B” -OR gate
5.A’ -INVERT gate
Boolean Algebra and DeMorgan’s Theorems
Boolean algebra can be used to formalize the combination of binary logic states.The fundamental relations are given in above tables.In these relations,A and B are binary quantities,that is ,they can be either logical true(T or 1) or logical false(F or 0).Most of these relations are obvious.Here are some of them:
AA=A; A+A=A; A+A’=1; AA’=0; A”=A.
Recall that the text sometimes uses an apostrophe for inversion (A’).We use overbar notation.
We can use algebraic expression to complete our definitions of the basic logic gates we began above.Note that the boolean operations of “multiplication” and “addition” are defined by truth tables for AND or OR gates.
We can make an INV gate from a 2-input NOR gate.Simply connect the two inputs of the NOR gate together.Algebraically ,if the two original NOR gate inputs are labbled B and C and they are combined to form A, then we have Q=(B+C)’=(A+A)’=A’,which is INV operation.
Note that an INV gate can not be made from OR or AND gates.For this reason the OR gate AND gates are not universal.So for example,no combination of AND gates can be combined to substitute for NOR gate.However the AND and NAND gates are universal.
Perhaps the most interesting if the boolean indentities are the two known as DeMorgan’s Theorems.
These expressions turn out to be quite useful, and we shall use them often.An example of algebraic logic manipulation follows.One is to show that XOR gatecan be composed of 4 NAND gates.We know that A XOR B =A’B+AB’. Since AA’=0 and BB’=0, we can add these ,rearrange,and apply the two DeMorgan relations to give
A XOR B=A(A’+B’)+B(A’+B’)=A(AB)’+B(AB)’=((A(AB)’)'(B(AB)’)’)’
The two DeMorgan expressions above can be envoked using gate symbol by following this prescription:change gate shape(AND ->OR) and invert all inputs and outputs.