Introduction In this article we will discuss the design of the arithmetic adder and subtracter circuits. These circuits will take input bits and produce two outputs sum and carry or difference and borrow. Half adder A half adder is a logic circuit that adds two bits at a time. These two bits may be the bits of two binary numbers or a bit of one number and a carry from previous stage. This circuit take two inputs and produces two outputs a Sum and a Carry. Truth table of a half adder The Sum output has two minterms and the carry output has one minterm. So the output equation in SOP form is Boolean equation are: Sum = A’B + AB’ Carry = AB The above equations do

# Digital Logic and Computer Design

The course on Digital Logic and Computer Design is designed for the students of B.Tech Computer Science and Engineering. It covers topics on Number systems, Boolean Algebra, Binary arithmetic, Simplification of Boolean functions, combinational and the sequential circuits, basic computer organization, computer arithmetic, input-output organization, modes of data transfer and the memory subsystem

# K-Map Redundant and Don’t Care

Introduction In Simplification using K-Map we used minterms or Maxterms to group 1's or 0's respectively to form pair, quad or octets. There may be situations where all the 1's which are part of one group are overlapped by other groups or certain input conditions may not be contributing in getting the minterms or maxterms of the output functions. In the following discussions, we will see how to account for such conditions. Redundant Groups A redundant group is one whose all the 1’s have been used or ovelapped by other groups. We can always eliminate such group. Let us take an example and understand how redundant group can be dropped thus not part of solution. Example: Simplify the SOP equation given by F(ABC) = ∑(2,3,5,7). Figure-1 There

# Simplification using K-Map

K-Map A K-Map is a pictorial representation of the Boolean expression. By entering the values of the minterm or Maxterm an SOP or POS equation can be very easily simplified. It is very easy tool for simplifying up to 5-variable Boolean equation, but as the variable increases, solution become tedious. To begin, we will learn how to draw a 2, 3 and 4 variable K-Map. Draw 2-Variable K-Map Draw 3-Variable K-Map Draw 4-Variable K-Map Simplification Using K-Map Simplifying 2-Variable SOP equation Ex-1: Simplifying F= A'B + AB Ex-2: Simplify A'B + AB + AB' Simplifying 3-Variable SOP equation Ex-3: Ex-4 Simplifying 4-Variable SOP Equation Ex-5 Simplify F(ABCD)=∑(m1, m3, m5, m7, m9, m10,m13, m15) Ans: D Practice Problem: Simplify F= A'B'+A'B+AB' Simplify F(WXY)=∑m(1,2,3,5,7) Simplify F(ABC)=∑m(2,3,4,5 ) Simplify F(ABC)=∑m(2,3,7) + d(1,5) Simplify F(ABCD)=∑m(2,3,5,7,10,15)+d(0, 4,9,13)

# Boolean Function

Boolean function A Boolean function is a Boolean algebraic equation derived generally from the word statement of the given problem or from the truth table. Such a function has two or more input variables. The function will produce a LOW or HIGH output when certain combinations of the input signal (binary value) is applied to the input variables. Two forms of Boolean functions Sum of Product Product of Sum Sum of Product: A sum of product (or SOP) equation is obtained by ORing the product or the minterms Minterm : A Minterm is also called as product term is one that produces a HIGH output when the values of the input variables are ANDed together. It is written for a ‘1’ or HIGH values in

# Floating Point Number representation

Representation of Real Numbers: Real numbers can be represented using the following two representations. · Fixed point Notation · Floating point notation Fixed point representation A fixed point number is one which is stored with fixed number of bits for integer part and also a fixed number of bits for the fraction part. It has the following parts · Integer Part · binary Point · Fraction Part IIIIIIIIIIIII . FFFFFF Suppose we have 8 bits storage to store a real number, of these 5 bits to store the integer part and 3 bits to store the fractional part as shown below: ( 1 0 1 0 1 . 1 0 1)2 = 1*24 +0*23 + 1*22 + 0*21 + 1*20 + 1*2-1 + 0*2-2 + 1*2-3 Smallest and largest Fixed point number With 8 bit