Chapter 2 Boolean algebra and Logic Gates


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1 Chapter 2 Boolean algebra and Logic Gates 2. Introduction In working with logic relations in digital form, we need a set of rules for symbolic manipulation which will enable us to simplify complex expressions and solve for unknowns. Originally, Boolean algebra which was formulated by George Boole, an English mathematician (85864) described propositions whose outcome would be either true or false. In computer work it is used in addition to describe circuits whose state can be either (true) or (false).using the relations defined in the AND, OR and NOT operation, a number of postulates are stated. In this chapter, we are going to learn What is called as Boolean algebra? Some basic theorems of Boolean algebra Basic logic gates used in logic circuits In what way Boolean algebra used in building the logic circuits 2.2 Definition of Boolean algebra In 854 George Boole introduced a systematic treatment of logic and developed for thispurpose an algebraic system now called Boolean algebra. Boolean algebra is an algebraic structure defined on a set of elements B togetherwith two binary operators + and.(dot) provided the following postulates aresatisfied: Let us assume that A and B are two logical variables, which can either be true or false. The + operator is the union or the OR operator. The is the intersection or the AND operator. The following are some of the laws of Boolean Algebra:. A+B=B+A 2. A B=B A 3. (A+B)+C = A+(B+C) 4. (A.B).C = (A.B).C 5. A+(B.C) = (A+B).(A+C) 6. A.(B+C) = (A.B)+(A.C) 7. A+=A 8. A+= 9. A+A=A. A.A=A. A+A = 2. A.A =
2 Where A is complement of A (Or A ). Venn diagrams can be used to illustrate the above. Figure Basic theorems of Boolean algebra From the postulates defined above we can prove some of the Boolean theorems which are often used in simplification of Boolean functions. Simplification of Boolean functions will reduce the number of gates required to build the logic circuit and thereby reduces the cost of the circuit. So we have to be familiar with the postulates and theorems. Duality theorem: It states that every algebraic expression remains valid if the operator and identity elements are interchanged. For example x+ = x is dual of x. = x Theorem (a): x+x = x Proof: x+x = (x+x). = (x+x).(x+x ) =x+x.x =x+ = x
3 Theorem (b): x.x = x Proof: x.x = (x.x)+ = (x.x)+(x.x ) =x.(x+x ) =x. = x Note that theorem (b) is dual of (a). Theorem 2(a): x+ = Proof: x+ = (x+). = (x+).(x+x ) =x+.x = x+x = Theorem 2(b): x. = (by duality principle) Theorem 3: (x ) = x Theorem 4(a): x+(y+z) = (x+y)+z Theorem 4(b): x.(y.z) = (x.y).z Theorem 5(a)DeMorgan s theorem: (x+y) = x.y Theorem 5(b)DeMorgan s theorem: (x.y) = x +y Theorem 6(a) absorption theorem: x+xy = x Proof: x+xy = x.+x.y =x.(+y) =x. =x Theorem 6(b) absorption theorem: x.(x+y) = x Proof: x.(x+y) = (x+).(x+y) =x+(.y) =x+ =x
4 2.4 Logic gates Basic logic gates All digital systems can be constructed by only three basic logic gates. These basic gates are called the NOT gate, AND gate and the OR gate. NAND and NOR are called as universal gates as using only NAND gates (or only NOR gates) we can construct the logic circuits. ExOR and ExNOR are two exclusive gates used in the logic circuits. NOT gate The NOT gate is a circuit which produces at its output the negated (inverted) version of its input logic. The circuit is also known as an inverter. If the input variable is A, the inverted output is written as A or A. AND gate Figure 2.2 The AND gate is a circuit which gives a high output (logic ) if all its inputs are high. A dot is used to indicate the AND operation. Figure 2.3
5 OR gate The OR gate is a circuit which gives a high output if one or more of its inputs are high. A plus sign (+) is used to indicate the OR operation. NAND gate Figure 2.4 The NAND gate is a NOTAND circuit which is equivalent to an AND circuit followed by a NOT circuit. The output of the NAND gate is high if any of its inputs is low. NOR gate Figure 2.5 The NOR gate is a NOTOR circuit which is equivalent to an OR circuit followed by a NOT circuit. The output of the NOR gate is low if any of its inputs is high.
6 Figure 2.6 ExOR gate The ExclusiveOR gate is a circuit which gives a high output if either of its two inputs is high, but not both. A encircled plus sign is used to indicate the ExOR operation. Figure Boolean functions and Expressions Boolean functions are formed by using binary variables. Binary variable means that a variable could take either or. So the output of Boolean function is also either or. Boolean expressions are generally consists of NOT, AND, OR operations. A Boolean expression can be implemented in practice by using the logic gates. General precedence of order is followed in evaluating the Boolean expression, i.e., Parenthesis, NOT, AND, OR. Let us consider a Boolean expression F = x.y +z. The variables in the expression are F, x, y and z. In this Boolean expression x, y and z are input variables and F is the variable used to specify the output. Now we are going to draw truth table for this expression.
7 Table 2. x y z y x.y F = x.y +z Three input variables x, y and z are written by possible combinations. As each variable can take either or, three variables can take 2 3 = 8 combinations starting from,,.. By using order of precedence rule, first calculate the y (NOT first), then x.y (AND next) and finally calculate (x.y )+z (OR function). Truth table is the table which consists of possible inputs and their corresponding output. So for Boolean expression F = x.y +z, the truth table is given as Table 2.2 Input Output x y z F = x.y +z
8 Note that this table contains only inputs (x, y and z) and output (F). Let us consider another Boolean expression F2 = (x+y ). We will construct truth table for this Boolean expression now. By order of precedence rule, first we have to compute (x+y ), then take the complement of (x+y ) to get output F. Table 2.3 x y y x+y F2 = (x+y ) So the truth table for the Boolean expression F2 = (x+y ) is given as Table 2.4 Input Output x y F2 So now we are able to construct truth table for given Boolean expression. Note that many Boolean expressions can give the same output. Two Boolean expressions that give the same output is said to be equivalent. For example consider the Boolean expression, F3 = x.y
9 Table 2.5 x y X F3 = x.y The truth table for the expression F3 = x.y is given as Table 2.6 Input Output x y F3 By comparing the truth table of F2 and F3, both Boolean expressions give the same output. So we can say that F2 = (x+y ) and F3 = x.y are equivalent. 2.6 Canonical and standard terms To conveniently specify the truth table in Boolean algebra, canonical and standard terms are used. As we have just discussed many Boolean expressions may give the same output. But while using canonical form of Boolean expression there is only one canonical form of Boolean expression to specify one truth table. To map the truth table into canonical Boolean expression, we have two standard forms. (i) the sum of minterms (product terms) (ii) the product of maxterms (sum terms). Let us know what is meant by minterms and maxterms.
10 Minterms Consider three input variables x, y and z. As binary variables can take either or, there could be eight possible combinations (2 3 = 8) i.e.,,,.. A zero is considered as complemented variable, one is considered as true variable. Thus 3 input variables have 8 minterms. x y z, x y z, x yz, x yz, xy z, xy z, xyz, xyz Minterms are also called as product terms as x,y and z are ANDed either in complemented or true form. Maxterms Note that here zero is considered as true variable, one is considered as complemented variable. Thus 3 input variables have 8 minterms. x+y+z, x+y+z, x+y +z, x+y +z, x +y+z, x +y+z, x +y +z, x +y +z Maxterms are also called as sum terms as x,y and z are ORed either in complemented or true form. Table 2.7 Input x Minterms Maxterms y z Term Designation Term Designation x y z m x+y+z M x y z m x+y+z M x yz m2 x+y +z M2 x yz m3 x+y +z M3 xy z m4 x +y+z M4 xy z m5 x +y+z M5 xyz m6 x +y +z M6 xyz m7 x +y +z M7 The above table shows minterms and maxterms for three input variables. In the same way two input variables will have 4 minterms and 4 maxterms. Four input variables will have 6 minterms and maxterms. So we know the minterms and maxterms now. Let us see how to map a truth table into Boolean expression using these minterms and maxterms. First consider the truth table with two input variables say A and B and an output variable X.
11 Table 2.8 Input A Output B X To specify this truth table in canonical form of Boolean expression using minterms, consider the output X wherever which is equal to one. The output is whenever A= and B= or A= and B=. If A= and B=, the corresponding minterm is A B. If A= and B=, the corresponding minterm is AB. So the Boolean expression is X = A B+AB This expression can also be written as X = m+m2 or X = (m,m2) or simply X= (,2). This is very convenient way of representing the truth table and it is unique. As output is expressed as sum of minterms, this type of expression is called as sum of product expression or SOP. The same truth table can be expressed using maxterms also. For maxterms we should consider the output X wherever it is equal to zero. The output X is zero if the input A= and B= or A= and B=. If A= and B=, the corresponding maxterm is (A+B). If A= and B=, the corresponding maxterms is (A +B ). So the Boolean expression is X = (A+B)(A +B ) This expression can also be written as X = M, M3 or X = (M,M3) or simply X= (,3). As output is expressed as product of maxterms, this type of expression is called as product of sum expression or POS. How to convert the Boolean algebra into Canonical forms There may be a question raised in your mind that how to convert any Boolean expression into canonical forms. It is very simple to convert any Boolean algebra into its equivalent canonical forms. The need for conversion is as already said it is convenient way to represent in canonical form and it is unique. Let us consider a Boolean expression F = A+B C. We need to convert this Boolean expression into its equivalent sum of product expression (SOP). By seeing the expression we can easily say that it has three inputs A, B and C and one output F. we will build a truth table first.
12 Table 2.9 A B C B B C F=A+B C From the table we can easily write F = A B C+AB C +AB C+ABC +ABC or F = m+m4+m5+m6+m7 or F = Σ(, 4, 5, 6, 7). Let consider another Boolean expression F = xy+x z and convert it into POS expression. Again draw the truth table Table 2. x y z X x z xy F=xy+x z From the table we can easily write F = (x+y+z)(x+y +z)(x +y+z)(x +y+z ) or F= M.M2.M4.M5 or F = Π(, 2, 4, 5).
13 How to convert SOP expression into POS and vice versa To convert SOP expression into POS and vice versa, writes the missing terms because POS and SOP are complement of each other. Consider the previous example F = xy+x z, and we have found that its POS expression is that F = Π(, 2, 4, 5). To convert this expression into its equivalent SOP form, find the missing terms. Missing terms are here, 3, 6, 7. So we can write SOP expression as F = Σ(, 3, 6, 7). 2.7 Implementation of Boolean function using logic gates Implementation of Boolean function using basic logic gates It is possible to implement any Boolean expression, using logic gates in practical. As a first case, we will construct the logic circuit for some Boolean expression using only basic gates like NOT, AND, OR gates. Later we will learn to implement the Boolean function using only NAND gates or using only NOR gates. Let consider a Boolean function F = x+y z. What are all the logic operations in this Boolean function? As order of precedence first NOT operation, then AND operation and OR operation. So we need a NOT gate, AND gate and OR gate to implement this function. Figure 2.8 First obtain the required complemented variable using NOT gate(s). In this particular example, we need complement of y only. Then using AND and OR gates we can complete this circuit.
14 Figure 2.9 Consider another Boolean function F2 = x y z+x yz+xy. As we have just seen, first obtain the required complement function. Note that for x and y only complemented variable needed, not for z. Figure 2.
15 Figure 2. Implementation of Boolean function using universal logic gates NAND implementation NAND and NOR gates are easier to fabricate and are the basic gates used in all IC digital logic families. So implementing Boolean function using NAND and NOR gates are necessary. First we implement basic logic gates NOT, AND and OR using only NAND gates. A one input NAND gate behaves like an inverter or NOT gate. Figure 2.2 As NAND function is complemented function of AND gate, to derive AND gate using NAND gates again it should be complemented. So connecting a NOT gate at the output of NAND gate will provide AND function. Figure 2.3 NOT gate in the output of NAND gate can be replaced by single input NAND gate as single input NAND gate will act as NOT gate or inverter.
16 Figure 2.4 OR gate can be implemeted by using NAND gates only by using DeMorgan s theorem. As per DeMorgan s theorem, (x+y) = x.y. So x+y = (x.y ). So first we need x and y which can be obtained by connecting a single input NAND gate to each input. Figure 2.5 Then x and y are connected to a two input NAND gate to obtain (x.y ) which is equal to OR function x+y. Figure 2.6 Sometimes this circuit is simply represented by using a bubble for inverter operation as shown below. Figure 2.7 To implement a Boolean function uisng only NAND gates Implement the Boolean function using basic logic gates i.e. NOT, AND, OR. Replace all NOT gates by single input NAND gates. Replace all the AND gates by NAND gates adding a bubble at the output of each AND gate. Replace all the OR gates by NAND gates adding a bubble at the inputs of each OR gate. If two bubbles are in the same line, then they will be get cancelled out naturally. If only one bubble is there in a line, then to compensate the invert operation add one more single input NAND gate in that line.
17 Example Consider a Boolean function F = AB+CD. We implement this Boolean function using only NAND gates. Let us draw the logic circuit using basic logic gates. Figure 2.8 Replace AND by NAND putting a bubble at the output of AND gate and replace OR gate by putting bubbles at the inputs of OR gate. Figure 2.9 As two bubbles are in the same lines, two invert functions will cancel each other. So we can replace all the gates by NAND gates. Figure 2.2
18 Example Consider another Boolean fucntion F = A(B+CD)+BC. We implement this Boolean function now using only NAND gates. Let us draw the logic circuit using basic logic gates first. Figure 2.2 First replace the NOT gate by single input NAND gate. Figure 2.22
19 Replace AND by NAND putting a bubble at the output of AND gate and replace OR gate by putting bubbles at the inputs of OR gate. Figure 2.23 As one bubble is there in two lines we have to add inverter to compensate it. Figure 2.24 Now all the gates can be replaced as NAND gates.
20 Figure 2.25 NOR implementation We implement basic logic gates NOT, AND and OR using only NOR gates. A one input NOR gate behaves like an inverter or NOT gate. Figure 2.26 As NOR function is complemented function of OR gate, to derive OR gate using NOR gates again it should be complemented. So connecting a NOT gate at the output of NOR gate will provide OR function. Figure 2.27 NOT gate in the output of NOR gate can be replaced by single input NOR gate as single input NOR gate will act as NOT gate or inverter.
21 Figure 2.28 AND gate can be implemeted by using NOR gates only by using DeMorgan s theorem. As per DeMorgan s theorem, (x.y) = x +y. So x.y = (x +y ). So first we need x and y which can be obtained by connecting a single input NOR gate to each input. Figure 2.29 Then x and y are connected to a two input NOR gate to obtain (x +y ) which is equal to AND function x.y. Figure 2.3 Sometimes this circuit is simply represented by using a bubble for inverter operation as shown below. Figure 2.3 To implement a Boolean function uisng only NOR gates Implement the Boolean function using basic logic gates i.e. NOT, AND, OR. Replace all NOT gates by single input NOR gates. Replace all the OR gates by NOR gates adding a bubble at the output of each OR gate. Replace all the AND gates by NOR gates adding a bubble at the inputs of each AND gate. If two bubbles are in the same line, then they will be get cancelled out naturally. If only one bubble is there in a line, then to compensate the invert operation add one more single input NOR gate in that line.
22 Example Consider a Boolean fucntion F = (A+B)(C+D)E. We implement this Boolean function using only NOR gates. Let us draw the logic circuit using basic logic gates first. Figure 2.32 Replace OR by NOR putting a bubble at the output of OR gate and replace AND gate by putting bubbles at the inputs of AND gate. Figure 2.33 As one bubble is there in one line we have to add inverter to compensate it. Figure 2.34
23 Now all the gates can be replaced as NOR gates. 2.8 Summary Figure 2.35 In this chapter we have defined the Boolean algebra by using some postulates. Boolean algebra postulates can be easily proved by using Venn diagram. Then some basic theorems like duality principle, DeMorgan s theorem and absorption theorem have been proved. These theorems are very much useful in the subsequent topics. NOT, AND and OR which are called as basic logic gates and NAND and NOR which are called as universal gates have been given along with their truth table, Boolean functions. Using these gates only we have to implement the digital system throughout this subject. Therefore learning these gates undoubtedly is important to further proceed. Boolean functions are implemented using logic gates. Canonical forms are unique way to represent a digital system and it is most convenient way also to represent the truth table. Finally we have learnt to construct the logic circuit using basic logic gates. Using only NAND or only NOR gates also we can build the logic circuits that is why NAND and NOR are called as universal gates. Review Questions. Demonstrate by means of truth tables the validity of the following identities: (a) DeMorgan's theorem for three variables: (xyz)' = x' + y' + z'. (b) The consensus theorem: xy + x z + yz = xy + x z. 2. Obtain the truth table of the following functions and express each function in sum of minterms and product of maxterms: (a) (xy + z)(y + xz) (b) (A' + B)(B' + C) (c) y'z + wxy' + wxz' + w'x'z 3. For the Boolean function F given in the truth table, express F in sum of minterms in algebraic form.
24 Table 2. x y z F 4. Express the following functions in sum of minterms and product of maxterms: (a) F(A,B, C,D) = B'D + A'D + BD (b) F(x, y, z) = (xy + z)(xz + y) 5. Convert the following to the other canonical form: (a) F(x, y, z) = Σ(, 3, 7) (b) F(A, B, C, D) = Π(,,2,3,4,6,2) 6. Draw the logic diagram corresponding to the following Boolean expressions (a) BC' + AB + ACD (b) (A + B)(C + D)(A' + B + D) (c) (AB + A'B')(CD' + CD) 7. Draw a NAND logic diagram that implements the complement of the following function: F(A, B, C, D)=Σ(,,2,3,4,8,9,2) 8. Draw a logic diagram using only twoinput NAND gates to implement the following expression: (AB + A'B')(CD' + C'D)
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