CHAPTER 3 LOGIC GATES & BOOLEAN ALGEBRA
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2 CHPTER 3 LOGIC GTES & OOLEN LGER C H P T E R O U T C O M E S Upon completion of this chapter, student should be able to: 1. Describe the basic logic gates operation 2. Construct the truth table for basic logic gates 3. Write the oolean expression 4. Implement the circuit from oolean expression. 5. pply De Morgan s theorems to simplify logic expressions. 1.1 INTRODUCTION Logic gate is the most basic type of digital circuit, which consists of two or more inputs and one output. gate can be used alone to perform a logic function. It can also be connected to several other gates to form a logic network. 1.2 SIC LOGIC GTES The basic gates are NOT, ND, OR, XOR, XNOR, NND & NOR. The symbols for these gates and their corresponding oolean expressions are briefly discussed in this chapter Inverter / NOT gate EHVIOR The output is the inverse of the input. SYMOLS Y = LGERICLLY Y =
3 TRUTH TLE Input, Output, Y TIMING DIGRM OR gate EHVIOR The output is 1 if any or all of the inputs are 1 SYMOL LGERICLLY = TRUTH TLE Input Output TIMING DIGRM = +
4 1.2.3 ND gate EHVIOUR The output is 1 only if all the inputs are 1 SYMOL LGERICLLY = = TRUTH TLE Input Output TIMING DIGRM =
5 1.2.4 NOR gate EHVIOR The Output is 1 only if all of the inputs are 0 SYMOL = LGERICLLY = TRUTH TLE Input Output TIMING DIGRM
6 1.2.5 NND gate EHVIOUR The output is 1 if any or all the inputs are 0. SYMOL = LGERICLLY = TRUTH TLE Input Output TIMING DIGRM
7 1.2.6 X-OR gate (Exclusive-OR gate) EHVIOUR The output is 1 only if an odd number of the inputs are 1 SYMOL LGERICLLY = = TRUTH TLE Input Output TIMING DIGRM
8 1.2.7 X-NOR gate (Exclusive-NOR gate) EHVIOUR The output is 1 only if an odd number of inputs are 0 SYMOL LEGRICLLY TRUTH TLE = = Input Output TIMING DIGRM
9 1.3 OOLEN EXPRESSIONS Implementing Circuit rom oolean Expression oolean expression is mathematics of logic. It is one of the most basic tools available to the logic designer and thus can be effectively used for simplification of complex logic expressions. The simplify logic circuit can be directly draw base on the final equation that obtained from this oolean expression. If an expression contains both ND and OR operation, the ND operations are performed first. Unless there are parentheses in the expression, where the operation inside the parentheses is to be performed first. Example 3-1: 1) Given = ( ) C, so the diagram is performed as: C + = +. C 2) Given = CD, the diagram is performed as: C D CD = + CD 3) Given = ( )( C), the diagram is performed as: C + + C = + + C
10 1.3.2 Evaluating Logic-Circuit Outputs Example 3.2: Given =0, =1, C=1, D=1 and the oolean expression is = C( D), determine the output for the expression. Solution = C( D) = 0 1 1(0 1) = 1(0) = Determining Output Level rom Circuit Diagram Example 3.3: Determining the output level from a circuit diagram, where the inputs are =0, =1, C=1,D=1 C D Solution
11 = OOLEN THEOREMS oolean theorems can be used to express logic circuit operations mathematically. It can help us to simplify logic expression. There are two methods to simplify a complicated logic circuit, which is: 1) oolean algebra & De Morgan s theorem 2) Karnaugh Map The most important of oolean algebra rules and laws are presented in the following section asic Rules of oolean lgebra Laws of oolean algebra
12 Commutative Laws ddition = Multiplication + = + = = ssociative Laws ddition +(+C) =(+)+C C + C +(+C) = C + (+)+C Distributive Law ( C) = C C + C (+C) = C C + C
13 1.4.2 Rules of oolean algebra (Single-variable theorems)
14 Table 3-2: Multivariable theorems (10a) + = + (10b) = (11) +(+C) = (+)+C (12) (C)=()C (13a) (+C) = +C (13b) (+)(C+D)=C+C+D+D (14) + = (15a) + =+ Proof it! (15b) + = + Example 3.4: 1) Proof the + = Solution + = (1+) = (1) theorem (6) = 2) Proof the (+)(+C) = + C Solution ( )( C) = C C = C C = ( 1 C) C = ( 1) C = ( 1 ) C = ( 1) C = C DeMorgan s Theorem DeMorgan s theorems are very useful in simplifying expressions for easy transfer back and forth from the product of variables to the sum of variables form. It allows for elimination of overbar(s) that are over several variables. The two theorems are:
15 irst theorem The inverted of a product of variables is equal to the sum of the inverted individually variable. = INPUTS OUTPUTS Second theorem The inverted of a sum of variables is equal to the product of the inverted individual variable. = INPUTS OUTPUTS
16 Example 3.5: pplying DeMorgan s theorem to the expression: XYZ = X Y Z. Solution XYZ = X Y Z X Y Z = X Y Z Example 3.6: pply DeMorgan s theorem to each of the following expressions: Solution (a) (c) (e) (b) C C (d) C (f) C (a) (c) (b) C C = C (d) (e) C (f) C = C pplying DeMorgans Theorems The following example illustrates the applications of De Morgan s theorems and oolean lgebra to the specific expression: Example 3.7:
17 Given C D( E ), simplify the expression. Solution: ssume each term as single variable Let C = X D ( E ) = Y Rewite the expressions. pply DeMorgan s Theorem Rewrite new expression. C D( E ) = X Y X Y = X Y ( C)( D( E )) Use rule 9 ( = ) to cancel the double bars over the left term. ( C) D( E ) = ( C)( D( E ) ) pplying DeMorgan s Theorem to the 2 nd term ( C)( D ( E ) ) Use rule 9 ( = ) to cancel the double bars over the E part of the term. C D E.answer Example 3.8: pply De Morgan s theorems to each of the following expressions: a) ( C) D (b) C DE (c) ( ) C D E Solution (a) ( C) D Let ( C) = X D = Y Rewrite; C D = XY pply DeMorgan s Theorem; XY = X Y C D = C D pply DeMorgan s Theorem to the 1 st term, C = C
18 Rewrite the expression, C D = C D (answer) b) C DE Let Rewrite; C = X DE = Y C DE = X Y pply DeMorgan s Theorem; X Y = X Y C DE = ( C)( DE ) pply DeMorgan s Theorem; C DE C D E = answer c) CD E CD E = CD E = CD E = C D E C D E = answer 1.5 SIMPLIICTION USING OOLEN LGER pply the laws, rules and theorems of oolean algebra to simplify general expressions. Example 3.9:
19 Using oolean algebra techniques, simplify this expression: Solution C C pply the distributive law to the 2 nd and 3 rd terms. pply rule 3 ( = ) to the 4 th term. pply rule 7 (+=) to the 1 st and 2 nd term pply rule 14 (+=) to the 3 rd and 4 th term pply rule 14 (+=) to the 1 st and 3 rd term C C C C C C C C answer Example 3.10: Simplify the following oolean Expression: (a) ( ( C D) ) C (b) C C Solution a) C D C C D C = C D C b) C C = CC DC C = C C = C = C answer C C = C C = C C = C C C = C C C = C C = C answer 1.6 Universality Of NND Gates & NOR Gates
20 ll oolean expressions consists of various combinations of the basic operations of OR, ND, and INVERTER. Therefore, any expression can be implement using combinations of these gates. NND Gates can be used to implement any oolean function = NOR Gates can be used to implement any oolean operation 1.7 IEEE/NSI Standard Logic Symbols
21 standard for logic symbols was developed in 1984; it is called the IEEE/NSI Standard for logic symbols. The IEEE/NSI standard uses rectangular symbols to represent all logic gates and circuits. special dependency notation inside the rectangular symbol indicates how the device outputs depend on the device inputs. Table 3-3: Standard logic symbols: (a) traditional; (b) IEEE/NSI
22 Problem 1. What is the oolean expression for the logic diagram as shown in ig. 3-1? C 2. Show the logic arrangements for implementing: (a) four-input NND gate using ND gates and NOT gates; (b) three-input NND gate using NND gates; (c) NOT circuit using EX-NOR gate. 3. Construct the truth table base on the diagram as shown in ig. 3-1? 4. Simplify the oolean expression obtain from question Given =0, =1, C=1and the oolean expression is base on question 3, determine the output for the expression.
23 6. Draw the circuits that will perform the functions described by both sides of + = by applying De Morgan's theorems, and also demonstrate the theorem is true using a truth table. 7. igure 3.2 shows the inputs waveforms for the two-input ND gate. Draw the output waveform,. igure Given the timing diagram in ig. 3-3, write out the truth table for the circuit responsible for it, the oolean equation describing its operation and draw the circuit diagram. C 9. Implement the oolean expression = C + C + C + C using XOR gates. 10. Implement the oolean expression = + C + C using NND gates only
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