×
DREAM IT. BELIEVE IT. ACHIEVE IT.
--Your friends at LectureNotes
Close

Switching Theory and Logic Design

by Jntu Heroes
Type: NoteInstitute: JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY Downloads: 216Views: 2938Uploaded: 8 months agoAdd to Favourite

Share it with your friends

Suggested Materials

Leave your Comments

Contributors

Jntu Heroes
Jntu Heroes
LECTURE NOTES ON SWITCHING THEORY AND LOGIC DESIGN II B. Tech I semester (JNTUH-R15) ELECTRONICS AND COMMUNICATION ENGINEERING
Switching Theory And Logic Design UNIT-I Number System and Boolean Algebra and Switching functions The Decimal Number system: The Decimal number system contains ten unique symbols. 0,1,2,3,4,5,6,7,8,9. Since Counting in decimal involves ten symbols its base or radix is ten. There is no symbol for its base. i.e, for ten .It is a positional weighted system i.e,the value attached to a symbol depends on its location w.r.t. the decimal point.In this system, any no.(integer, fraction or mixed) of any magnitude can be rep. by the use of these ten symbols only. Each symbol in the no. is called a Digit. The leftmost digit in any no.rep ,which has the greatest positional weight out of all the digits present in that no. is called the MSD (Most Significant Digit) and the right most digit which has the least positional weight out of all the digits present in that no. is called the LSD(Least Significant Digit).The digits on the left side of the decimal pt. form the integer part of a decimal no. & those on the right side form the fractional part.The digits to the right of the decimal pt have weights which are negative powers of 10 and the digits to the left of the decimal pt have weights are positive powers of 10. The value of a decimal no.is the sum of the products of the digit of that no. with their respective column weights. The weights of each column is 10 times greater than the weight of unity or 1010.The first digit to the right of the decimal pt. has a weight of 1/10 or 10-1.for the second 1/100 & for third 1/1000.In general the value of any mixed decimal no. is dn dn-1 dn-2 ………d1 d0.d-1 d-2 d-3 …….d-k is given by (dn x10n)+(dn-1 x10 n-1)+ ………(d1 x101)+(d0 x101)+(d-1 x102)(d-2 x103) ……. 9’s & 10’s Complements: It is the Subtraction of decimal no.s can be accomplished by the 9‘s & 10‘s compliment methods similar to the 1‘s & 2‘s compliment methods of binary . the 9‘s compliment of a decimal no. is obtained by subtracting each digit of that decimal no. from 9. The 10‘s compliment of a decimal no is obtained by adding a 1 to its 9‘s compliment. Example: 9‘s compliment of 3465 and 782.54 is 9999 -3465 ---------6534 999.99 -782.54 ----------217.45 ------------------ --------------------
10‘s complement of 4069 is 9999 - 4069 ---------5930 +1 ---------5931 ----------9’s compliment method of subtraction: To perform this, obtain the 9‘s compliment of the subtrahend and it to the minuend now call this no. the intermediate result .if there is a carry to the LSD of this result to get the answer called end around carry.If there is no carry , it indicates that the answer is negative & the intermediate result is its 9‘s compliment. Example: Subtract using 9‘s comp (1)745.81-436.62 (2)436.62-745.82 436.62 745.81 -436.62 -745.81 ------------------309.19 -309.19 ------------------745.81 436.62 +563.37 9‘s compliment of 436.62 +254.18 --------------------1309.18 Intermediate result 690.80 +1 end around carry ----------309.19 ------------If there is ono carry indicating that answer is negative . so take 9‘s complement of intermesiate result & put minus sign (-) result should ne -309.19 If carry indicates that the answer is positive +309.19 10’s compliment method of subtraction: To perform this, obtain the 10‘s compliment of the subtrahend& add it to the minuend. If there is a carry ignore it. The presence of the carry indicates that the answer is positive, the result is the answer. If there is no carry, it indicates that the answer is negative & the result is its 10‘s compliment. Obtain the 10‘s compliment of the result & place negative sign infront to get the answer.
Example: (a)2928.54-41673 2928.54 -0416.73 ---------2511.81 ----------2928.54 +9583.27 ---------12511.81 10‘s compliment of 436.62 ignore the carry (b)416.73-2928.54 0416.73 -2928.54 ----------2511.81 --------0416.73 +7071.46 -----------7488.19 The Binary Number System: It is a positional weighted system. The base or radix of this no. system is 2 Hence it has two independent symbols. The basic itself can‘t be a symbol. The symbol used are 0 and 1.The binary digit is called a bit. A binary no. consist of a sequence of bits each of which is either a 0 or 1. The binary point seperates the integer and fraction parts. Each digit (bit) carries a weight based on its position relative to the binary point. The weight of each bit position is on power of 2 greater than the weight of the position to its immediate right. The first bit to the left of the binary point has a weight of 20 & that column is called the Units Column.The second bit to the left has a weight of 21 & it is in the 2‘s column & the third has weight of 22& so on.The first bit to the right of the binary point has a weight of 2-1 & it is said to be in the ½ ‗s column , next right bit with a weight of 2-2 is in ¼‘s column so on..The decimal value of the binary no. is the sum of the products of all its bits multiplied by the weight of their respective positions. In general , binary no. wioth an integer part of (n+1) bits & a fraction parts of k bits can be dn dn-1 dn-2 ………d1 d0.d-1 d-2 d-3 …….d-k In decimal equivalent is (dn x2n)+(dn-1 x2n-1)+ ………(d1 x21)+(d0 x20)+(d-1 x2-1)(d-2 x2-2) ……. The decimal equivalent of the no. system dn dn-1 dn-2 ………d1 d0.d-1 d-2 d-3 …….d-k in any system with base b is (dn xbn)+(dn-1 xbn-1)+ ………(d1 xb1)+(d0 xb0)+(d-1 xb-1)(d-2 xb-2) ……. The binary no. system is used in digital computers because the switching circuits used in these computers use two-state devices such as transistors , diodes etc. A transistor can be OFF or ON a switch can be OPEN or CLOSED , a diode can be OFF or ON etc( twopossible states). These two states represented by the symbols 0 & 1 respectively.

Lecture Notes