EC1312 Digital Logic Circuits B.E Question Bank : niceindia.com
Name of the College : Noorul Islam College of Engineering
University : Anna University
Degree : B.E
Department : Electrical & Electronic Engineering
Subject Code/Name : EC 1312 – Digital Logic Circuits
Document Type : Question Bank
Website : niceindia.com
Download Model/Sample Question Paper : https://www.pdfquestion.in/uploads/niceindia.com/3000-EC_1312_Digital_Logic_Circuits_.pdf
NICE Digital Logic Circuits Question Paper
1) Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction (a) X-Y and (b) Y – X using 2’s complements.
Related : Noorul Islam College of Engineering CS1629 Network Security M.E Question Bank : www.pdfquestion.in/3107.html
a) X = 1010100 :
2’s complement of Y = + 0111101 ————–
Sum = 10010001
Discard end carry
Answer : X – Y = 0010001
b) Y = 1000011 :
2’s complement of X = + 0101100
—————
Sum = 1101111
There is no end carry,
Therefore the answer is Y-X = -(2’s complement of 1101111) = -0010001
2). Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction (a) X -Y and (b) Y – X using 1’s complements.
a). X – Y = 1010100 – 1000011 :
X = 1010100
1’s complement of Y = + 0111100
————–
Sum = 10010000
End -around carry = + 1
————–
Answer : X – Y = 0010001
b). Y – X = 1000011 – 1010100 :
Y = 1000011
1’s complement of X = + 0101011
———–
Sum = + 1101110
There is no end carry.
Therefore the answer is Y – X = -(1’s complement of 1101110) = -0010001
3). What is meant by parity bit :
A parity bit is an extra bit included with a message to make the total number of 1’s either even or odd. Consider the following two characters and their even and odd parity :
With even parity With odd parity
ASCII A = 1000001 01000001 11000001
ASCII T = 1010100 11010100 01010100
In each case we add an extra bit in the left most position of the code to produce an even number of 1’s in the character for even parity or an odd number of 1’s in the character for odd parity. The parity bit is helpful in detecting errors during the transmission of information from one location to another.
4).What are registers :
register is a group of binary cells. A register with n cells can store any discrete quantity of information that contains n bits. The state of a register is an n-tuple number of 1’s and 0’s, with each bit designating the state of one cell in the register.
5). What is meant by register transfer? :
A register transfer operation is a basic operation in digital systems. It consists of transfer of binary information from one set of registers into another set of registers. The transfer may be direct from one register to another, or may pass through data processing circuits to perform an operation.
6). Define binary logic? :
Binary logic consists of binary variables and logical operations. The variables are designated by the alphabets such as A, B, C, x, y, z, etc., with each variable having only two distinct values : 1 and 0. There are three basic logic operations: AND, OR, and NOT.
7). Define logic gates? :
Logic gates are electronic circuits that operate on one or more input signals to produce an output signal. Electrical signals such as voltages or currents exist throughout a digital system in either of two recognizable values. Voltage- operated circuits respond to two separate voltage levels that represent a binary variable equal to logic 1 or logic 0.
8).Define duality property :
Duality property states that every algebraic expression deducible from the postulates of Boolean algebra remains valid if the operators and identity elements are interchanged. If the dual of an algebraic expression is desired, we simply interchange OR and AND operators and replace 1’s by 0’s and 0’s by 1’s.
9).Find the complement of the functions F1 = x’yz’ + x’y’z and F2 = x(y’z’ +yz). By applying De Morgan’s theorem as many times as necessary :
F1’ = (x’yz’ + x’y’z)’ = (x’yz’)’(x’y’z)’ = (x + y’ + z)(x + y +z’)
F2’ = [x(y’z’ + yz)]’ = x’ + (y’z’ + yz)’
= x’ + (y’z’)’(yz)’
= x’ + (y + z)(y’ + z’)
10).Find the complements of the functions F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz). by taking their duals and complementing each literal :
F1 = x’yz’ + x’y’z
The dual of F1 is (x’ + y + z’)(x’ + y’ + z) Complementing each literal : (x + y’ + z)(x + y + z’)
F2 = x(y’z’ + yz).
The dual of F2 is x + (y’ + z’)(y + z). Complement of each literal : x’ + (y + z)(y’ + z’)
11).State De Morgan’s theorem :
De Morgan suggested two theorems that form important part of Boolean algebra.
They are,
1) The complement of a product is equal to the sum of the complements. (AB)’ = A’ + B’
2) The complement of a sum term is equal to the product of the complements. (A + B)’ = A’B’