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# Important Short Question and Answers: Digital Electronics

Basic Electrical and electronics - Digital Electronics - Important Short Question and Answers: Digital Electronics

DIGITAL ELECTRONICS

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.

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

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:

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).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.

5).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.

6).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.

7).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')

8).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')

9).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'

10).Reduce A.A'C

A.A'C = 0.c            [A.A' = 1]

= 0

11). Reduce A(A + B)

A(A + B) = AA + AB

= A(1 + B)          [1 + B = 1]

= A.

12.            Reduce A'B'C' + A'BC' + A'BC

A'B'C' + A'BC' + A'BC = A'C'(B' + B) + A'B'C

= A'C' + A'BC         [A + A' = 1]

= A'(C' + BC)

= A'(C' + B)            [A + A'B = A + B]

13.) Reduce AB + (AC)' + AB'C(AB + C)

AB + (AC)' + AB'C(AB + C) = AB + (AC)' + AAB'BC + AB'CC

= AB + (AC)' + AB'CC [A.A' = 0]

= AB + (AC)' + AB'C    [A.A = 1]

= AB + A' + C' =AB'C  [(AB)' = A' + B']

= A' + B + C' + AB'C    [A + AB' = A + B]

= A' + B'C + B + C'       [A + A'B = A + B]

= A' + B + C' + B'C

=A' + B + C' + B'

=A' + C' + 1

= 1    [A + 1 =1]

14. Simplify the following expression Y = (A + B)(A + C' )(B' + C' )

Y = (A + B)(A + C' )(B' + C' )

= (AA' + AC +A'B +BC )(B' + C')            [A.A' = 0]

= (AC + A'B + BC)(B' + C' )

= AB'C + ACC' + A'BB' + A'BC' + BB'C + BCC'

= AB'C + A'BC'

15).Simplify the following using De Morgan's theorem [((AB)'C)'' D]'

[((AB)'C)'' D]' = ((AB)'C)'' + D'                                     [(AB)' = A' + B']

= (AB)' C + D'

= (A' + B' )C + D'

16.Show that (X + Y' + XY)( X + Y')(X'Y) = 0

(X + Y' + XY)( X + Y')(X'Y) = (X + Y' + X)(X + Y' )(X' + Y)         [A + A'B = A + B]

= (X + Y' )(X + Y' )(X'Y)        [A + A = 1]

= (X + Y' )(X'Y)   [A.A = 1]

= X.X' + Y'.X'.Y

= 0    [A.A' = 0]

17).Prove that ABC + ABC' + AB'C + A'BC = AB + AC + BC

ABC + ABC' + AB'C + A'BC =AB(C + C') + AB'C + A'BC

=AB + AB'C + A'BC

=A(B + B'C) + A'BC

=A(B + C) + A'BC

=AB + AC + A'BC

=B(A + C) + AC

=AB + BC + AC

=AB + AC +BC             ...Proved

18).Convert the given expression in canonical SOP form Y = AC + AB + BC

Y = AC + AB + BC

=AC(B + B' ) + AB(C + C' ) + (A + A')BC

=ABC + ABC' + AB'C + AB'C' + ABC + ABC' + ABC

=ABC + ABC' +AB'C + AB'C'                  [A + A =1]

19).Convert the given expression in canonical POS form Y = ( A + B)(B + C)(A + C)

Y = ( A + B)(B + C)(A + C)

= (A + B + C.C' )(B + C + A.A' )(A + B.B' + C)

= (A + B + C)(A + B + C' )(A + B +C)(A' + B +C)(A + B + C)(A + B' + C)

[A + BC = (A + B)(A + C)   (Distributive law]

= (A + B + C)(A + B + C')(A' + B + C)(A' + B + C)(A + B' + C)

20)                       Write down the steps in implementing a Boolean function with levels of NAND Gates?

Simplify the function and express it in sum of products.

Draw a NAND gate for each product term of the expression that has at least two literals.

The inputs to each NAND gate are the literals of the term. This constitutes a group of first level gates. Draw a single gate using the AND-invert or the invert-OR graphic symbol in the second level, with inputs coming from outputs of first level gates.

A term with a single literal requires an inverter in the first level. How ever if the single literal is complemented, it can be connected directly to an input of the second level NAND gate.

21) Give the general procedure for converting a Boolean expression in to multilevel NAND diagram?

Draw the AND-OR diagram of the Boolean expression.

Convert all AND gates to NAND gates with AND-invert graphic symbols. Convert all OR gates to NAND gates with invert-OR graphic symbols.

Check all the bubbles in the same diagram. For every bubble that is not compensated by another circle along the same line, insert an inverter or complement the input literal.

22) What are combinational circuits?

A combinational circuit consists of logic gates whose outputs at any time are determined from the present combination of inputs. A combinational circuit performs an operation that can be specified logically by a set of Boolean functions. It consists of input variables, logic gates, and output variables.

23) Give the design procedures for the designing of a combinational circuit.

The procedure involves the following steps,

From the specification of the circuit, determine the required number of inputs and outputs and assign a symbol to each.

Derive the truth table that defines the required relationships between inputs and outputs. Obtain the simplified Boolean functions for each output as a function of the input variables.

Draw the logic diagram and verify the correctness of the design.

A combinational circuit that performs the addition of two bits is called a half adder. A half adder needs two binary inputs and two binary outputs. The input variables designate the augend and addend bits; the output variables produce the sum and carry

A combinational circuit that performs the adtion of three bits is a full adder.It consists of three inputs and two outputs.

The carry into sign bit position and the carry out of the sign bit position. If these two carries are not equal, an overflow has occurred.

26.            Represent binary number 1101 - 101 in power of 2 and find its decimal equivalent

N = 1 x 2 3 + 1 x 2 2 + 0 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 0 x 2 -2 + 1 x 2 -3

= 13.625 10

27.            What are the different classification of binary codes?

1. Weighted codes

2. Non - weighted codes

3. Reflective codes

. Sequential codes

5. Alphanumeric codes

6. Error Detecting and correcting codes.

28.            Write the names of basic logical operators.

1. NOT / INVERT

2. AND

3. OR

29.            Simplify the following expression y = (A + B) (A = C) (B + C)

= (A A + A C + A B + B C) (B + C) = (A C + A B + B C) (B + C)

= A B C + A C C + A B B + A B C + B B C + B C C = A B C = A B C

30.            Show that the NAND connection is not associative

The NAND connection is not associative says that A . B . C A . B. C

A . B + C A + B C AB + C A + BC

31.            What is a Logic gate?

Logic gates are the basic elements that make up a digital system. The electronic gate is a circuit that is able to operate on a number of binary inputs in order to perform a particular logical function.

32.            Write the names of Universal gates.

1. NAND gate

2. NOR gate

33. Why are NAND and NOR gates known as universal gates?

The NAND and NOR gates are known as universal gates, since any logic function can be implemented using NAND or NOR gates.

34. Define combinational logic

When logic gates are connected together to produce a specified output for certain specified combinations of input variables, with no storage involved, the resulting circuit is called combinational logic.

35.            Explain the design procedure for combinational circuits

¢  The problem definition

¢  The determination of number of available input variables & required O/P variables.

¢  Assigning letter symbols to I/O variables

¢  Obtain simplified boolean expression for each O/P.

¢  Obtain the logic diagram.

The logic circuit which performs the addition of two bits is a half adder.

The circuit which performs the addition of three bits is a full adder.

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Basic Electrical and electronics : Digital Electronics : Important Short Question and Answers: Digital Electronics |