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JNTU B.Tech II DIGITAL LOGIC DESIGN Previous Question Paper (October - 2020 - R15)

B.Tech | II | II-I | R15 | October - 2020

Answer PaperPaper Code: 123CTMaximum Marks: 75Time: 2 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Write about error detecting codes and error correcting codes used in digital systems.

8 marks

Answer

Error detecting and correcting codes are redundant bits added to digital data to ensure reliable transmission and storage. Error detecting codes allow the receiver to determine if an error occurred during transmission, but they cannot identify the specific bit in error. Common examples include Parity bits, which add a single bit to make the total number of 1s even or odd, and Checksums or Cyclic Redundancy Checks (CRC), which use polynomial division to verify data integrity.

Error correcting codes go further by allowing the receiver to detect and automatically fix errors without retransmission. This is achieved by adding sufficient redundant bits to create a unique distance between valid code words, known as the Hamming distance. The most common example is the Hamming code, which uses multiple parity bits placed at power-of-two positions to identify and correct single-bit errors.

b

Convert (i) 68BE16 to Binary and then Octal (ii) Convert 34510 to binary and hexadecimal.

7 marks

Answer

(i) Conversion of 68BE16:

To convert Hexadecimal to Binary, replace each hex digit with its 4-bit binary equivalent: 6=0110, 8=1000, B=1011, E=1110. Thus, 68BE16 = 01101000101111102.

To convert to Octal, group the binary digits into sets of three starting from the right: 011 010 001 011 111 0. Adding leading zeros gives 011 010 001 011 111 010, which corresponds to 3213728.

(ii) Conversion of 34510:

To convert to Binary, perform repeated division by 2: 345/2 = 172 R 1, 172/2 = 86 R 0, 86/2 = 43 R 0, 43/2 = 21 R 1, 21/2 = 10 R 1, 10/2 = 5 R 0, 5/2 = 2 R 1, 2/2 = 1 R 0, 1/2 = 0 R 1. Reading remainders from bottom to top: 1010110012.

To convert to Hexadecimal, divide by 16: 345/16 = 21 R 9, 21/16 = 1 R 5, 1/16 = 0 R 1. Reading remainders from bottom to top: 15916.

215 marks
a

Describe the basic theorems and properties of Boolean algebra.

9 marks

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b

Express the following function in sum of minterms and product of maxterms: F(A,B,C,D) = B'D + A'D + BD

6 marks

Answer

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3

Implement the following Boolean function F, together with the don't care conditions d, using no more than two NOR gates.

F (A,B,C,D) = Σ (0, 1, 2, 9, 11)
d (A,B,C,D) = Σ (8, 10, 14, 15)

Assume that both the normal and complement inputs are available.

15 marks

Answer

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4

Simplify the following Boolean functions using a four variable Karnaugh map method and implement the simplified function using NOR gates.

F(A,B,C,D) = Σ(2, 4, 5, 6, 7, 10, 13, 15)
15 marks

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515 marks
a

Discuss the analysis and design procedure and draw the logic diagram of a 4 bit BCD subtractor.

9 marks

Answer

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b

Implement the following Boolean function with a multiplexer:

F (A,B,C,D) = Σ (0, 1, 3, 4, 8, 9, 15)
6 marks

Answer

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6

Design a combinational circuit that converts a 4 bit gray code to 4 bit binary converter using XOR gates and explain its operation with examples.

15 marks

Answer

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715 marks
a

When does a race condition occur in asynchronous sequential circuits? Explain the noncritical and critical race conditions.

6 marks

Answer

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b

Draw the logic diagram of a 4 bit binary ripple down counter using flip-flops that trigger on the negative edge of the clock and explain its operation with truth table and timing diagram.

9 marks

Answer

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815 marks
a

Explain the memory hierarchy in terms of nature, capacity, access times.

6 marks

Answer

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b

Explain in detail about sequential memories with examples.

9 marks

Answer

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