JNTU

JNTU B.Tech II SWITCHING THEORY AND LOGIC DESIGN Previous Question Paper (March - 2021 - R13)

B.Tech | II | I | R13 | March - 2021

Answer PaperPaper Code: 123BU/113BUMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Solve for x: (i) (367)8=(x)2 (ii) (378.93)10 = (x)8 (iii) (B9F.AE)16 = (x)8 (iv) (16)10 = (100)x

7 marks

Answer

  1. (367)8: Each octal digit is represented by 3 bits. 3=011, 6=110, 7=111. Thus, (011110111)2.
  2. (378.93)10: For the integer part, 378/8=47 R 2, 47/8=5 R 7, 5/8=0 R 5. Result: 572. For the fractional part, 0.93×8=7.44 (7), 0.44×8=3.52 (3), 0.52×8=4.16 (4). Result: (572.734)8.
  3. (B9F.AE)16: Convert to binary first: B=1011, 9=1001, F=1111, A=1010, E=1110. Binary: 101110011111.10101110. Grouping into 3 bits from the radix point: 101 110 011 111 . 101 011 100. Result: (5637.534)8.
  4. (16)10 = (100)x: 16 = 1×x2 + 0×x1 + 0×x0. Therefore, x2 = 16, which gives x = 4.
b

Convert (163.875)10 to binary, octal, hexadecimal.

8 marks

Answer

To convert (163.875)10:

  • Binary: Integer 163 = 101000112. Fractional 0.875×2=1.75 (1), 0.75×2=1.5 (1), 0.5×2=1.0 (1). Result: (10100011.111)2.
  • Octal: Integer 163/8=20 R 3, 20/8=2 R 4, 2/8=0 R 2. Integer part is 243. Fractional 0.875×8=7.0. Result: (243.7)8.
  • Hexadecimal: Integer 163/16=10 R 3. 10 is A. Integer part is A3. Fractional 0.875×16=14.0. 14 is E. Result: (A3.E)16.
215 marks
a

Prove the following Boolean theorems: (i) AB+A'C = (A+C)(A'+B) (ii) AB+A'C+BC = AB+A'C

7 marks

Answer

Sign in to view the complete answer paper.

b

Simplify the following Boolean expressions: (i) ABC+AB'+ABC' (ii) ACD+A'BCD.

8 marks

Answer

Sign in to view the complete answer paper.

315 marks
a

Minimize the following expressions using K-map and realize using NAND Gates: f = Σm(0,1,4,5,6,7,9,11,15) + d(10,14)

7 marks

Answer

Sign in to view the complete answer paper.

b

Minimize the following expression using K-map and realize using NOR Gates: f = ΠM(1,4,5,11,12,14) . d(6,7,15).

8 marks

Answer

Sign in to view the complete answer paper.

415 marks
a

Explain the differences between a MUX and a DEMUX. Realize 16-input multiplexer by cascading of two 8-input multiplexers 74151.

7 marks

Answer

Sign in to view the complete answer paper.

b

Realize the function f(A,B,C,D) = Π(1,4,6,10,14) . d(0,8,11,15) using: (i) 16:1 MUX (ii) 8:1 MUX

8 marks

Answer

Sign in to view the complete answer paper.

515 marks
a

Is an X-OR function a threshold function? Justify.

7 marks

Answer

Sign in to view the complete answer paper.

b

Realize the following function using a PROM of size 8x3: F = Σm(1,2,4,6).

8 marks

Answer

Sign in to view the complete answer paper.

615 marks
a

Draw and explain 4-bit universal shift register.

7 marks

Answer

Sign in to view the complete answer paper.

b

Explain the differences between asynchronous and synchronous counters. Design a MOD-6 ripple counter.

8 marks

Answer

Sign in to view the complete answer paper.

715 marks
a

What are the Moore and Melay machines? Compare them.

7 marks

Answer

Sign in to view the complete answer paper.

b

Explain the procedure for state minimization using merger graph and merger table.

8 marks

Answer

Sign in to view the complete answer paper.

815 marks
a

Differentiate between an ASM chart and a conventional flow chart.

7 marks

Answer

Sign in to view the complete answer paper.

b

Explain in detail the ASM technique of designing a sequential circuit.

8 marks

Answer

Sign in to view the complete answer paper.

About this paper

This page contains the question paper of SWITCHING THEORY AND LOGIC DESIGN from JNTU for B.Tech MECT-MCT-MEM II I, March - 2021, R13.

For the solved question paper with answers, switch to the Answer Paper tab.

Other papers in SWITCHING THEORY AND LOGIC DESIGN

Loading Question Papers

Fetching data from database...