1
Find the octal equivalent of hexadecimal numbers AB.CD

2 M

2
State and prove the consensus theorem.

2 M

3
Implement the function G=Σm(0,3) using a 2×4 decoder.

2 M

4
Draw the circuit for 2-to-1 line multuplexer.

2 M

5
Write the characteristics table and equation of JK flip flop.

2 M

6
Write any two applications of shift register.

2 M

7
Define Race condition.

2 M

8
What are the types of hazards?

2 M

9
What is memory decoding?

2 M

10
Define ASIC.

2 M

Answer any one question from Q11 (a) & Q11 (b)

11 (a)
Simplify the following functions, using K-map techniques

(i) G=Π

(ii) f(W,X,Y,Z)=Σm(0, 7, 8, 9, 10, 12)+ Σd(2, 5, 13)

(i) G=Π

_{M}(0, 1, 3, 7, 9, 11)(ii) f(W,X,Y,Z)=Σm(0, 7, 8, 9, 10, 12)+ Σd(2, 5, 13)

16 M

11 (b)
Minimize the expression using Quine McCluskey (Tabulation) method F=Σm (0,1,9,15,24,29,30)+Σd(8,11,31)

16 M

Answer any one question from Q12 (a) & Q12 (b)

12 (a)
Design a circuit that converts 8421 BCD code to Excess-3 code.

16 M

12 (b)
Implement the following Boolean function using 8 to 1. Multiplexer F(A, B, C, D)= A'BD'+ACD+B'CD+A'C'D. Also implement the function using 16 to 1 multiplexer.

16 M

Answer any one question from Q13 (a) & Q13 (b)

13 (a)
Implement T flip flop using D flip flop and JK flip flop using D flip flop.

16 M

13 (b)
Design a synchronous conunter which counts in the sequence 000, 001, 010, 011, 100, 101, 111, 000 using D-FF.

16 M

Answer any one question from Q14 (a) & Q14 (b)

14 (a)
Explain the steps for the design of asynchronous sequential circuits.

16 M

14 (b)
Implement the switching function F=?m(1,3,5,7,8,9,14,15) by a static hazard free two level AND OR gate network.

16 M

15 (a)
Implement the following function using PLA

A(x,y,z)=Σm(1,2,4,6)

B(x,y,z)=Σm(0,1,6,7)

C(x,y,z)=Σm(2,6)

A(x,y,z)=Σm(1,2,4,6)

B(x,y,z)=Σm(0,1,6,7)

C(x,y,z)=Σm(2,6)

16 M

15 (b)
The following messages have been coded in the even parity Hamming code and transmitted through a noisy channel. Decode the messages, assuming that at most a single error has occurred in each codeword.

(i) 1001001

(ii) 0111001

(iii) 1110110

(iv) 0011011.

(i) 1001001

(ii) 0111001

(iii) 1110110

(iv) 0011011.

16 M

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