Thursday, 25 May 2017

bit102 smu bsc it spring 2017 (jul/aug 2017 exam) Ist sem assignment

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DRIVE SPRING 2017
PROGRAM Bachelor of Science in Information Technology - B.SC(IT)
SEMESTER I
SUBJECT CODE & NAME BIT102 – Digital Electronics


Assignment Set -1

1 Convert the following hexadecimal numbers to base 10:
a. 145
b. A2C1

Answer: a. 14516 in Hexadecimal number system and we want to translate it into Decimal.

14516 = 1*16^2+4*16^1+5*16^0

= 256+64+5



2 Construct XOR and NAND Gate using NOR Gate.

Answer: a)


3 Simplify f (a, b, c, d) = Σm(0, 2, 4, 6, 7, 8, 9, 11, 12, 14)

Answer: The given function is:
f(a.b.c.d)= ∑m(0,2,4, 6,7,8,9,11,12,14)
This is a four variable function so the k


Assignment Set -2

1 Write a short note on J-K Master Slave Flip-Flop.

Answer: J-K Master Slave Flip Flop
The logic symbol for the master-slave flip-flop only indicates the initial inputs to the master and the outputs from the slave as



2 Explain the working of 4 bit Johnson counter with the help of neat diagram.

Answer: A Johnson counter is constructed using serial-in and serial-out (SISO) shift register. The output of the shift register is connected back to the input after passing it through an inverter. Depending on the initial bit pattern stored in the shift register, the shift register content changes for every clock and the bit pattern gets repeated after 2n clocks, where n is the number of bits in the shift register. These are called “walking ring” counters and have specialist applications like digital-to-analog converters (DAC) etc. Many user applications require counter outputs in the form of decimal digits in place of binary format. One way to get it is by using a decoder that decodes the binary output and converts it


3 Explain BCD adder and BCD subtractor.

Answer: BCD Adder
The addition in BCD can be performed in three steps: 1) Add the numbers in binary. 2) Check if the number is greater than 9. 3) Then convert the sum of two numbers to BCD number by adding correction value. The correction value will be equal to 6 if the sum output in step 1 is greater than 9, else the correction value will be zero. For example: 8 + 6 = 14 = [1000] + [0110] = [0000 1110] in binary. However, in BCD, per \



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