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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
May2017
ASSESSMENT_CODE BCA114_May2017
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
496303_1
What is D flip-flop? Write its Truth table.
QUESTION_TEXT
D flip-flop definition:
Table of D flipflop:
D flip-flop or Delay flip flop is a sequential circuit whose output takes the value of input aft
SCHEME OF
EVALUATION
Truth table of D flip-flop
Clock
Rising edge
Rising edge
Non-Rising
D
0
1
X
Q
0
1
constant
Qprev
X
X
(2.5 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496304_2
List out any five theorems in Boolean Algebra.
QUESTION_TEXT
The important theorems are:
Theorem-1: X + X = X
SCHEME OF EVALUATION
Theorem-2: X • X = X
Theorem-3: X + 0 = X
Theorem-4: X • 1 = X
Theorem-5: X • 0 = 0
Theorem-6: X + 1 = 1
Theorem-7: (X + Y)’ = X’ • Y’
Theorem-8: (X • Y)’ = X’ + Y’
Theorem-9: X + X•Y = X
Theorem-10: X •(X + Y) = X
Theorem-11: X + X’Y = X+Y
Theorem-12: X’ • (X + Y’) = X’Y’
Theorem-13: XY + XY’ = X
Theorem-14: (X’+Y’) • (X’ + Y) = Y’
Theorem-15: X + X’ = 1
Theorem-16: X • X’ = 0
(5 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496305_3
QUESTION_TEXT
Define the term Switch. Why are mechanical switches replaced by
electronic switching devices?
Definition –
(2 marks)
SCHEME OF
EVALUATION
Reasoning answer
We can define a switch in terms of its operation and use. In general, a
switch is a device that can be used to either establish or remove (also
called make or break) connections between at least two points in an
electric or electronic circuit.
Mechanical switches are replaced by electronic switching devices
because it can be automated and intelligently controlled.
(3 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496306_4
What is 300-baud modem and 2400 bit/sec modem?
QUESTION_TEXT
If a modem transmitted 1-bit every baud, it’s a 300bit/sec modem or
300-baud modem.
(2.5 mark)
SCHEME OF
EVALUATION
A modem that switches its state 600 times/sec and for each baud if it
transmits 4 bits, it is said to be a 2400 bit/sec modem.
(2.5 mark)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496307_5
QUESTION_TEXT
What are the differences between deep space telecom modems and
landline modems for digital modulation?
Differences between deep space telecom modems and landline
modems for digital modulation include:
SCHEME OF
EVALUATION
 digital modulation formats that have high doppler immunity are
typically used
(2 marks)
 waveform complexity tends to be low-typically binary phase shift
keying
(1 mark)
 Error correction varies mission to mission, but is typically much
stronger than most landline modems.
(2 marks)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496308_6
Write short notes on Pulse Width Modulator.
QUESTION_TEXT
SCHEME OF
EVALUATION
Pulse Width Modulator is the simplest type of DAC. In this method, a
stable current or voltage is switched into a low-pass analog filter and
duration is determined by the digital input code. This is a commonly used
technique in speed control of electric motor and is now commonly used in
high fidelity audio applications. (5 marks
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496309_7
Give an overview about design of Modulo n-counter
QUESTION_TEXT
The input lines……0 and N-1 (1 mark)
In the first design step 1……development of counter only (2
marks)
SCHEME OF
EVALUATION
The second design step2…….course a function of N (2 marks)
The third design step 3 …….a sequence of impulses (2 marks)
The fourth design step 4 is……. By a master-slave flip flop(2
marks)
Note……work properly together (1 mark)
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
QUESTION_TEXT
496310_8
Explain the concept of signed magnitude representation, 1’s Complement and 2’s compleme
system.
The signed magnitude representation is one of the methods to represent the negative and po
system. It is a regular binary representation with addition of one more bit in the MSB side re
negative or positive. The MSB “0” represent the positive value and MSB “1” represent the n
The range for this representation is: - 2n-1 -1 to + 2n-1 -1
The drawbacks of this representation are: It has both positive and negative zero, and comple
computation.
marks)
1’s Complement
SCHEME OF
EVALUATION
The 1’s complement is slightly different from the signed magnitude number. All binary bits
represented along with the sign bit.
The range function for this representation is: - 2n-1 -1 to + 2n-1 -1
The drawbacks of this representation are: It has both positive and negative zero and comple
computation.
marks)
2’s Complement
The 2’s Complement is asymmetric system with additional bit is required to represent any n
are complemented and added 1 with it. The range function for this representation is: - 2n-1
1.
marks)
QUESTION_TYPE DESCRIPTIVE_QUESTION
QUESTION_ID
496311_9
Explain Excess-3 binary coded decimal (XS-3) in brief.
QUESTION_TEXT
SCHEME OF
EVALUATION
Excess-3 binary coded decimal (XS-3), also called biased representation or Excess-N, is a n
computers that uses a pre-specified number N as a biasing value. It is a way to represent val
and negative numbers. In XS-3, numbers are represented as decimal digits, and each digit is
value plus 3 (the "excess" amount):
The smallest binary number represents the smallest value. (i.e. 0 - Excess Value). The great
largest value. (i.e. 2N - Excess Value - 1)
Decimal
Binary
Decimal
Binary
Decimal
Bin
-3
0000
1
0100
5
1000
-2
0001
2
0101
6
1001
-1
0010
3
0110
7
1010
0
0011
4
0111
8
1011
To encode a number such as 127, then, one simply encodes each of the decimal digits as abo
The primary advantage of XS-3 coding over BCD coding is that a decimal number can be n
as easily as a binary number can be ones' complemented; just invert all bits.
Adding Excess-3 works on a different algorithm than BCD coding or regular binary number
together, the result is not an XS-3 number. For instance, when you add 1 and 0 in XS-3 the
order to correct this problem, when you are finished adding each digit, you have to subtract
decimal 10 and add three if the number is greater than or equal to decimal 10.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
496312_10
QUESTION_TEXT
SCHEME OF
EVALUATION
Write the steps to simplify the given Boolean expression using QuineMcCluskey method.
The following steps are followed to simplify the given Boolean
expression using Quine-McCluskey method.
Step 1: If the function is not given in minterm form, then translate the
decimal values to minterms.
Step 2: If the function is not given in binary minterm form, then translate
the decimal values to binary notation of minterms.
Step 3: Minterms are grouped depending upon number of one’s they have
and then enter in table form.
Step 4: The minterms in the adjacent blocks are compared to determine
the minterms which are differed by only one bit. Replace the missing
literal by – and place the minterms in the next column. The minterms in
the present column which are combined are placed with a check mark
Step 5: The minterms in the adjacent blocks are compared to determine
the minterms which are differed by only one bit. Note -’s must line up.
Replace the missing literal by – and place the minterms in the next
column. The minterms in the present column which are combined are
placed with a check mark.
Step 6: The prime implicant chart is to be formed with minterms in a row
and prime implicants in column. When there is intersection of minterm
and the prime implicant, the  placed
Step 7: Select prime implicants for minterms with only one  in a column.
Step 8: Repeat step 8 for minterms with only two  in a column. If the
minterms of those colums are already included, then not required to be
considered.
Step 9: Note the final expression from the table and Translate to literal
notation.
(Each step carries 1 mark step 5 contains 2 marks)