Zeal Education Society’s

ZEAL POLYTECHNIC, PUNE.

NARHE │PUNE -41 │ INDIA

SECOND YEAR (SY)

DIPLOMA IN COMPUTER ENGINEERING
SCHEME: I SEMESTER: III

NAME OF SUBJECT: DIGITAL TECHNIQUE

Subject Code: 22320

MSBTE QUESTION PAPERS & MODEL ANSWERS

1. MSBTE WINTER-18 EXAMINATION
2. MSBTE SUMMER-19 EXAMINATION
3. MSBTE WINTER-19 EXAMINATION

Page 1 of 10
 22320
11819
3 Hours / 70 Marks Seat No.

Instructions – (1) All Questions are Compulsory.

(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.

Marks

1. Attempt any FIVE of the following: 10

a) Write the radix of binary, octal, decimal and hexadecimal
number system.
b) Draw the circuit diagram for AND and OR gates using diodes.
c) Write simple examples of boolean expression for SOP and POS.
d) State the necessity of multiplexer.
e) Draw logic diagram of T flip-flop and give its truth table.
f) Define modulus of a counter. Write the numbers of flip flops
required for Mod-6 counter.
g) State function of preset and clear in flip flop.

P.T.O.
22320 [2]
Marks
2. Attempt any THREE of the following: 12
a) Draw the block diagram of Programmable Logic Array.
b) Convert -
(i) (255)10 = ( ? )16 = ( ? )8
(ii) (157)10 = ( ? )BCD = (? ) Excess3
c) Draw the symbol, truth table and logic expression of any
one universal logic gate. Write reason why it is called
universal gate.
d) Minimize the following expression using K-Map.
f (A, B, C, D) = Σm (0, 1, 2, 4, 5, 7, 8, 9, 10)

3. Attempt any THREE of the following: 12

a) Compare TTL and CMOS logic families on the basis of
following:
(i) Propagation delay
(ii) Power Dissipation
(iii) Fan-out
(iv) Basic gate
b) Describe the function of Full Adder Circuit using its truth table,
K-Map simplification and logic diagram.
c) Realize the basic logic gates, NOT, OR and AND gates using
NOR gates only.
d) Describe the working of JK flip-flop with its truth table and
logic diagram.

4. Attempt any THREE of the following: 12

a) Draw and explain working of 4 bit serial Input parallel Output
shift register.
b) Draw 16:1 MUX tree using 4:1 MUX.
c) Calculate analog output of 4 bit DAC for digital input 1101.
Assume VFS = 5V.
d) State De Morgan’s theorem and prove any one.
e) Design one digit BCD Adder using IC 7483.
22320 [3]
Marks
5. Attempt any TWO of the following: 12
a) Subtract using 2’s compliment method
(35)10 – (5)10
b) Design a 4 bit synchronous counter and draw its logic diagram.
c) Describe the working of successive Approximation ADC. Define
Resolution and conversion time associated with ADC.

6. Attempt any TWO of the following: 12

a) Design 4 bit Binary to Gray code converter.
b) Compare the following (Any three points)
(i) Volatile with Non-volatile memory
(ii) SRAM with DRAM memory
c) Give block schematic of decade counter IC 7490. Design
Mod-7 counter using this IC.
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MODEL ANSWER
WINTER– 18 EXAMINATION
Subject Title: Digital Techniques Subject Code: 22320
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for
subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures
drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and
there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q. Sub Answer Marking

No. Q.N. Scheme
Q.1 Attempt any FIVE of the following : Total Marks
10
a) Write the radix of binary,octal,decimal and hexadecimal number system. 2M
Ans: Radix of: Binary – 2 ½ M each

Octal - 8

Decimal - 10

Hexadecimal -16

b) Draw the circuit diagram for AND and OR gates using diodes. 2M

Ans: 1 M each
Diode AND gate :Diode OR gate :

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c) Write simple example of Boolean expression for SOP and POS. 2M

Ans: SOP form: 1 M each

(any proper
example can
be
considered)
POS form:

d) State the necessity of multiplexer. 2M

Ans: Necessity of Multiplexer:

2 M(any two
 It reduces the number of wires required to pass data from source to proper
destination. points)

 For minimizing the hardware circuit.

 For simplifying logic design.

 In most digital circuits, many signals or channels are to be transmitted,

and then it becomes necessary to send the data on a single line
simultaneously.

 Reduces the cost as sending many signals separately is expensive and

requires more wires to send.

e) Draw logic diagram of T flip-flop and give its truth table. 2M

Ans: Note: Diagram Using logic gates with proper connection also can be 1M (any one
consider. diagram)
Logic Diagram:

1M

OR

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Truth Table:

Input Output Operation Performed

Tn Qn+1
0 Qn No change

1 Qn Toggle

f) Define modulus of a counter. Write the numbers of flip flops required for 2M
Mod-6 counter.

Ans:  Modulus of counter is defined as number of states/clock the counter Definition:

countes. 1M
 The numbers of flip flops required for Mod-6 counter is 3. No. of FF-
1M
g) State function of preset and clear in flip flop. 2M

Ans:  In the flip flop , when the power is switched on, the state of the circuit 1 M for each
is uncertain i.e. may be Q = 1 or Q = 0. function (
 Hence, the function of preset is to set a flip flop i.e. Q = 1andthe table is
optional)
function of clear is to clear a flip flop i.e. Q = 0.

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Q2 Attempt any THREE of the following : 12-Total

Marks
4M
a) Draw the block diagram of Programmable Logic Array.
Ans: Diagram :- 4M

Block diagram of Programmable Logic Array

OR

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b) Convert – 4M
(255)10 = (?)16 = (?)8
(157)10 = (?)BCD = (?) Excess3
Ans: (i) (255)10 = (FF)16 = (377)8

(255)10 = (FF)16 1M

(255)10 = (377)8
1M

(ii) (157)10 = (000101010111)BCD = (010010001010) Excess3

(157)10 = (000101010111)BCD

1M

(000101010111)BCD = (010010001010) Excess3

1M

c) Draw the symbol, truth table and logic expression of any one universal 4M
logic gate. Write reason why it is called universal gate.
Ans: (Note: Any one universal gate has to be considered.)
Universal Gates: NAND or NORSymbol:
1M

Truth table:
1M

Logic expression:
1M

NAND and NOR gates are called as “Universal Gate” as it is possible to

implement any Boolean expression using these gates. 1M

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d) Minimize the following expression using K-Map. 4M

ƒ(A, B, C, D) = ∑m (0, 1, 2, 4, 5, 7, 8, 9, 10)
Ans: 1M–
drawing k
map
1M–
Representin
g function in
k map
1M–
Grouping
1M – Final
expression

OR

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Q. 3 Attempt any THREE: 12-Total

Marks
a) Compare TTL and CMOS logic families on the basis of following: 4M
(i) Propagation delay
(ii) Power Dissipation
(iii) Fan-out
(iv) Basic gate

Ans: NOTE :- ( Relevant points of comparison- 1 M for each point) 1 Marks

each point
Parameter CMOS TTL
Propagation delay 70-105 nsec/more than 10 nsec/Less than
TTL CMOS
Power Dissipation Less 0.1 mW/Less than More 10 mW/ More
TTL than CMOS
Fan-out 50/More than TTL 10/Less than CMOS

Basic gate NAND/NOR NAND

b) Describe the function of full Adder Circuit using its truth table, K-Map 4M
simplification and logic diagram.
Ans: ( Diagram- 1M,Truth table-1M, K-map- 1M,Logic diagram-1 M)

A full adder is a combinational logic circuit that performs addition between

three bits, the two input bits A and B, and carry C from the previous bit.

Block diagram : 1M

1M

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Truth Table :

1M

K-Map :-

1M

Logic Diagram:

(Note: Logic Diagram using basic or universal gate also can be consider)

S= (A ⊕ B) ⊕ Cin

Carry= (AB + ACin +BCin)

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c) Realize the basic logic gates, NOT, OR and AND gates using NOR gates 4M
only.
Ans:
( NOT GATE USING NOR GATE:1 M ) 1M

where, X = A NOR A
x=Ā

(AND GATE USING NOR GATE:1.5 MARKS)

1.5M
___
Q=Ā+̅B = Ā+̅B
___
________________

=A.B
= A.B

(OR GATE USING NOR GATE:1.5 MARKS)

1.5 M

Q=A+B
=A+B

d) Describe the working of JK flip-flop with its truth table and logic diagram. 4M
Ans: (Diagram-2 M,Working-1M,Truth table-1M)

Truth Table :- 1M

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Diagram :-

2M

Working :-

The JK flip flop is basically a gated SR flip-flop with the addition of a clock
input circuitry that prevents the illegal or invalid output condition that can
occur when both inputs S and R are equal to logic level “1”. Due to this
additional clocked input, a JK flip-flop has four possible input combinations, 1M
“logic 1”, “logic 0”, “no change” and “toggle”.

Both the S and the R inputs of the previous SR bistable have now been
replaced by two inputs called the J and K inputs, respectively after its inventor
Jack Kilby. Then this equates to: J = S and K = R.
The two 2-input AND gates of the gated SR bistable have now been replaced
by two 3-input NAND gates with the third input of each gate connected to the
outputs at Q and Q. This cross coupling of the SR flip-flop allows the
previously invalid condition of S = “1” and R = “1” state to be used to produce
a “toggle action” as the two inputs are now interlocked.
If the circuit is now “SET” the J input is inhibited by the “0” status
of Q through the lower NAND gate. If the circuit is “RESET” the K input is
inhibited by the “0” status of Q through the upper NAND gate. As Q and Q are
always different we can use them to control the input. When both
inputs J and K are equal to logic “1”, the JK flip flop toggles

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Q. 4 A) Attempt any THREE of the following: 12-Total

Marks
a) Draw and explain working of 4 bit serial Input parallel Output shift 4M
register.
Ans: (Diagram:2M,Explaination:2M)

Diagram :-
2M

Explaination :-
If a logic “1” is connected to the DATA input pin of FFA then on the first
clock pulse the output of FFA and therefore the resulting QA will be set HIGH
to logic “1” with all the other outputs still remaining LOW at logic “0”. 2M
Assume now that the DATA input pin of FFA has returned LOW again to logic
“0” giving us one data pulse or 0-1-0.
The second clock pulse will change the output of FFA to logic “0” and the
output of FFBand QB HIGH to logic “1” as its input D has the logic “1” level
on it from QA. The logic “1” has now moved or been “shifted” one place along
the register to the right as it is now at QA.
When the third clock pulse arrives this logic “1” value moves to the output
of FFC ( QC ) and so on until the arrival of the fifth clock pulse which sets all
the outputs QA to QD back again to logic level “0” because the input
to FFA has remained constant at logic level “0”.
The effect of each clock pulse is to shift the data contents of each stage one
place to the right, and this is shown in the following table until the complete
data value of 0-0-0-1 is stored in the register. This data value can now be read
directly from the outputs of QA to QD.
Then the data has been converted from a serial data input signal to a parallel
data output. The truth table and following waveforms show the propagation of
the logic “1” through the register from left to right as follows.

Basic Data Movement Through A Shift Register

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Clock Pulse No QA QB QC QD

0 0 0 0 0

1 1 0 0 0

2 0 1 0 0

3 0 0 1 0

4 0 0 0 1

5 0 0 0 0

b) Draw 16:1 MUX tree using 4:1 MUX. 4M

Ans: Diagram :-

4M

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c) Calculate analog output of 4 bit DAC for digital input 1101. 4M

Assume VFS = 5V.
Ans: (Formula- 1M, Correct problem solving- 3M)

Formula :- 1M

VR = VFS

3M

= 5(1x2-1 + 1x2-2+0x2-3+1x2-4)
= 5(0.5+O.25+0+0.0625)
= 4.0625 Volts

OR
𝑏3 𝑏2 𝑏1 𝑏0
VFS = VR . ( + + + )
2 4 8 16
2 Marks for
Note – (Since VR is not given find VR) VR and 2
marks for
Full Scale o/p mean Vo

b3 b2 b1 b0 = 1111

VFS = 5V
1 1 1 1
5 = VR . ( 2 + + + )
4 8 16

VR = 5.33

For digital i/p b3 b2 b1 b0 = 1101

1 1 0 1
V0 = 5.33 ( 2 + + + )
4 8 16

V0 = 4.33V

d) State De Morgan’s theorem and prove any one. 4M

Ans: (Each State and proof using table- 2M each)

2M

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2M

OR

e) Design one digit BCD Adder using IC 7483

Ans: (Diagram:4M)

(Note: Labeled combinational circuit can be drawn using universal gate 4M

also)

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Q.5 Attempt any TWO of the following : Total Marks

12
a) Subtract using 2’s compliment method 6M
(35)10 – (5)10
Ans: Each step 3
Step 1 – Obtain binary equivalent of (35)10& (5)10 & then take 2’s compliment Marks
of (5)10 .
i.e. (35)10 = (100011)2
(5)10 = (101)2

2’s compliment of (5)10 = (000101)2 = 111010  1’s compliment

+ 1
----------------------------------
(111011)2 2’s Compliment

Step -2 :

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Now add (100011)2 and (111011)2

100011
+ 111011
----------------------------
1 011110
Carry is generated so answer is in positive form, so will
discard the carry generated
Therefore final answer will be (011110)2 = (30)2

b) Design a 4 bit synchronous counter and draw its logic diagram. 6M

Ans: State Table:

2M-State
table

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Kmap:
2M-Kmap

2M-Logic
Logic Diagram:
Diagram

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c) Describe the working of Successive Approximation ADC. Define 6M

Resolution and conversion time associate with ADC.
Ans: Circuit Diagram:

2 Marks
Diagram

When the start signal goes low the successive approximation register
SAR is cleared and output voltage of DAC will be 0V. When start goes high 2 Marks
Explanation
the conversion starts.
After starts, during first clock pulse the control circuit set MSB bit so
SAR output will be 1000 0000. This is connected as input to DAC so output of
DAC is (analog output) compared with Vin input voltage. If VDAC is more than
Vin the comparator output –Vsat, if VDAC is less than Vin, the comparator output
is +Vsat.
1 Marks
If output of DAC i.e. VDAC is + Vsat (i.e unknown analog input Each
voltage Vin> VDAC) then MSB bit is kept set, otherwise it is reset.
Consider MSB is set so SAR will contain 1000 0000.
The next clock pulse will set next bit i.e D6 a digital output of 1100 0000. The
output voltage of DAC i.e VDAC is compared with Vin, if it is + Vsatthe D6 bit is
kept as it is, but if it is –Vsat the D6 bit reset.
The process of checking and taking decision to keep bit set or to reset is
continued upto D0.
Then the DAC input will be digital data equal to analog input.
When the conversation if finished the control circuits sends out an end
of conversion signal and data is locked in buffer register

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Resolution: The voltage input change necessary for a one bit change in the
output is called resolution.
Conversion Time: The conversion time is the time required for conversion
from an analog input voltage to the stable digital output

OR

Circuit Diagram:

2 Marks
Diagram

Explanation: 2 Marks
Explanation
DAC= Digital to Analog converter
EOC= End of conversion
SAR =Succesive approximation register
S/H= Sample and hold circuit
Vin= input voltage
Vref= reference voltage
The successive approximation Analog to Digital converter circuit typically
consisting of four sub circuits-
1. A sample and hold circuit to acquire the input voltage Vin.
2. An analog voltage comparator that compares Vin to the output of internal
DAC and outputs the result of comparison to successive approximation
register(SAR).
3. SAR sub circuits designed to supply an approximate digital code of Vin to
the internal DAC.
4. An internal reference DAC that supplies the comparator with an analog
voltage equivalent of digital code output of SAR for comparison with Vin.

The successive approximation register is initialized so that most significant bit

(MSB) is equal to digital 1. This code is fed into DAC which the supplies the
analog equivalent of this digital code Vref/2 into the comparator circuit for the
comparison with sampled input voltage. If this analog voltage exceeds Vin the
comparator causes the SAR to reset the bit, otherwise a bit is left as 1. Then the

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next bit is set to 1 and the same test is done continuing this binary search until
every bit in the SAR has been tested. The resulting code is the digital
approximation of the sampled input voltage and is finally output by DAC at
end of the conversion (EOC).

Resolution and conversion time associate with ADC-

Resolution:
It is the maximum number of digital output codes.
Resolution= 2^n
(OR) 1 Marks
It is defined as the ratio of change in the value of input analog voltage required each
to change the digital output by 1 LSB.

Conversion time:
The time difference between two instants i.e. 'to' where SOC signal is given as
input to the ADC and 't1' where EOC signal we get as output from ADC. it
should be small as possible.

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Q.6 Attempt any TWO of the following: Total Marks

12
a) Design 4 bit Binary to Gray code converter. 6M
Ans: 2M for truth
Truth Table for 4 bit Binary to Gray code converter table
Binary Input Gray output
B3 B2 B1 B0 G3 G2 G1 G0 1/2m for
0 0 0 0 0 0 0 0 each output
0 0 0 1 0 0 0 1 equation
0 0 1 0 0 0 1 1 2M for
0 0 1 1 0 0 1 0 realization
0 1 0 0 0 1 1 0 using gates
0 1 0 1 0 1 1 1
0 1 1 0 0 1 0 1
0 1 1 1 0 1 0 0
1 0 0 0 1 1 0 0
1 0 0 1 1 1 0 1
1 0 1 0 1 1 1 1
1 0 1 1 1 1 1 0
1 1 0 0 1 0 1 0
1 1 0 1 1 0 1 1
1 1 1 0 1 0 0 1
1 1 1 1 1 0 0 0

K-MAP FOR G3:

G3=B3

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K-MAP FOR G2:

= B3 XOR B2

K-MAP FOR G1:

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= B1 XOR B2

K-MAP FOR G0:

= B1 XOR B0

Diagram for 4 bit Binary to Gray code converter:

Note: Realization of output equations can be done using Basic or Universal

gates

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b) Compare the following (Any three points) 6M

i) Volatile with Non-volatile memory
ii) SRAM with DRAM memory

Ans:
Parameter Volatile memory Non-Volatile memory
definition Memory required Memory that will keep Any 3points
electrical power to keep storing its information (each 1
information stored is without the need of mark)
called volatile memory electrical power is
called nonvolatile
memory.
classification All RAMs ROMs, EPROM,
magnetic memories
Effect of power Stored information No effect of power
is retained only as on stored
long as power is on. information
applications For temporary For permanent
storage storage of
information

2. SRAM with DRAM memory

Parameter SRAM DRAM

Circuit configuration Each SRAM cell is Each cell is one
a flip flop MOSFET & a capacitor
Bits stored In the form of In the form of charges
voltage
No of components per More Less
cell
Storage capacity Less More
Refreshing It does not require It require refreshing.
refreshing
Cost It is expensive It is cheaper
Speed It is faster It is slower
comparatively

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c) Give block schematic of decade counter IC 7490. Design Mod-7 counter 6M

using this IC.
Ans:
1. block schematic of decade counter IC 7490-
2M block
schematic

Mod-7 means states are from o,1,2,3,4,5,6,0

Therefore we have to reset counter IC 7490 when QD,QC,QB,QA=0111
Design reset logic:
Output of reset circuit should be HIGH because R0(1) and R0(2) are active high inputs.
Therefore reset logic output should be low for states 0 to 6.

Output should be HIGH for states 7 onwards.

Truth table & K-map:

Truth
Table-1M
Kmap-1M
Logical Dig-
2M

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Expression for Y:

Y= QC QB QA + QD

Circuit is-

Logic Diagram:

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 22320
21819
3 Hours / 70 Marks Seat No.

Instructions – (1) All Questions are Compulsory.

(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Use of Non-programmable Electronic Pocket
Calculator is permissible.
(7) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.

Marks

1. Attempt any FIVE of the following: 10

a) List the binary, octal and hexadecimal numbers for decimal
no. 0 to 15.
b) Define fan-in and fan-out of a gate.
c) Compare between synchronous and asynchronous counter (any
two points).
d) State two specification of DAC.
e) Write the gray code to given no. (⊥⊥0⊥)2 = (?) Gray.
f) Define encoder, write the IC number of IC used as decimal to
BCD encoder.
g) Draw the logical symbol of EX-OR and EX-NOR gate.

P.T.O.
22320 [2]
Marks
2. Attempt any THREE of the following: 12
a) Convert:
(i)(AD92 . BC A)16 = (?)10 = (?)8 = (?)2
b) Simplify the following and realize it
Y = A + A BC + A B C + ABC + A B
c) Explain the flowing characteristics w.r.t logic families:
(i) Noise margin
(ii) Power dissipation
(iii) Figure of merit
(iv) Speed of operation
d) Draw logic diagram of half adder circuit.

3. Attempt any THREE of the following: 12

a) Draw the circuit of successive approximation type ADC and
explain it’s working.
b) Describe the operation of R-5 flip-flop using NAND gates only.
c) Give classification of memory and compare RAM and ROM
(any four points).
d) State the applications of shift register.

4. Attempt any THREE of the following: 12

a) Subtract the given number using 2’s complement method:
(i) (⊥⊥0⊥⊥)2 – (⊥⊥⊥00)2
(ii)(⊥0⊥0)2 – (⊥0⊥)2
b) State De-Morgan’s theorem and prove any one.
c) Compare between PLA and PAL.
d) Reduce the following expression using K-map and implement it
F (A, B, C, D) = p M (1,3,5,7,8,10,14)
e) Describe the working of J-K flip- flop and state the race
around condition.
22320 [3]
Marks
5. Attempt any TWO of the following: 12
a) Design BCD to seven segment decoder using IC 7447 with its
truth table.
b) Describe the working of 4 bit universal shift register.
c) Design basic logic gates using NAND and NOR gate.

6. Attempt any TWO of the following: 12

a) Design a mod-6 Asynchronous counter with truth-table and
logic.
b) Design ⊥:8 demultiplexer using 1:4 demultiplexer.
c) Draw the circuit diagram of 4 bit R-2R ladder DAC and
obtain its output voltage expression.
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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in themodel answer
scheme.
2) The model answer and the answer written by candidate may vary but the examiner may tryto assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given moreImportance (Not
applicable for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in thefigure. The
figures drawn by candidate and model answer may vary. The examiner may give credit for anyequivalent
figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constantvalues
may vary and there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer
based on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.

Q. Sub Answers Marking

No. Q. N. Scheme

1 (A) Attempt any FIVE of the following: 10- Total

Marks

(a) List the binary,octal and hexadecimal numbers for decimal no. 0 to 15 2M

Ans: 2M
DECIMAL BINARY OCTAL HEXADECIMAL
0 0000 0 0
1 0001 1 1
2 0010 2 2
3 0011 3 3
4 0100 4 4
5 0101 5 5
6 0110 6 6
7 0111 7 7
8 1000 10 8

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Subject Name: Digital technique Model Answer Subject Code: 22320
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

(b) Define fan-in and fan-out of a gate. 2M

Ans: Fan-in is a term that defines the maximum number of digital inputs that a single logic gate 1M
can accept. Most transistor-transistor logic ( TTL ) gates have one or two inputs, although
some have more than two. A typical logic gate has a fan-in of 1 or 2.

Fan-out is a term that defines the maximum number of digital inputs that the output of a
single logic gate can feed. Most transistor-transistor logic ( TTL ) gates can feed up to 10 1M
other digital gates.

(c) Compare between synchronous and asynchronous counter (any two points). 2M

Ans:

Any two

1M
for each
compari
son

Page 2/
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Subject Name: Digital technique Model Answer Subject Code: 22320

(d) State two specification of DAC. 2M

Ans: 1.Resolution: Any

Resolution is defined as the ratio of change in analog output voltage resulting from a two,
change of 1 LSB at the digital input VFS is defined as the full scale analog output
voltage i.e. the analog output voltage when all the digital input with all digits 1. 1M for
Resolution = VFS /(2n −1) each
2. Accuracy:
Accuracy indicates how close the analog output voltage is to its theoretical value. It indicates
the deviation of actual output from the theoretical value. Accuracy depends on the accuracy
of the resistors used in the ladder, and the precision of the reference voltage used. Accuracy
is always specified in terms of percentage of the full scale output that means maximum
output voltage
3. Linearity:
The relation between the digital input and analog output should be linear.
However practically it is not so due to the error in the values of resistors used for the
resistive networks.
4. Temperature sensitivity:
The analog output voltage of D to A converter should not change due to changes in
temperature.
But practically the output is a function of temperature. It is so because the resistance values
and OPAMP parameters change with changes in temperature.
5. Settling time:
The time required to settle the analog output within the final value, after the change in
digital input is called as settling time.
The settling time should be as short as possible.
6. Long term drift
Long term drift are mainly due to resistor and semiconductor aging and can affect all the
characteristics.
Characteristics mainly affected are linearity, speed etc.
7. Supply rejection
Supply rejection indicates the ability of DAC to maintain scale, linearity and other important
characteristics when the supply voltage is varied.
Supply rejection is usually specified as percentage of full scale change at or near full scale
voltage at 25oe
8. Speed:
It is defined as the time needed to perform a conversion from digital to analog. It is also
defined as the number of conversions that can be performed per second.

Page 3/
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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
e) Write the gray code to given no.(1101)2 =(?) Gray. 2M

Ans: 2M

(1101)2 = (1011) Gray

f) Define encoder, write the IC number of IC used asdecimal l to BCD encoder. 2M

Ans: An encoder is a device or circuit that converts information from one format or code to Definati
another, for the purpose of standardization, speed or compression. on-1M

Decimal to BCD encoder IC- 74147 IC-1M

g) Draw the logical symbol ofEX-OR and EX-NOR gate. 2M

Ans:

EX-OR-
1M

EX-NOR-
1M

Q. Sub Answers Marking

No. Q. N. Scheme

2 Attempt any THREE of the following: 12- Total

Marks

Page 4/
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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
a) Convert: 4M

(i) (AD92.BCA)16= (?)10 = (?)8 = (?)2

Ans: (AD92.BCA)16 1.5M

= (10 × 16³) + (13 × 16²) + (9 × 16¹) + (2 × 16⁰) + (11 × 16⁻¹) + (12 × 16⁻²) + (10 × 16⁻³)

= 40960 + 3328 + 144 + 2 + 0.6857 + 0.046875 + 0.00244

= (44434.7368)10 1M

1.5M
(AD92.BCA)16 =(1010 1101 1001 0010.1011 1100 1010)2

(AD92.BCA)16 = (1010 1101 1001 0010.1011 1100 1010)2

=(001 010 110 110 010 010.101 111 001 010)2

=(126622.5712)8

Note: any other method can be considered.

b) Simplify the following and realize it 4M

Y=A+ C+ + ABC+

Ans: Y=A+ C+ + ABC+ 4M

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320

c) Explain the following characteristics w.r.t. logic families : 4M

(i) Noise margin

(ii) Power dissipation
(iii) Figure of merit
(iv) Speed of operation

Ans: Noise margin indicates the amount to noise voltage circuit can tolerate at its input for 1M each
definitio
both logic 1 and logic0.
n

Power Dissipation: It is the amount of power dissipated in an IC.

Figure of Merit: It is defined as the product of propagation delay and power dissipated by

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
the gate.

Speed of Operation: Speed of a logic circuit is determined by the time between the
application of input and change in the output of the circuit.

d) Draw logic diagram of half adder circuit 4M

Ans: 4M

OR

Note: logic diagram using NAND/NOR also can be considered.

Q. Sub Answers Marking

No. Q. N. Scheme

3 Attempt any THREE of the following : 12- Total

Marks

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
a) Draw the circuit of successive approximation type ADC and explain its working 4M

Ans:

Diagram

2M

The successive approximation A/D converter is as shown in fig. An analog voltage (Va) is
constantly compared with voltage Vi, using a comparator. The output produced by
comparator (Vo) is applied to an electronic Programmer.
If Va=Vi, then Vo=0 & then no conversion is required. The programmer displays the value of
Vi in the form of digital O/P.
But if Va Vi, then the O/P is changed by the programmer.
Explanat
If Va> Vi, then value of Vi is increased by 50% of earlier value.
ion 2M
But if Va< Vi, then value of Vi is decreased by 50% of earlier value.

This new value is converted into analog form, by D/A converter so as to compare it with Va
again. This procedure is repeated till we get Va=Vi. As the value of Vi is changed successively,
this method is called as successive-approximation A/D converter.

b) Describe the operation of R-S flip flop using NAND gates only . 4M

Ans:

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Subject Name: Digital technique Model Answer Subject Code: 22320

Logical
Diagram
2M

Description/explanation-

When clock = 0, the outputs of NAND gates 3 and 4 will be forced to be 1 irrespective of the
values of S and R. That means R’= S’ = 1.Hence the outputs of basic SR/F/F i.e. Q n+1 and Explanat
will not change. Thus if clock = 0, then there is no change in the output of the ion 2M
clocked SR flip-flop.
Explanat
Case I : S = R = 0, clock = 1: No change ion
If S=R=0 then outputs of NAND gate 3 and 4 are forced to become 1. without
Hence R' and S' both will be equal to 1. Since R' and S' are the inputs of the basic S – R flip- clock
flop using NAND gates. There will be no change in the state of outputs. pulse
must
Case II : S =1, R = 0, clock = 1: Set also be
Now S=0, R=1 and a positive going edge is applied to the clock consider
Output of NAND 3 i.e. R’ = 0 and output of NAND 4 i.e. S’ = 1. ed
Hence output of SR flip-flop is Q n+1 = 1 and = 0.
This is the set condition.

Case III : S =0, R = 1, clock = 1: Reset

Now S=0, R=1 and a positive edge is applied to the clock input.
Since S=0, output of NAND – 3 i.e. R´= 1. And as R’ = 1 and clock = 1 the output of NAND-4
i.e. S´ = 0. Hence output of SR flip-flop is Q n+1 = 0 and = 1.
This is the reset condition.

Case IV : S =1, R = 1, clock = 1: Undefined/ forbidden

As S=1, R=1 and clock = 1, the outputs of NAND gates 3 and 4 both are 0 i.e. S' = R'=0. So
both the outputs Q n+1 = 1 and
Hence output is Undefined/ forbidden.

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
CLK INPUTS OUTPUTS REMARK
S R Qn+1
0 X X Qn No change
1 0 0 Qn No change
1 0 1 0 1 Reset
1 1 0 1 0 Set
1 1 1 ? ? Forbidden
c) Give classification of memory and compare RAM and ROM (any four points) 4M

Ans: classification of memory Classific

ation
2M

Consider
even if
Seconda
ry
memory
is not
written

Comparison between RAM and ROM

RAM RAM

1. Temporary Storage. 1.Permanent Storage.

2 .Store data in MBs. 2.Store data in GBs.

3. Volatile . 3.Non-Volatile

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
4. Writing data is Faster. 4.Writing data is Slower.

Compari
son 2M

d) State the applications of shift register. 4M

Ans: 1] Shift register is used as Parallel to serial converter, which converts the parallel data into Each
serial data. It is utilized at the transmitter section after Analog to Digital Converter (ADC) Applicati
block. on 1M

2] Shift register is used as Serial to parallel converter, which converts the serial data into Any
parallel data. It is utilized at the receiver section before Digital to Analog Converter (DAC) other
block. relevant
applicati
3] Shift register along with some additional gate(s) generate the sequence of zeros and on must
ones. Hence, it is used as sequence generator. be
consider
4] Shift registers are also used as counters. There are two types of counters based on the ed
type of output from right most D flip-flop is connected to the serial input. Those are Ring
counter and Johnson Ring counter.

Q. Sub Answers Marking

No. Q. N. Scheme

4 Attempt any THREE of the following : 12- Total

Marks

(a) Subtract the given number using 2’s compliment method: 4M

(i) (11011)2 – (11100)2

(ii) (1010)2 - (101)2

Ans: i) Subtract (11011)2 – (11100)2using 2’s complement binary arithmetic.

Solution:

(11011)2 – (11100)2
Now,
2’s complement of (11100)2= 1’s complement of (11100)2+1 2’s
1’s complement of (11100)2 = (00011)2 comple
ment

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Subject Name: Digital technique Model Answer Subject Code: 22320
2’s complement = 00011+1 = 00100 1M

Therefore, 1 1 0 1 1
+
0 0 1 0 0

1 1 1 1 1

There is no carry it indicates that results is negative and in 2’s complement form i.e.(11111)2.
Therefore, for getting true value i.e.(+1) take 2’s complement of (11111) is
1’s complement + 1 Final
= 00000 + 1 Answer-
Ans= (00001)2 1M

Ans: (11011)2 – (11100)2 = 2’s complement of (11111)2 = (-1)10

ii) Subtract (1010)2 - (101)2using 2’s complement binary arithmetic.

2’s complement of (0101)2 = 1’s complement of (0101)2 +1

1’s complement of (0101)2 = (1010)2
2’s complement = 1010+1 = 1011 2’s
Therefore, 1 0 1 0 comple
+ ment
1 0 1 1 1M

1 0 1 0 1

There is carry ignore it, which indicates that results is positive i.e.(+5)
= (0101)2
Ans: (1010)2 - (101)2 = (0101)2= (+5)10 Final
Answer-
1M

(b) Stare De-Morgan’s theorem and prove any one 4M

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
Ans: De Morgan´s 1st Theorem: Stateme
It states that the compliment of sum is equal to the product of the compliment of nts-1M
individual variables. each

Anyone
Proof: proof -
2M

A B A+B

0 0 1 1 0 1 1
0 1 1 0 1 0 0
1 0 0 1 1 0 0
1 1 0 0 1 0 0

De Morgan´s 2nd Theorem:

It states that the compliment of product is equal to the sum of the compliments of
individual variables.

Proof:

A.B
A B

0 0 1 1 0 1 1
0 1 1 0 0 1 1
1 0 0 1 0 1 1
1 1 0 0 1 0 0

(c) 4M

Compare between PLA and PAL.

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
Ans: PLA PAL Any four
4 points-
1) Both AND and OR arrays are 1) OR array is fixed and AND array is 1M each
programmable programmable.

2) Costliest and complex than PAL 2) Cheaper and simpler

3) AND array can be programmed to 3) AND array can be programmed to

get desired minterms. get desired minterm.

4) Large number of functions can be 4) Provides the limited number of

implemented. functions.

5) Provides more programming 5) Offers less flexibility, but more likely

flexibility. used.

(d) Reduce the following expression using K-map and implement it 4M

F(A,B,C,D ) = M (1,3,5,7,8,10,14)

Ans: Kmap-
1M

Pairs-
1.5M

Final
Ans-
1.5M

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320

(e) Describe the working of J-K flip-flop and state the race around condition. 4M

Ans: Diagram
-1.5M

Working
-1.5M

State-
1M

The clock signal is applied to CLK input.

IF CLK =0 than F/F is disabled and O/P Q and do not change

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Subject Name: Digital technique Model Answer Subject Code: 22320
If CLK= 1 and J=K=O then the output Q and will not change their state.
If J=0 and K= 1 then JK flip flop will reset and Q= 0 & =1
If J=1 and K=0 then output will be set and Q=1 & =0
If J= K=1 then Q & outputs are inverted and FF will toggle
Race Around condition:
Race around condition occurs in J K Flip-flop only when J=K=1 and clock/enable is high (logic
1) as shown below-

In JK Flip-flop when J=K=1 and when clock goes high, output should toggle (change to
opposite state), but due to multiple feedback, output changes/toggles many times till the
clock/enable is high.
Thus toggling takes place more than once, called as racing or race around condition.

Q. Sub Answers Marking

No. Q. N. Scheme

5. Attempt any TWO of the following: 12- Total

Marks

a) Design BCD to seven segment decoder using IC 7447 with its truth table. 6M

Page 16/
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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
Ans: Note: Any one type of display shall be considered Explaina
1. BCD to 7 segment decoder is a combinational circuit that accepts 4 bit BCD input and tion 2M
generates appropriate 7 segment output.
2. In order to produce the required numbers from 0 to 9 on the display the correct
combination of LED segments need to be illuminated.
3. A standard 7 segment LED display generally has 8 input connections, one from each LED
segment & one that acts as a common terminal or connection for all the internal segments Circuit
4. Therefore there are 2 types of display 1. Common Anode Display 2. Common Cathode Diagram
Display : 2M
Common Anode Display

Truth
Table
2M

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Subject Name: Digital technique Model Answer Subject Code: 22320
Common Cathode Display:

b) Describe the working of 4 bit universal shift register. 6M

Ans:

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Subject Name: Digital technique Model Answer Subject Code: 22320
Circuit
Diagram
3M

Working
3M

Fig:4 bit universal shift register

Working:
1. PARALLEL LOAD: When mode control (M) is connected to logic 1, AND gates 2, 4, 6, 8 will
be enables and AND gates 1, 3,5,7, will be disabled . The 4-bit binary data will be loaded
parallel. The clock-2 input will be applied to the flip-flops , since M= 1, AND gates -10 is
enabled and gate-9 is disabled. Input will transfer parallel data to QA to QD outputs.
2. SHIFT RIGHT: When mode control (M) is connected to logic 0, AND gates 1,3,5,7 will be
enabled and gates 2, 4,,6, 8,will be disabled. The data will be shifted serially. The clock -1,
input will be applied to the flip-flops, Since M = 0, AND gates - 9 is enabled, and gates -10 is
disabled. The data is shifted serially to right from QA to QD.
3. SHIFT LEFT: When mode control (M) is connected to logic 1, AND gates 2,4,6,8 will be
enabled. This mode permits parallel loading of the resister and shift -left operation. . The
shift -left operation can be accomplished by connecting the output of each flip flop to the
parallel input of the previous flip- flop and serial input is applied at the input.

c) Design basic logic gates using NAND and NOR gate. 6M

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SUMMER-19 EXAMINATION
Subject Name: Digital technique Model Answer Subject Code: 22320
Ans: Each
Gate
Design
1 Marks

NOT gate using NAND =

OR gate using NOR gate:

Expression for OR gate is Y= =A+B

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Subject Name: Digital technique Model Answer Subject Code: 22320

AND gate using NOR gate:

Expression for AND gate is Y = = . = A.B (Applying De Morgan‟s theorem)

NOT gate using NOR Y= =

Q. Sub Answers Marking

No. Q. N. Scheme

6. Attempt any TWO of the following : 12- Total

Marks

a) Design a mod-6 Asynchronous counter with truth-table and logic. 6M

Ans: MOD 6 asynchronous counter will require 3 flip flops and will count from 000 to 101. Rest of
the states are invalid. To design the combinational circuit of valid states, following truth
table and K-map is drawn:

Truth
Table
2M

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From the above truth table, we draw the K-maps and get the expression for the MOD 6 Logic
asynchronous counter. Diagram
2M

Fig: K-map for above truth table

Thus reset logic is OR of complemented forms of QC and QB. This will be given to the reset
inputs of the counter so that as soon as count 110 reaches, the counter will reset. Thus the
counter will count from 000 to 101. The implementation of the designed MOD 6
asynchronous counter is shown below:
Circuit
Diagram
2M

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Fig: Circuit diagram of MOD 6 asynchronous counter

b) Design 1:8 de multiplexer using 1:4 de multiplexer 6M

Ans:

Design
3M

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Truth
Table
3M

Fig:1:8 Demultiplexer using 1:4 demultiplexer

Fig: Truth Table of 1:8 Demultiplexer .

c) Draw the circuit diagram of 4 bit R-2R ladder DAC and obtain its output voltage expression 6M

Ans:

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Subject Name: Digital technique Model Answer Subject Code: 22320
2M

2M

Fig 1: 4 bit R-2R ladder DAC

2M

2M

Fig 2:Simplified circuit diagram of Fig 1

Therefore output analog voltage V0 is given by,

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Subject Name: Digital technique Model Answer Subject Code: 22320

2M

Page 26/
 22320
11920
3 Hours / 70 Marks Seat No.

Instructions – (1) All Questions are Compulsory.

(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.

Marks

1. Attempt any FIVE of the following: 10

a) Convert (D8F)16 into binary and octal.
b) Draw Symbol, Truth Table and logic equation of Ex-OR gate.
c) State the DeMorgan's Theorems.
d) Convert the following expression into standard SOP form.
Y = AB + AC + BC
e) Draw symbol and write truth table of D and T Flip Flop.
f) Write down number of flip flops are required to count 16
clock pulses.
g) List the types of DAC

P.T.O.
22320 [2]
Marks
2. Attempt any THREE of the following: 12
a) Perform the subtraction using 2'S Complement methods.
(52)10 – (65)10
b) Simplify the following Boolean Expression and Implement using
logic gate.
ABC D + ABCD + ABCD + ABCD
c) Minimize the four variable logic function using K map.
f(A,B,C,D) = ∑m(0, 1, 2, 3, 5, 7, 8, 9, 11, 14)
d) Implement the following functions using demultiplexer.
f1 = ∑m (0, 2, 4, 6)
f2 = ∑m (1, 3, 5)

3. Attempt any THREE of the following: 12

a) Realize the following logic expressions using only NAND gates.
(i) OR
(ii) AND
(iii) NOT
b) Draw binary to gray converter and write its truth table.
c) Describe the working of JK flip flop with truth table and
logic diagram.
d) Describe the working of 4 bit SISO (serial in serial out)
Shift Register with diagram and waveform if input is 01101.

4. Attempt any THREE of the following: 12

a) Design a full Adder using Truth Table and K-map.
b) Describe the working of ring counter using D flip flop with
diagram and waveforms.
c) Draw block diagram of programmable logic Array.
d) Compare the following:
(i) Volatile with Non Volatile.
(ii) EPROM with EEPROM.
e) Describe the working principle of successive approximation ADC.
22320 [3]
Marks
5. Attempt any TWO of the following: 12
a) (i) Convert the following binary number (11001101)2 into
Gray Code and Excess-3 Code.
(ii) Perform the BCD Addition.
(17)10 + (57)10

(iii) Perform the binary addition.

(10110 • 110)2 + (1001 • 10)2
b) Design a 4bit ripple counter using JK flip flop, with truth
table and waveforms.
c) Calculate the analog output for 4 bit weighted register type
DAC for inputs
(i) 1011
(ii) 1001
Assume (Vfs) full scale range of voltage is 5V

6. Attempt any TWO of the following: 12

a) Compare TTL, CMOS and ECL logic family on the following
points.
(i) Basic Gates
(ii) Propogation delay
(iii) Fan out
(iv) Power Dissipation
(v) Noise immunity
(vi) Speed Power Product.
b) Design a BCD adder using IC 7483.
c) Design a 3 bit synchronous counter using JK FlipFlop.
 22320
11920
3 Hours / 70 Marks
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WINTER – 19EXAMINATIONS

Subject Name: Digital Techniques Model Answer Subject Code: 22320

Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in themodel answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may tryto assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given moreImportance (Not applicable for
subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in thefigure. The figures drawn
by candidate and model answer may vary. The examiner may give credit for anyequivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constantvalues may vary and
there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q. Sub Answer Marking

No. Q. N. Scheme

10-Total
Q.1 Attempt any FIVE of the following:
Marks

a) Convert (D8F) 16into binary and octal. 2M

Ans: 1M

1M

b) Draw symbol, Truth table and logic equation of Ex-OR gate. 2M

Ans: ½M

½M
Logic Equation = A ̅ + ̅B OR
Truth Table:
Inputs Output 1M
A B Y

0 0 0

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0 1 1

1 0 1

1 1 0

c) State the DeMorgan’s Theorems. 2M

Ans: De Morgan‟s 1st 1st-1M

Theorem complement of sum is equal to product of their individual complements. 2nd -1M
OR ̅̅̅̅̅̅̅̅ = ̅̅̅ ● ̅
De Morgan‟s 2nd theorem
Complement of product is equal to sum of their individual complements.
OR ̅̅̅̅̅̅̅ = ̅ + ̅
d) Convert the following expression into standard SOP form. 2M
Y= AB+ A̅ + BC
Ans: Y = AB+ A ̅ + BC 2M
Total variable ABC
1st Product term = AB ( C is missing)
2nd Product term = A ̅ ( B is missing)
3rd Product term = BC ( A is missing)
Y= AB●1 + A ̅ ●1 + BC● 1
Y= AB(C+ ̅ ) A ̅ (B+ ̅ ) + BC(A+ ̅)

Y= ABC + AB ̅ + AB ̅ + A ̅ ̅ +ABC + ̅BC

Y= ABC + AB ̅ + A ̅ ̅ + ̅BC Standard SOP Form
e) Draw symbol and write truth table of D and T Flip Flop. 2M

Ans: (Note: Symbol with other triggering method also can be consider) 1M
Symbol

1M
Truth
table

f) Write down number of flip flops are required to count 16 clock pulses. 2M

Ans: No of states= no.of clock pulses = 16 2M

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2n =m
n = no.of flip flops requried
m= no.of states
2n = 16
n=4
4 flip flops are required to count 16 clock pulse.
g) List the types of DAC 2M

Ans: 1) Binary weighted DAC 1M each

2) R –2R ladder network DAC

12-Total
Q.2 Attempt any THREE of the following:
Marks

Perform the subtraction using 2’S Complement methods.

a) 4M
(52)10 – (65)10
Ans: Conversio
n-1M each

Complime
nt-1M

Final
answer-
1M

b) Simplify the following Boolean Expressionand Implement using logic gate. 4M

AB ̅ ̅ + AB ̅ D + ABC ̅ + ABCD
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Ans: 2M

2M

c) Minimize the four variable logic function using K map. 4M

F(A,B,C,D) = ∑m(0,1,2,3,5,7,8,9,11,14)
Ans: Kmap
with place
value-1M

Pair-1M

Answer-
2M

Implement the following function using demultiplexer.

d) f1 = ∑m(0,2,4,6) 4M
f2 = ∑m(1,3,5)

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Ans: 4M

12-Total
Q.3 Attempt any THREE of the following:
Marks

Realize the following logic expression using only NAND gates.

a) (i) OR 4M
(ii) AND
(iii)NOT
Ans: (i)OR 1½ M

(ii)AND

1½ M

(ii)NOT
1M

(out put A bar)

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b) Draw binary to gray converter and write its truth table. 4M

Ans: Truth Table for 4 bit Binary to Gray code converter 2M Truth
Binary Input Gray Output table
B3 B2 B1 B0 G3 G2 G1 G0
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 1
0 0 1 0 0 0 1 1
0 0 1 1 0 0 1 0
0 1 0 0 0 1 1 0
0 1 0 1 0 1 1 1 Note:
0 1 1 0 0 1 0 1
0 1 1 1 0 1 0 0 Kmap is
1 0 0 0 1 1 0 0 optional
1 0 0 1 1 1 0 1
1 0 1 0 1 1 1 1
1 0 1 1 1 1 1 0
1 1 0 0 1 0 1 0
1 1 0 1 1 0 1 1
1 1 1 0 1 0 0 1
1 1 1 1 1 0 0 0
K-MAP FOR G3:

2M
Logical
diagram

G3=B3
K-MAP FOR G2

G2 = ̅̅̅̅B2 + ̅̅̅̅B3
=B3 XOR B2
K-MAP FOR G1:
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G0 = ̅̅̅̅B0+B1̅̅̅̅
= B1 XOR BO

(Note: Realization of output equation can be done Basic or Universal)

c) Describe the working of JK flip flop with truth table and logic Diagram. 4M

Ans: logic Diagram: 1M

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1M

Working:
The JK flip flop is basically a gated SR flip-flop with the addition of a clock input circuitry 2M
that prevents the illegal or invalid output condition that can occur when both inputs S and R
are equal to logic level “1”. Due to this additional clocked input, a JK flip-flop has four
possible input combinations, “logic 1”, “logic 0”, “no change” and “toggle”.
Both the S and the R inputs of the previous SR bistable have now been replaced by two
inputs called the J and K inputs, respectively after its inventor Jack Kilby. Then this equates
to: J = S and K = R.
The two 2-input AND gates of the gated SR bistable have now been replaced by two 3-
input NAND gates with the third input of each gate connected to the outputs at Q and Q.
This cross coupling of the SR flip-flop allows the previously invalid condition of S = “1”
and R = “1” state to be used to produce a “toggle action” as the two inputs are now
interlocked.
If the circuit is now “SET” the J input is inhibited by the “0” status
Of Q through the lower NAND gate. If the circuit is “RESET” the K input is inhibited by
the “0” status of Q through the upper NAND gate. As Q and Q are always different we can
use them to control the input. When both
inputs J and K are equal to logic “1”, the JK flip flop toggles
Describe the working of 4 bit SISO (serial in serial out) shift register with diagram
d) 4M
and waveform if input is 01101.

Ans: Diagram:(use SR or JK or D type flip flop) 1M

Working:
The DATA leaves the shift register one bit at a time in a serial pattern, hence the 1½ M
name Serial-in to Serial-Out Shift Register or SISO.
The SISO shift register is one of the simplest of the four configurations as it has only three
connections, the serial input (SI) which determines what enters the left hand flip-flop, the
serial output (SO) which is taken from the output of the right hand flip-flop and the
sequencing clock signal (Clk). The logic circuit diagram below shows a generalized serial-
in serial-out shift register, Output of FFA is Q4,FFB Q3,FFC Q2 and FFD is Q1
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Waveform:(Input is 01101) 1½ M

Q.4 Attempt any THREE of the following : 12-Total

Marks

a) Design a full Adder using Truth Table and K-map. 4M

Ans: A full adder is a combinational logic circuit that performs addition between three bits, the
two input bits A and B, and carry C from the previous bit.

Truth Table:

Truth
table 1½
M

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1M

Logical diagram:

1½ M

b) Describe the working of ring counter using D flip flop with diagram and waveforms. 4M

Ans: Diagram:1
Diagram:
½M

Waveforms:
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Waveform
:1½ M

Working:
The ring counter is a cascaded connection of flip flops, in which the output of last flip flop
is connected to input of first flip flop. In ring counter if the output of any stage is 1, then its
reminder is 0. The Ring counters transfers the same output throughout the circuit.
That means if the output of the first flip flop is 1, then this is transferred to its next stage i.e.
2nd flip flop. By transferring the output to its next stage, the output of first flip flop
becomes 0. And this process continues for all the stages of a ring counter. If we use n flip
flops in the ring counter, the „1‟ is circulated for every n clock cycles.
Explainati
on:1 M
c) Draw block diagram of programmable logic Array. 4M

Ans: Diagram: 4M

d) Compare the following:

(i) Volatile with Non Volatile. 4M
(ii) EPROM with EEPROM.
(i)Volatile with Non Volatile. 2M (Any
Parameter Volatile memory Non-Volatile memory two point
definition Memory required electrical power to Memory that will keep 1 M each)
keep information stored is called storing its information
Ans: volatile memory without the need of
electrical power is called
nonvolatile memory.
classification All RAMs ROMs, EPROM, magnetic
memories
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Effect of power Stored information is retained only as No effect of power on stored
long as power is on. information
applications For temporary storage For permanent storage of
information
1M(Any
ii)EPROM with EEPROM.
Parameter EPROM EEPROM. two point
Stands for Erasable Programable Read- Electrically Erasable each)
Only Memory. Programmable Read-Only
Memory.
Basic Ultraviolet Light is used to EEPROM contents are
erase the content of erased using electrical
EPROM. signal.
Appearance EPROM has a transparent EEPROM are totally
quartz crystal window at the encased in an opaque plastic
top. case.
Technology EPROM is modern version EEPROM is the modern
of PROM. version of EPROM.
e) Describe the working principal of successive approximation ADC. 4M
Ans: Note: Other relevant diagram and explanation also can be considered.
Diagram:

2M

Working:
The successive approximation A/D converter is as shown in fig. An analog voltage (Va) is
constantly compared with voltage Vi, using a comparator. The output produced by
comparator (Vo) is applied to an electronic Programmer. If Va=Vi, then Vo=0 & then no
conversion is required. The programmer displays the value of Vi in the form of digital O/P.
But if Va Vi, then the O/P is changed by the programmer. If Va> Vi, then value of Vi is
increased by 50% of earlier value. But if Va< Vi, then value of Vi is decreased by 50% of
earlier value. 2M
This new value is converted into analog form, by D/A converter so as to compare it with Va
again. This procedure is repeated till we get Va=Vi. As the value of Vi is changed
successively, this method is called as successive-approximation A/D converter.

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OR

When the starts signal goes low the successive approximation register SAR is cleared and
output voltage of DAC will be 0v. When start goes high the conversion starts.
After starts, during first clock pulse the control circuit set MSB bit so SAR output wiil be
1000 0000. This is connected as input to DAC so output of DAC is compared with Vin
input voltage. If VDAC is more than Vin the comparator output –Vsat, if VDAC is less than
Vin, the comparator output is +Vsat.
If output of DAC i.e. VDAC is +Vsat (i.e. unknown analog input voltage Vin> VDAC) then
MSB bit is kept set, otherwise it is reset.
Consider MSB is set so SAR will contain 1000 0000.
The next clock pulse will set next bit i.e. D6 bit is kept as it is, but if it –Vsat the D6 bit
reset. The process of checking and taking decision to keep bit set or to reset is continued
upto D0. Then the DAC input will be digital data equal to analog input.
When the conversion is finished the control circuits sends out an end of conversion signal
and data is locked in buffer register.

Q.5 Attempt any TWO of the following : 12- M

(i)Convert the following binary number (11001101)2 into Gray Code and Excess-3
(a) 2M
Code.

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Ans: 1M each
conversion

(ii)Perform the BCD Addition. 2M

(17)10 + (57)10
Ans: (17)10 0001 0111
(57)10 + 0101 0111 -------(1/2 M)
0110 1110
Valid Invalid
BCD BCD ----------(1/2 M)
ADD 0110 TO Invalid BCD
½ Each
1 11
step
0110 1110
+ 0000 0110
01110100 -----------(1/2 M)
7 4
-----------(1/2 M)

(iii)Perform the binary addition.

2M
(10110 ● 110)2 + (1001 ● 10)2

Ans: 10110.110)2 – (1001.10)2 = (100000.010)2 2M

11111
10110.110
+ 1001.10
100000.010
(b) Design a 4bit ripple counter using JK flip flop, with truth table and waveforms. 6M

Ans: 2M
Circuit Diagram:

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Truth Table:
2M

Timing Diagram / Waveforms:

2M

Calculate the analog output for 4 bit weighted register type DAC for inputs
(c) (i) 1011 6M
(ii) 1001
Assume (Vfs) full scale range of voltage is 5V
Ans: Given: 3M each
VR = Vfs = 5V
Formula Used:
Vo = - VR (B1.2-1 + B2.2-2 + B3.2-3 + B4.2-4 )
1. 1011
Vo = - VR (B1.2-1 + B2.2-2 + B3.2-3 + B4.2-4 )

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= - 5 (1*1/2 + 0 + 1*1/23 + 1 *1/24 )
= - 5 (1*1/2 + 1*1/8 + 1 *1/16)
= - 5 (0.5 + 0.125 + 0.0625) = 3.4375V
Vo = 3.4375 V
2. 1001
Vo = - VR (B1.2-1 + B2.2-2 + B3.2-3 + B4.2-4 )
= - 10 (1*1/2 + 0 + 0 + 1 *1/24 )
= - 10 (1*1/2 + 0 + 0 + 1 *1/16)
= - 10 (0.5 + 0.0625) = 2.8125V
Vo = 2.8125 V
12-Total
Q.6 Attempt any TWO of the following:
Marks

Compare TTL, CMOS and ECL logic family on the following points.
(i) Basic Gates
(ii) Propogation dealy
(a) (iii)Fan out 6M
(iv) Power Dissipation
(v) Noise immunity
(vi) Speed power product
Ans:
1M Each
Parameter TTL CMOS ECL
parameter

Basic gates NAND NOR/NAND OR/NOR

Propagation delay 10 70-105 2

Fan out 10 50 25

Power Dissipation 10mW 1.01mW 40-55mW

Noise Immunity 0.2V 5V 0.25V

Speed Power 100 0.7 100

Product

(b) Design a BCD adder using IC 7483. 6M

Ans: (Note: Labeled combinational circuit can be drawn using universal gate also)
1) To implement BCD adder we require:
• 4-bit binary adder for initial addition
• Logic circuit to detect sum greater than 9
• One more 4-bit adder to add 0110201102 in the sum if sum is greater than 9 or carry is 1
2) The logic circuit to detect sum greater than 9 can be determined by simplifying the

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Boolean expression of given truth Table.

Truth
Table: 2M

K-Map:
1M

3) Y=1 indicates sum is greater than 9. We can put one more term, C_out in the above
expression to check whether carry is one.
4) If any one condition is satisfied we add 6(0110) in the sum.
5) With this design information we can draw the block diagram of BCD adder, as shown in
figure below.

Circuit
Diagram:
3M

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(c) Design a 3 bit synchronous counter using JK Flip Flop. 6M

Ans: 1) Step1: 2M
Construct JK state table with corresponding excitation table:
Output State Transitions

Present Next state

State
Q2 Q1 Q0 Flip-flop inputs
Q2 Q1 Q0
J2 K2 J1 K1 J0 K0
000 001 0X 0X 1X

001 010 0X 1X X1

010 011 0X X0 1X

011 100 1X X1 X1

100 101 X0 0X 1X

101 110 X0 1X X1

110 111 X0 X0 1X

111 000 X1 X1 X1

State Table and Corresponding Excitation Table (d=don't care)

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Step 2:
Build Karnaugh Map or Kmap for each JK inputs:

2M

Step3:
Draw the complete design as below:
2M

Note: It can also be designed using any other Flip Flop.

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