project.name

Naman Agrawal

Member since: 1 year

Educational Institution: Not Entered

Country: Not Entered

T1 25 L43+L44 22BCE2904 POS F=πM( 1,3,5,6,7,8,10,13,15) = (A'+B+D)(A+B'+C')(A+D')(B'+D')

T1 25 L43+L44 22BCE2904 POS F=πM( 1,3,5,6,7,8,10,13,15) = (A'+B+D)(A+B'+C')(A+D')(B'+D')
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cat2 b2 q1

cat2 b2 q1
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CAT2 B2 Q3

CAT2 B2 Q3
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CAT 2 B2 Q3

CAT 2 B2 Q3
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CAT1 B2 Q2

CAT1 B2 Q2
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CAT1 B2 Q3

CAT1 B2 Q3
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CAT B1 Q1

CAT B1 Q1
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CAT 2 B1 Q3

CAT 2 B1 Q3
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Half Adder

Half Adder
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Full Adder

Full Adder
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Half Subtractor

Half Subtractor
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Full Subtractor

Full Subtractor
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3*3 Multiplier

3*3 Multiplier
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T1 25 L43+L44 22BCE2904 Naman Agrawal F’=Σm( 1,3,5,6,7,8,10,13,15 ) = AB’D’+A’BC+A’D+BD

T1 25 L43+L44 22BCE2904 Naman Agrawal F’=Σm( 1,3,5,6,7,8,10,13,15 ) = AB’D’+A’BC+A’D+BD
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T1 25 L43+L44 22BCE2904 POS NOR F=πM( 1,3,5,6,7,8,10,13,15) = (A'+B+D)(A+B'+C')(A+D')(B'+D')

T1 25 L43+L44 22BCE2904 POS NOR F=πM( 1,3,5,6,7,8,10,13,15) = (A'+B+D)(A+B'+C')(A+D')(B'+D')
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Q7

Q7
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Q16

Q16
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Pos Canonical Expression in MUX

Pos Canonical Expression in MUX
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Circuit diagram of 2:1 MUX Circuit using SOP Equation --- NAND circuit

Circuit diagram of 2:1 MUX Circuit using SOP Equation --- NAND circuit
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B1 CAT1 Q2

B1 CAT1 Q2
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CAT2 B2 Q2

CAT2 B2 Q2
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CAT2 B2 Q2

CAT2 B2 Q2
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cat2 b2 q5

cat2 b2 q5
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Circuit diagram for 3 bit even parity checker

Circuit diagram for 3 bit even parity checker
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Decoder 22BCE2904

Decoder 22BCE2904
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(x'y'z')+(xyz')Decoder Based design 22BCE2904

(x'y'z')+(xyz')Decoder Based design 22BCE2904
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(x'y'z')+(x+y+z') 22BCE2904 Multiplexer

(x'y'z')+(x+y+z') 22BCE2904 Multiplexer
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1 BIT alu

1 BIT alu
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30_22BCE3117_1Bit_ALU

30_22BCE3117_1Bit_ALU
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4 BIT ALU

4 BIT ALU
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FS 8:1 MUX 22BKT0156

FS 8:1 MUX 22BKT0156
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F POS decoder

F POS decoder
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Ha decoder

Ha decoder
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FADECODER 22BKT0156

FADECODER 22BKT0156
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HS using Decoder

HS using Decoder
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FA using Decoder

FA using Decoder
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HS using Decoder

HS using Decoder
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HS usiing Decoder

HS usiing Decoder
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f sop using decoder

f sop using decoder
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HS 4X1 MUX 22BKT0156

HS 4X1 MUX 22BKT0156
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HS 4X1 MUX 22BKT0156

HS 4X1 MUX 22BKT0156
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Half Subtractor MUX

Half Subtractor MUX
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T2 HA 4:1 MUX

T2 HA 4:1 MUX
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FS 4:1 MUX 22BKT0156

FS 4:1 MUX 22BKT0156
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Half Subtractor using MUX

Half Subtractor using MUX
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Untitled

Untitled
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NAND logic circuit of 2:4 decoder 03 22BCE0220 L43+L44

NAND logic circuit of 2:4 decoder 03 22BCE0220 L43+L44
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NAND logic circuit of 2:4 decoder 25 22BCE2904L43+L44

NAND logic circuit of 2:4 decoder 25 22BCE2904L43+L44
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Untitled

Untitled
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1:2 Decoder using Mux

1:2 Decoder using Mux
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SOPBCD3321

SOPBCD3321
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DA Q3 POS (OAI)

DA Q3 POS (OAI)
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BCDMUX

BCDMUX
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BCDPOSOAI

BCDPOSOAI
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DA QUES 3 DECODER

DA QUES 3 DECODER
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BCDDecoder

BCDDecoder
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DA Q3 NOR

DA Q3 NOR
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8:1 MUX USING 4:1 25 22BCE2904

8:1 MUX USING 4:1 25 22BCE2904
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Untitled

Untitled
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4:1 MUX using 2:1 MUX 22BCE2904 25

4:1 MUX using 2:1 MUX 22BCE2904 25
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8:1 MUX using 4:1 MUX & 2:1 MUX

8:1 MUX using 4:1 MUX & 2:1 MUX
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Question 1 DSD

Question 1 DSD
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Bidirectional Shift Register

Bidirectional Shift Register
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Bidirectional shift register 22BCE2904 24

Bidirectional shift register 22BCE2904 24
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Ring Counter 22BCE2904 24

Ring Counter 22BCE2904 24
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Johnson 24 22BCE2904

Johnson 24 22BCE2904
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CAT2 B2 Q5

CAT2 B2 Q5
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Decoder

Decoder
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3 Bit Even Parity Checker 22BCE2904 Naman Agrawal

3 Bit Even Parity Checker 22BCE2904 Naman Agrawal
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1:2 Decoder using Mux

1:2 Decoder using Mux
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CAT2 B1 Q5

CAT2 B1 Q5
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question 8 part A

question 8 part A
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CAT1B1Q3

CAT1B1Q3
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CAT 1 Q3 B1

CAT 1 Q3 B1
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CAT 1 Q4 B1

CAT 1 Q4 B1
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cat1 b1 q5

cat1 b1 q5
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CAT1 B1 Q5

CAT1 B1 Q5
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Cat2 B2 Q1

Cat2 B2 Q1
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CAT2 B2 Q4

CAT2 B2 Q4
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CAT2 B2 Q4

CAT2 B2 Q4
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B1 CAT1 Q1

B1 CAT1 Q1
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B1 CAT1 Q1

B1 CAT1 Q1
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CAT1 B2 Q1

CAT1 B2 Q1
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cat1b2q1

cat1b2q1
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cat1b2q3

cat1b2q3
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B2 Q4 CAT1

B2 Q4 CAT1
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cat1b2q5

cat1b2q5
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CAT-I Q4 B2

CAT-I Q4 B2
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cat2 b1 q2

cat2 b1 q2
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CAT2 B1 Q4

CAT2 B1 Q4
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cat2b1q4

cat2b1q4
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CAT2 B1 Q3

CAT2 B1 Q3
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CAT1 B2 Q5

CAT1 B2 Q5
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cat2b1q1

cat2b1q1
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CAT2 B1 Q5

CAT2 B1 Q5
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DA QUES3 MUX

DA QUES3 MUX
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SOPBCD

SOPBCD
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DA Q3 NOR

DA Q3 NOR
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DA QUES 3 DEMUX

DA QUES 3 DEMUX
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DA Q3 NAND

DA Q3 NAND
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BCDNOR

BCDNOR
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BCDDEmux

BCDDEmux
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BCDNAND

BCDNAND
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T1 25 L43+L44 22BCE2904 POS NAND F(0,2,4,9,11,12,14)=A’B’D’+A’C’D’+AB’D+ABD’

T1 25 L43+L44 22BCE2904 POS NAND F(0,2,4,9,11,12,14)=A’B’D’+A’C’D’+AB’D+ABD’
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T1 25 L43+L44 22BCE2904 SOP NAND F’=Σm( 1,3,5,6,7,8,10,13,15 ) = AB’D’+A’BC+A’D+BD

T1 25 L43+L44 22BCE2904 SOP NAND F’=Σm( 1,3,5,6,7,8,10,13,15 ) = AB’D’+A’BC+A’D+BD
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T1 25 L43+L44 22BCE2904 POS F’=πM( 0,2,4,9,11,12,14) = (A+B+D)(A+C+D)(A’+B+D’)(A’+B’+D)

T1 25 L43+L44 22BCE2904 POS F’=πM( 0,2,4,9,11,12,14) = (A+B+D)(A+C+D)(A’+B+D’)(A’+B’+D)
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4 bit Johnson Counter

4 bit Johnson Counter
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(x+y')(x'+y)(z')Multiplexer based design 22BCE2904

(x+y')(x'+y)(z')Multiplexer based design 22BCE2904
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Circuit diagram of 2:1 MUX Circuit using SOP Equation -- AOI circuit

Circuit diagram of 2:1 MUX Circuit using SOP Equation -- AOI circuit
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AOI logic circuit of 4:2 encode25 22BCE2904 L43+L44

AOI logic circuit of 4:2 encode25 22BCE2904 L43+L44
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NAND logic circuit of 4:2 encoder 03 22BCE0220 L43+L44

NAND logic circuit of 4:2 encoder 03 22BCE0220 L43+L44
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30_22BCE3117_4Bit ALU

30_22BCE3117_4Bit ALU
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7 Seg Decoder

7 Seg Decoder
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Circuit diagram of 2:1 MUX Circuit using POS Equation --- NOR circuit

Circuit diagram of 2:1 MUX Circuit using POS Equation --- NOR circuit
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AOI logic circuit of 2:4 decoder 03 22BCE0220 L43+L44

AOI logic circuit of 2:4 decoder 03 22BCE0220 L43+L44
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Circuit diagram of 2:1 MUX Circuit using POS Equation --- OAI circuit

Circuit diagram of 2:1 MUX Circuit using POS Equation --- OAI circuit
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Half Adder using Mux

Half Adder using Mux
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4 Bit Adder

4 Bit Adder
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Synchronous 4 bit up down counter

Synchronous 4 bit up down counter
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T1 25 L43+L44 22BCE2904 Naman Agrawal SOP Circuit F=Σm( 0,2,4,9,11,12,14 )=A’B’D’+A’C’D’+AB’D+ABD’

T1 25 L43+L44 22BCE2904 Naman Agrawal SOP Circuit F=Σm( 0,2,4,9,11,12,14 )=A’B’D’+A’C’D’+AB’D+ABD’
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AOI logic circuit of 4:2 encoder 03 22BCE0220 L43+L44

AOI logic circuit of 4:2 encoder 03 22BCE0220 L43+L44
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cat2b1q2

cat2b1q2
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T1 25 L43+L44 22BCE2904 Naman Agrawal POS F’=πM( 0,2,4,9,11,12,14) = (A+B+D)(A+C+D)(A’+B+D’)(A’+B’+D)

T1 25 L43+L44 22BCE2904 Naman Agrawal POS F’=πM( 0,2,4,9,11,12,14) = (A+B+D)(A+C+D)(A’+B+D’)(A’+B’+D)
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