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CS302 Digital Logic Design Assignment 01 Spring 2021 Solution / Discussion

CS302

Question no 1

Perform the following arithmetic operations on given data, you can use indirect conversion or repeated division methods only. Convert the result into Caveman Number system.

 

Solution:

     5

      -4602031

 Reminder

     5

         920406

     1

     5

           184081

     1

     5

      36816

   1

     5

      7363

    1

     5

       1472

    3

     5

        294

    2

     5

          58

    4

     5

           11

    3

 

            2

    1

 

 

Convert this base 5 number to Caveman Number System

 

Question No.2

Convert   into the Grey Code.

Solution:

Convert the hexadecimal number  to binary by replacing the each hexadecimal symbol with appropriate four bits binary number.

Step 1:

Hexa Number

    F

E

4

5

E

A

Binary number

1111

1110

0100

0101

1110

1010

 

Step 2:

Binary

Number

1111

1110

0100

0101

1110

1010

Gray Code

1000

1001

0110

0111

1001

1111

 

So, the Grey Code is = 10001001011001111001111

Question No.2 (Part B)

  • Find the equivalence Excesss-7 code for the number

Solution

Convert octal number to Excess 7

Octal Number

     7

     6

     3

    7

    2

Adding 8

      8

     8

     8

    8

     8

We get

      15

     14

     11

    15

     10

Convert into binary each bit

     1111

    1110

    1011

  1111

     1010

 

Equivalent Excess-7 code for the number is equal to (11111110101111111010)

 

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CS302 Digital Logic Design Assignment 01 Spring 2021 Solution.

Click on the below link to download the file

CS302-Assignment-1-Solution-Spring-2021.docx

CS302 Assignment 1 Solution Spring 2021 ||Digital Logic Design|| 100% Correct Solution

 

Solution Assignment No.302

 

Question no 1

Perform the following arithmetic operations on given data, you can use indirect conversion or repeated division methods only. Convert the result into Caveman Number system.

 

Solution:

     5

      -4602031

 Reminder

     5

         920406

     1

     5

           184081

     1

     5

      36816

   1

     5

      7363

    1

     5

       1472

    3

     5

        294

    2

     5

          58

    4

     5

           11

    3

 

            2

    1

 

 

Convert this base 5 number to Caveman Number System

 

Question No.2

Convert   into the Grey Code.

Solution:

Convert the hexadecimal number  to binary by replacing the each hexadecimal symbol with appropriate four bits binary number.

Step 1:

Hexa Number

    F

E

4

5

E

A

Binary number

1111

1110

0100

0101

1110

1010

 

Step 2:

Binary

Number

1111

1110

0100

0101

1110

1010

Gray Code

1000

1001

0110

0111

1001

1111

 

So, the Grey Code is = 10001001011001111001111

Question No.2 (Part B)

  • Find the equivalence Excesss-7 code for the number

Solution

Convert octal number to Excess 7

Octal Number

     7

     6

     3

    7

    2

Adding 8

      8

     8

     8

    8

     8

We get

      15

     14

     11

    15

     10

Convert into binary each bit

     1111

    1110

    1011

  1111

     1010

 

Equivalent Excess-7 code for the number is equal to (11111110101111111010)

 

CS302 Assignment No. 1

Question :1

(ΩΩ∑↑↑∆>)C + (↑>Ω∆∆Ω)C = (?)C = (?)H

3

3

0

4

4

1

2

 

4

2

3

1

1

3

 

 

3304412

3*56 + 3*55 + 0*54 + 4*53 + 4*52 + 1*51 + 2*50

46875 + 9375 + 0 + 500 + 100 + 5 + 2

56857

423113

4*55 + 2*54 + 3*53 + 1*52 + 1*51 + 3*50

12500 + 1250 + 375 + 25 + 5 + 3

14158

56857 + 14158

(71015)10

5

71015

5

14203 – 0

5

2840 – 3

5

568 – 0

5

113 – 3

5

22 – 3

 

4 – 2

 

 

(71015)10 = (↑>ΩΩ∑Ω∑)c

16

71015

16

4438 – 7

16

277 – 6

16

17 – 5

 

1 – 1

         

     (71015)10 = (11567)H

 

  1. (FDEA2)H – (8989F)H + (767657)O = (?)H

 

F

D

E

A

2

15

13

14

10

2

 

8

9

8

9

F

8

9

8

9

15

7

6

7

6

5

7

7

6

7

6

5

7

 

 

    

      151314102

      1*168 + 5*167 + 1*166 + 3*165 + 1*164 + 4*163 + 1*162 + 0*161 + 2*160

      4294967296 + 1342177280 + 16777216 + 3145728 + 65536 + 16384 + 256 + 0 + 2

      5657149698

      898915

      8*165 + 9*164 + 8*163 + 9*162 + 1*161 + 5*160

      8388608 + 589824 + 32768 + 2304 + 16 + 5

      9013525

      767657

      7*85 + 6*84 + 7*83 + 6*82 + 5*81 + 7*80

      229376 + 24576 + 3584 + 384 + 40 + 7

      257967

(5657149698)10 – (9013525)10 + (257967)10 = (?)H

      (5648136173)10 + (257967)10 = (?)H

      (5648394140)10 = (?)H

     

16

5648394140

16

353024633 – 12

16

22064039 – 9

16

1379002 – 7

16

86187 – 10

16

5386 – 11

16

336 – 10

16

21 – 0

 

1 – 5

     

      (5648394140)10 = (150ABA79C)H

 

Question: 2

  1. -831.696972

2

831

2

415 – 1

2

27 – 1

2

13 – 1

2

6 – 1

2

3 – 0

 

1 – 1

 

-831 = -1101111

                                                                                           

.696972*2 = 1.393944

.393944*2 = 0.787888

.787888*2 = 1.575776

.575776*2 = 1.151552

.151552*2 = 0.303104

 

.696972 = .10110

 

-831.696972 = (-1101111.10110) 

 

  1. (01000011111111111010100010101011)2 = (?)10

 

0*231 + 1*230 + 0*229 + 0*228 + 0*227 + 0*226 + 1*225 + 1*224 + 1*223 + 1*222 + 1*221 + 1*220 +

1*219 + 1*218 + 1*217 + 1*216 + 1*215 + 0*214 + 1*213 + 0*212 + 1*211 + 0*210 + 0*29 + 0*28 + 1*27

+ 0*26 + 1*25 + 0*24 + 1*23 + 0*22 + 1*21 + 1*20

 

0 + 1073741824 + 0 + 0 + 0 + 0 + 33554432 + 16777216 + 8388608 + 4194304 + 2097152 + 1048576

+ 524288 + 262144 + 131072 + 65536 + 32768 + 0 + 8192 + 0 + 2048 + 0 + 0 + 0 + 128 + 0 + 32 + 0

+ 8 + 0 + 2 + 1

 

1140828331

 

(01000011111111111010100010101011)2 = (1140828331)10

 

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