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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.

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

 Ω Ω ∑ ↑ ↑ ∆ > 3 3 0 4 4 1 2

 ↑ > Ω ∆ ∆ Ω 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