Question

4. Solve the following conversions: a. (0126F9D4)16 ? in base 8 b.(11100010)2 ? in base 16 5. Show the 1's complement representation on 1 Byte for the following signed decimal numbers: b.57 a.-100 6. Show the 2's complement representation on 1 Byte for the following signed decimal numbers: b. 57 a.-100
4-6 with explained work

Answer 1

  

| 40) (0126F 984) 16 = ? in base & First this 16 base number convert the 10 base number and then we convert the 10 base number to 8 base number (0126F9 pubian ( 216 +( Je 16 base to lo base .. 0x 167+ 1x168+2x163 + 6x16" +Fx16² +9x1220x 16 + 4+1 = 16777216+2097152 +393216+61448 +2304+208+ 4 2 19331540 10 base to 8 base :- Quotient Remainder 19331540-8=2416442 -2416442 + 8 = 302055 302055 8 37756 37756 = 8 = 4719 4719 8 = 589 589 +8 = 73 73 + 8 = 9 - 9+8 = 1 - → 4 1+8 -0 Answer!-(11157 47 24), ( TH

4.6) (11100010)2 = ? in base 16 First we convert to the no in 10 base then we convert to base no to 16 base. (11100010)2=( 7,0 =(. Sie 2 base to 10 baseir 1x2?+ 12®+ 1x 78 +0x2?+ 0x2 + orz' \x2'tox2 = 128 +64 +32+2 = 226 10 base no to 16 baser 226 +16 - Quotient Quotient 14 Remainder 14 - 16 = 0 141 € ...Answer - (E2), (Here' 145 €)

5) a) -100 If + signed no 21100 then most significant 21500 bit is o 2725 o and 2012- 11 If - signed no 21670 then most significant 213 +0 bit is 1 21 +1 0 71 Signed magnitude ! - | 1 1 1 1101010101 signed a's complement ! 11010 11110 111 ] 6) a) signed as complement - Here signed I's complement +1 .1001 1 0 1 1 10011100 Answer! 11010 111 110101

50 57 2157 2728511 214 o 217 70 2 13 1 2 I 1 o >1 Signed Magnitude 010 111111010111 For positive no l's complement and 2's complement are same as Signed Maynitude, Here no change occoured. so signed I's complement :- 6) b) signed as complement in 0101 111]110 1011)