Consider the following 8bit hexadecimal numbers the second c
Solution
Range of 8-bit unsigned integers : 0 to (28 -1) = 0 to 255
Range of 8-bit signed integers(in 2\'s complement) : 0 to 127 (as two\'s complement can only represent positive integers).
(a) 0xC2
unsigned : C216 = 12 *161 + 2 * 160 = 192 + 2 = 19410
signed : C216 = 1100 0010 => negative number as most significant number is 1
its 1\'s complement = 0011 1101
2\'s complement = 1\'s complement + 1 = 0011 1110 = 62
(b) 0x0A
unsigned : 0A16 = 0*161 + 10 * 160 = 0 +10 = 1010
signed : 0A16 = 0000 1010 => positive number as most significant number is 0, so its value is 10.
(c) 0xEB
unsigned : EB16 = 14 *161 + 11 * 160 = 224 + 11 = 23510
signed : EB16 = 1110 1011 => negative number as most significant number is 1
its 1\'s complement = 0001 0100
2\'s complement = 1\'s complement + 1 = 0001 0101 = 21
(d) 0xCD
unsigned : CD16 = 12 *161 + 13 * 160 = 192+ 13 = 20510
signed : CD16 = 1100 1101 => negative number as most significant number is 1
its 1\'s complement = 0011 0010
2\'s complement = 1\'s complement + 1 = 0011 0011 = 51

