Complete the following 1. Assurning 6 bit two\'s complernent storage, corivert e
ID: 3745072 • Letter: C
Question
Complete the following 1. Assurning 6 bit two's complernent storage, corivert each of the following to base ten 0 110101 o 011011 e 100110 2 Convert each of the following to 8 bit two's complement -35 0 77 o 26 3 What are the largest and smallest numbers that can be stored 8 bit two's complement? 4 Convert each of the following from excess eight notation to the equivalent decimal value: o 1010 o 1101 0 0111 5. Convert each of the following base ten numbers to the equivalent excess eight notation: o -5 6. What are the largest and smalles numbers that can be stored in excess eight notation?Explanation / Answer
IN 2's complement form, if a number starts with 1--> it is a negative number, positive otherwise.
for 110101
-> invert (logically toggle the bits) the number -> 001010
-> add 1 in the LSB ->001011
-> convert to decimal with a negative sign -> -11
011011 -> its a positive number -> simply convert to decimal -> 27
100110 -> negative number
-> inverted -> -26
2) Similarly, the logic gets reversed for 2's complement form: (invert and add 1 to binary form of number's absolute value)
-35 -> 11011101
77 -> 01001101
-26 -> 11100110
3) If the maximum allowed bits in a decimal storage are N then:
-> Maximum number = 2(N-1) - 1
-> Minimum number = - 2(N-1)
Hence for 8 bit representation it is 127 and -128
4) If a N bit binary number is in excess 8 notation -> simply convert it to decimal and subtract subtract 2(N-1) from it. Hence:
1010 -> 10 -> 2
1101 -> 13 -> 5
0111 -> 7 -> -1
5) Similarly, reversing the logic (assuming 4 bits in representation)
4 -> 12 -> 1100
-5 -> 3 -> 0011
6 -> 14 -> 1110
6) The range of values for the excess 8 notation is: 7 to -8
For N bit representation:
-> Maximum number = 2(N-1) - 1
-> Minimum number = - 2(N-1)
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