st 6-bit binary number that can be represented with What is the largest and smal
ID: 3880712 • Letter: S
Question
st 6-bit binary number that can be represented with What is the largest and smalle (e) unsigned numbers? (b) two's complement numbers? (e) sign/magnitude numbers? 2. 3. Convert the following two's complement binary numbers to decimal numbers. a. 110101 b. 01101010 c. 10110101 4. Convert the following decimal numbers to 6-bit two's complement binary numbers and add them. Write the results in 6-bit two's complement binary number format and decimal number format. Indicate whether or not the sum overflows a 6-bit result. a. 14 +12 b. 30 +2 c. 7 +20 d. -18-17 5. What is the signed/magnitude and two's complement range of numbers that can be represented with: a. 16 bits b. 32 bitsExplanation / Answer
2.)
a.) Unassigned Number
Largest : 011111
Smalles : 000000
b.) 2's Complement Number
Largest = [2^(61)1] = [2^5 - 1] = [31] = 2's complement = 011111
Smallest = [(2^(61))] = [-2^5] = [-32] = 2's complement = 100000
c.) Singed / magnitude numbers
Largest = 31
Smallest = -32
3.) Convert 2's complement binary numbers to decimal
a. 110101 = -11
Since it starts with 1 , it is -ve number.
change 1 -> 0 and 0 -> and add 1
001010 + 1 = 001011 = (0 * 2^5) + (0*2^4) + (1 * 2^3) + (0 * 2^2) + (1 * 2^1) + (1 * 2^0) = 0 + 0 + 8 + 0 + 2 + 1 = 11
Since it is - ve so answer is -11.
b.) 01101010 = 106
Since it starts with 0, it is +ve number, so simply covert to decimal
01101010 = (0 * 2^7) + (1 * 2^6) + (1 * 2^5) + (0* 2^4) +(1 * 2^3) + (0 * 2^2) + (1 * 2^1) + (0 * 2^0) =
0 + 64 + 32 + 0 + 8 + 0 + 2 + 0 = 106
c.) 10110101
Since it starts with 1 , it is -ve number.
change 1 -> 0 and 0 -> and add 1
01001010+ 1 = 01001011= (0 * 2^7) + (1*2^6) + (0 * 2^5) + (0 * 2^4) + (1 * 2^3) + (0 * 2^2) + (1 * 2^1) + (1 * 2^0)= 0 + 64 + 0 + 0 + 8 + 0 + 2 + 1 = 75
Since it is - ve so answer is -75.
5.)
In general, the representation range in 2's-complement is:
[(2^(n1)),2^(n1)1]
a.) 16 bits = [(2^(161)),2^(161)1] = [-2^15, 2^15 - 1] = [-32768, 32767]
b.) 32 bits = [(2^(321)),2^(321)1] = [-2^31, 2^31 - 1] = [- 2147483648, 2147483647]
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