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Radda [10]
3 years ago
8

If the floating-point number representation on a certain system has a sign bit, a 3-bit exponent, and a 4-bit significand. What

is the smallest positive decimal number that can be stored on this system if the storage is normalized? (Assume no bits are implied, there is no biasing, exponents use two's complement notation, and exponents of all zeros and all ones are allowed.) Note: no leading/trailing zero's.
Computers and Technology
1 answer:
vekshin13 years ago
7 0

Answer:

Normalization of storage is a typical method of storing the floating point number by shifting the decimal after the first figure of the number such as 1101.101 is normalized to 1.101101x23.

If the number that is in hovering point representation has 1 sign bit, 3-bit exponent with a 4-bit significant:

whenever the storage is normalized, then the biggest positive floating spot number in 2`s and the complement notation is 0.11112 x 23 = 111.12 =7.5

If the storage is returned to normal, then the minimum positive floating point number is 0.12 x 2-4 =0.000012 =1/32 = 0.03125.

Explanation:

Whenever the floating figure is keyed into the computer memory, then the first bit will be the sign bit, the next 8 bits are for exponent and 23 bits are used for storing significand. The array of exponents is from -127 to 128. While Exponent 127 stands for 0 and positive figures can be represented by values bigger than 127. The biggest floating point number will be represented as 0.111111.... 1111x 211111111

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