Consider a 7 bit floating point representation based on the IEEE floating point

Question

Consider a 7 bit floating point representation based on the IEEE floating point format with one sign bit, three exponent bits, and three fraction bits.

What is the bias in this representation?

What is the smallest denormal number in this representation?

sign bit:

exponent bits:

mastissa bits:

What is the smallest normal number in this representation?

sign bit:

exponent bits:

mastissa bits:

What is the largest denormal number in this representation?

sign bit:

exponent bits:

mastissa bits:

What is the largest normal number in this representation?

sign bit:

exponent bits:

mastissa:

What is the number 1 in this representation?

sign bit:

exponent bits:

mastissa:

Explanation / Answer

Answer for Given Question:

Given that 7 bit floating point number based on IEEE floating point format.

Sign Bits - 1
Exponent Bits - 3
Fraction Bits - 3

s(1 bit) exp(3 bit) frac(3 bits)

Part 1:
Bias is: where k is exp bits
bias = 2^(k-1) - 1 = 2^(3-1) - 1= 3

Part 2:
smallest normal number in this representation?

V = (-1)^s ×M×2^E, s=0; Positive value
E = exp-bias = (001) base 2-2=1-2=1
M=1+frac×2^-n = 1+(000) ×2^-3 = 1.0

sign bit: 0
exponent bits: 001
mantissa bits: 000

Part 3: Largest denormalized value
s = 0, exp =000, frac = 111
M=frac ×2^3
= 0.11,

E=1-bias=1-3=2;
V = M×2^E
=0.11×2^-2

Part 4:
largest normalized positive value
s = 0, exp!=000 and exp!=111, exp=(110)^2=4+2=6
E=exp-bias=6-3=3, frac = (111)2
M = 1 + frac ×2^-4 =
=1+0.11=1.11=
= (1.11) ×2^3

Part 5: Floating-point representation of 1:
V = 1.0= (-1)^0 ×1.0×2^-0
s = 0, M = 1.0, E=0
E = 0 = exp - bias => exp = bias = 3 = (011)2
M = 1 + frac×2^-2 => frac = (000) base 2
=1.0 => frac = (000)2

sign bit : 0
exponent bit : 011
mantissa : 000

Answer for Given Question:

Given that 7 bit floating point number based on IEEE floating point format.

Sign Bits - 1
Exponent Bits - 3
Fraction Bits - 3

s(1 bit) exp(3 bit) frac(3 bits)

Part 1:
Bias is: where k is exp bits
bias = 2^(k-1) - 1 = 2^(3-1) - 1= 3

Part 2:
smallest normal number in this representation?

V = (-1)^s ×M×2^E, s=0; Positive value
E = exp-bias = (001) base 2-2=1-2=1
M=1+frac×2^-n = 1+(000) ×2^-3 = 1.0

sign bit: 0
exponent bits: 001
mantissa bits: 000

Part 3: Largest denormalized value
s = 0, exp =000, frac = 111
M=frac ×2^3
= 0.11,

E=1-bias=1-3=2;
V = M×2^E
=0.11×2^-2

Part 4:
largest normalized positive value
s = 0, exp!=000 and exp!=111, exp=(110)^2=4+2=6
E=exp-bias=6-3=3, frac = (111)2
M = 1 + frac ×2^-4 =
=1+0.11=1.11=
= (1.11) ×2^3

Part 5: Floating-point representation of 1:
V = 1.0= (-1)^0 ×1.0×2^-0
s = 0, M = 1.0, E=0
E = 0 = exp - bias => exp = bias = 3 = (011)2
M = 1 + frac×2^-2 => frac = (000) base 2
=1.0 => frac = (000)2

sign bit : 0
exponent bit : 011
mantissa : 000

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