1) IEEE 754 Floating Point Standard: This standard is used to represent floating-point numbers (real numbers) digitally in computer memory. The standard provides two representations which have different accuracies: double precision and single precision. > Double Precision- 8bytes = 64 bits o Bit 63 (one bit): Sign (0=positive, 1=negative) o Bits 62 to 52 (11 bits): Exponent, biased by 1023 o Bits 51 to 0 (52 bits): Fraction f of the number 1.f Single Precision – 4bytes = 32 bits o Bit 31 (one bit): Sign (0-positive, 1-negative) o Bits 30 to 23 (8 bits): Exponent, biased by 127 o Bits 22 to 0 (23 bits): Fraction f of the number 1.f Procedure to Find the Double-Precision Representation of a Number x: Let the representation be (b63, b62, ..., b, b,, bo) with b63 being the most-significant bit and bo being the least significant bit. Sign Bit: x20 (0, l1, x< 0 Ехрonent: Replace x with |x], i.e. ignore the sign of x. We need to find an integer e such that Ose< 2047 and The value of e can be found as [log2(x)] The bits (b62, b61, ., bs2) are the binary representation of f – 1023. Fraction: The fraction part f is the first 52 bits of the binary representation of x2e – 1. It can be found as follows: y+x2-e Repeat until 52 bits are obtained { y+ 2y if y >1 { set new bit as 1 y+ y-1
1) IEEE 754 Floating Point Standard: This standard is used to represent floating-point numbers (real numbers) digitally in computer memory. The standard provides two representations which have different accuracies: double precision and single precision. > Double Precision- 8bytes = 64 bits o Bit 63 (one bit): Sign (0=positive, 1=negative) o Bits 62 to 52 (11 bits): Exponent, biased by 1023 o Bits 51 to 0 (52 bits): Fraction f of the number 1.f Single Precision – 4bytes = 32 bits o Bit 31 (one bit): Sign (0-positive, 1-negative) o Bits 30 to 23 (8 bits): Exponent, biased by 127 o Bits 22 to 0 (23 bits): Fraction f of the number 1.f Procedure to Find the Double-Precision Representation of a Number x: Let the representation be (b63, b62, ..., b, b,, bo) with b63 being the most-significant bit and bo being the least significant bit. Sign Bit: x20 (0, l1, x< 0 Ехрonent: Replace x with |x], i.e. ignore the sign of x. We need to find an integer e such that Ose< 2047 and The value of e can be found as [log2(x)] The bits (b62, b61, ., bs2) are the binary representation of f – 1023. Fraction: The fraction part f is the first 52 bits of the binary representation of x2e – 1. It can be found as follows: y+x2-e Repeat until 52 bits are obtained { y+ 2y if y >1 { set new bit as 1 y+ y-1
Chapter3: Data Representation
Section: Chapter Questions
Problem 23VE: ___________ occurs when the result of an arithmetic operation exceeds the number of bits available...
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![1) IEEE 754 Floating Point Standard:
This standard is used to represent floating-point numbers (real numbers) digitally in
computer memory. The standard provides two representations which have different
accuracies: double precision and single precision.
> Double Precision – 8bytes = 64 bits
o Bit 63 (one bit): Sign (0=positive, 1=negative)
o Bits 62 to 52 (11 bits): Exponent, biased by 1023
o Bits 51 to 0 (52 bits): Fraction f of the number 1.f
Single Precision – 4bytes = 32 bits
o Bit 31 (one bit): Sign (0-positive, 1=negative)
o Bits 30 to 23 (8 bits): Exponent, biased by 127
o Bits 22 to 0 (23 bits): Fraction f of the number 1.f
Procedure to Find the Double-Precision Representation of a Number x:
Let the representation be (b63, b62, ..., b, b,, bo) with b63 being the most-significant
bit and bo being the least significant bit.
Sign Bit:
(0,
x20
l1,
x< 0
Еxрonent:
Replace x with |x], i.e. ignore the sign of x. We need to find an integer e such that
Ose< 2047
and
The value of e can be found as
log:(x)]
The bits (b62, b61, ., b52) are the binary representation of f – 1023.
Fraction:
The fraction part f is the first 52 bits of the binary representation of x2 – 1. It can
be found as follows:
y +x2-e
Repeat until 52 bits are obtained
{
y+ 2y
if y >1
{
set new bit as 1
y + y-1
else
set new bit as 0
EE 426
Homework 2
Spring 2020/2021
a) Find the double-precision representation of –3.85 x 10°.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde5481a5-1a58-4e06-afd6-7fa7e832b3d0%2F61d9ecf2-2c59-49fc-a7da-8a13f600b2eb%2F5brnvwo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1) IEEE 754 Floating Point Standard:
This standard is used to represent floating-point numbers (real numbers) digitally in
computer memory. The standard provides two representations which have different
accuracies: double precision and single precision.
> Double Precision – 8bytes = 64 bits
o Bit 63 (one bit): Sign (0=positive, 1=negative)
o Bits 62 to 52 (11 bits): Exponent, biased by 1023
o Bits 51 to 0 (52 bits): Fraction f of the number 1.f
Single Precision – 4bytes = 32 bits
o Bit 31 (one bit): Sign (0-positive, 1=negative)
o Bits 30 to 23 (8 bits): Exponent, biased by 127
o Bits 22 to 0 (23 bits): Fraction f of the number 1.f
Procedure to Find the Double-Precision Representation of a Number x:
Let the representation be (b63, b62, ..., b, b,, bo) with b63 being the most-significant
bit and bo being the least significant bit.
Sign Bit:
(0,
x20
l1,
x< 0
Еxрonent:
Replace x with |x], i.e. ignore the sign of x. We need to find an integer e such that
Ose< 2047
and
The value of e can be found as
log:(x)]
The bits (b62, b61, ., b52) are the binary representation of f – 1023.
Fraction:
The fraction part f is the first 52 bits of the binary representation of x2 – 1. It can
be found as follows:
y +x2-e
Repeat until 52 bits are obtained
{
y+ 2y
if y >1
{
set new bit as 1
y + y-1
else
set new bit as 0
EE 426
Homework 2
Spring 2020/2021
a) Find the double-precision representation of –3.85 x 10°.
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