IBM 7090 Instruction-Reference page 71

Fixed-Point Binary
(4) 000100
(11) 001001
Floating Point
.1 x 2 011
.1001 x 2
100
Because the 7090 works in binary, all floating-point numbers will be to the base 2.
Therefore, to represent a float ing-point number in the computer, there is no need to
carry the base along with the number. This cuts our need to representing the fraction
and the exponent. The exponent is represented in positions (1-8) of the word and is
now called the characteristic. The fraction is contained in positions (9-35). The
binary point is to the left of the 9 bit. The sign position is used to sign the fraction.
Word layout takes this format:
S
1 ----------- 8 .9---------------------35
Characteristic
Fraction
The value of the number in the characteristic field signifies the exponent and its
sign. The characteristic is derived by adding 200 8 to the exponent. If the character-
istic is 200 8 the exponent is zero. If the number is 201 to 377, the exponent is posi-
tive. If it is 0 to 177, the exponent is negative. The following chart gives examples
of exponential numbers and their floating point representation:
Exponential Floating Point
Binary
S
1 - 8 9 - 35
+
.1 x 2 011
+
10000011 10000----0
-.01 x 2 001
10000001 0100-----0
+ •
1 x 2- 011
+
01111101 1000-----0
Normal and Unnormal Forms
A floating-point number is said to be in normal form when the digit immediately to
the right of the point is a significant bit (1). If the number is a zero, it is said to be in
unnormal form. The exception to this rule is a normal zero: a normal zero is a
floating-point number whose characteristic and fraction are both zero.
To go along with the two types of numbers, the instructions are also divided into two
categories, normal and unnormal. The difference in computer operation is that the
normal instructions always attempt to produce a normal answer and the unnormal in-
structions do not.
Arithmetic of Floating Point
Addition of floating-point numbers is done by adding the fractions of floating-point
numbers which have equal characteristics. The characteristics are set equal pre-
ceding the addition by placing the number with the smallest characteristic in the AC.
The fraction is then shifted right, and for each right shift one is added to the character-
istic. When the characteristic of the AC equals the characteristic of the SR, shifting
stops, and the fractions of the AC and SR are added. Bits shifted out of AC(35) enter
MQ(9). The sum appears in the AC and forms the most significant part of the answer.
The least significant part is the bits that were shifted into the MQ. The MQ character-
istic is set 2710 less than the AC characteristic to complete an unnormalized floating
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