1. ECE 103 Engineering Programming Chapter 3 Numbers
Herbert G. Mayer, PSU
Status 6/20/2016
Initial content copied verbatim from
ECE 103 material developed by
Professor Phillip PSU ECE

2. Syllabus Binary Numbers Bitwise Operations Other Base Representations
Positive and Negative Integers
Floating Point

3. Binary Numbers Bit  Smallest unit of information (binary digit)
A single bit has two distinct states:
0 (logical False)
1 (logical True)
A binary number consists of n bits grouped together.
LSB
MSB
bn-1bn-2…b1b0 = bn-12n-1 + bn-22n-2 + … + b121 + b020 2

4. Table 1: Given n bits, the number of possible states = 2n- 10
1024 20 1 2 11
2048 21 4 12
4096 22 3 8 13
8192 23 16 14
16384 24 5 32 15
32768 : 6 64
65536 30 7
128 17
131072 31
256 18
262144 9
512 19
524288 3

5. Convert from binary to its equivalent base 10 value  Expand the powers of two.
Example: What is in decimal?
11102 = (123) + (122) + (121) + (020)
= (18) + (14) + (12) + (01)
= 1410
Convert from base 10 to its equivalent binary value  Successively divide by two; keep track of remainders.
Example: What is 1410 in binary?
Base 10
Remainder
14 / 2 = 7
7 / 2 = 3 1
3 / 2 = 1
1 / 2 = 0
Read the remainders backwards. Hence,
1410 = 11102 4

6. Bitwise Operations 1 1 1 1 1 A B A & B A B A | B A B A ^ B A ~A1 A B
A | B 1 A B
A ^ B 1 A ~A 1
Bitwise Complement
Bitwise AND
Bitwise OR
Bitwise XOR A B
A + B
Carry 1
Bitwise Addition 5

7. Logic operations are done a bit at a time (unary operator) or a pair of bits at a time (binary operator).
Example:
~1011 = Complement
1010 & 1100 = Bitwise AND
1010 | 1100 = Bitwise OR
1010 ^ 1100 = Bitwise XOR
■ ■ ■ ■ 6

8. Table 2: 4-bit positive integer conversion table
Other Base Representations
Octal (base 8  0, …, 7)
Hexadecimal (base 16  0, …, 9, A, B, C, D, E, F)
Table 2: 4-bit positive integer conversion table
Dec
Bin
Oct
Hex
0000 8
1000 10 1
0001 9
1001 11 2
0010
1010 12 A 3
0011
1011 13 B 4
0100
1100 14 C 5
0101
1101 15 D 6
0110
1110 16 E 7
0111
1111 17 F 7

9. Converting from binary to its equivalent hex: 1) Separate binary value into 4-bit groups 2) Replace each group by its hex value
Example:
= = AC4416
Converting from hex to its equivalent binary: Replace each hex value by its 4-bit binary value.
= 6B1316 = 8

10. Positive and Negative Integer Numbers
Integers are exactly representable in base 2
Table 3: 4-bit positive only values 0 to 15
If only positive integers and zero are needed, then all of the bits in the binary representation are available to express the value.
Given : n bits
Range: 0 to 2n – 1
Base 10
Base 2
0000 8
1000 1
0001 9
1001 2
0010 10
1010 3
0011 11
1011 4
0100 12
1100 5
0101 13
1101 6
0110 14
1110 7
0111 15
1111 9

11. Table 4: 4-bit positive and negative values -8 to +7
For negative integers, the most significant bit (MSB) of the binary value is reserved as a sign bit.
Negative values are expressed as 2’s complement.
Table 4: 4-bit positive and negative values -8 to +7
If both positive and negative integers are needed, the maximum positive value is reduced by a factor of two.
Given : n bits
Range: –2n-1 to 2n-1 – 1
Base 10
Base 2
(2's comp)
0000 -1
1111 1
0001 -2
1110 2
0010 -3
1101 3
0011 -4
1100 4
0100 -5
1011 5
0101 -6
1010 6
0110 -7
1001 7
0111 -8
1000 10

12. Floating Point Numbers
Floating point is used to express real-valued numbers. There is an implicit decimal point.
Example:
–634.9
Example: In scientific notation format (base 10)
–6.349 × 102
mantissa
sign
base
exponent 11

13. IEEE 754 single-precision (32-bit) standard s e1e2…e8 b1b2…b23
Binary can be used to represent floating point values, but usually only as an approximation.
IEEE 754 single-precision (32-bit) standard
s e1e2…e8 b1b2…b23
1 bit
Sign
0→ +
1→ –
8 bits
Interpreted as unsigned integer e'
23 bits
Interpreted as a base 2 value defined as
m' = 0.b1b2…b23 = b b22-2 +…+ b232-23
if e' ≠ 0 then FP number = (-1)s × (1 + m') × 2e'-127
if e' = 0 then FP number = (-1)s × m' × 2-126 12

14. Example: IEEE 754 single precision (32-bit)
The more bits available, the more precise the mantissa and the larger the exponent range.
See:
s = 0
e' = 17210
m' = =
Number = (-1)s × (1 + m') × 2e'-127 = × 245 ≈ × 1013 13