Week 5 - Number Systems.pdf

Number Systems
Decimal (base 10) {0 1 2 3 4 5 6 7 8 9}
o Place value gives a logarithmic representation
of the number
o Ex. 4378 means
 4 X 103 = 4000
 3 X 102 = 300
 7 X 101 = 70
 8 X 100 = 8
o The place also gives the exponent of the base

Example
• 432,600
4 3 2 6 0 0
105
104
103
100
101
102
Powers of ten:
100 = 1 102 = 100 104 = 10000
101 = 10 103 = 1000 105 = 100000

Binary (base 2) {0 1}
Binary Decimal
0 0
1 1
10 2
11 3
100 4
101 5
110 6
111 7
1000 8
1001 9
1010 10

Example
1 1 0 1 1 0 0 1
27
26
25
20
21
22
24 23

Decimal Equivalent
 1101 1001
1 X 27 = 128
+ 1 X 26 = 64
+ 0 X 25 = 0
+ 1 X 24 = 16
+ 1 X 23 = 8
+ 0 X 22 = 0
+ 0 X 21 = 0
+ 1 X 20 = 1
217
Notice how powers of two
stand out:
20 = 1
21 = 10
22 = 100
23 = 1000

Decimal to Binary Conversion
 Ex. 575
o Find the largest power of two less than the number
o 29 = 512
o Subtract that power of two from the number
o 575 – 512 = 63
o Repeat steps 1 and 2 for the new result until you reach zero.
o 25 = 32 63 – 32 = 31
o 24 = 16 31 – 16 = 15
o 23 = 8 15 – 8 = 7
o 22 = 4 7 – 4 = 3
o 21 = 2 3 – 2 = 1
o 20 = 1 1 – 1 = 0
o Construct the number
o 1000111111

Another Example
144
o 27 = 128 144 – 128 = 16
o 24 = 16 16 – 16 = 0
Result 10010000

Hexadecimal (base 16)
 {0 1 2 3 4 5 6 7 8 9 A B C D E F}
 Assignments Dec Hex Dec Hex
0 0 8 8
1 1 9 9
2 2 10 A
3 3 11 B
4 4 12 C
5 5 13 D
6 6 14 E
7 7 15 F

Example
163
162
160
161
3 B 6 E
3 X 163 = 12288
11 X 162 = 2816
6 X 161 = 96
14 X 160 = 14
 15214

Hexadecimal is Convenient for
Binary Conversion
Binary Hex Binary Hex
0 0 1001 9
1 1 1010 A
10 2 1011 B
11 3 1100 C
100 4 1101 D
101 5 1110 E
110 6 1111 F
111 7 1 0000 10
1000 8  Nibble

Binary to Hex Conversion
 Group binary number by fours (nibbles)
o 1101 1001 0110
 Convert each nibble into hex equivalent
o 1101 1001 0110
D 9 6

Decimal to Hex Conversion
 Ex. 284
o 162 = 256 284 – 256 = 28
o 161 = 16 28 - 16 = 12 (Hex C)
o Result 1 1 C

Another Example with an Extension
 1054
o 162 = 256
 But we have several multiples of 256 in 1054
o 1054/256 = 4.12 take integer part
o This eliminates 4*256 = 1024
 1054 – 1024 = 30
o 161 = 16 30 – 16 = 14 (Hex E)
o Result 4 1 E

Truth Table
Binary Decimal Hexadecimal
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Truth Table
Binary Decimal Hexadecimal
0000 0 0
0001 1 1
0010 2 2
0011 3 3
0100 4 4
0101 5 5
0110 6 6
0111 7 7
1000 8 8
1001 9 9
1010 10 A
1011 11 B
1100 12 C
1101 13 D
1110 14 E
1111 15 F

Practice
Convert 212 decimal to binary
o 212 – 27 = 84
o 84 – 26 = 20
o 20 – 24 = 4
o 4 – 22 = 0
o Result: 1101 0100

More Practice
Convert 1101 0010 binary to hex
o 0010 = 2
o 1101 = 13 = D
o Result D2

Notation
Some books use a subscript to denote the
base.
o Ex: 1210 = 12 decimal
o 1216 = 12 hex = 18 decimal

Transistors as Switches
• VBB voltage controls whether the transistor
conducts in a common base configuration.
• Logic circuits can be built

AND
In order for current to flow, both switches
must be closed
¤ Logic notation AB = C
(Sometimes AB = C)
A B C
0 0 0
0 1 0
1 0 0
1 1 1

OR
Current flows if either switch is closed
¤ Logic notation A + B = C
A B C
0 0 0
0 1 1
1 0 1
1 1 1

Properties of AND and OR
Commutation
o A + B = B + A
o A  B = B  A
Same as
Same as

Commutation Circuit
A + B B + A
A  B B  A

Associative Property
A + (B + C) = (A + B) + C
A  (B  C) = (A  B)  C
=

Distributive Property
A + B  C = (A + B)  (A + C)
A + B  C
A B C Q
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1

Distributive Property
(A + B)  (A + C)
A B C Q
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1

Binary Addition
A B S C(arry)
0 0 0 0
1 0 1 0
0 1 1 0
1 1 0 1
Notice that the carry results are the same as AND
C = A  B

Inversion (NOT)
A Q
0 1
1 0
Logic: A
Q 

Exclusive OR (XOR)
Either A or B, but not both
This is sometimes called the
inequality detector, because the
result will be 0 when the inputs are the
same and 1 when they are different.
The truth table is the same as for
S on Binary Addition. S = A  B
A B S
0 0 0
1 0 1
0 1 1
1 1 0

Getting the XOR
A B S
0 0 0
1 0 1
0 1 1
1 1 0
Two ways of getting S = 1
B
A
or
B
A 


Circuit for XOR
Accumulating our results: Binary addition is the
result of XOR plus AND
B
A
B
A
B
A 





Half Adder
Called a half adder because we haven’t allowed for any carry bit
on input. In elementary addition of numbers, we always need to
allow for a carry from one column to the next.
18
25
4
3 (plus a carry)

Full Adder
INPUTS OUTPUTS
A B CIN COUT S
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1

Chaining the Full Adder
Possible to use the same
scheme for subtraction by
noting that
A – B = A + (-B)

Binary Counting
Use 1 for ON
Use 0 for OFF
= 00101011
So our example has 25 + 23 + 21 + 20 = 32 + 8 + 2 + 1 = 43

Counting in Binary
1 1 11 1011 21 10101
2 10 12 1100 22 10110
3 11 13 1101 23 10111
4 100 14 1110 24 11000
5 101 15 1111 25 11001
6 110 16 10000 26 11010
7 111 17 10001 27 11011
8 1000 18 10010 28 11100
9 1001 19 10011 29 11101
10 1010 20 10100 30 11110

NAND (NOT AND)
A B Q
0 0 1
0 1 1
1 0 1
1 1 0
B
A
Q 


NOR (NOT OR)
A B Q
0 0 1
0 1 0
1 0 0
1 1 0
B
A
Q 


DeMorgan’s Theorem
A NAND gate is equivalent to an inversion followed by an OR
A NOR gate is equivalent to an inversion followed by and AND

DeMorgan Truth Table
A B
0 0 1 1 1 1
0 1 1 1 0 0
1 0 1 1 0 0
1 1 0 0 0 0
NAND NOR

Exclusive NOR
A B Q
0 0 1
0 1 0
1 0 0
1 1 1
Equality Detector
B
A
Q 


Summary
Summary for all 2-input gates
Inputs Output of each gate
A B AND NAND OR NOR XOR XNOR
0 0 0 1 0 1 0 1
0 1 0 1 1 0 1 0
1 0 0 1 1 0 1 0
1 1 1 0 1 0 0 1

Logic Gates and Symbols
AND
NAND

More Gates and Symbols
NOR
NOT
OR

Week 5 - Number Systems.pdf

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