CSE140: Components and Design Techniques
for Digital Systems
Introduction
Tajana Simunic Rosing
1
Welcome to CSE 140!
•
•
•
•
•
•
•
•
Course time: T/Th 2-3:20pm, WLH 2205
Discussion session: M 11-11:50pm
11-11:50pm, WLH 2205
Instructor: Tajana Simunic Rosing
• Email: tajana@ucsd.edu; please put CSE140 in the subject line
• Ph. 858 534-4868
• Office Hours: Tu/Th 1-2pm,
1-2pm CSE 2118
Instructor’s Assistant: Sheila Manalo
– Email: shmanalo@ucsd.edu
– Phone: (858) 534-8873
TA: Raid Ayoub
– Email: rayoub@cs.ucsd.edu
– Office hours: WF: 10:00am-12:00pm
TA: Chun Chen Liu
– Email: chl084@ucsd.edu
chl084@ucsd edu
– Office hours: TBA
Class Website:
– http://www.cse.ucsd.edu/classes/sp08/cse140/
Grades: http://webct.ucsd.edu
http://webct ucsd edu
2
Course Description
p
• Prerequisites:
– CSE 20 or Math 15A, and CSE 30.
– CSE 140L must be taken concurrently
• Objective:
– Introduce digital components and system design concepts
• Grading
– Homeworks (~8): 15%
• Lowest grade on the HW will be dropped
• HW picked up at beginning of the class, 0 pts if late
– Two midterms: 25% each
• No makeup exams; exceptions only for
– Documented illness (signed doctor’s statement), death in the family
– Final exam: 35%
• Regrade
g
requests:
q
turn in a written request
q
at the end of the
class where your work (HW or exam) is returned
3
Textbook and Recommended Readings
g
• Required textbook:
– Contemporary Logic Design
by R. Katz & G. Borriello
•
Recommended textbook:
– Digital Design by F. Vahid
•
Lecture slides are derived from
the slides designed for both
books
4
Whyy Studyy Digital
g Design?
g
• Look “under the hood” of computers
– Become a better p
programmer
g
when
aware of hardware resource issues
• Everyday devices becoming digital
– Enables:
• Better devices: Better sound recorders,
cameras, cars, cell phones, medical
devices,...
• New devices: Video games, PDAs, ...
– Known as “embedded systems”
• Thousands of new devices every year
• Designers needed: Potential career
Satellites
DVD
players
Portable
music players
Cell phones
1995
1999
1997
Video
recorders
Cameras
2001
2003
Musical
instruments
TVs
2005
2007
???
5
Implementing Digital Systems: Programming
Microprocessors Vs
Vs. Designing Digital Circuits
Desired motion-at-night detector
I0
I1
I2
I3
I4
I5
I6
I7
Programmed Custom designed
microprocessor
digital circuit
P0
P1
P2
P3
P4
P5
P6
P7
6
Digital Design: When Microprocessors Aren’t Good
Enough
Execution time
•
With microprocessors so easy,
cheap, and available, why
design a digital circuit?
– Microprocessor may be too
slow
– Or too big, power hungry, or
costly
Sample digital camera task execution times (in
seconds) on a microprocessor versus a digital
circuit:
Task
Microprocessor
5
0.1
Compress
8
0.5
1
(a)
Memory
Image Sensor
0.8
Microp
processor
(Read,
Compress,
and Store)
Read
circuit
Compress
circuit
(b)
Store
circuit
Memory
Custom
Digital Circuit
Read
Store
Image Sensor
Image Sensor
(c)
Memory
Read
i it
circuit
Compress
circuit
i it
Microprocessor
(Store)
7
What will we learn in this class?
• The language of logic design
– Boolean algebra,
algebra logic minimization,
minimization state
state, timing
timing, CAD tools
• The concept of state in digital systems
– analogous to variables and program counters in software systems
• How to specify/simulate/compile/realize our designs (140L)
–
–
–
–
hardware description languages
tools to simulate the workings of our designs
logic compilers to synthesize the hardware blocks of our designs
mapping onto programmable hardware
• Contrast with software design
– sequential and parallel implementations
– specify algorithm as well as computing/storage resources it will use
8
A qquick historyy lesson
• 1850: George Boole invents Boolean algebra
– maps logical propositions to symbols
– permits manipulation of logic statements using mathematics
• 1938: Claude Shannon links Boolean algebra to switches
– his Masters’
Masters thesis
• 1945: John von Neumann develops the first stored program
computer
– its switching elements are vacuum tubes (a big advance from relays)
• 1946: ENIAC . . . The world’s first completely electronic computer
– 18,000 vacuum tubes
– several hundred multiplications per minute
• 1947: Shockley, Brittain, and Bardeen invent the transistor
– replaces vacuum tubes
– enable integration of multiple devices into one package
– gateway to modern electronics
9
What is logic
g design?
g
• What is design?
• What is logic design?
10
Computation:
p
abstract vs. implementation
p
•
•
•
•
Up to now, computation has been abstract (paper, programs)
This class is about physically implementing computation using physical
devices that use voltages to represent digital values
Basic units of computation:
– Representation:
0, 1
– Assignment
x=y
– Data operations
x+y-5
– Control
• Sequential statements
a;b;c
• Conditionals
if x==1 then y
• Loops
for (i=1; i<10; i++)
• Procedures
a; proc (…); b;
We will study how each of these can be implemented in hardware
11
Outline
•
What we will cover in the next few lectures:
– Number representations
• Analog vs.
vs Digital
• Digital representations:
– Binary
– Hexadecimal
– Octal
O t l
• Binary addition and subtraction
• Binary multiplication, division
– Switches, MOS transistors, Logic gates
•
•
•
•
What is a switch
How a transistor operates
Building logic gates out of transistors
Building larger functions from logic gates
– Boolean algebra
• Properties
• How Boolean algebra can be used to design logic circuits
12
CSE140: Components and Design Techniques
for Digital Systems
Number representations &
Binary arithmetic
Tajana Simunic Rosing
13
What Does “Digital”
g
Mean?
• Analog signal
• Digital signal
– Infinite possible values
• E
Ex: voltage
l
on a wire
i
created by microphone
– Finite possible values
• E
Ex: button
b tt pressed
d on a
keypad
Sound waves
move the
th
membrane,
1
which moves
the magnet,
microphone
i
h
2
3
4
2
digital
signal
valuee
which creates
current in the nearby wire
Possible values:
1 00 11.01,
1.00,
01 22.0000009,
0000009
... infinite possibilities
time
valuee
analog
signal
4
3
2
1
0
Possible values:
0 1,
0,
1 2,
2 3,
3 or 4.
4
That’s it.
time
14
Digital
g Signals
g
with Onlyy Two Values: Binaryy
One Binary digIT is a BIT
Popular because it is easy storing a single
bit, and simple to operate using two values
(0,1) with transistors
valuee
• Binary digital signal -- only two
possible values
1
0
time
15
Digitized version enables
near-perfect save/cpy/trn.
– “Sample” voltage at
particular rate, save
sample using bit encoding
– Voltage levels still not kept
perfectly
– But we can distinguish 0s
from 1s
Let bit encoding be:
1 V: “01”
Digitized signal not
2 V: “10”
perfect re-creation,
but higher sampling
3 V: “11”
rate
t andd more bits
bit per
encoding brings closer.
time
a2d
1 digitized signal
0
time
01 10 11 10 11
3
2
1 received signal
0
time
How fix -- higher,
g
lower, ?
01 10 11 10 11
Volts
•
original signal
lengthy traansmission
(e.g, celll phone)
– Voltage levels not
saved/copied/transmitted
perfectly
3
2
1
0
lengthy trransmission
(e.g, ceell phone)
Analog signal (e.g., audio)
may lose quality
V
Volts
•
Volts
Example
p of Digitization
g
Benefit
1
0
time
Can fix -- easily distinguish 0s
and 1s, restore
d2a
3
2
1
0
time
16
Digitized
g
Audio: Compression
p
Benefit
• Digitized audio can be
compressed
– e.g., MP3s
– A CD can hold about 20
songs
g uncompressed,
p
but about 200
compressed
Example compression scheme:
00 --> 0000000000
01 -->
> 1111111111
1X --> X
0000000000 0000000000 0000001111 1111111111
• Compression also done
on digitized pictures
(jpeg), movies (mpeg),
and more
• Digitization has many
other benefits too
17
How Do We Encode Data as Binary for Our Digital
System?
analog
phenomena
button
•
sensors and
other inputs
electric
digital
signal
data
A2D
digital
data
– Button: not pressed (0),
pressed (1)
•
digital
data
D2A
electric
signal
actuators and
other outputs
Some inputs inherently
digital
– Just need encoding in
binary
– e.g., multi-button
lti b tt input:
i
t
encode red=001, blue=010,
...
Digital System
digital
data
Some inputs inherently
binary
•
Some inputs analog
– Need analog-to-digital
conversion
0
red
1
blue
green
black
0 0 0
red
blue
green
black
0 0 1
red
blue
green
black
0 1 0
air
33
degrees
temperature
sensor
0 0 1 0 0 0 0 1
18
How to Encode Text: ASCII, Unicode
• ASCII: 7- (or 8-) bit
g of each letter,,
encoding
number, or symbol
• Unicode: Increasingly
pop lar 16
popular
16-bit
bit bit encoding
– Encodes characters from
various world languages
Symbol
Encoding
Symbol
Encoding
R
S
T
L
N
E
0
1010010
1010011
1010100
1001100
1001110
1000101
0110000
0101110
0001001
r
s
t
l
n
e
9
1110010
1110011
1110100
1101100
1101110
1100101
0111001
0100001
0100000
.
<tab>
!
<space>
What does this ASCII bit sequence represent?
1010010 1000101 1010011 1010100
19
How to Encode Numbers: Binaryy Numbers
• Each position represents a
quantity; symbol in position
means how many of that
quantity
– Base ten (decimal)
• Ten symbols: 0, 1, 2, ..., 8, and 9
• More than 9 -- next position
5
2
3
104 103 102 101 100
– So each position power of 10
• N
Nothing
thi special
i l about
b tb
base 10 -used because we have 10
fingers
– Base two ((binary)
y)
• Two symbols: 0 and 1
• More than 1 -- next position
24
23
1
0
1
22
21
20
– So each position power of 2
20
How to Encode Numbers: Binaryy Numbers
• Working with binary numbers
– In base ten, helps to know
powers of 10
• one, ten, hundred, thousand, ten
th
thousand,
d ...
29
28
27
26
25
24
23
22
21
20
512 256 128 64 32
16
8
4
2
1
– In base two, helps to know
powers of 2
21
Converting from Decimal to Binary Numbers: Subtraction
Method (Easy for Humans)
•
Goal
– Get the binary weights to add up to
the decimal quantity
• Work from left to right
•
32 16 8
4
2
1
Subtraction method:
– To make the job easier, we just
subtract a selected binary weight from
the (remaining) quantity
• Then
Then, we have a new remaining
quantity, and we start again (from the
present binary position)
• Stop when remaining quantity is 0
22
Converting from Decimal to Binary Numbers: Division
Method (Good for Computers)
•
Divide decimal number by 2 and insert remainder into new binary
number.
– Continue dividing quotient by 2 until the quotient is 0.
•
Example: Convert decimal number 12 to binary
Decimal
ec a Number
u be
Binary
a y Number
u be
23
Bases Sixteen & Eight
g
164
163
•
8
A
F
162
161
160
8
A
F
1000 1010 1111
hex
binary
hex
bina ry
0
1
2
3
4
5
6
7
0000
0001
0010
0011
0100
0101
0110
0111
8
9
A
B
C
D
E
F
1000
1001
1010
1011
1100
1101
1110
1111
•
Base sixteen
– nice because each position represents four
base two positions
– Used as compact means to write binary
numbers
– Basic digits: 0-9, A-F
– Known as hexadecimal, or just hex
Base eight
– Used in some digital designs
– Each position represents three base two
positions
iti
– Basic digits: 0-7
Write 11110000 in hex
Write 11110000 in octal
24
Sign
g and magnitude
g
• One bit dedicate to sign (positive or negative)
– sign: 0 = positive (or zero), 1 = negative
• Rest represent the absolute value or magnitude
– three low order bits: 0 (000) thru 7 (111)
• Range for n bits
–7
– +/– 2n–1 –1 (two representations for 0)
• Cumbersome addition/subtraction
– mustt compare magnitudes
it d
to determine sign of result
–6
–5
+0
+1
1111 0000
1110
0001
1101
0010
+2
–4 1100
0011 +3
–3 1011
0100 +4
0101 +5
–2 1010
1001
0110
+6
–1
1000 0111
0
–0
+7
25
1s complement
p
•
•
•
If N is a positive number, then the negative of N (its 1s complement or
N' ) is N' = (2n– 1) – N (n= total # bits); higher order -> sign bit
• Shortcut: bit wise complement (7->-7 0111->1000
Subtraction implemented by 1s complement and then addition
Problem: two representations
p
of zero
–0
–1
2
–2
+0
+1
1111 0000
1110
0001
1101
0010
+2
–3 1100
0011 +3
1011
0100 +4
–4
–5
1010
0101
1001
0110
1000 0111
+6
–6
–7
+7
+5
26
2s complement
p
•
1s complement with negative numbers shifted one position clockwise
– only one representation for 0, high-order bit -> sign bit, more negative #s
•
If N iis a positive
iti number,
b th
then th
the negative
ti off N (it
(its 2s
2 complement
l
t or
N* ) is N* = 2n – N
– Shortcut: bit-wise complement plus 1
– 7 ->
> -7
7 : 0111 ->
> 1000 + 1 = 1001 ((-7)
7)
– -7 -> 7: 1001 -> 0110 + 1 = 0111 (7)
–1
–2
–3
+0
+1
1111 0000
1110
0001
1101
0010
+2
–4 1100
0011 +3
–5 1011
0100 +4
–6
1010
0101
1001
0110
1000 0111
+6
–7
–8
+5
+7
27
2s complement
p
addition and subtraction
• Simple addition and subtraction
– makes 2s complement the unanimous choice for integers in computers
• Check overflow conditions:
– Add two positive numbers and get a negative, or two negative and get
positive
4
–4
+3
+ (– 3)
7
–7
4
–4
–3
+3
1
–1
28
Overflow
• Sometimes result can’t be represented with given number
of bits
– Either too large magnitude of positive or negative
– e.g., 4-bit two’s complement addition of 0111+0001 (7+1=8). But
4-bit
4
bit two’s
two s complement can
can’tt represent number >7
• 0111+0001 = 1000 WRONG answer, 1000 in two’s complement is 8, not +8
– Adder/subtractor should indicate when overflow has occurred
occurred, so
result can be discarded
29
Detectingg Overflow: Method 1
•
Assuming 4-bit two’s complement numbers, can detect overflow by
detecting when the two numbers’ sign bits are the same but are
different from the result’s
result s sign bit
– If the two numbers’ sign bits are different, overflow is impossible
• Adding a positive and negative can’t exceed largest magnitude positive or
negative
•
Simple circuit
– overflow = a3’b3’s3 + a3b3s3’
– Include “overflow” output bit on adder/subtractor
sign bits
0 1 1 1
1 1 1 1
1 0 0 0
+0 0 0 1
+1 0 0 0
+0 1 1 1
1 0 0 0
overflow
(a)
0 1 1 1
overflow
(b)
1 1 1 1
no overflow
(c)
If the numbers’ sign bits have the same value, which
differs from the result’s sign bit, overflow has occurred.
30
Detectingg Overflow: Method 2
•
•
Even simpler method: Detect difference between carry-in to sign bit
and carry-out from sign bit
Yields simpler circ
circuit:
it o
overflow
erflo = c3 xor
or c4
1 1 1
0 1 1 1
0 0 0
1 1 1 1
0 0 0
1 0 0 0
+0 0 0 1
+1 0 0 0
+0 1 1 1
0 1 0 0 0
10 1 1 1
01 1 1 1
overflow
((a))
overflow
((b))
no overflow
((c))
If the carry into the sign bit column differs from the
carry out of that column, overflow has occurred.
31
Multiplication
p
• Generalized representation of multiplication by hand
32
Division
•
Repeated subtraction
– Set quotient to 0
– Repeat while dividend >=
divisor
• Subtract divisor from dividend
• Add 1 to quotient
– When dividend < divisor:
• Reminder = dividend
• Quotient is correct
Example:
• Dividend: 101; Divisor: 10
Dividend
101 -
Quotient
0 +
10
1
11 -
1 +
10
1
1
10
33
Summaryy of number representation
p
•
Conversion between basis
–
–
–
–
•
Decimal
Bi
Binary
Octal
Hex
Addition & subtraction in binary
– Overflow detection
•
Multiplication
– Partial products
• For demo see:
http://courses.cs.vt.edu/~cs1104/BuildingBlocks/multiply.010.html
•
Division
– Repeated subtraction
• For demo see:
http://courses.cs.vt.edu/~cs1104/BuildingBlocks/Binary.Divide.html
34
CSE140: Components and Design Techniques
for Digital Systems
Combinational circuit building blocks:
Switches and gates
Tajana Simunic Rosing
35
Combinational Circuit Introduction
Digital
g
circuit
• We’ll start with a simple form of circuit:
– Combinational circuit
• A digital circ
circuit
it whose
hose o
outputs
tp ts depend solel
solely on
the present combination of the circuit inputs’
values
• Built out of simple components: switches and
gates
1
a
b
Combinational
0
1
F
digital circuit
1
a
b
Sequential
0
?
F
g
circuit
digital
36
Switches
• Electronic switches are the basis of
g
circuits
binaryy digital
– Electrical terminology
• Voltage: Difference in electric potential
between two p
points
–
45A
4.5
9V
4.5 A
+
– Analogous to water pressure
• Current: Flow of charged particles
2 ohms
– Analogous to water flow
• Resistance: Tendency of wire to resist
current flow
– Analogous to water pipe diameter
9V
0V
4.5 A
• V = I * R (Oh
(Ohm’s
’ L
Law))
37
Switches
•
A switch has three parts
control
input
– Source input, and output
“off”
• Current wants to flow from source
input to output
source
input
– Control input
• Voltage that controls whether that
current can flow
•
The amazing shrinking switch
–
–
–
–
output
control
input
source
input
1930s: Relays
1940s: Vacuum tubes
1950s: Discrete transistor
1960s: Integrated circuits (ICs)
“ ”
“on”
output
(b)
• Initially just a few transistors on IC
• Then tens, hundreds, thousands...
relay
discrete
transistor
vacuum tube
IC
quarter
(to see the relative size)
38
The CMOS Circuit
• CMOS circuit
– Consists of N and PMOS transistors
– Both N and PMOS are similar to basic switches
– Rp ~ 3 Rn => PMOS in series is much slower than NMOS
A positive
voltage here
here...
...attracts electrons here,
turning the channel
between source and drain
into a conductor.
nMOS
gate
1
0
gate
oxide
source
IC package
drain
conducts
does not
conduct
1
0
pMOS
(a)
IC
gate
Silicon -- not quite a conductor or insulator:
Semiconductor
does not
conduct
conducts
39
MOS networks
X
what is the
relationship
between x and y?
3v
x
Y
0v
y
0 volts
3 volts
3 volts
0 volts
40
Two input
p networks
X
Y
3
3v
Z1
0v
what is the
relationship
between x, y and z?
x
X
Y
y
z1
z2
l
0 volts 0 volts
0 volts 3 volts
3v
3 volts 0 volts
Z2
3 volts 3 volts
0v
41
Relatingg Boolean Algebra
g
to Digital
g Design
g
Boolean
algebra
(mid-1800s)
Boole’s intent: formalize
human thought
For telephone
Switches
(1930s) switching and other
NOT
Symbol
F
x
0
1
Shannon (1938)
x
F
y
F
1
0
x
0
0
1
1
electronic uses
Showed application
of Boolean algebra
to design of switchbased circuits
AND
x
x
Truth table
OR
1
y
0
1
0
1
F
0
1
1
1
x
0
0
1
1
Digital design
x
F = x’
F’
F= x or y
0
x
y
0
1
1
Implement logic operators using transistors
F
0
0
0
1
1
0
•
y
0
1
0
1
1
y
Transistor
x
circuit
i
it
F
y
y
x
– Call those implementations logic gates
F=x & y
F’
y
0
x
0
42
NOT/OR/AND Logic
g Gate Timingg Diagrams
g
1
1
1
x
x
0
1
x
0
y
y
1
0
1
0
1
F
0
F
time
0
1
F
0
0
time
time
43
More Gates
NAND
NOR
XOR
x
x
F
y
1
NAND
y
F
x
0
0
1
1
•
•
•
•
y
0
1
0
1
F
1
1
1
0
x
0
0
1
1
y
0
1
0
1
F
1
0
0
0
x
0
0
1
1
y
0
1
0
1
F
0
1
1
0
x
0
0
1
1
y
0
1
0
1
F
1
0
0
1
NOR
x
y
x
F
y
1
XNOR
F
x
y
x
y
0
0
NAND: Opposite of AND (“NOT AND”)
NOR: Opposite of OR (“NOT
( NOT OR
OR”))
XOR: Exactly 1 input is 1, for 2-input
XOR. (For more inputs -- odd number
of 1s)
XNOR: Opposite of XOR (“NOT XOR”)
44
CMOS Example
• The rules:
– NMOS connects to GND,, PMOS to power
p
supply
pp y Vdd
– Duality of NMOS and PMOS
– Rp ~ 3 Rn => PMOS in series is much slower than NMOS in series
• Implement Z using
sing CMOS
CMOS: Z = (A + BC)’
45
Another CMOS Example
• Implement F using CMOS:
F=A*(B+C)
46
Completeness
p
of NAND
• Any logic function can be implemented using just NAND
gates. Likewise for NOR. Why?
g
y
– Need AND, OR and NOT
47
Buildingg Circuits Usingg Gates
• Motion-in-dark example
48
Example: Converting a Logic Equation to a Circuit
of Logic Gates
• Convert the following equation to logic gates:
F = a AND NOT( b OR NOT(c) )
49
Another example
• More than one way to map expressions to gates
g , Z = A’ • B’ • ((C + D)) = ((A’ • ((B’ • ((C + D)))
)))
e.g.,
– Use only two input gates, or using also three input gates
50
Waveform view of logic
g functions
• Just a sideways truth table
– but note how edges
g don’t line up
p exactly
y
– it takes time for a gate to switch its output!
time
change in Y takes time to "propagate" through gates
51
Example:
p Seat Belt Warningg Light
g System
y
•
•
Design circuit for warning light
S
Sensors
– s=1: seat belt fastened
– k=1: key inserted
– p=1: person in the seat
•
Capture logic equation
– What are conditions for warning
light
g t to go o
on?
•
Convert equation to circuit
52
Convertingg to Logic
g Equations
q
a
• Q1. A fire sprinkler system should spray water if high
y
is set to enabled.
heat is sensed and the system
– Let variable h represent “high heat is sensed,” e represent
“enabled,” and F represent “spraying water.”
– Answer: F =
• Q2. A car alarm should sound if the alarm is enabled,
and either the car is shaken or the door is opened.
– Let a represent “alarm is enabled,” s represent “car is shaken,” d
represent “door is opened,” and F represent “alarm sounds.”
– Answer: F =
53
CSE140: Components and Design Techniques
for Digital Systems
Boolean algebra
Tajana Simunic Rosing
54
Boolean Algebra
g
and its Relation to Digital
g Circuits
•
•
•
To understand the benefits of “logic gates” vs. switches, we should first
understand Boolean algebra
“Traditional” algebra
– Variable represent real numbers
– Operators
p
operate
p
on variables,, return real numbers
An algebraic structure consists of
– a set of elements B
– binary operations { + , • }
– and a unary operation { ’ }
– such that the following axioms hold:
1. the set B contains at least two elements: a, b
2. closure:
a + b is in B
a • b is in B
3. commutativity: a + b = b + a
a•b=b•a
4. associativity: a + (b + c) = (a + b) + c
a • (b • c) = (a • b) • c
5. identity:
a+0=a a•1=a
6. distributivity: a + (b • c) = (a + b) • (a + c) a • (b + c) = (a • b) + (a • c)
55
7. complementarity:
a + a’ = 1 a • a’ = 0
Boolean algebra
g
• Boolean Algebra
–
–
–
–
B = {{0,, 1}}
Variables represent 0 or 1 only
Operators return 0 or 1 only
Basic operators
a
0
0
1
1
b
0
1
0
1
AND
0
0
0
1
a
0
0
1
1
• • is logical AND: a AND b returns 1 only when both a=1 and b=1
• + is logical OR: a OR b returns 1 if either (or both) a=1 or b=1
• ’ is logical NOT: NOT a returns the opposite of a (1 if a=0
a=0, 0 if a=1)
– All algebraic axioms hold
a
0
1
b
0
1
0
1
NOT
1
0
56
OR
0
1
1
1
Axioms and theorems of Boolean algebra
g
•
•
•
•
•
•
•
identity
1. X + 0 = X
null
2. X + 1 = 1
idempotency:
3. X + X = X
involution:
4. (X’)’ = X
complementarity:
5. X + X’ = 1
commutativity:
y
6. X + Y = Y + X
associativity:
7. ((X + Y)) + Z = X + ((Y + Z))
1D. X • 1 = X
2D. X • 0 = 0
3D. X • X = X
5D. X • X’ = 0
6D. X • Y = Y • X
7D. ((X • Y)) • Z = X • ((Y • Z))
57
Axioms and theorems of Boolean algebra (cont’d)
•
•
•
•
•
distributivity:
8. X • (Y + Z) = (X • Y) + (X • Z)
Z)
uniting:
9. X • Y + X • Y’ = X
absorption:
10. X + X • Y = X
11. (X + Y’) • Y = X • Y
factoring:
12. (X + Y) • (X’ + Z) =
X • Z + X’ • Y
concensus:
13. (X • Y) + (Y • Z) + (X’ • Z) =
X • Y + X’ • Z
8D. X + (Y • Z) = (X + Y) • (X +
9D. (X + Y) • (X + Y’) = X
10D. X • (X + Y) = X
11D. (X • Y’) + Y = X + Y
12D. X • Y + X’ • Z =
(X + Z) • (X’ + Y)
13D. (X + Y) • (Y + Z) • (X’ + Z) =
(X + Y) • (X’ + Z)
58
Axioms and theorems of Boolean algebra (cont’d)
• de Morgan’s:
g
14. (X + Y + ...)’ = X’ • Y’ • ...14D. (X • Y • ...)’ = X’ + Y’ +
...
• generalized
li d d
de M
Morgan’s:
’
15. f’(X1,X2,...,Xn,0,1,+,•) = f(X1’,X2’,...,Xn’,1,0,•,+)
• establishes relationship between • and +
59
Axioms and theorems of Boolean algebra (cont’d)
• Duality
– a dual of a Boolean expression is derived by replacing
• by +, + by •, 0 by 1, and 1 by 0, and leaving variables unchanged
– any theorem that can be proven is thus also proven for its dual!
– a meta-theorem (a theorem about theorems)
• duality:
16. X + Y + ... ⇔ X • Y • ...
• generalized duality:
17. f (X1,X2,...,Xn,0,1,+,•) ⇔ f(X1,X2,...,Xn,1,0,•,+)
• Different
Diff
t th
than d
deMorgan’s
M
’ L
Law
– this is a statement about theorems
– this is not a way to manipulate (re-write) expressions
60
Provingg theorems
• Using the axioms of Boolean algebra:
– e.g.,
g , prove
p
the theorem:
distributivity (8)
p
y (5)
( )
complementarity
identity (1D)
– e.g., prove the theorem:
identity (1D)
distributivity (8)
identity (2)
identity (1D)
X • Y + X • Y’ = X
X • Y + X • Y’
X • (Y
( + Y’))
X • (1)
X+X•Y
X
X
X
X
+ X•Y
•1 + X•Y
• (1 + Y)
• (1)
= X • (Y + Y’)
= X • ((1))
= X9
= X
=
=
=
=
X
X
X
X
•1 + X•Y
• (1 + Y)
• (1)
9
61
Provingg theorems example
p
• Prove the following using the laws of Boolean algebra:
– ((X • Y)) + (Y
( • Z)) + (X’
( • Z)) = X • Y + X’ • Z
(X • Y) + (Y • Z) + (X’ • Z)
identity
(X • Y) + (1) • (Y • Z) + (X’ • Z)
complementarity
(X • Y) + (X’ + X) • (Y • Z) + (X’ • Z)
distributivity
(X • Y) + (X’ • Y • Z) + (X • Y • Z) + (X’ • Z)
commutativity
(X • Y) + (X • Y • Z) + (X’ • Y • Z) + (X’ • Z)
factoring
(X • Y) • (1 + Z) + (X’ • Z) • (1 + Y)
null
(X • Y) • (1) + (X’ • Z) • (1)
identity
(X • Y) + (X’ • Z) 9
62
Provingg theorems (p
(perfect induction))
• Using perfect induction (complete truth table):
– e.g.,
g , de Morgan’s:
g
(X + Y)’ = X’ • Y’
NOR is equivalent to AND
with inputs complemented
X
0
0
1
1
Y
0
1
0
1
X
X’
1
1
0
0
Y
Y’
1
0
1
0
(X + Y)
Y)’ X
X’ • Y
Y’
(X • Y)’ = X’ + Y’
NAND is equivalent to OR
with inputs complemented
X
0
0
1
1
Y
0
1
0
1
X’
1
1
0
0
Y’
1
0
1
0
(X • Y)’
X’ + Y’
63
A simple
p example:
p 1-bit binaryy adder
Cout Cin
• Inputs: A, B, Carry-in
• Outputs: Sum,
Sum Carry-out
Carry out
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
Cin Cout S
0
1
0
1
0
1
0
1
A
B
Cin
A
B
A
B
A
B
A
B
A
B
S
S
S
S
S
S
Cout
64
Boolean Algebra
g
Operator
p
Precedence
•
Evaluate the following Boolean equations, assuming a=1, b=1, c=0, d=1.
– Q1. F = a * b + c.
• Answer: F =
– Q2. F = ab + c.
• Answer: F =
– Q3. F = ab’.
a
• Answer: F =
– Q4. F = (ac)’.
• Answer: F =
– Q5. F = (a + b
b’)) * c + d
d’..
• Answer: F =
65
Example
p that Applies
pp Boolean Algebra
g
Properties
p
•
Want automatic door opener
circuit (e.g., for grocery store)
– Output: f=1 opens door
– Inputs:
•
Found inexpensive chip that
computes:
• f = c’hp + c’hp’ + c’h’p
– Can we use it?
• p=1: person detected
• h=1:
h 1 switch
it h fforcing
i h
hold
ld open
• c=1: key forcing closed
– Want open door when
• h=1 and c=0,, or
• h=0 and p=1 and c=0
66
Boolean Algebra:
g
Additional Properties
p
Aircraft lavatory sign example
•
Behavior
•
•
•
Three lavatories, each with
sensor (a, b, c), equals 1 if
door locked
Light “Available” sign (S) if
any lavatory available
•
Alternative:
• Instead of lighting “Available
“Available,”” light
“Occupied”
Equation and circuit
67
Summaryy
• What we covered thus far:
–
–
–
–
Number representations
p
Switches, MOS transistors, Logic gates
Boolean algebra
Supplement: NMOS and PMOS transistor characteristics
• What is next:
– Combinatorial logic:
• Representations
• Minimization
• Implementations
68