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Logic GatesLogic GatesCS/APMA 202, Spring 2005CS/APMA 202, Spring 2005
Rosen, section 10.3Rosen, section 10.3 Aaron Bloomfield Aaron Bloomfield
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Review of Boolean algebraReview of Boolean algebra
Not is a horizontal bar above the number Not is a horizontal bar above the number 0 = 10 = 1 1 = 01 = 0
Or is a plusOr is a plus 0+0 = 00+0 = 0 0+1 = 10+1 = 1 1+0 = 11+0 = 1 1+1 = 11+1 = 1
And is multiplication And is multiplication 0*0 = 00*0 = 0 0*1 = 00*1 = 0 1*0 = 01*0 = 0
1*1 = 11*1 = 1
_ _
_ _
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Quick surveyQuick survey
I understand the basics of BooleanI understand the basics of Boolean
algebraalgebra
a)a)
Absolutely!Absolutely!
b)b) More or lessMore or less
c)c) Not reallyNot really
d)d) Boolean what?Boolean what?
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Today’s demotivatorsToday’s demotivators
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Basic logic gatesBasic logic gates
NotNot
And And
Or Or
NandNand
Nor Nor
Xor Xor
x x
x y
xy x y
xyz
z x +y x y
x y
x +y +z
z
x y
xy
x +y x y
x ⊕y x y
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Rosen, §10.3 question 1Rosen, §10.3 question 1
Find the output of the following circuitFind the output of the following circuit
Answer: ( Answer: ( x+y x+y )y)y Or (Or ( x x ∨∨y y ))∧¬∧¬yy
x y x +y
y
( x +y )y
__ __
x y
y
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x
y
Rosen, §10.3 question 2Rosen, §10.3 question 2
Find the output of the following circuitFind the output of the following circuit
Answer: xy Answer: xy Or Or ¬¬((¬¬ x x ∧¬∧¬y y ) ≡) ≡ x x ∨∨y y
x
y
x y x y
_ _ _ _ ___ ___
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Quick surveyQuick survey
I understand how to figure out whatI understand how to figure out what
a logic gate doesa logic gate does
a)a)
Absolutely!Absolutely!
b)b) More or lessMore or less
c)c) Not reallyNot really
d)d) Not at allNot at all
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Rosen, §10.3 question 6Rosen, §10.3 question 6
Write the circuits for the followingWrite the circuits for the following
Boolean algebraic expressionsBoolean algebraic expressions
a)a)
x x
++
y y
x
y
x
y
x
y
__ __
x x +y
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x
y
x
y
x
y
Rosen, §10.3 question 6Rosen, §10.3 question 6
Write the circuits for the followingWrite the circuits for the following
Boolean algebraic expressionsBoolean algebraic expressions
b)b)
(( x x
++
y y )) x x
_______ _______
x
y
x +y x +y ( x +y ) x
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Writing xor using and/or/notWriting xor using and/or/not
p p ⊕⊕ qq ≡≡ (( p p ∨∨ qq)) ∧∧¬(¬( p p ∧∧ qq))
x x ⊕⊕ yy ≡≡ (x + y)((x + y)( xy xy ))
x y x ⊕y
1 1 0
1 0 1
0 1 1
0 0 0
x
y
x +y
xy xy
( x +y )(xy)
____ ____
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Quick surveyQuick survey
I understand how to write a logicI understand how to write a logic
circuit for simple Boolean formulacircuit for simple Boolean formula
a)a) Absolutely!Absolutely!
b)b) More or lessMore or less
c)c) Not reallyNot really
d)d) Not at allNot at all
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Converting decimal numbers toConverting decimal numbers to
binarybinary
5353 = 32 + 16 + 4 + 1= 32 + 16 + 4 + 1
= 2= 255 + 2+ 244 + 2+ 222 + 2+ 200
= 1*2= 1*255 + 1*2+ 1*244 + 0*2+ 0*233 + 1*2+ 1*222 + 0*2+ 0*211 + 1*2+ 1*200
= 110101 in binary= 110101 in binary= 00110101 as a full byte in binary= 00110101 as a full byte in binary
211= 128 + 64 + 16 + 2 + 1211= 128 + 64 + 16 + 2 + 1
= 2= 277 + 2+ 266 + 2+ 244 + 2+ 211 + 2+ 200
= 1*2= 1*277 + 1*2+ 1*266 + 0*2+ 0*255 + 1*2+ 1*244 + 0*2+ 0*233 + 0*2+ 0*222 ++
1*21*211 + 1*2+ 1*200
= 11010011 in binary= 11010011 in binary
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Converting binary numbers toConverting binary numbers to
decimaldecimalWhat is 10011010 in decimal?What is 10011010 in decimal?1001101010011010 = 1*2= 1*277 + 0*2+ 0*266 + 0*2+ 0*255 + 1*2+ 1*244 + 1*2+ 1*233 ++
0*20*222 + 1*2+ 1*211 + 0*2+ 0*200
= 2= 277 + 2+ 244 + 2+ 233 + 2+ 211
= 128 + 16 + 8 + 2= 128 + 16 + 8 + 2= 154= 154
What is 00101001 in decimal?What is 00101001 in decimal?
00101001 = 0*200101001 = 0*277 + 0*2+ 0*266 + 1*2+ 1*255 + 0*2+ 0*244 + 1*2+ 1*233 ++0*20*222 + 0*2+ 0*211 + 1*2+ 1*200
= 2= 255 + 2+ 233 + 2+ 200
= 32 + 8 + 1= 32 + 8 + 1
= 41= 41
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A bit of binary humor A bit of binary humor
Available for $15 at Available for $15 athttp://www.thinkgeek.com/http://www.thinkgeek.com/tshirts/frustrations/5aa9/tshirts/frustrations/5aa9/
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Quick surveyQuick survey
I understand the basics of I understand the basics of
converting numbers betweenconverting numbers between
decimal and binarydecimal and binary
a)a) Absolutely!Absolutely!
b)b) More or lessMore or less
c)c) Not reallyNot reallyd)d) Not at allNot at all
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How to add binary numbersHow to add binary numbers
Consider adding two 1-bit binary numbersConsider adding two 1-bit binary numbers x x andand y y 0+0 = 00+0 = 0
0+1 = 10+1 = 1
1+0 = 11+0 = 1
1+1 = 101+1 = 10
Carry isCarry is x x AND AND y y
Sum isSum is x x XORXOR y y
The circuit to compute this is called a half-adder The circuit to compute this is called a half-adder
x y Carry Sum
0 0 0 00 1 0 1
1 0 0 1
1 1 1 0
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The half-adder The half-adder
Sum =Sum = x x XORXOR y y
Carry =Carry = x x AND AND y y
x y Sum
Carry
x y Sum
Carry
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Using half addersUsing half adders
We can then use a half-adder to computeWe can then use a half-adder to compute
the sum of two Boolean numbersthe sum of two Boolean numbers
1 1 0 0
+ 1 1 1 0
010?
001
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Quick surveyQuick survey
I understand half addersI understand half adders
a)a) Absolutely!Absolutely!
b)b) More or lessMore or lessc)c) Not reallyNot really
d)d) Not at allNot at all
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The full adder The full adder
The “HA” boxes are half-addersThe “HA” boxes are half-adders
HAX
Y
S
C
HAX
Y
S
C
x
y
c
c
s
HAX
Y
S
C
HAX
Y
S
C
x
y
c
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The full adder The full adder
The full circuitry of the full adder The full circuitry of the full adder
x
y
s
c
c
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Adding bigger binary numbers Adding bigger binary numbers
Just chain full adders together Just chain full adders together
HAX
Y
S
C
FAC
Y
X
S
C
FAC
Y
X
S
C
FAC
Y
X
S
C
x 1y 1
x 2 y 2 x 3y 3
x 0
y 0 s0
s1
s2
s3c
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Adding bigger binary numbers Adding bigger binary numbers
A half adder has 4 logic gates A half adder has 4 logic gates
A full adder has two half adders plus a OR gate A full adder has two half adders plus a OR gate Total of 9 logic gatesTotal of 9 logic gates
To addTo add nn bit binary numbers, you need 1 HA andbit binary numbers, you need 1 HA andnn-1 FAs-1 FAs
To add 32 bit binary numbers, you need 1 HATo add 32 bit binary numbers, you need 1 HAand 31 FAsand 31 FAs Total of 4+9*31 = 283 logic gatesTotal of 4+9*31 = 283 logic gatesTo add 64 bit binary numbers, you need 1 HATo add 64 bit binary numbers, you need 1 HAand 63 FAsand 63 FAs Total of 4+9*63 = 571 logic gatesTotal of 4+9*63 = 571 logic gates
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Quick surveyQuick survey
I understand (more or less) aboutI understand (more or less) about
adding binary numbers using logicadding binary numbers using logic
gatesgates
a)a) Absolutely!Absolutely!
b)b) More or lessMore or less
c)c)
Not reallyNot reallyd)d) Not at allNot at all
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More about logic gatesMore about logic gates
To implement a logic gate in hardware,To implement a logic gate in hardware,
you use a transistor you use a transistor
Transistors are all enclosed in an “IC”, or Transistors are all enclosed in an “IC”, or
integrated circuitintegrated circuit
The current Intel Pentium IV processorsThe current Intel Pentium IV processors
have 55 million transistors!have 55 million transistors!
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Pentium math error 1Pentium math error 1
Intel’s PentiumsIntel’s Pentiums(60Mhz – 100 Mhz)(60Mhz – 100 Mhz)had a floating pointhad a floating pointerrorerror
Graph of z = y/xGraph of z = y/x
Intel reluctantlyIntel reluctantly
agreed to replaceagreed to replacethem in 1994them in 1994
Graph from http://kuhttp.cc.ukans.edu/cwis/units/IPPBR/pentium_fdiv/pentgrph.htmlGraph from http://kuhttp.cc.ukans.edu/cwis/units/IPPBR/pentium_fdiv/pentgrph.html
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Flip-flopsFlip-flops
Consider the following circuit:Consider the following circuit:
What does it do?What does it do?
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MemoryMemory
A flip-flop holds a single bit of memory A flip-flop holds a single bit of memory The bit “flip-flops” between the two NANDThe bit “flip-flops” between the two NAND
gatesgates
In reality, flip-flops are a bit moreIn reality, flip-flops are a bit more
complicatedcomplicated Have 5 (or so) logic gates (transistors) per flip-Have 5 (or so) logic gates (transistors) per flip-
flopflop
Consider a 1 Gb memory chipConsider a 1 Gb memory chip
1 Gb = 8,589,934,592 bits of memory1 Gb = 8,589,934,592 bits of memory That’s about 43 million transistors!That’s about 43 million transistors!
In reality, those transistors are split into 9In reality, those transistors are split into 9
ICs of about 5 million transistors eachICs of about 5 million transistors each
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Quick surveyQuick survey
I felt I understood the material in thisI felt I understood the material in this
slide set…slide set…
a)a) Very wellVery well
b)b) With some review, I’ll be goodWith some review, I’ll be good
c)c) Not reallyNot really
d)d) Not at allNot at all
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Quick surveyQuick survey
The pace of the lecture for this slide The pace of the lecture for this slide
set was…set was…
a)a) FastFast
b)b) About rightAbout right
c)c) A little slowA little slow
d)d) Too slow Too slow
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End of lecture on 27 January 2005End of lecture on 27 January 2005
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