1 - UBC Computer Science

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CPSC 121: Models of Computation
2011 Winter Term 1
Building & Designing
Sequential Circuits
Steve Wolfman, based on notes by
Patrice Belleville and others
1
Outline
• Prereqs, Learning Goals, and Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
2
Learning Goals: Pre-Class
We are mostly departing from the readings
to talk about a new kind of circuit on our
way to a full computer: sequential circuits.
The pre-class goals are to be able to:
– Trace the operation of a deterministic finitestate automaton (represented as a diagram)
on an input, including indicating whether the
DFA accepts or rejects the input.
– Deduce the language accepted by a simple
DFA after working through multiple example
inputs.
3
Learning Goals: In-Class
By the end of this unit, you should be able
to:
– Translate a DFA to a corresponding
sequential circuit, but with a “hole” in it for the
circuitry describing the DFA’s transitions.
– Describe the contents of that “hole” as a
combinational circuitry problem (and therefore
solve it, just like you do other combinational
circuitry problems!).
– Explain how and why each part of the
resulting circuit works.
4
Where We Are in
The Big Stories
Theory
How do we model
computational systems?
Now: With our powerful
modelling language (pred
logic), we can prove
things like universality
of NOR gates for
combinational circuits.
Our new model (DFAs) are
sort of full computers.
Hardware
How do we build devices to
compute?
Now: Learning to build a
new kind of circuit with
memory that will be
the key new feature
we need to build fullblown computers!
(Something you’ve seen
in lab from a new angle.) 5
Outline
• Prereqs, Learning Goals, and Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
6
Problem: Push-Button Light Switch
Problem: Design a circuit to control a light
so that the light changes state any time its
“push-button” switch is pressed.
(Like the light switches in Dempster: Press and release,
and the light changes state. Press and release again,
and it changes again.)
?
7
A Light Switch “DFA”
?
pressed
light
off
light
on
pressed
This Deterministic Finite Automaton (DFA) isn’t really about
8
accepting/rejecting; its current state is the state of the light.
Problem:
Light Switch
?
Problem: Design a circuit to control a light so that the light
changes state any time its “push-button” switch is
pressed.
Identifying inputs/outputs: consider Which are most useful for
these possible inputs and outputs: this problem?
Input1: the button was pressed
Input2: the button is down
Output1: the light is on
Output2: the light changed states
a.
b.
c.
d.
e.
Input1 and Output1
Input1 and Output2
Input2 and Output1
Input2 and Output2
None of these
9
COMPARE TO:
?
Lecture 2’s Light Switch
Problem: Design a circuit to control a light so that the
light changes state any time its switch is flipped.
Identifying inputs/outputs: consider Which are most useful for
these possible inputs and outputs: this problem?
Input1: the switch flipped
Input2: the switch is on
Output1: the light is on
Output2: the light changed states
a.
b.
c.
d.
e.
Input1 and Output1
Input1 and Output2
Input2 and Output1
Input2 and Output2
None of these
10
Outline
• Prereqs, Learning Goals, and !Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
11
Departures from
Combinational Circuits
MEMORY:
We need to “remember”
the light’s state.
EVENTS:
We need to act on
a button push rather
than in response
to an input value.
12
Problem:
How Do We Remember?
We want a circuit that:
• Sometimes… remembers its current state.
• Other times… loads a new state to remember.
Sounds like a choice.
13
What circuit element do we have for modelling choices?
Worked Problem:
“Muxy Memory”
How do we use a mux to store a bit of memory?
We choose to remember on a control value of 0
and to load a new state on a 1.
???
0
output
new data
1
control
14
Worked Problem:
Muxy Memory
How do we use a mux to store a bit of memory?
We choose to remember on a control value of 0
and to load a new state on a 1.
old output (Q’)
0
output (Q)
new data (D)
1
control (G)
15
This violates our basic combinational constraint: no cycles.
Truth Table for “Muxy Memory”
Fill in the MM’s truth table:
G
0
0
D
0
0
Q'
0
1
a.
Q
0
1
b.
Q
0
1
c.
Q
0
1
d.
Q
0
0
0
0
1
1
1
0
0
1
0
0
1
0
0
1
0
0
1
0
0
1
1
1
1
1
0
1
1
1
0
1
1
0
1
0
1
1
F
F
1
1
0
1
e.
None
of
these
16
Worked Problem:
Truth Table for “Muxy Memory”
Worked Problem: Write a truth table for the
MM:
G
D
Q'
Q
0
0
0
0
0
1
0
1
0
0
1
1
1
1
0
0
0
1
0
1
0
1
0
0
1
1
0
1
1
1
1
1
17
Like a “normal” mux table, but what happens when Q'  Q?
Worked Problem:
Truth Table for “Muxy Memory”
Worked Problem: Write a truth table for the
MM:
G
D
Q'
Q
0
0
0
0
0
1
0
1
0
0
1
1
1
1
0
0
0
1
0
1
0
1
0
0
1
1
0
1
1
1
1
1
18
Q' “takes on” Q’s value at the “next step”.
“D Latch” Symbol + Semantics
We call a “muxy memory” a “D latch”.
When G is low, the latch maintains its memory.
When G is high, the latch loads a new value from D.
old output (Q’)
0
output (Q)
new data (D)
1
control (G)
19
D Latch Symbol + Semantics
When G is low, the latch maintains its memory.
When G is high, the latch loads a new value from D.
old output (Q’)
0
output (Q)
new data (D)
1
control (G)
20
D Latch Symbol + Semantics
When G is low, the latch maintains its memory.
When G is high, the latch loads a new value from D.
new data (D)
D
Q
control (G)
output (Q)
G
21
Hold On!!
Why does the D Latch have two inputs and one output
when the mux inside has THREE inputs and one
output?
a. The D Latch is broken as is; it should have three
inputs.
b. A circuit can always ignore one of its inputs.
c. One of the inputs is always true.
d. One of the inputs is always false.
e. None of these (but the D Latch is not broken as is).
22
Using the D Latch
for Circuits with Memory
Problem: What goes in the cloud? What do
we send into G?
D
Q
??
G
Combinational
Circuit to calculate
next state
input
23
We assume we just want Q as the output.
Using the D Latch
for Our Light Switch
Problem: What do we send into G?
D
Q
??
current light state
G
no (0 bit) input
a. T if the button is down, F if it’s up.
b. T if the button is up, F if it’s down.
c. Neither of these.
output
24
A Timing Problem:
We Need EVENTS!
Problem: What do we send into G?
D
“pulse”
when
button is
pressed
button
pressed
Q
current light state
G
output
As long as the button is down, D flows to Q
flows through the NOT gate and back to25D...
which is bad!
A Timing Problem, Metaphor
(from MIT 6.004, Fall 2002)
What’s wrong with this
tollbooth?
P.S. Call this a “bar”, not a “gate”,
or we’ll tie ourselves in (k)nots.
26
A Timing Solution, Metaphor
(from MIT 6.004, Fall 2002)
Is this one OK?
27
A Timing Problem
Problem: What do we send into G?
D
“pulse”
when
button is
pressed
button
pressed
Q
current light state
G
output
As long as the button is down, D flows to Q
flows through the NOT gate and back to28D...
which is bad!
A Timing Solution (Almost)
D
D
Q
G
Q
G
Never raise both “bars” at
the same time.
button
pressed
output
29
A Timing Solution
D
D
Q
G
??
Q
G
output
The two latches are never enabled at the same time (except
for the moment needed for the NOT gate on the left to
compute, which is so short that no “cars” get through).
30
A Timing Solution
D
D
Q
G
button
press
signal
Q
G
output
button
pressed
31
Button/Clock is LO
(unpressed)
1
1
D
1
LO
D
Q
Q
G
0
G
0
output
Why isn’t the right latch’s Q output 1?
a. An error: it should be.
b. The G input is 0.
c. Not enough time for the D input to propagate.
32
d. I don’t know.
Button goes HI
(is pressed)
1
1
D
1
HI
D
Q
Q
G
0
G
1
output
33
Propagating signal..
left NOT, right latch
1
1
D
0
HI
D
Q
Q
G
1
G
1
output
34
Propagating signal..
right NOT (steady state)
0
1
D
0
HI
D
Q
Q
G
1
G
1
output
35
Button goes LO (released)
0
1
D
0
LO
D
Q
Q
G
1
G
0
output
36
Propagating signal..
left NOT
0
1
D
1
LO
D
Q
Q
G
1
G
0
output
37
Propagating signal..
left latch (steady state)
0
0
D
1
LO
D
Q
Q
G
1
G
0
output
And, we’re done with one cycle.
38
How does this compare to our initial state?
Master/Slave D Flip-Flop
Symbol + Semantics
When CLK goes from low to high, the flip-flop loads a new
value from D.
Otherwise, it maintains its current value.
new
data
(D)
D
D
Q
control
or
“clock”
signal
(CLK)
G
Q
output
(Q)
G
39
Master/Slave D Flip-Flop
Symbol + Semantics
When CLK goes from low to high, the flip-flop loads a new
value from D.
Otherwise, it maintains its current value.
new
data
(D)
D
D
Q
control
or
“clock”
signal
(CLK)
G
Q
output
(Q)
G
40
Master/Slave D Flip-Flop
Symbol + Semantics
When CLK goes from low to high, the flip-flop loads a new
value from D.
Otherwise, it maintains its current value.
new
data
(D)
D
D
Q
control
or
“clock”
signal
(CLK)
G
Q
output
(Q)
G
41
Master/Slave D Flip-Flop
Symbol + Semantics
When CLK goes from low to high, the flip-flop loads a new
value from D.
Otherwise, it maintains its current value.
new data
D
Q
output
clock signal
The clock and data inputs are SWAPPED
42
compared to Logisim’s symbol; sorry!
Outline
• Prereqs, Learning Goals, and !Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
43
Branch Prediction Machine
An “if” in hardware is called a “branch”.
Computers “pre-execute” instructions... but to
pre-execute a “branch”, they must guess
whether to execute the then or else part.
Here’s one reasonable guess: whatever the
last branch did, the next one will, too.
Why? In recursion, how often do we hit the
base case vs. the recursive case?
44
Implementing the Branch
Predictor (First Pass)
Here’s the corresponding DFA. (Instead of
accept/reject, we care about the current
state.)
not
taken
taken
taken
yes
not
taken
no
Experiments show it generally works well to add “inertia”
45
so that it takes two “wrong guesses” to change the prediction…
Implementing the Branch
Predictor (Final Version)
Here’s a version that takes two wrong
guesses in a row to admit it’s wrong:
taken
taken
YES!
not
taken
yes?
taken
taken
not
taken
no?
not
taken
NO!
not
taken
Can we build a branch prediction circuit?
46
Abstract Template
for a DFA Circuit
Each time the clock “ticks” move from one
state to the next.
clock
input
store
current
state
compute
next
state
47
Template for a DFA Circuit
Each time the clock “ticks” move from one
state to the next.
D
Q
CLK
Combinational
circuit to calculate
next state/output
input
Each of these lines (except the clock) may carry multiple bits;
48
the D flip-flop may be several flip-flops to store several bits.
How to Implement a DFA:
Slightly Harder
(1) Number the states and figure out b: the number of
bits needed to store the state number.
(2) Lay out b D flip-flops. Together, their memory is
the state as a binary number.
(3) Build a combinational circuit that determines the
next state given the input and the current state.
(4) Use the next state as the new state of the flip-flops.
49
How to Implement a DFA:
Slightly Easier MUX Solution
(1) Number the states and figure out b: the number of
bits needed to store the state number.
(2) Lay out b D flip-flops. Together, their memory is
the state as a binary number.
(3) For each state, build a combinational circuit
that determines the next state given the input.
(4) Send the next states into a MUX with the current
state as the control signal: only the appropriate
next state gets used!
(5) Use the MUX’s output as the new state of the flipflops.
With a separate circuit for each
50
state, they’re often very simple!
Implementing the Predictor
Step 1
taken
taken
not
taken
YES!
3
yes?
2
taken
not
taken
What is b (the number of 1-bit flip-flops
needed to represent the state)?
a. 0, no memory needed
b. 1
c. 2
d. 3
e. None of these
taken
no?
1
not
taken
NO!
0
not
taken
As always, we use numbers
to represent the inputs:
taken = 1
51
not taken = 0
Just Truth Tables...
taken
taken
Reminder: taken = 0
not taken = 1
YES!
3
take
taken
no?
1
not
taken
yes?
2
Current State
input
New state
0
0
0
0
0
0
0
1
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
not
taken
not
taken
not
taken
NO!
0
What’s in this row?
a. 0 0
b. 0 1
c. 1 0
d. 1 1
e. None of these.
52
Just Truth Tables...
taken
taken
Reminder: taken = 0
not taken = 1
YES!
3
take
taken
no?
1
not
taken
yes?
2
Current State
input
New state
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
0
1
1
1
1
1
not
taken
not
taken
not
taken
NO!
0
53
Implementing the Predictor:
Step 3
We always use this pattern.
In this case, we need two flip-flops.
DD
Q
Q
CLK
Combinational
circuit to calculate
next state/output
input
54
Let’s switch to Logisim schematics...
Implementing the Predictor:
Step 3
DD
Q
Q
CLK
??
state0
input
???
state1
55
Implementing the Predictor:
Step 3
state0
???
state1
input
56
Implementing the Predictor:
Step 5 (easier than 4!)
state0
The MUX “trick” here is much
like in the ALU from lab!
state1
57
input
Implementing the Predictor:
Step 4
state0
state1
In state number 0, what’s the new value of state0?
a. input
b. ~input
c. 1
d. 0
58
e. None of these.
input
Implementing the Predictor:
Step 4
state0
state1
In state number 1, what’s the new value of state0?
a. input
b. ~input
c. 1
d. 0
59
e. None of these.
input
Implementing the Predictor:
Step 4
state0
state1
In state number 2, what’s the new value of state0?
a. input
b. ~input
c. 1
d. 0
60
e. None of these.
input
Implementing the Predictor:
Step 4
state0
state1
In state number 3, what’s the new value of state0?
a. input
b. ~input
c. 1
d. 0
61
e. None of these.
input
Just Truth Tables...
state0
state1
input
state0'
state1'
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
0
1
1
1
1
1
62
input
Implementing the Predictor:
Step 4
state0
state1
63
input
Implementing the Predictor:
Step 4
In state number 0, what’s the new value of state1?
a. input
b. ~input
c. 1
d. 0
e. None of these.
state1
64
input
Implementing the Predictor:
Step 4
state0
state1
TADA!!
65
input
You can often simplify,
but that’s not the point.
state0
state1
66
input
Just for Fun: Renaming to
Make a Pattern Clearer...
taken
taken
YES!
3
take
taken
no?
1
not
taken
yes?
2
pred
last
input
pred'
last'
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
0
1
1
1
1
1
not
taken
not
taken
not
taken
NO!
0
67
Outline
• Prereqs, Learning Goals, and !Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
68
Steps to Build a Powerful Machine
(1) Number the states and figure out b: the number of
bits needed to store the state number.
(2) Lay out b D flip-flops. Together, their memory is
the state as a binary number.
(3) For each state, build a combinational circuit
that determines the next state given the input.
(4) Send the next states into a MUX with the current
state as the control signal: only the appropriate
next state gets used!
(5) Use the MUX’s output as the new state of the flipflops.
With a separate circuit for each
69
state, they’re often very simple!
How Powerful Is a DFA?
How does a DFA compare to a modern
computer?
a. Modern computer is more powerful.
b. DFA is more powerful.
c. They’re the same.
70
Where We’ll Go From Here...
We’ll come back to DFAs again later in
lecture.
In lab you have been and will continue to
explore what you can do once you have
memory and events.
And, before long, how you combine these
into a working computer!
Also in lab, you’ll work with a widely used
representation equivalent to DFAs:
regular expressions.
71
Outline
• Prereqs, Learning Goals, and !Quiz Notes
• Problems and Discussion
– A Pushbutton Light Switch
– Memory and Events: D Latches & Flip-Flops
– General Implementation of DFAs
(with a more complex DFA as an example)
– How Powerful are DFAs?
• Next Lecture Notes
72
Learning Goals: In-Class
By the end of this unit, you should be able
to:
– Translate a DFA into a sequential circuit that
implements the DFA.
– Explain how and why each part of the
resulting circuit works.
73
Next Lecture Learning Goals:
Pre-Class
By the start of class, you should be able to:
– Convert sequences to and from explicit formulas
that describe the sequence.
– Convert sums to and from summation/”sigma”
notation.
– Convert products to and from product/”pi”
notation.
– Manipulate formulas in summation/product
notation by adjusting their bounds, merging or
splitting summations/products, and factoring out
values.
74
Next Lecture Prerequisites
See the Mathematical Induction Textbook
Sections at the bottom of the “Textbook
and References” page.
Complete the open-book, untimed quiz on
Vista that’s due before the next class.
75
snick

snack
More problems to solve...
(on your own or if we have time)
76
Problem: Traffic Light
Problem: Design a DFA with outputs to
control a set of traffic lights.
77
Problem: Traffic Light
Problem: Design a DFA with outputs to control a set
of traffic lights.
Thought: try allowing an output that sets a timer
which in turn causes an input like our “button
press” when it goes off.
Variants to try:
- Pedestrian cross-walks
- Turn signals
- Inductive sensors to indicate presence of cars
- Left-turn signals
78
PROBLEM: Does this accept
the empty string?
a,b,c
b
1
b,c
a
2
a
3
c
79
“Moore Machines” 
DFAs whose state matters
A “Moore Machine” is a DFA that’s not about
accepting or rejecting but about what state
it’s in at a particular time.
Let’s work with a specific example...
80
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