### STATE DIAGRAM AND STATE TABLES

```STATE DIAGRAM AND
STATE TABLES
DeGroat, ECE, OSU
2
Problem Statement specifies the desired
relationship between the input and output
sequences. Sometimes called the specification.
 First step is to translate this specification into a
state table or state graph.
 In the HDL world, there is a style that allows
creation of the next state specification that does
not require either a state graph or state table.

9/2/2012 – ECE 3561 Lect
6
DERIVATION OF STATE GRAPHS
STATE DIAGRAM

State transition diagram


a circle: a state
a directed lines connecting the circles: the transition
between the states

Each directed line is labeled “inputs/outputs”
state: A B
input: x
FLIP-FLOP INPUT EQUATIONS
 The
part of circuit that
generates the inputs to
flip-flops

Also called excitation
functions

DA = Ax +Bx

DB = A'x
 The
Ax
Ax +Bx
Bx
A 'x
output equations

to fully describe the
sequential circuit

y = (A+B)x'
A+B
ANALYSIS WITH D FLIP-FLOPS
The input equation
 DA=A⊕x⊕y
 The state equation
 A(t+1)=A⊕x⊕y

ANALYSIS WITH JK FLIP-FLOPS


Determine the flip-flop input function in terms of
the present state and input variables
Used the corresponding flip-flop characteristic
table to determine the next state
Fig. 5-18
Sequential circuit
with
JK flip-flop
JA = B
KA= Bx'
JB = x '
KB = A'x + Ax '
STATE TABLE FOR FIG. 5-18
JA = B, KA= Bx'
JB = x ', KB = A'x + Ax '
State Transition Diagram for Fig. 5-18

Method 1
The characteristic equation of JK FF is
A(t  1)  JA  K A
B(t  1)  JB  K B
State equation for A and B :
A(t  1)  BA  ( Bx) A  AB  AB  Ax
,
,
B(t  1)  xB  ( A  x)B  Bx  ABx  ABx
State Transition Diagram for Fig. 5-18

Method 2
x
AB00
A(t +1)
01
11
10
0
0
1
0
1 AB’
1
0
1
1
1
A’B
Ax
Using K-map, we also can derive A(t+1).
A(t +1)=A ’B+AB ’+Ax
ANALYSIS WITH T FLIP-FLOPS

The characteristic equation
 Q(t+1)= T⊕Q = TQ'+T'Q

The specification

DeGroat, ECE, OSU
11
The circuit will examine a string of 0’s and 1’s applied
serially, once per clock, to the X input and produce a
1 only when the prescribed input sequence occurs.
Any sequence ending in 101 will produce and output
of Z=1 coincident with the last 1 input. The circuit
does not reset when a 1 output occurs so when ever a
101 is in the data stream a 1 is output coincident
with the last 1.
9/2/2012 – ECE 3561 Lect
6
A SEQUENCE DETECTOR EXAMPLE
9/2/2012 – ECE 3561 Lect
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GENERAL FORM OF THE CIRCUIT

The circuit has the general form
X – serial input stream
 Z – serial output stream
 Clk – the clock

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DeGroat, ECE, OSU


Here meaning of starting state can be
The system has been reset and this is the
initial state
 A sequence of 2 or more 0’s has been received

DeGroat, ECE, OSU
13
Choose a starting state and a meaning for that
state. The starting state is typically a reset
state.
9/2/2012 – ECE 3561 Lect
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START CONSTRUCTION OF THE GRAPH.

DeGroat, ECE, OSU
14
Meaning – a sequence of 0…01 has been received
when coming from state S0
 Meaning – the first 1 has been received.

9/2/2012 – ECE 3561 Lect
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
What happens when in S1
DeGroat, ECE, OSU
15
A 0 input causes transition to a new state S2 with
new meaning
 A 1 keeps you in S1 where the first 1 of a possible 101
sequence has occurred.

9/2/2012 – ECE 3561 Lect
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TRANSITIONS FROM S1

State S2 – what is the meaning of being here?

DeGroat, ECE, OSU
16
When transition is from S1 it means we have receive
an input stream of xxx10.
9/2/2012 – ECE 3561 Lect
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STATE S2

Are currently in S2

A 1 arrives and now have a sequence of 101

Action – Output a 1 and have the first 1 of a
new sequence, i.e., transition to S1
A 0 arrives – now have a sequence of 100

Action – Move back to state S0 where you do
not even have the start of a sequence, i.e.,
one or more 0 inputs.
DeGroat, ECE, OSU
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
9/2/2012 – ECE 3561 Lect
6
TRANSITIONS FROM S2

The now completed
diagram
state
9/2/2012 – ECE 3561 Lect
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THE FULL STATE DIAGRAM
18

This can now be used to generate a state table –
more on that later
DeGroat, ECE, OSU

DeGroat, ECE, OSU
19
Problem Statement: The circuit has the same
form as before and shown below. The circuit will
detect input sequences that end in 010 or 1001.
When a sequence is detected the output Z is 1,
otherwise Z is 0.
9/2/2012 – ECE 3561 Lect
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ANOTHER EXAMPLE
DeGroat, ECE, OSU
20
The RESET state – have no inputs yet
 Then if you have a 0 input the output is 0 –
transition to S1
 If you have a 1 input the output is 0 and
transition to S2

9/2/2012 – ECE 3561 Lect
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THE INITIAL STATE
9/2/2012 – ECE 3561 Lect
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MEANING OF STATES
S0 – Reset
 S1 – 0 but not 10
 S4 – 1 but not 01

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DeGroat, ECE, OSU
DeGroat, ECE, OSU
22
Add S2 having meaning that a 01 sequence has
 Add S3 having meaning that the sequence 10 has

9/2/2012 – ECE 3561 Lect
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MORE STATES
S0 – Reset
 S1 – 0 but not 10
 S2 – Sequence of 01
 S3 – Sequence of 10
 S4 – 1 but not 01

9/2/2012 – ECE 3561 Lect
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MEANING OF STATES AFTER S2 S3
23
DeGroat, ECE, OSU
DeGroat, ECE, OSU
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In S2 (01) and get a 0 – Transition to S3 (10) –
output a 1
 In S3 (10) and get a 1 – Transition to S2 (01)

9/2/2012 – ECE 3561 Lect
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CONSIDER INPUTS WHEN IN S2, S3

S5 – Have received input sequence 100
9/2/2012 – ECE 3561 Lect
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DeGroat, ECE, OSU
9/2/2012 – ECE 3561 Lect
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WHEN IN S5

In S5
Input of a 1 means you have had
a input of 1001 so transition to
S2 as the input sequence now
ends in 01 while Z is 1.
26

DeGroat, ECE, OSU
9/2/2012 – ECE 3561 Lect
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Complete the transitions not
yet covered
 Each state should have an
output transition for both a 0
and a 1.

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DeGroat, ECE, OSU
9/2/2012 – ECE 3561 Lect
6
THE MEANING OF THE STATES
S0 – Reset
 S1 – 0 (but not 10)
 S2 – Sequence of 01
 S3 – Sequence of 10
 S4 – 1 (but not 01)
 S5 – Sequence of 100

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DeGroat, ECE, OSU

Guidelines for Construction of State Graphs






DeGroat, ECE, OSU
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
First, construct some sample input and output sequences
to make sure you understand the problem (ref slides 5 and
13)
Determine under what conditions the circuit is in reset
state.
If only one or two sequences lead to a 1 output construct a
partial state graph.
OR determine what sequences or groups of sequences must
be remembered
When adding transitions see if you transition to a defined
state or a new state is to be added
Make sure all state have a transition for both a 0 and a 1
but only 1!
Add annotation or create a table to expound the meaning of
each state.
9/2/2012 – ECE 3561 Lect
6
GUIDELINES
MOORE MACHINE
MEALY MACHINE VS. MOORE
MACHINE
Modern Design
Register-transfer-level block diagram
C: Combinational circuit
S: Sequential circuit
C
Q
S
D
S
.
.
C
C
Q
D
S
Control Unit
C
C
Datapath
FSM DESIGN
D or JK or T ??
current state register
next
state
In
？
DX
clk
QXcurrent
state
Out
QY
？
DY
clk
next state logic
output logic
34
The same sequence detector to detect a sequence
ending in 101 but this time a Moore machine
implementation.
 Moore machine implementation is much the
same except that the output designation is now
indicated within the state.

9/2/2012 – ECE 3561 Lect
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CONSIDER THE SEQUENCE DETECTOR
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ne
DeGr
oat,
ECE,
35
S0 –a state where you have received a non
middle 0 or a long string of 0s. Output is 0.
 Output is indicated within the state not on the
transition.

9/2/2012 – ECE 3561 Lect
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START IN S0
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Joan
ne
DeGr
oat,
ECE,
36
On a 0 you stay in state 1
 On a 1 you transition to state S1. Meaning of S1
– have the 1st 1 of the sequence

9/2/2012 – ECE 3561 Lect
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TRANSITIONS FORM STATE 1
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ne
DeGr
oat,
ECE,
37
On a 1 have the first 1 of a sequence – stay in S1.
 On a 0 now have a sequence that ends in 10 so
define a new state S2 and transition to it.

9/2/2012 – ECE 3561 Lect
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TRANSITION FROM S1
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Joan
ne
DeGr
oat,
ECE,
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S2 has meaning that you have an input sequence
that ends in 10 so far.
 Transitions from S2

9/2/2012 – ECE 3561 Lect
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STATE S2
0 input – Back to S0
 1 input – Valid sequence

go to new state S3

which outputs a 1

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ne
DeGr
oat,
ECE,
39
S3 – have received input sequence that ends in
101.
 Next input

9/2/2012 – ECE 3561 Lect
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STATE S3
0 – end of seq

(10 so back to S2)
 1 – back to S1

(11 so 1st 1)

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Joan
ne
DeGr
oat,
ECE,

Easy to convert state graph to state table


40

Moore machine
note output is function
of the state
9/2/2012 – ECE 3561 Lect
7
STATE TABLE FROM STATE GRAPH
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Joan
ne
DeGr
oat,
ECE,

Mealy machine state graph and state table
41
In Mealy machine the output is a function of the

state and the current input

9/2/2012 – ECE 3561 Lect
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CONTRAST THIS TO MEALY MACHINE
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Joan
ne
DeGr
oat,
ECE,

Detect the sequences 010 and 1001 and on those
output a 1.
9/2/2012 – ECE 3561 Lect
7
NOW ON TO THE OTHER EXAMPLE
42

Starting state on reset is S0

On a 0 transition to S1 - output 0


Have a first 0
On a 1 transition to S3 - output 0

Have a first 1
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Joan
ne
DeGr
oat,
ECE,

State S1 have the first 0 of a possible 010
43
On a 1 now have 01

Transition to a new state S2/0 with meaning that
you have 01
 On a 0 stay in S1/0

9/2/2012 – ECE 3561 Lect
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IN S1/0
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Joan
ne
DeGr
oat,
ECE,

44
S2/0 has meaning that you have 01
so far
9/2/2012 – ECE 3561 Lect
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FROM S2/0
Input is a 0 – Need a new state S4
with meaning that you have received
010 (so output is a 1) and have a 10
for a start of that string.
 Input is a 1 so the input is 011 – Go
to S3 where as this is the first 1.

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Joan
ne
DeGr
oat,
ECE,

S3/0 has meaning that you have the first 1 of the
1001 sequence.
45
Input is a 0 – Go to S5 – meaning have 10
 Input is a 1 – stay in S3

9/2/2012 – ECE 3561 Lect
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FROM S3/0
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Joan
ne
DeGr
oat,
ECE,

46
sequence has been 010 so
far.
9/2/2012 – ECE 3561 Lect
7
Input is a 0 – Now have 100
– Need a new state with this
meaning – S6/0
 Input is a 1 – Now have 101
so go back to S2/0

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Joan
ne
DeGr
oat,
ECE,

S5/0 means you have 10 so far
47
Input is a 0 – transition to S6/0 – have 100 so far
 Input is a 1 – now have 101 or the 01 which is the
meaning of S2/0

9/2/2012 – ECE 3561 Lect
7
TRANSITIONS FROM S5/0
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Joan
ne
DeGr
oat,
ECE,

Input is a 1 so have 1001 –
a new state S7/1 to signal
the sequence 1001.
 Input is a 0 so have 1000
and back to S1 as you have
a first 0.
48
S6/0 has meaning that you
have a sequence of 100 so
far
9/2/2012 – ECE 3561 Lect
7
STATE S6/0

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Joan
ne
DeGr
oat,
ECE,



Input is a 0 so have 010 –
go to S4/1
Input is a 1 so have 011 –
go to S3 as you have a
first 1.
49
S7 has meaning of 1001
so you also have the 01
for the start of that
sequence
9/2/2012 – ECE 3561 Lect
7
FROM S7/1
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Joan
ne
DeGr
oat,
ECE,
9/2/2012 – ECE 3561 Lect
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THE STATE TABLE FOR EACH

For the Mealy Machine
50
Present State
S0
S1
S2
S3
S4
S5
X=0
S1
S1
S3
S5
S3
S1
NEXT STATE
X=1
S4
S2
S4
S2
S4
S2
OUTPUT
X=0 X=1
0
0
0
0
1
0
0
0
0
0
0
1
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Joan
ne
DeGr
oat,
ECE,

Present State
S0
S1
S2
S3
S4
S5
S6
S7
Next State X=0
S1
S1
S4
S5
S6
S6
S1
S4
Next State X=1
S3
S2
S3
S3
S2
S2
S7
S3
Output Z
0
0
0
0
1
0
0
1
51
The state table for the Moore
machine – output is associated
with the state.
9/2/2012 – ECE 3561 Lect
7
FOR THE MOORE MACHINE
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Joan
ne
DeGr
oat,
ECE,
52
The next step to implementation is state
assignment
 In state assignment the binary code for each
state is chosen.

9/2/2012 – ECE 3561 Lect
7
THE NEXT STEP
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Joan
ne
DeGr
oat,
ECE,
53
Choosing one state assignment versus another
can have significant implications for circuit
implementation.
 But first – how do you reduce the number of
states in the state table?

9/2/2012 – ECE 3561 Lect
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EFFECT OF CHOOSING STATE ASSIGNMENT
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Joan
ne
DeGr
oat,
ECE,

Programmed Example 14.2
9/2/2012 – ECE 3561 Lect
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EXAMPLE THAT HAS SINK STATE
54
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Joan
ne
DeGr
oat,
ECE,

The start of the state graph
9/2/2012 – ECE 3561 Lect
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INITIAL STATES
55
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Joan
ne
DeGr
oat,
ECE,

More states
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STEP 2
56
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Joan
ne
DeGr
oat,
ECE,
9/2/2012 – ECE 3561 Lect
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COMPLETE STATE GRAPH
57
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Joan
ne
DeGr
oat,
ECE,

From the state graph the state table can be
generated
9/2/2012 – ECE 3561 Lect
7
CORRESPONDING STATE TABLE
58
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Joan
ne
DeGr
oat,
ECE,
59
Have covered state graphs for Mealy and Moore
machines
 Have covered how to transition from state graphs
to state tables.

9/2/2012 – ECE 3561 Lect
7
LECTURE SUMMARY
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ne
DeGr
oat,
ECE,
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