1. STATE DIAGRAM AND STATE TABLES

2. Derivation of State Graphs
9/2/2012 – ECE 3561 Lect 6
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.
Copyright Joanne DeGroat, ECE, OSU

3. 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

4. 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 output equations
to fully describe the sequential circuit
y = (A+B)x'
Ax +Bx Ax Bx
A 'x
A+B

5. Analysis with D flip-flops
The input equation
DA=A⊕x⊕y
The state equation
A(t+1)=A⊕x⊕y

6. 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 '

7. State Table for Fig. 5-18
JA = B, KA= Bx'
JB = x ', KB = A'x + Ax '

8. Method 1 State Transition Diagram for Fig. 5-18
The characteristic equation of JK FF is
State equation for A and B : ,

9. Method 2 State Transition Diagram for Fig. 5-18
x AB 1
AB’
A(t +1)
A’B Ax
Using K-map, we also can derive A(t+1).
A(t +1)=A ’B+AB ’+Ax

10. Analysis with T Flip-Flops
The characteristic equation
Q(t+1)= T⊕Q = TQ'+T'Q

11. A Sequence Detector Example
9/2/2012 – ECE 3561 Lect 6
The specification
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.
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12. General Form of the circuit
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The circuit has the general form
X – serial input stream
Z – serial output stream
Clk – the clock
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13. Start construction of the graph.
9/2/2012 – ECE 3561 Lect 6
Choose a starting state and a meaning for that state. The starting state is typically a reset state.
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
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Add the next state
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Now add state S1
Meaning – a sequence of 0…01 has been received when coming from state S0
Meaning – the first 1 has been received.
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Transitions from S1
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What happens when in S1
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.
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State S2
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State S2 – what is the meaning of being here?
When transition is from S1 it means we have receive an input stream of xxx10.
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Transitions from S2
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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.
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The full state diagram
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The now completed state diagram
This can now be used to generate a state table – more on that later
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Another example
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Problem Statement: The circuit has the same form as before and shown below. The circuit will detect input sequences that end in 010 or When a sequence is detected the output Z is 1, otherwise Z is 0.
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The initial state
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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
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Meaning of states
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S0 – Reset
S1 – 0 but not 10
S4 – 1 but not 01
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More states
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Add S2 having meaning that a 01 sequence has been received.
Add S3 having meaning that the sequence 10 has been received
Copyright Joanne DeGroat, ECE, OSU

23. Meaning of states after S2 S3
9/2/2012 – ECE 3561 Lect 6
S0 – Reset
S1 – 0 but not 10
S2 – Sequence of 01
S3 – Sequence of 10
S4 – 1 but not 01
Copyright Joanne DeGroat, ECE, OSU

24. Consider inputs when in S2, S3
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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)
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Add a new state S5
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S5 – Have received input sequence 100
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When in S5
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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.
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Add other transitions
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Complete the transitions not yet covered
Each state should have an output transition for both a 0 and a 1.
Copyright Joanne DeGroat, ECE, OSU

28. The meaning of the states
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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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Guidelines
9/2/2012 – ECE 3561 Lect 6
Guidelines for Construction of State Graphs
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.
Copyright Joanne DeGroat, ECE, OSU

30. MOORE MACHINE

31. Mealy Machine vs. Moore Machine

32. Modern Design Register-transfer-level block diagram Datapath
C: Combinational circuit
S: Sequential circuit C S
Control Unit
Datapath D Q .

33. FSM Design D or JK or T ?? current state register QX DX Out In clk ？
next
state QX
current
state DX
Out In
clk ？ QY ？ DY
clk
output logic
next state logic

34. Consider the sequence detector
9/2/2012 – ECE 3561 Lect 7
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.
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Start in S0
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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.
Copyright Joanne DeGroat, ECE, OSU

36. Transitions form state 1
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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
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Transition from S1
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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.
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State S2
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S2 has meaning that you have an input sequence that ends in 10 so far.
Transitions from S2
0 input – Back to S0
1 input – Valid sequence
go to new state S3
which outputs a 1
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State S3
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S3 – have received input sequence that ends in 101.
Next input
0 – end of seq
(10 so back to S2)
1 – back to S1
(11 so 1st 1)
Copyright Joanne DeGroat, ECE, OSU

40. State Table from State Graph
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Easy to convert state graph to state table
Moore machine
note output is function
of the state
Copyright Joanne DeGroat, ECE, OSU

41. Contrast this to Mealy Machine
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Mealy machine state graph and state table
In Mealy machine the output is a function of the
state and the current input
Copyright Joanne DeGroat, ECE, OSU

42. Now on to the other example
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Detect the sequences 010 and 1001 and on those output a 1.
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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In S1/0
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State S1 have the first 0 of a possible 010
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
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From S2/0
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S2/0 has meaning that you have 01 so far
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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From S3/0
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S3/0 has meaning that you have the first 1 of the sequence.
Input is a 0 – Go to S5 – meaning have 10
Input is a 1 – stay in S3
Copyright Joanne DeGroat, ECE, OSU

46. Add transitions from S4/1
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S4/1 had meaning that the sequence has been 010 so far.
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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Transitions from S5/0
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S5/0 means you have 10 so far
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
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State S6/0
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S6/0 has meaning that you have a sequence of 100 so far
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.
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From S7/1
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S7 has meaning of so you also have the 01 for the start of that sequence
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.
Copyright Joanne DeGroat, ECE, OSU

50. The state table for each
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For the Mealy Machine
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For the Moore machine
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The state table for the Moore machine – output is associated with the state.
Present State
Next State X=0
Next State X=1
Output Z S0 S1 S3 S2 S4 S5 S6 1 S7
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The next step
9/2/2012 – ECE 3561 Lect 7
The next step to implementation is state assignment
In state assignment the binary code for each state is chosen.
Copyright Joanne DeGroat, ECE, OSU

53. Effect of choosing state assignment
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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?
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54. Example that has sink state
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Programmed Example 14.2
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Initial states
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The start of the state graph
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Step 2
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More states
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Complete state graph
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58. Corresponding State Table
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From the state graph the state table can be generated
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Lecture summary
9/2/2012 – ECE 3561 Lect 7
Have covered state graphs for Mealy and Moore machines
Have covered how to transition from state graphs to state tables.
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