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**STATE DIAGRAM AND STATE TABLES**

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**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

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**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

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**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

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**Analysis with D flip-flops**

The input equation DA=A⊕x⊕y The state equation A(t+1)=A⊕x⊕y

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**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 '

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

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**Method 1 State Transition Diagram for Fig. 5-18**

The characteristic equation of JK FF is State equation for A and B : ,

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**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

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**Analysis with T Flip-Flops**

The characteristic equation Q(t+1)= T⊕Q = TQ'+T'Q

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**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. Copyright Joanne DeGroat, ECE, OSU

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**General Form of the circuit**

9/2/2012 – ECE 3561 Lect 6 The circuit has the general form X – serial input stream Z – serial output stream Clk – the clock Copyright Joanne DeGroat, ECE, OSU

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**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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Add the next state 9/2/2012 – ECE 3561 Lect 6 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Transitions from S1 9/2/2012 – ECE 3561 Lect 6 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

State S2 9/2/2012 – ECE 3561 Lect 6 State S2 – what is the meaning of being here? When transition is from S1 it means we have receive an input stream of xxx10. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Transitions from S2 9/2/2012 – ECE 3561 Lect 6 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

The full state diagram 9/2/2012 – ECE 3561 Lect 6 The now completed state diagram This can now be used to generate a state table – more on that later Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Another example 9/2/2012 – ECE 3561 Lect 6 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

The initial state 9/2/2012 – ECE 3561 Lect 6 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

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

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

More states 9/2/2012 – ECE 3561 Lect 6 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

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**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

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**Consider inputs when in S2, S3**

9/2/2012 – ECE 3561 Lect 6 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) Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Add a new state S5 9/2/2012 – ECE 3561 Lect 6 S5 – Have received input sequence 100 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

When in S5 9/2/2012 – ECE 3561 Lect 6 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Add other transitions 9/2/2012 – ECE 3561 Lect 6 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

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**The meaning of the states**

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) S5 – Sequence of 100 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

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

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MOORE MACHINE

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**Mealy Machine vs. Moore Machine**

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**Modern Design Register-transfer-level block diagram Datapath**

C: Combinational circuit S: Sequential circuit C S Control Unit Datapath D Q .

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**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

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**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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Start in S0 9/2/2012 – ECE 3561 Lect 7 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

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**Transitions form state 1**

9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Transition from S1 9/2/2012 – ECE 3561 Lect 7 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

State S2 9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

State S3 9/2/2012 – ECE 3561 Lect 7 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

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**State Table from State Graph**

9/2/2012 – ECE 3561 Lect 7 Easy to convert state graph to state table Moore machine note output is function of the state Copyright Joanne DeGroat, ECE, OSU

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**Contrast this to Mealy Machine**

9/2/2012 – ECE 3561 Lect 7 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

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**Now on to the other example**

9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

In S1/0 9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

From S2/0 9/2/2012 – ECE 3561 Lect 7 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

From S3/0 9/2/2012 – ECE 3561 Lect 7 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

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**Add transitions from S4/1**

9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Transitions from S5/0 9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

State S6/0 9/2/2012 – ECE 3561 Lect 7 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. Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

From S7/1 9/2/2012 – ECE 3561 Lect 7 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

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**The state table for each**

9/2/2012 – ECE 3561 Lect 7 For the Mealy Machine Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

For the Moore machine 9/2/2012 – ECE 3561 Lect 7 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 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

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

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**Effect of choosing state assignment**

9/2/2012 – ECE 3561 Lect 7 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? Copyright Joanne DeGroat, ECE, OSU

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**Example that has sink state**

9/2/2012 – ECE 3561 Lect 7 Programmed Example 14.2 Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Initial states 9/2/2012 – ECE 3561 Lect 7 The start of the state graph Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Step 2 9/2/2012 – ECE 3561 Lect 7 More states Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

Complete state graph 9/2/2012 – ECE 3561 Lect 7 Copyright Joanne DeGroat, ECE, OSU

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**Corresponding State Table**

9/2/2012 – ECE 3561 Lect 7 From the state graph the state table can be generated Copyright Joanne DeGroat, ECE, OSU

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**Copyright 2012 - Joanne DeGroat, ECE, OSU**

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. Copyright Joanne DeGroat, ECE, OSU

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