1 Synchronous Sequential Circuit Analysis
1 Synchronous Sequential Circuit Analysis. 2 Synchronous Sequential Circuit State Memory – A set of n edge-triggered flip-flops that store the current.
Dec 22, 2015
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Synchronous Sequential Circuit Analysis
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Synchronous Sequential Circuit
• State Memory – A set of n edge-triggered flip-flops that store the current state of the machine
– All flip-flops are triggered from the same master clock signal
– All change state together
• Combinational circuit
– Next state logic
– Output logic – Mealy and Moore
Combinational
circuit
Inputs
State Memory
Outputs
Clock
CurrentState
NextState
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Mealy Model
next state = F (current state, inputs)
outputs = G (current state, inputs)
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Moore Model
next state = F (current state, inputs)
outputs = G (current state)
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Analysis - Goals
• Characterize as Mealy or Moore machine
• Determine next state equations, i.e., find the function F
– next state = F (current state, inputs)
• Determine output equations
– Meally: outputs = F (current state, inputs), or
– Moore: outputs = F (current state)
• Express as machine behavior
– State table, or
– State diagram
• Formulate English description of machine behavior
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An example sequential circuit
• A sequential circuit with two JK flip-flops
• State or memory: Q1Q0
• One input: X; One output: Z
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State table of example circuit
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1
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Output Equations
• From the diagram, you can see that
Z = Q1Q0X
Mealy model circuit !!!
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 00 0 1 00 1 0 00 1 1 01 0 0 01 0 1 01 1 0 01 1 1 1
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Next State Equations – Q(t+1)
• Find the flip-flop input equations/excitation equations
• Substitute excitation equations in the flip-flop’s characteristic equation
J1 = X’ Q0
K1 = X + Q0
J0 = X + Q1
K0 = X’
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Next State Equations – Q(t+1)
• Next state equations:
– Q1(t+1) = K1’Q1(t) + J1Q1’(t)
= (X + Q0(t))’ Q1(t) + X’ Q0 (t) Q1’(t)
= X’ (Q0(t)’ Q1(t) + Q0(t) Q1(t)’)
= X’ (Q0(t) Q1(t))
– Q0(t+1) = K0’Q0(t) + J0Q0’(t)
= X Q0(t) + (X + Q1(t)) Q0’(t)
= X + Q0(t)’ Q1(t)
• Excitation equations:
– J1 = X’ Q0 and K1 = X + Q0
– J0 = X + Q1 and K0 = X’
• Characteristic equation of the JK flip-flop:
– Q(t+1) = K’Q(t) + JQ’(t)
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State Table & Next State Equations
• Q1(t+1) = X’ (Q0(t) Q1(t))
– Q1=0, Q0=0, X= 0 => Q1(t+1)= 0
• Q0(t+1) = X + Q0(t)’ Q1(t)
– Q1=0, Q0=0, X= 0 => Q0(t+1)= 0
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 00 0 1 00 1 0 00 1 1 01 0 0 01 0 1 01 1 0 01 1 1 1
0 0
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State Table & Next State Equations
• Q1(t+1) = X’ (Q0(t) Q1(t))
– Q1=0, Q0=1, X= 1 => Q1(t+1)= 0
• Q0(t+1) = X + Q0(t)’ Q1(t)
– Q1=0, Q0=1, X= 1 => Q0(t+1)= 1
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 00 0 1 00 1 0 00 1 1 01 0 0 01 0 1 01 1 0 01 1 1 1
0 0
0 1
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State Table & Next State Equations
• Q1(t+1) = X’ (Q0(t) Q1(t))
• Q0(t+1) = X + Q0(t)’ Q1(t)
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 0 0 00 0 1 0 1 00 1 0 1 0 00 1 1 0 1 01 0 0 1 1 01 0 1 0 1 01 1 0 0 0 01 1 1 0 1 1
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State Table & Characteristic Table
• The general JK flip-flop characteristic equation is:
Q(t+1) = K’Q(t) + JQ’(t)
• We can also determine the next state for each input/current state combination directly from the characteristic table
J K Q(t+1) Operation
0 0 Q(t) No change0 1 0 Reset1 0 1 Set1 1 Q’(t) Complement
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State Table & Characteristic Table
• With these equations, we can make a table showing J1, K1, J0 and K0
for the different combinations of present state Q1Q0 and input X
J1 = X’ Q0 J0 = X + Q1
K1 = X + Q0 K0 = X’
Present State I nputs Flip-flop I nputsQ1 Q0 X J 1 K1 J 0 K0
0 0 0 0 0 0 10 0 1 0 1 1 00 1 0 1 1 0 10 1 1 0 1 1 01 0 0 0 0 1 11 0 1 0 1 1 01 1 0 1 1 1 11 1 1 0 1 1 0
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State Table & Characteristic Table
Present State I nputs FF I nputs Next State Q1 Q0 X J 1 K1 J 0 K0 Q1 Q0
0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 0
J K Q(t+1)
0 0 Q(t) 0 1 0 1 0 1 1 1 Q’(t)
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State Table & Characteristic Table
Present State I nputs FF I nputs Next State Q1 Q0 X J 1 K1 J 0 K0 Q1 Q0
0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 0
J K Q(t+1)
0 0 Q(t) 0 1 0 1 0 1 1 1 Q’(t)
0
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A different look
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 0 0 00 0 1 0 1 00 1 0 1 0 00 1 1 0 1 01 0 0 1 1 01 0 1 0 1 01 1 0 0 0 01 1 1 0 1 1
Present State Q1 Q0
Next StateOutput Z
Input X= 0
Input X= 1
X= 0 X= 1
0 0 0 0 0 1 0 0
0 1 1 0 0 1 0 0
1 0 1 1 0 1 0 0
1 1 0 0 0 1 0 1
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State diagrams (Mealy model)
Present State I nputs Next State OutputsQ1 Q0 X Q1 Q0 Z
0 0 0 0 0 00 0 1 0 1 00 1 0 1 0 00 1 1 0 1 01 0 0 1 1 01 0 1 0 1 01 1 0 0 0 01 1 1 0 1 1
• We can also represent the state table graphically with a state diagram
• A diagram corresponding to our example state table is shown below
00 01
1011
1/0
0/00/0
0/0
0/0 1/0
1/01/1
input output
state
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Sizes of state diagrams
00 01
1011
1/0
0/00/0
0/0
0/0 1/0
1/01/1
• Always check the size of your state diagrams
– If there are n flip-flops, there should be 2n nodes in the diagram
– If there are m inputs, then each node will have 2m outgoing arrows
• In our example,
– We have two flip-flops, and thus four states or nodes.
– There is one input, so each node has two outgoing arrows.
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Another Mealy Circuit
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Excitation Equations
• DD00 = EN’ = EN’ QQ00 + EN + EN QQ00’’
• DD11 = EN’ = EN’ QQ11 + EN + EN QQ11’ ’ QQ00 + EN + EN QQ11 QQ00’’
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Next State/Output Equations
• QQ00(t+1) = D(t+1) = D00 = EN’ = EN’ QQ00 + EN + EN QQ00’’
• QQ11(t+1) = D(t+1) = D11 = EN’ = EN’ QQ11 + EN + EN QQ11’ ’ QQ00 + EN + EN QQ11 QQ00’’
• MAX= EN MAX= EN QQ11 QQ00
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Mealy State Table
Present State Q1 Q0
Next StateOutput MAX
Input EN= 0
Input EN= 1
X= 0 X= 1
0 0 0 0 0 1 0 0
0 1 0 1 1 0 0 0
1 0 1 0 1 1 0 0
1 1 1 1 0 0 0 1
• QQ00(t+1) = D(t+1) = D00 = EN’ = EN’ QQ00 + EN + EN QQ00’’
• QQ11(t+1) = D(t+1) = D11 = EN’ = EN’ QQ11 + EN + EN QQ11’ ’ QQ00 + EN + EN QQ11 QQ00’’
• MAX= EN MAX= EN QQ11 QQ00
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Mealy State Diagram
Present State Q1 Q0
Next StateOutput MAX
Input EN= 0
Input EN= 1
X= 0 X= 1
0 0 0 0 0 1 0 0
0 1 0 1 1 0 0 0
1 0 1 0 1 1 0 0
1 1 1 1 0 0 0 1
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Moore Circuit
X
Remove input connection to output logic => Moore machine
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Next State/Output Equations
• QQ00(t+1) = D(t+1) = D00 = EN’ = EN’ QQ00 + EN + EN QQ00’’
• QQ11(t+1) = D(t+1) = D11 = EN’ = EN’ QQ11 + EN + EN QQ11’ ’ QQ00 + EN + EN QQ11 QQ00’’
• MAX= MAX= QQ11 QQ00
X
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Moore State Table
Present State Q1 Q0
Next State
Output MAX
Input EN= 0
Input EN= 1
0 0 0 0 0 1 0
0 1 0 1 1 0 0
1 0 1 0 1 1 0
1 1 1 1 0 0 1
• QQ00(t+1) = D(t+1) = D00 = EN’ = EN’ QQ00 + EN + EN QQ00’’
• QQ11(t+1) = D(t+1) = D11 = EN’ = EN’ QQ11 + EN + EN QQ11’ ’ QQ00 + EN + EN QQ11 QQ00’’
• MAX= MAX= QQ11 QQ00
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Moore State Diagram
Present State Q1 Q0
Next StateOutput MAX
Input EN= 0
Input EN= 1
X= 0 X= 1
0 0 0 0 0 1 0 0
0 1 0 1 1 0 0 0
1 0 1 0 1 1 0 0
1 1 1 1 0 0 0 1
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State Transitions
• MAX : Output of the Mealy circuit
• MAXS : Output of the Moore circuit
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Sequential circuit analysis summary
• To analyze sequential circuits, you have to:
– Find Boolean expressions for the outputs of the circuit and the
flip-flop inputs
– Use these expressions to fill in the output and flip-flop input columns
in the state table
– Finally, use the characteristic equation or characteristic table of theflip-flop to fill in the next state columns.
• The result of sequential circuit analysis is a state table or a state
diagram describing the circuit
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