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Fundamentals of Computer SystemsFinite State Machines
Stephen A. Edwards
Columbia University
Spring 2012
Finite State Machine Components
CL
CL
CLOCK
Next State Current State
Current State
Inputs
Next State
Current State
Inputs?
Outputs
Moore and Mealy Machines
Next StateLogic
Output Logic
CLK
NextState
CurrentState
Inputs Outputs
The Moore Form:
Outputs are a function of only the current state.
Moore and Mealy Machines
Next StateLogic
Output Logic
CLK
NextState
CurrentState
Inputs Outputs
The Mealy Form:
Outputs may be a function of both the current state andthe inputs.
A mnemonic: Moore machines often have more states.
Mealy Machines are the Most General
CLCurrentState
Next State
Inputs Outputs
CLK
Another, equivalent way of drawing Mealy Machines
This is exactly the synchronous digital logic paradigm
Moore vs. Mealy FSMs
Alyssa P. Hacker has a snail thatcrawls down a paper tape with1’s and 0’s on it. The snail smileswhenever the last four digits ithas crawled over are 1101.Design Moore and Mealy FSMs ofthe snail’s brain.
State Transition Diagrams: Looking for “1101”
S00
Moore Machine: States indicate output
S0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
1
0
Moore Machine: States indicate output
S0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
Moore Machine: States indicate output
S0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
01
0
1
0
Moore Machine: States indicate output
S0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
01
0
1
0
Moore Machine: States indicate output
S0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
01
0
1
0
Moore Machine: States indicate output
S0 S11/0
0/0
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
01
0
1
0
Moore Machine: States indicate output
S0 S1 S2 S31/0
0/0
1/0 0/0
1/1
Mealy Machine: Arcs indicate input/output
State Transition Diagrams: Looking for “1101”
S00
S10
S20
S30
S41
1
0
1 0 1
01
0
1
0
Moore Machine: States indicate output
S0 S1 S2 S31/0
0/0
1/0 0/0
1/10/0
1/0
0/0
Mealy Machine: Arcs indicate input/output
Moore Machine
Next StateS A S’
S0 0 S0S0 1 S1S1 0 S0S1 1 S2S2 0 S3S2 1 S2S3 0 S0S3 1 S4S4 0 S0S4 1 S2
OutputS Y
S0 0S1 0S2 0S3 0S4 1
CLK
CLK
CLK
S′0
S′1
S′2
A
S0
S1
S2Y
Moore Machine
Next StateS A S’
000 0 000000 1 001001 0 000001 1 010010 0 011010 1 010011 0 000011 1 100100 0 000100 1 010
OutputS Y
000 0001 0010 0011 0100 1
CLK
CLK
CLK
S′0
S′1
S′2
A
S0
S1
S2Y
Mealy Machine
S A S’ Y
S0 0 S0 0S0 1 S1 0S1 0 S0 0S1 1 S2 0S2 0 S3 0S2 1 S2 0S3 0 S0 0S3 1 S1 1
CLK
CLK
S′1
S′0
A
S0
S1
Y
Mealy Machine
S A S’ Y
00 0 00 000 1 01 001 0 00 001 1 10 010 0 11 010 1 10 011 0 00 011 1 01 1
CLK
CLK
S′1
S′0
A
S0
S1
Y
More Intuitive Solutions using Shift Registers
A
Y
CLK
Mealy Form: Output Depends on Input Immediately
A
Y
CLK
Moore Form: Output Depends Only on State
FSM Example: A Traffic Light Controller
C
C
This controls a trafficlight at the intersectionof a busy highway and a farmroad. Normally, the highwaylight is green but if a sensordetects a car on the farmroad, the highway light turns
yellow then red. The farm road light then turns greenuntil there are no cars or after a long timeout. Then, thefarm road light turns yellow then red, and the highwaylight returns to green. The inputs to the machine arethe car sensor, a short timeout signal, and a longtimeout signal. The outputs are a timer start signal andthe colors of the highway and farm road lights.Source: Mead and Conway, Introduction to VLSI Systems, 1980, p. 85.
State Transition Diagram for the TLC
HGH : GF : R
HYH : YF : R
C+ L/TCL/T
FGH : RF : G
FYH : RF : Y
S/T
S/T
CL/TC+ L/T
S/T
S/T
Inputs:C: Car sensorS: Short TimeoutL: Long Timeout
Outputs:T: Timer ResetH: Highway colorF: Farm road color
S H F
HG G RHY Y RFG R GFY R Y
State Transition Diagram for the TLC
HGH : GF : R
HYH : YF : R
C+ L/TCL/T
FGH : RF : G
FYH : RF : Y
S/T
S/T
CL/TC+ L/T
S/T
S/T
Inputs:C: Car sensorS: Short TimeoutL: Long Timeout
Outputs:T: Timer ResetH: Highway colorF: Farm road color
S H F
HG G RHY Y RFG R GFY R Y
State Transition Diagram for the TLC
HGH : GF : R
HYH : YF : R
C+ L/TCL/T
FGH : RF : G
FYH : RF : Y
S/T
S/T
CL/TC+ L/T
S/T
S/T
Inputs:C: Car sensorS: Short TimeoutL: Long Timeout
Outputs:T: Timer ResetH: Highway colorF: Farm road color
S H F
HG G RHY Y RFG R GFY R Y
State Transition Diagram for the TLC
HGH : GF : R
HYH : YF : R
C+ L/TCL/T
FGH : RF : G
FYH : RF : Y
S/T
S/T
CL/TC+ L/T
S/T
S/T
S C S L T S’
HG 0 X X 0 HGHG X X 0 0 HGHG 1 X 1 1 HYHY X 0 X 0 HYHY X 1 X 1 FGFG 1 X 0 0 FGFG 0 X X 1 FYFG X X 1 1 FYFY X 0 X 0 FYFY X 1 X 1 HG
Inputs:C: Car sensorS: Short TimeoutL: Long Timeout
Outputs:T: Timer ResetH: Highway colorF: Farm road color
S H F
HG G RHY Y RFG R GFY R Y
State and Output EncodingS C S L T S’
HG 0 X X 0 HGHG X X 0 0 HGHG 1 X 1 1 HYHY X 0 X 0 HYHY X 1 X 1 FGFG 1 X 0 0 FGFG 0 X X 1 FYFG X X 1 1 FYFY X 0 X 0 FYFY X 1 X 1 HG
S H F
HG G RHY Y RFG R GFY R Y
First idea: use a binaryencoding:
HG 00HY 01FG 10FY 10
G 00Y 01R 10
State and Output EncodingS C S L T S’
00 0 X X 0 0000 X X 0 0 0000 1 X 1 1 0101 X 0 X 0 0101 X 1 X 1 1010 1 X 0 0 1010 0 X X 1 1110 X X 1 1 1111 X 0 X 0 1111 X 1 X 1 00
S H F
00 00 1001 01 1010 10 0011 10 01
T = S1 S0CL+ S1S0S +
S1S0(C+ L) + S1S0S
S′1 = S1 S0S+ S1 S0 +
S1 S0S
S′0 = S1 S0CL+ S1 S0 S +
S1 S0(C+ L) + S1S0S
H1 = S1
H0 = S1 S0
F1 = S1
F0 = S1 S0
State and Output EncodingS C S L T S’
00 0 X X 0 0000 X X 0 0 0000 1 X 1 1 0101 X 0 X 0 0101 X 1 X 1 1010 1 X 0 0 1010 0 X X 1 1110 X X 1 1 1111 X 0 X 0 1111 X 1 X 1 00
S H F
00 00 1001 01 1010 10 0011 10 01
T = S1 S0CL+ S0S +
S1S0(C+ L)
S′1 = S0S+ S1 S0
S′0 = S1 S0CL +
S0 S+ S1 S0(C+ L)
H1 = S1
H0 = S1 S0
F1 = S1
F0 = S1 S0
State and Output EncodingS C S L T S’
00 0 X X 0 0000 X X 0 0 0000 1 X 1 1 0101 X 0 X 0 0101 X 1 X 1 1010 1 X 0 0 1010 0 X X 1 1110 X X 1 1 1111 X 0 X 0 1111 X 1 X 1 00
S H F
00 00 1001 01 1010 10 0011 10 01
T = S0�
S1CL+ S1(C+ L)�
+
S0S
S′1 = S0S+ S1 S0
S′0 = S0�
S1CL + S1(C+ L)�
+
S0 S
H1 = S1
H0 = S1 S0
F1 = S1
F0 = S1 S0
State and Output Encoding
10
10
10
10
CL
C
LS1
T
S′1
S′0
S
SS1
S
S0
S1
F1
S0H0
S1
H1
S0F0
T = S0�
S1CL+ S1(C+ L)�
+
S0S
S′1 = S0S+ S1 S0
S′0 = S0�
S1CL + S1(C+ L)�
+
S0 S
H1 = S1
H0 = S1 S0
F1 = S1
F0 = S1 S0
State and Output EncodingS C S L T S’
HG 0 X X 0 HGHG X X 0 0 HGHG 1 X 1 1 HYHY X 0 X 0 HYHY X 1 X 1 FGFG 1 X 0 0 FGFG 0 X X 1 FYFG X X 1 1 FYFY X 0 X 0 FYFY X 1 X 1 HG
S H F
HG G RHY Y RFG R GFY R Y
Second idea: use a one-hotencoding:
HG 0001HY 0010FG 0100FY 1000
G 001Y 010R 100
State and Output EncodingS C S L T S’
0001 0 X X 0 00010001 X X 0 0 00010001 1 X 1 1 00100010 X 0 X 0 00100010 X 1 X 1 01000100 1 X 0 0 01000100 0 X X 1 10000100 X X 1 1 10001000 X 0 X 0 10001000 X 1 X 1 0001
S H F
0001 001 1000010 010 1000100 100 0011000 100 010
T = S0CL+ S1S +
S2(C+ L) + S3S
S′3 = S2(C+ L) + S3S
S′2 = S1S+ S2(C+ L)
S′1 = S0CL+ S1S
S′0 = S0(CL) + S3S
HR = S2 + S3
HY = S1
HG = S0
FR = S0 + S1
FY = S3
FG = S2
State and Output Encoding
S′0
S′1
S′2
S′3
S0
HG
S1
HY
FR
HR
S
S2
FG
C
L
S3
FY
T = S0CL+ S1S +
S2(C+ L) + S3S
S′3 = S2(C+ L) + S3S
S′2 = S1S+ S2(C+ L)
S′1 = S0CL+ S1S
S′0 = S0(CL) + S3S
HR = S2 + S3
HY = S1
HG = S0
FR = S0 + S1
FY = S3
FG = S2
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