Synchronous
Sequential Circuit
Chapter 5
Analysis
Analysis
Analyze existing circuits to determine
their function
Synchronous sequential circuit
The basic modeling structure
Two circuit models,Mealy model & Moore model
Three approach used to describe the circuit:
Logic function equation
State table
State diagram
Universal model
System
output
Variables
O0
Om
Er
Excitation
Variables
E0
Combinational
logic
System
Input
Variables
I0
In
S0
Sx
State
Variables
Memory
Flip-
flops
CLOCK
Analysis principle
Determine the system variables,input,state and
output.
Assign names to the variables if they are not clear
from the logic diagram.
Determine the flip-flop type,Write the
characteristics equations.
Write the excitation equations by inspection from
the logic diagram.
Write the output variable equations,Determine
whether the circuit is a Mealy or a Moore machine.
Analysis principle
Mealy model machine
E = f( I,St )
St+1 = f( St,E ) flip-flop characteristics equation
O = g( I,St )
Moore model machine
E = f( I,St )
St+1 = f( St,E ) flip-flop characteristics equation
O = g(St )
Write the next state equations for each state
variable,using the flip-flop characteristics
equation and the circuit excitation equations,or
construct the excitation table from the
excitation equations.
Construct a transition table,Identify all of the
states possible for a given number of state
variables.
Assign symbols to the states and construct a
state table or state diagram.
When possible,construct a timing diagram.
Analysis principle
Logic circuit
Output variable equations
&Excitation equations
Next-state truth table Next-state equations
flip-flop
Characteristic
table State diagram &State table
Describe the circuit by a timing diagram and statement
Flip-flop
Characteristic
equation
table algebra
Analysis flowchart
Example 6.2
Start/stop sensor
START
STARTN
STOP
STOPN
t1 t2
L
Vprojectile = L/ (t2-t1)
Start/ stop sensors
As the bullet passes over the start sensor,the
start signal (SRT) is generated.
When the same bullet passes over the stop
sensor,the stop signal (STP) is generated.
Example 6.2
Start/stop sensor
START (SRT)
STARTN
STOP (STP)
STOPN
Example 6.2
BCD counter,
t2-t1
measure the elapsed time of the
bullet directly by counting the
number of clock pulse,
Example 6.2
BCD counter
Clock input
Clock generator,2MHz Square-wave
Clock frequency divider,1MHz Square-
wave
Output,LED display
Enable input,Reset/latch
Sequential circuit controller
SRT,SPT
Sequential circuit controller
The external input variables,SRT,SPT
The external output variables
Clock enable,CEN (U3D)
Reset enable,CRST (U2A)
Latch enable,CLTCH (U2B)
The excitation variables
J1,K1 (U1A)
J2,K2 (U6A)
The state variables
F1,F1’ (U1A)
F2,F2’ (U6A)
Example 6.2
Sequential circuit controller
The external output variables
Clock enable,CEN =F2·F1 (U3D)
Reset enable,CRST =F2’·F1 (U2A)
Latch enable,CLTCH =F2·F1’ (U2B)
The excitation variables
J1=SRT’·F1,K1=STP·F2 (U1A)
J2=SRT’·F1,K2=STP’F1’ (U6A)
The state variables
F1t+1=J1F1’+K1’F1= SRT’·F2’·F1+STP·F2+F2·F1
F1t+1(F2,F1,SRT,STP)=∑m(4,5,9,11,12,13,14,15)
F2T+1=J2F2’+K2’F2= SRT’·F2’·F1’+STP’·F1+F2’·F1
F2t+1(F2,F1,SRT,STP)=∑m(2,3,4,5,6,7,12,14)
SRT’ SRT
D/
A/
C/
B/
STP
STP’ SRT’
SRT’
SRT
STP
CRST
CEN
CLTCH
A,No bullet passes over the start sensor.
B,A bullet is passing over the start sensor.
C,The bullet passed over the start sensor,and is on the
itinerary between the start sensor and the stop sensor.
D,The bullet is passing over the stop sensor.
A,The bullet passed over the stop sensor and no bullet
passes over the start sensor.
cp
Determine flip-flop type:
negative edge triggered J-K
Determine circuit model,Moore
Determine variables
1 input variable,X
1 output variable,Y1,Y2
2 state variables,Y1,Y2
4 excitation variables,J1K1,J2K2
J1 =1 ; K1 =1
J2 = X⊕ Y1 ; K2 = X⊕ Y1
Analysis example1
J1K1
Y1
J2K2
Y2
=1 1
X
Next-state Truth Table
Input Present
state
Excitation
J2 K2 J1 K1
Next
state
X Y2 Y1 Y2t+1 Y1t+1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
1
0
0
1
0
1
1
0
1
0
J1 =1
K1 =1
J2 = X⊕ Y1
K2 = X⊕ Y1
Excitation
equation
Analysis example1
Present
state Excitation
Next
state
Yt
0
0
0
0
1
1
1
1
J K
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
Yt+1
0
0
1
1
1
0
1
0
Flip-flop
Characteristic Table
Next_state Truth Table
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
1
0
0
1
0
1
1
0
1
0
Input Present
state
Excitation
J2 K2 J1 K1
Next
state
X Y2 Y1 Y2t+1 Y1t+1
1
0
1
0
1
0
1
0
0
1
1
0
1
0
0
1
Analysis example1
Present
state
Next state/output
Y2 Y1
Y2t+1 Y1t+1
0
0
1
1
0
1
0
1
1
0
1
0
0
1
1
0
X = 0 X = 1
1
0
1
0
1
0
0
1
State Table
Y2Y1 X
0 0
1 01 1
0
0
0
0
1
1
1
1
0 1
State Diagram
Synchronous Modulo-4 up-down
counter
Analysis example1
Statement of the logic Circuit
Synchronous Modulo-4 up-down counter
Timing
Diagram
Negative edge
trigger clock
Analysis example1
1 2 3 4 5 6 7 8 9
1 1 1 1 0 0 0 0 0
0 1 1 0 0 0 1 1 0
0 1 0 1 0 1 0 1 0
0 0 0 1 0 0 0 1 0
X
CP
Y2
Y1
Z
Analysis example2
Output equation
Z=XY2Y1’
Excitation equation
T2=((Y2Y1’)’((Y2⊕ Y1)X)’)’
=Y2Y1’+(Y2⊕ Y1)X
T1=Y1⊕ X
Flip-flop characteristics
equation
Yt+1=Yt⊕ T
T2
Y2 Y2’
T1
Y1 Y1’
X
Z
CP
Mealy
Next-state equation
Y1t+1=Y1⊕ T1=Y1⊕ Y1⊕ X=X
Y2t+1=Y2⊕ T2=Y2⊕ (Y2Y1’+(Y2⊕ Y1)X)
Y2⊕ (Y2Y1’+(Y2⊕ Y1)X)
Analysis example2
Present
state
Next state/output
Y2 Y1
Y2t+1 Y1t+1/Z
0
0
1
1
0
1
0
1
0
0
0
0
0
0
0
1
X = 0 X = 1
/0
/0
/0
/0
1
1
1
1
0
1
0
1
/0
/0
/1
/0
0
0
1
1
0
0
1
1
0
0
0
0
1
1
1
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
,1101” sequence detector
The last bit,1” can overlap with the
next,1101” sequence
Analysis example2
10
00
11
01
0/0
1/0
0/0
1/1
0/0
0/0 1/0
1/0 D
A
C
B
0/0
1/0
0/0
1/1
0/0
0/0 1/0
1/0
positive edge
trigger clock
Analysis example2
1 2 3 4 5 6 7 8 9
1 1 1 0 1 1 0 1 0
1 1 1 0 1 1 0 1 0
0 1 1 1 0 1 1 0 0
0 0 0 0 0 0 0
X
CP
Y2
Y1
Z 1
1 1
1
Analysis example3
Output equation
Z=Y2Y1’
Excitation equation
D2=X’Y1
D1=X+Y2’Y1
Flip-flop characteristics
equation
Yt+1=D
D2
Y2 Y2’
D1
Y1 Y1’
X
Z
CP
Moore
Next-state equation
Y2t+1=D2=X’Y1
Y1t+1=D1=X+Y2’Y1
Analysis example3
Present
state
Next state
Y2 Y1
Y2t+1Y1t+1
0
0
1
1
0
1
0
1
0
1
0
0
0
1
0
1
X=0 X=1
1
1
1
1
0
0
0
0
Output
Z
0
0
1
0
,100” sequence detector
Analysis example3
D/1
A/0
C/0
B/0
0
1
1
0
0 01
1
10/1
00/0
11/0
01/0
0
1
1
0
0 01
1
Analysis example3
1 2 3 4 5 6 7 8 9
1 1 0 0 0 1 0 0 1
0 0 1 1 0 0 1 1 0
1 1 1 0 0 1 1 0 1
0 0 0 1 0 0 0
X
CP
Y2
Y1
Z 1
Sequential Circuit
Chapter 5
Analysis
Analysis
Analyze existing circuits to determine
their function
Synchronous sequential circuit
The basic modeling structure
Two circuit models,Mealy model & Moore model
Three approach used to describe the circuit:
Logic function equation
State table
State diagram
Universal model
System
output
Variables
O0
Om
Er
Excitation
Variables
E0
Combinational
logic
System
Input
Variables
I0
In
S0
Sx
State
Variables
Memory
Flip-
flops
CLOCK
Analysis principle
Determine the system variables,input,state and
output.
Assign names to the variables if they are not clear
from the logic diagram.
Determine the flip-flop type,Write the
characteristics equations.
Write the excitation equations by inspection from
the logic diagram.
Write the output variable equations,Determine
whether the circuit is a Mealy or a Moore machine.
Analysis principle
Mealy model machine
E = f( I,St )
St+1 = f( St,E ) flip-flop characteristics equation
O = g( I,St )
Moore model machine
E = f( I,St )
St+1 = f( St,E ) flip-flop characteristics equation
O = g(St )
Write the next state equations for each state
variable,using the flip-flop characteristics
equation and the circuit excitation equations,or
construct the excitation table from the
excitation equations.
Construct a transition table,Identify all of the
states possible for a given number of state
variables.
Assign symbols to the states and construct a
state table or state diagram.
When possible,construct a timing diagram.
Analysis principle
Logic circuit
Output variable equations
&Excitation equations
Next-state truth table Next-state equations
flip-flop
Characteristic
table State diagram &State table
Describe the circuit by a timing diagram and statement
Flip-flop
Characteristic
equation
table algebra
Analysis flowchart
Example 6.2
Start/stop sensor
START
STARTN
STOP
STOPN
t1 t2
L
Vprojectile = L/ (t2-t1)
Start/ stop sensors
As the bullet passes over the start sensor,the
start signal (SRT) is generated.
When the same bullet passes over the stop
sensor,the stop signal (STP) is generated.
Example 6.2
Start/stop sensor
START (SRT)
STARTN
STOP (STP)
STOPN
Example 6.2
BCD counter,
t2-t1
measure the elapsed time of the
bullet directly by counting the
number of clock pulse,
Example 6.2
BCD counter
Clock input
Clock generator,2MHz Square-wave
Clock frequency divider,1MHz Square-
wave
Output,LED display
Enable input,Reset/latch
Sequential circuit controller
SRT,SPT
Sequential circuit controller
The external input variables,SRT,SPT
The external output variables
Clock enable,CEN (U3D)
Reset enable,CRST (U2A)
Latch enable,CLTCH (U2B)
The excitation variables
J1,K1 (U1A)
J2,K2 (U6A)
The state variables
F1,F1’ (U1A)
F2,F2’ (U6A)
Example 6.2
Sequential circuit controller
The external output variables
Clock enable,CEN =F2·F1 (U3D)
Reset enable,CRST =F2’·F1 (U2A)
Latch enable,CLTCH =F2·F1’ (U2B)
The excitation variables
J1=SRT’·F1,K1=STP·F2 (U1A)
J2=SRT’·F1,K2=STP’F1’ (U6A)
The state variables
F1t+1=J1F1’+K1’F1= SRT’·F2’·F1+STP·F2+F2·F1
F1t+1(F2,F1,SRT,STP)=∑m(4,5,9,11,12,13,14,15)
F2T+1=J2F2’+K2’F2= SRT’·F2’·F1’+STP’·F1+F2’·F1
F2t+1(F2,F1,SRT,STP)=∑m(2,3,4,5,6,7,12,14)
SRT’ SRT
D/
A/
C/
B/
STP
STP’ SRT’
SRT’
SRT
STP
CRST
CEN
CLTCH
A,No bullet passes over the start sensor.
B,A bullet is passing over the start sensor.
C,The bullet passed over the start sensor,and is on the
itinerary between the start sensor and the stop sensor.
D,The bullet is passing over the stop sensor.
A,The bullet passed over the stop sensor and no bullet
passes over the start sensor.
cp
Determine flip-flop type:
negative edge triggered J-K
Determine circuit model,Moore
Determine variables
1 input variable,X
1 output variable,Y1,Y2
2 state variables,Y1,Y2
4 excitation variables,J1K1,J2K2
J1 =1 ; K1 =1
J2 = X⊕ Y1 ; K2 = X⊕ Y1
Analysis example1
J1K1
Y1
J2K2
Y2
=1 1
X
Next-state Truth Table
Input Present
state
Excitation
J2 K2 J1 K1
Next
state
X Y2 Y1 Y2t+1 Y1t+1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
1
0
0
1
0
1
1
0
1
0
J1 =1
K1 =1
J2 = X⊕ Y1
K2 = X⊕ Y1
Excitation
equation
Analysis example1
Present
state Excitation
Next
state
Yt
0
0
0
0
1
1
1
1
J K
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
Yt+1
0
0
1
1
1
0
1
0
Flip-flop
Characteristic Table
Next_state Truth Table
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
1
0
0
1
0
1
1
0
1
0
Input Present
state
Excitation
J2 K2 J1 K1
Next
state
X Y2 Y1 Y2t+1 Y1t+1
1
0
1
0
1
0
1
0
0
1
1
0
1
0
0
1
Analysis example1
Present
state
Next state/output
Y2 Y1
Y2t+1 Y1t+1
0
0
1
1
0
1
0
1
1
0
1
0
0
1
1
0
X = 0 X = 1
1
0
1
0
1
0
0
1
State Table
Y2Y1 X
0 0
1 01 1
0
0
0
0
1
1
1
1
0 1
State Diagram
Synchronous Modulo-4 up-down
counter
Analysis example1
Statement of the logic Circuit
Synchronous Modulo-4 up-down counter
Timing
Diagram
Negative edge
trigger clock
Analysis example1
1 2 3 4 5 6 7 8 9
1 1 1 1 0 0 0 0 0
0 1 1 0 0 0 1 1 0
0 1 0 1 0 1 0 1 0
0 0 0 1 0 0 0 1 0
X
CP
Y2
Y1
Z
Analysis example2
Output equation
Z=XY2Y1’
Excitation equation
T2=((Y2Y1’)’((Y2⊕ Y1)X)’)’
=Y2Y1’+(Y2⊕ Y1)X
T1=Y1⊕ X
Flip-flop characteristics
equation
Yt+1=Yt⊕ T
T2
Y2 Y2’
T1
Y1 Y1’
X
Z
CP
Mealy
Next-state equation
Y1t+1=Y1⊕ T1=Y1⊕ Y1⊕ X=X
Y2t+1=Y2⊕ T2=Y2⊕ (Y2Y1’+(Y2⊕ Y1)X)
Y2⊕ (Y2Y1’+(Y2⊕ Y1)X)
Analysis example2
Present
state
Next state/output
Y2 Y1
Y2t+1 Y1t+1/Z
0
0
1
1
0
1
0
1
0
0
0
0
0
0
0
1
X = 0 X = 1
/0
/0
/0
/0
1
1
1
1
0
1
0
1
/0
/0
/1
/0
0
0
1
1
0
0
1
1
0
0
0
0
1
1
1
1
0
1
0
1
0
1
0
1
0
0
1
1
0
0
1
1
,1101” sequence detector
The last bit,1” can overlap with the
next,1101” sequence
Analysis example2
10
00
11
01
0/0
1/0
0/0
1/1
0/0
0/0 1/0
1/0 D
A
C
B
0/0
1/0
0/0
1/1
0/0
0/0 1/0
1/0
positive edge
trigger clock
Analysis example2
1 2 3 4 5 6 7 8 9
1 1 1 0 1 1 0 1 0
1 1 1 0 1 1 0 1 0
0 1 1 1 0 1 1 0 0
0 0 0 0 0 0 0
X
CP
Y2
Y1
Z 1
1 1
1
Analysis example3
Output equation
Z=Y2Y1’
Excitation equation
D2=X’Y1
D1=X+Y2’Y1
Flip-flop characteristics
equation
Yt+1=D
D2
Y2 Y2’
D1
Y1 Y1’
X
Z
CP
Moore
Next-state equation
Y2t+1=D2=X’Y1
Y1t+1=D1=X+Y2’Y1
Analysis example3
Present
state
Next state
Y2 Y1
Y2t+1Y1t+1
0
0
1
1
0
1
0
1
0
1
0
0
0
1
0
1
X=0 X=1
1
1
1
1
0
0
0
0
Output
Z
0
0
1
0
,100” sequence detector
Analysis example3
D/1
A/0
C/0
B/0
0
1
1
0
0 01
1
10/1
00/0
11/0
01/0
0
1
1
0
0 01
1
Analysis example3
1 2 3 4 5 6 7 8 9
1 1 0 0 0 1 0 0 1
0 0 1 1 0 0 1 1 0
1 1 1 0 0 1 1 0 1
0 0 0 1 0 0 0
X
CP
Y2
Y1
Z 1
