sequential logic circuit design. 1 sequential circuits combinational circuit memory elements inputs...
TRANSCRIPT
![Page 1: SEQUENTIAL LOGIC CIRCUIT DESIGN. 1 Sequential Circuits Combinational Circuit Memory Elements Inputs Outputs Asynchronous Synchronous Combinational](https://reader033.vdocuments.us/reader033/viewer/2022061422/5697c0241a28abf838cd4dcf/html5/thumbnails/1.jpg)
SEQUENTIAL LOGIC
CIRCUIT DESIGN
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22
Sequential CircuitsSequential Circuits
CombinationalCircuit
MemoryElements
Inputs Outputs
Asynchronous
Synchronous
CombinationalCircuit
Flip-flops
Inputs Outputs
Clock
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33
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0
0
1
0
0
0 1 Q = Q0
Initial Value
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44
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1
1
0
0
0
1 0 Q = Q0
Q = Q0
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55
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0
0
1
1
0
1 Q = 0
Q = Q0
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66
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 11
0
1
0
0 1Q = 0
Q = Q0
Q = 0
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77
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0
0
1
0
1
1 0
Q = 0
Q = Q0
Q = 1
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88
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1
1
0
0
1
1 0
Q = 0
Q = Q0
Q = 1
Q = 1
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99
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1 1 01 1 0
0
1
1
1
0 0
Q = 0
Q = Q0
Q = 1
Q = Q’
0
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1010
LatchesLatches
SR Latch
R
S
Q
Q
S R Q0 Q Q’
0 0 0 0 10 0 1 1 00 1 0 0 10 1 1 0 11 0 0 1 01 0 1 1 01 1 0 0 01 1 1
1
0
1
1
0 0
Q = 0
Q = Q0
Q = 1
Q = Q’
0
Q = Q’
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1111
LatchesLatches
SR Latch
R
S
Q
Q
S R Q
0 0 Q0
0 1 0
1 0 1
1 1 Q=Q’=0
No change
Reset
Set
Invalid
S
R
Q
Q
S R Q
0 0 Q=Q’=1
0 1 1
1 0 0
1 1 Q0
Invalid
Set
Reset
No change
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1212
LatchesLatches
SR Latch
R
S
Q
Q
S R Q
0 0 Q0
0 1 0
1 0 1
1 1 Q=Q’=0
No change
Reset
Set
Invalid
S’ R’ Q
0 0 Q=Q’=1
0 1 1
1 0 0
1 1 Q0
Invalid
Set
Reset
No change
S
R
Q
Q
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1313
Controlled LatchesControlled Latches
SR Latch with Control Input
C S R Q
0 x x Q0
1 0 0 Q0
1 0 1 0
1 1 0 1
1 1 1 Q=Q’
No change
No change
Reset
Set
Invalid
S
R
Q
Q
S
R
C
S
RQ
QS
R
C
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1414
Controlled LatchesControlled Latches
D Latch (D = Data)
C D Q
0 x Q0
1 0 0
1 1 1
No change
Reset
Set
S
R
Q
Q
D
C
C
Timing Diagram
D
Q
t
Output may change
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1515
Controlled LatchesControlled Latches
D Latch (D = Data)
C D Q
0 x Q0
1 0 0
1 1 1
No change
Reset
Set
C
Timing Diagram
D
Q
Output may change
S
R
Q
Q
D
C
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1616
Flip-FlopsFlip-Flops
Controlled latches are level-triggered
Flip-Flops are edge-triggered
C
CLK Positive Edge
CLK Negative Edge
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1717
Flip-FlopsFlip-Flops
Master-Slave D Flip-Flop
D Latch(Master)
D
C
QD Latch(Slave)
D
C
Q QD
CLK CLK
D
QMaster
QSlave
Looks like it is negative edge-triggered
Master Slave
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1818
Flip-FlopsFlip-Flops
Edge-Triggered D Flip-Flop
D
CLK
Q
Q
D Q
Q
D Q
Q
Positive Edge
Negative Edge
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1919
Flip-FlopsFlip-Flops
JK Flip-Flop
D Q
Q
Q
QCLK
J
K
J Q
QK
D = JQ’ + K’Q
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2020
Flip-FlopsFlip-Flops
T Flip-Flop
D = TQ’ + T’Q = T Q
J Q
QK
T D Q
Q
T
D = JQ’ + K’QT Q
Q
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2121
Flip-Flop Characteristic TablesFlip-Flop Characteristic Tables
D Q
Q
D Q(t+1)0 01 1
Reset
Set
J K Q(t+1)0 0 Q(t)0 1 01 0 11 1 Q’(t)
No change
Reset
Set
Toggle
J Q
QK
T Q
Q
T Q(t+1)0 Q(t)1 Q’(t)
No change
Toggle
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2222
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
D Q
Q
D Q(t+1)0 01 1
Q(t+1) = D
J K Q(t+1)0 0 Q(t)0 1 01 0 11 1 Q’(t)
Q(t+1) = JQ’ + K’Q
J Q
QK
T Q
Q
T Q(t+1)0 Q(t)1 Q’(t)
Q(t+1) = T Q
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2323
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 00 1 11 0 01 0 11 1 01 1 1
No change
Reset
Set
Toggle
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2424
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 01 0 11 1 01 1 1
No change
Reset
Set
Toggle
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2525
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 01 1 1
No change
Reset
Set
Toggle
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2626
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 0
No change
Reset
Set
Toggle
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2727
Flip-Flop Characteristic EquationsFlip-Flop Characteristic Equations
Analysis / Derivation
J Q
QK
J K Q(t) Q(t+1)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 0
K
0 1 0 0J 1 1 0 1
Q
Q(t+1) = JQ’ + K’Q
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2828
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Reset
D Q
Q
R
Reset
R’ D CLK Q(t+1)
0 x x 0
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2929
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Reset
D Q
Q
R
Reset
R’ D CLK Q(t+1)
0 x x 0
1 0 ↑ 0
1 1 ↑ 1
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3030
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0D Q
Q
CLR
Reset
PR
Preset
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3131
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0
0 1 x x 1
D Q
Q
CLR
Reset
PR
Preset
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3232
Flip-Flops with Direct InputsFlip-Flops with Direct Inputs
Asynchronous Preset and Clear
PR’ CLR’ D CLK Q(t+1)
1 0 x x 0
0 1 x x 1
1 1 0 ↑ 0
1 1 1 ↑ 1
D Q
Q
CLR
Reset
PR
Preset
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3333
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
The State
● State = Values of all Flip-Flops
Example
A B = 0 0
D Q
Q
CLK
D Q
Q
A
B
y
x
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3434
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Equations
D Q
Q
CLK
D Q
Q
A
B
y
x
A(t+1) = DA
= A(t) x(t)+B(t) x(t)
= A x + B x
B(t+1) = DB
= A’(t) x(t)
= A’ x
y(t) = [A(t)+ B(t)] x’(t)
= (A + B) x’
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3535
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Table (Transition Table)
D Q
Q
CLK
D Q
Q
A
B
y
x
A(t+1) = A x + B x
B(t+1) = A’ x
y(t) = (A + B) x’
Present State
InputNext State
Output
A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1
t+1 tt
0 0 00 1 00 0 11 1 00 0 11 0 00 0 11 0 0
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3636
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Table (Transition Table)
D Q
Q
CLK
D Q
Q
A
B
y
x
A(t+1) = A x + B x
B(t+1) = A’ x
y(t) = (A + B) x’
Present State
Next State Output
x = 0 x = 1 x = 0 x = 1
A B A B A B y y0 0 0 0 0 1 0 00 1 0 0 1 1 1 01 0 0 0 1 0 1 01 1 0 0 1 0 1 0
t+1 tt
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3737
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
State Diagram
D Q
Q
CLK
D Q
Q
A
B
y
x
0 0 1 0
0 1 1 1
0/0
0/1
1/0
1/0
1/0
1/0 0/10/1
AB input/output
Present State
Next State Output
x = 0 x = 1 x = 0 x = 1
A B A B A B y y
0 0 0 0 0 1 0 0
0 1 0 0 1 1 1 0
1 0 0 0 1 0 1 0
1 1 0 0 1 0 1 0
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3838
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
D Flip-Flops
Example: D Q
Q
x
CLK
yA
Present State
InputNext State
A x y A0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1
01101001
0 100,11 00,11
01,10
01,10
A(t+1) = DA = A x y
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3939
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
JK Flip-Flops
Example:
J Q
QK
CLK
J Q
QK
x
A
B
JA = B KA = B x’
JB = x’ KB = A x
A(t+1) = JA Q’A + K’A QA
= A’B + AB’ + AxB(t+1) = JB Q’B + K’B QB
= B’x’ + ABx + A’Bx’
Present State
I/PNext State
Flip-FlopInputs
A B x A B JA KA JB KB
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 1 0
0 0 0 1
1 1 1 0
1 0 0 1
0 0 1 1
0 0 0 0
1 1 1 1
1 0 0 0
0 1
0 0
1 1
1 0
1 1
1 0
0 0
1 1
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4040
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
JK Flip-Flops
Example:
J Q
QK
CLK
J Q
QK
x
A
BPresent State
I/PNext State
Flip-FlopInputs
A B x A B JA KA JB KB
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0 1 0
0 0 0 1
1 1 1 0
1 0 0 1
0 0 1 1
0 0 0 0
1 1 1 1
1 0 0 0
0 1
0 0
1 1
1 0
1 1
1 0
0 0
1 1
0 0 1 1
0 1 1 0
1 0 1
0
1
00
1
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4141
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
T Flip-Flops
Example:
TA = B x TB = xy = A B
A(t+1) = TA Q’A + T’A QA
= AB’ + Ax’ + A’BxB(t+1) = TB Q’B + T’B QB
= x B
A
B
T Q
QR
T Q
QR
CLK Reset
xy
Present State
I/PNext State
F.FInputs
O/P
A B x A B TA TB y
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 1
1 0
1 0
1 1
1 1
0 0
0
0
0
0
0
0
1
1
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4242
Analysis of Clocked Sequential CircuitsAnalysis of Clocked Sequential Circuits
T Flip-Flops
Example:
A
B
T Q
QR
T Q
QR
CLK Reset
xy
Present State
I/PNext State
F.FInputs
O/P
A B x A B TA TB y
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 0
1 1
0 0
0 1
0 1
1 0
1 0
1 1
1 1
0 0
0
0
0
0
0
0
1
1
0 0 0 1
1 1 1 0
0/01/0
0/0
1/0
1/0
1/1
0/00/1
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4343
Mealy and Moore ModelsMealy and Moore Models
The Mealy model: the outputs are functions of both the present state and inputs (Fig. 5-15).
● The outputs may change if the inputs change during the clock pulse period.
♦ The outputs may have momentary false values unless the inputs are synchronized with the clocks.
The Moore model: the outputs are functions of the present state only (Fig. 5-20).
● The outputs are synchronous with the clocks.
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4444
Mealy and Moore ModelsMealy and Moore Models
Fig. 5.21 Block diagram of Mealy and Moore state machine
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4545
Mealy and Moore ModelsMealy and Moore Models
Present State
I/PNext State
O/P
A B x A B y0 0 0 0 0 00 0 1 0 1 00 1 0 0 0 10 1 1 1 1 01 0 0 0 0 11 0 1 1 0 01 1 0 0 0 11 1 1 1 0 0
MealyMealy
For the same statestate,the outputoutput changes with the inputinput
Present State
I/PNext State
O/P
A B x A B y0 0 0 0 0 00 0 1 0 1 00 1 0 0 1 00 1 1 1 0 01 0 0 1 0 01 0 1 1 1 01 1 0 1 1 11 1 1 0 0 1
MooreMoore
For the same statestate,the outputoutput does not change with the inputinput
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4646
Moore State DiagramMoore State Diagram
State / Output
0 0 / 0 0 1 / 0
1 1 / 1 1 0 / 0
0
1
1
1
00
01
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4747
State Reduction and AssignmentState Reduction and Assignment
State Reduction Reductions on the number of flip-flops and the number of gates.
● A reduction in the number of states may result in a reduction in the number of flip-flops.
● An example state diagram showing in Fig. 5.25.
Fig. 5.25 State diagram
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4848
State ReductionState Reduction
● Only the input-output sequences are important.
● Two circuits are equivalent
♦ Have identical outputs for all input sequences;
♦ The number of states is not important.
Fig. 5.25 State diagram
State:a a b c d e f f g f g a
Input:0 1 0 1 0 1 1 0 1 0 0
Output:
0 0 0 0 0 1 1 0 1 0 0
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4949
Equivalent states
● Two states are said to be equivalent
♦ For each member of the set of inputs, they give exactly the same output and send the circuit to the same state or to an equivalent state.
♦ One of them can be removed.
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5050
Reducing the state table
● e = g (remove g);
● d = f (remove f);
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5151
● The reduced finite state machine
State:a a b c d e d d e d e a
Input:0 1 0 1 0 1 1 0 1 0 0
Output:
0 0 0 0 0 1 1 0 1 0 0
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5252
● The unused states are treated as don't-care condition fewer combinational gates.
Fig. 5.26 Reduced State diagram
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5353
State AssignmentState Assignment
State Assignment
To minimize the cost of the combinational circuits.
● Three possible binary state assignments. (m states need n-bits, where 2n > m)
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5454
● Any binary number assignment is satisfactory as long as each state is assigned a unique number.
● Use binary assignment 1.
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5555
Design Procedure Design Procedure
Design Procedure for sequential circuit
● The word description of the circuit behavior to get a state diagram;
● State reduction if necessary;
● Assign binary values to the states;
● Obtain the binary-coded state table;
● Choose the type of flip-flops;
● Derive the simplified flip-flop input equations and output equations;
● Draw the logic diagram;
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5656
Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits
Example:
Detect 3 or more consecutive 1’s
S0 / 0 S1 / 0
S3 / 1 S2 / 0
0
1
1
0 0
1
0
1
State A BS0 0 0
S1 0 1S2 1 0
S3 1 1
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5757
Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits
Example:
Detect 3 or more consecutive 1’s
Present State
InputNext State
Output
A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1
0 0 00 1 00 0 01 0 00 0 01 1 00 0 11 1 1
S0 / 0 S1 / 0
S3 / 1 S2 / 0
0
1
1
0 0
1
0
1
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5858
Design of Clocked Sequential CircuitsDesign of Clocked Sequential Circuits
Example:
Detect 3 or more consecutive 1’s
Present State
InputNext State
Output
A B x A B y0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1
0 0 00 1 00 0 01 0 00 0 01 1 00 0 11 1 1
A(t+1) = DA (A, B, x)
= ∑ (3, 5, 7)
B(t+1) = DB (A, B, x)
= ∑ (1, 5, 7)
y (A, B, x) = ∑ (6, 7)
Synthesis using DD Flip-Flops
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5959
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
DA (A, B, x) = ∑ (3, 5, 7)
= A x + B x
DB (A, B, x) = ∑ (1, 5, 7)
= A x + B’ x
y (A, B, x) = ∑ (6, 7)
= A B
Synthesis using DD Flip-Flops
B
0 0 1 0
A 0 1 1 0x B
0 1 0 0
A 0 1 1 0xB
0 0 0 0
A 0 0 1 1x
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6060
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with DD F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
DA = A x + B x
DB = A x + B’ x
y = A B
Synthesis using DD Flip-Flops
D Q
Q
A
CLK
x
BD Q
Q
y
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6161
Flip-Flop Excitation TablesFlip-Flop Excitation Tables
Present State
Next State
F.F.Input
Q(t) Q(t+1) D0 00 11 01 1
Present State
Next State
F.F.Input
Q(t) Q(t+1) J K0 00 11 01 1
0 0 (No change)0 1 (Reset)
0 x
1 x
x 1
x 0
0
1
0
1
1 0 (Set)1 1 (Toggle)0 1 (Reset)1 1 (Toggle)0 0 (No change)1 0 (Set)
Q(t) Q(t+1) T0 00 11 01 1
0
1
1
0
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6262
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
Present State
InputNext State
Flip-FlopInputs
A B x A B JA KA JB KB
0 0 0 0 00 0 1 0 10 1 0 0 00 1 1 1 01 0 0 0 01 0 1 1 11 1 0 0 01 1 1 1 1
0 x0 x0 x1 xx 1x 0x 1x 0
JA (A, B, x) = ∑ (3)dJA (A, B, x) = ∑ (4,5,6,7)KA (A, B, x) = ∑ (4, 6)dKA (A, B, x) = ∑ (0,1,2,3)JB (A, B, x) = ∑ (1, 5)dJB (A, B, x) = ∑ (2,3,6,7)KB (A, B, x) = ∑ (2, 3, 6)dKB (A, B, x) = ∑ (0,1,4,5)
Synthesis using JKJK F.F.
0 x1 xx 1x 10 x1 xx 1x 0
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6363
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with JKJK F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
JA = B x KA = x’
JB = x KB = A’ + x’
Synthesis using JKJK Flip-Flops
B
0 0 1 0
A x x x xx
B
x x x x
A 1 0 0 1x
B
0 1 x x
A 0 1 x xx
B
x x 1 1
A x x 0 1x
CLK
J Q
QK
x
A
B
J Q
QK y
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6464
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
Present State
InputNext State
F.F.Input
A B x A B TA TB
0 0 0 0 00 0 1 0 10 1 0 0 00 1 1 1 01 0 0 0 01 0 1 1 11 1 0 0 01 1 1 1 1
00011010
Synthesis using TT Flip-Flops01110110
TA (A, B, x) = ∑ (3, 4, 6)TB (A, B, x) = ∑ (1, 2, 3, 5, 6)
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6565
Design of Clocked Sequential Circuits with Design of Clocked Sequential Circuits with TT F.F. F.F.
Example:
Detect 3 or more consecutive 1’s
TA = A x’ + A’ B x
TB = A’ B + B x
Synthesis using TT Flip-Flops
B
0 0 1 0
A 1 0 0 1x
B
0 1 1 1
A 0 1 0 1x
A
B
y
T Q
Q
x
CLK
T Q
Q