Digital Design Synchronous Sequential Logic 5j195q
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3i3n4
Overview 26281t
& View Digital Design Synchronous Sequential Logic as PDF for free.
More details 6y5l6z
- Words: 3,585
- Pages: 61
Sequential Circuits Asynchronous Inputs
Combinational Circuit
Outputs Memory Elements
Synchronous Inputs
Outputs
Combinational Circuit Clock
Flip-flops
2 / 60
Latches SR Latch
S R Q0 0 0 0
0
0
0
1
Q 0
Q’ 1
Q = Q0
Initial Value 3 / 60
Latches SR Latch
S R Q0 0 0 0 0 0 1
0
1
0
0
Q 0 1
Q’ 1 0
Q = Q0 Q = Q0
4 / 60
Latches SR Latch
1
0
0
1
S 0 0 0
R 0 0 1
Q0 0 1 0
Q 0 1 0
Q’ 1 0 1
Q = Q0 Q=0
5 / 60
Latches SR Latch
1
1
0
0
S 0 0 0 0
R 0 0 1 1
Q0 0 1 0 1
Q 0 1 0 0
Q’ 1 0 1 1
Q = Q0 Q=0 Q=0
6 / 60
Latches SR Latch
0
1
0
S 0 0 0 0 1
R 0 0 1 1 0
Q0 0 1 0 1 0
Q 0 1 0 0 1
Q’ 1 0 1 1 0
Q = Q0 Q=0 Q=1
1
7 / 60
Latches SR Latch
0
1
1
S 0 0 0 0 1 1
R 0 0 1 1 0 0
Q0 0 1 0 1 0 1
Q 0 1 0 0 1 1
Q’ 1 0 1 1 0 0
Q = Q0 Q=0 Q=1 Q=1
0
8 / 60
Latches SR Latch
1
1
0
S 0 0 0 0 1 1 1
R 0 0 1 1 0 0 1
Q0 0 1 0 1 0 1 0
Q 0 1 0 0 1 1 0
Q’ 1 0 1 1 0 0 0
Q = Q0 Q=0 Q=1 Q = Q’
10
9 / 60
Latches SR Latch
1
1
10
S 0 0 0 0 1 1 1 1
R 0 0 1 1 0 0 1 1
Q0 0 1 0 1 0 1 0 1
Q 0 1 0 0 1 1 0 0
Q’ 1 0 1 1 0 0 0 0
Q = Q0 Q=0 Q=1 Q = Q’ Q = Q’
0
10 / 60
Latches SR Latch
S R
Q Q0
No change Reset 0 Set 1 Invalid Q=Q’=0
0 0 1 1
0 1 0 1
S 0 0 1 1
Q R Invalid 0 Q=Q’=1 Set 1 1 Reset 0 0 Q0 No change 1 11 / 60
Latches SR Latch
S R 0 0 1 1
0 1 0 1
S’ R’ 0 0 1 1
0 1 0 1
Q Q0
No change Reset 0 Set 1 Invalid Q=Q’=0 Q Invalid Q=Q’=1 Set 1 Reset 0 Q0 No change 12 / 60
Controlled Latches SR Latch with Control Input
C S R 0 1 1 1 1
x 0 0 1 1
x 0 1 0 1
Q Q0 Q0 0 1 Q=Q’
No change No change Reset Set Invalid
13 / 60
Controlled Latches D Latch (D = Data)
Timing Diagram C D Q
C D 0 x 1 0 1 1
Q Q0 0 1
t No change Reset Set
Output may change
14 / 60
Controlled Latches D Latch (D = Data)
Timing Diagram C D Q
C D 0 x 1 0 1 1
Q Q0 0 1
No change Reset Set
Output may change
15 / 60
Flip-Flops Controlled latches are level-triggered C
Flip-Flops are edge-triggered CLK
Positive Edge
CLK
Negative Edge
16 / 60
Flip-Flops Master-Slave D Flip-Flop D
D C
D Latch (Master)
Q
D C
D Latch (Slave) Master
CLK
Q
Q
Slave
CLK D
Looks like it is negative edge-triggered
QMaster QSlave 17 / 60
Flip-Flops Edge-Triggered D Flip-Flop D
Q Q
Positive Edge D
Q Q
Negative Edge
18 / 60
Flip-Flops JK Flip-Flop
J
Q
K
Q
D = JQ’ + K’Q 19 / 60
Flip-Flops T Flip-Flop
T
J K
Q
D
T
Q
Q
D = JQ’ + K’Q D = TQ’ + T’Q = T Q
Q
T
Q Q 20 / 60
Flip-Flop Characteristic Tables D
Q Q
J
Q
K
Q
T
Q Q
D 0 1 J 0 0 1 1 T 0 1
Q(t+1) 0 1
Reset Set
K Q(t+1) 0 Q(t) 1 0 0 1 1 Q’(t)
No change Reset Set Toggle
Q(t+1) Q(t) Q’(t)
No change Toggle 21 / 60
Flip-Flop Characteristic Equations D
Q Q
J
Q
K
Q
T
Q Q
D 0 1 J 0 0 1 1
Q(t+1) 0 1 K Q(t+1) 0 Q(t) 1 0 0 1 1 Q’(t)
T 0 1
Q(t+1) Q(t) Q’(t)
Q(t+1) = D
Q(t+1) = JQ’ + K’Q
Q(t+1) = T Q 22 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 1 1 0 0 0 1 1 0 1 1
No change Reset Set Toggle
23 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 1 0 1 1
No change Reset Set Toggle
24 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1
No change Reset Set Toggle
25 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0
No change Reset Set Toggle
26 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J
Q
K
Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0
K 0 J 1
1 1
0 0
0 1
Q
Q(t+1) = JQ’ + K’Q 27 / 60
Flip-Flops with Direct Inputs Asynchronous Reset
D
Q
R’ 0
D CLK Q(t+1) x x 0
Q R Reset
28 / 60
Flip-Flops with Direct Inputs Asynchronous Reset
D
Q Q R
R’ 0 1 1
D CLK Q(t+1) x x 0 ↑ 0 0 ↑ 1 1
Reset
29 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0
Q CLR Reset
30 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0 x 0 1 x 1
Q CLR Reset
31 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q Q CLR
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0 x 0 1 x 1 ↑ 1 1 0 0 ↑ 1 1 1 1
Reset
32 / 60
Analysis of Clocked Sequential Circuits The State ● State = Values of all Flip-Flops
Example AB=00
33 / 60
Analysis of Clocked Sequential Circuits State Equations
A(t+1) = DA = A(t) x(t)+B(t) x(t) =Ax+Bx B(t+1) = DB = A’(t) x(t) = A’ x y(t) = [A(t)+ B(t)] x’(t) = (A + B) x’
34 / 60
Analysis of Clocked Sequential Circuits State Table (Transition Table) Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1 t
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1 t+1
B 0 1 0 1 0 0 0 0
Output
y 0 0 1 0 1 0 1 0 t
A(t+1) = A x + B x B(t+1) = A’ x y(t) = (A + B) x’
35 / 60
Analysis of Clocked Sequential Circuits State Table (Transition Table) Present State
A 0 0 1 1 t
B 0 1 0 1
Next State
Output
x=0
x=1 x=0 x=1
A 0 0 0 0
A 0 1 1 1
B 0 0 0 0
t+1
B 1 1 0 0
y 0 1 1 1
y 0 0 0 0 t
A(t+1) = A x + B x B(t+1) = A’ x y(t) = (A + B) x’
36 / 60
Analysis of Clocked Sequential Circuits State Diagram
Present State
input/output
AB
0/0
1/0 0/1
00
Next State x=0
Output
x=1
x=0
x=1
A B 0 0
A B A B 0 0 0 1
y
y
0
0
0 1 1 0 1 1
0 0 0
1 1 1
0 0 0
0 0 0
1 1 1
1 0 0
10
0/1 1/0
0/1 01
1/0 11
1/0
37 / 60
Analysis of Clocked Sequential Circuits D Flip-Flops
Example: Present Input State
A 0 0 0 0 1 1 1 1
x 0 0 1 1 0 0 1 1
y 0 1 0 1 0 1 0 1
Next State
A 0 1 1 0 1 0 0 1
x y
D CLK
Q
A
Q
A(t+1) = DA = A x y 01,10 00,11
0
1
00,11
01,10 38 / 60
Analysis of Clocked Sequential Circuits JK Flip-Flops
Example: Present Next I/P State State A B x A B 0 0 0 0 1
Flip-Flop Inputs JA KA JB KB 0 0 1 0
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
1
1
0
0
1
1
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
1
0
1
1
0
0
0
0
0
1
1
0
0
0
1
1
1
1
1
1
1
1
1
1
0
0
0
JA = B JB = x’
KA = B x’ KB = A x
A(t+1) = JA Q’A + K’A QA = A’B + AB’ + Ax B(t+1) = JB Q’B + K’B QB = B’x’ + ABx + A’Bx’
39 / 60
Analysis of Clocked Sequential Circuits JK Flip-Flops
Example: Present Next I/P State State A B x A B 0 0 0 0 1
Flip-Flop Inputs JA KA JB KB 0 0 1 0
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
1
1
0
0
1
1
1
0
1
0
0
1
1
0
0
1
1
1
0
0
1
1
0
1
1
0
0
0
0
0
1
1
0
0
0
1
1
1
1
1
1
1
1
1
1
0
0
0
1
0
1 11
00
0
0 01
0 10
1
1 40 / 60
Analysis of Clocked Sequential Circuits T Flip-Flops
Example: Present Next F.F I/P O/P State State Inputs A B x A B TA TB y 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
1
0
0
1
0
0
0
0
1
1
1
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
1
TA = B x TB = x y =AB A(t+1) = TA Q’A + T’A QA = AB’ + Ax’ + A’Bx B(t+1) = TB Q’B + T’B QB =xB
41 / 60
Analysis of Clocked Sequential Circuits T Flip-Flops
Example: Present Next F.F I/P O/P State State Inputs A B x A B TA TB y 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
1
0
0
1
0
0
0
0
1
1
1
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
1
0/0
0/0 00
1/0
01
1/1 0/1
1/0 11
10
1/0
0/0 42 / 60
Mealy and Moore Models Mealy Present State A B 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
I/P x 0 1 0 1 0 1 0 1
Next O/P State A B y 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 1 0 0
For the same state, state the output changes with the input
Moore Present State A B 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
I/P x 0 1 0 1 0 1 0 1
Next O/P State A B y 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1
For the same state, state the output does not change with the input 43 / 60
Moore State Diagram State / Output 0
0 1 00/0
01/0
1
1
11/1
10/0 1
0
0 44 / 60
Timing Diagram 0 A x
y
1 00/0 0
B
01/0 1
0
11/1 No effect
0
10/0
1
1
CLK x State
A
0
0
1
0
0
0
B
0
1
0
0
1
1
y
45 / 60
Timing Diagram 1/0
0/0
00
A x
y B
0/0
01
1/1
1/0
0/0 11
0/0
10 1/1
CLK x State
A
1
B
0
y
46 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s 1
0 S0 / 0
S1 / 0 0
0
1
0 S3 / 1
1
S2 / 0
State A B S0 0 0 S1 0 1 S2
1 0
S3
1 1
1 47 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
Output
y 0 0 0 0 0 0 1 1
1
0 S0 / 0
S1 / 0 0
0
0
S3 / 1 1
1 S2 / 0
1 48 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
Output
y 0 0 0 0 0 0 1 1
Synthesis using D Flip-Flops
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)
49 / 60
Design of Clocked Sequential Circuits with D F.F. Example: Detect 3 or more consecutive 1’s Synthesis using D Flip-Flops
DA (A, B, x) = ∑ (3, 5, 7) = Ax + Bx
B 0 0 1 0 A 0 1 1 0 x
DB (A, B, x) = ∑ (1, 5, 7) = A x + B’ x y (A, B, x) = ∑ (6, 7) = AB
B 0 0 0 0 A 0 0 1 1 x
B 0 1 0 0 A 0 1 1 0 x
50 / 60
Design of Clocked Sequential Circuits with D F.F. Example: Detect 3 or more consecutive 1’s Synthesis using D Flip-Flops
DA = A x + B x DB = A x + B’ x y = AB
51 / 60
Flip-Flop Excitation Tables Present Next State State
Q(t) Q(t+1) 0 0 0 1 1 0 1 1
F.F. Input
D 0 1 0 1
Present Next State State
F.F. Input
Q(t) Q(t+1) J K 0 x 0 0 0 1 1 x 1 0 x 1 1 1 x 0 Q(t) Q(t+1) 0 0 0 1 1 0 1 1
0 0 (No change) 0 1 (Reset) 1 0 (Set) 1 1 (Toggle) 0 1 (Reset) 1 1 (Toggle) 0 0 (No change) 1 0 (Set)
T 0 1 1 0 52 / 60
Design of Clocked Sequential Circuits with JK F.F. Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
Flip-Flop Inputs
B JA KA JB KB 0 0 x 0 x 1 0 x 1 x 0 0 x x 1 0 1 x x 1 0 x 1 0 x 1 x 0 1 x 0 x 1 x 1 1 x 0 x 0
Synthesis using JK F.F. 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) 53 / 60 dKB (A, B, x) = ∑ (0,1,4,5)
Design of Clocked Sequential Circuits with JK F.F. Example: Detect 3 or more consecutive 1’s Synthesis using JK Flip-Flops
JA = B x
KA = x’
JB = x
KB = A’ + x’
B
B
0 0 1 0 A x x x x x B
x x x x A 1 0 0 1 x B
0 1 x x A 0 1 x x x
x x 1 1 A x x 0 1 x
54 / 60
Design of Clocked Sequential Circuits with T F.F. Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
F.F. Input
TA 0 0 0 1 1 0 1 0
TB 0 1 1 1 0 1 1 0
Synthesis using T Flip-Flops TA (A, B, x) = ∑ (3, 4, 6) TB (A, B, x) = ∑ (1, 2, 3, 5, 6)
55 / 60
Design of Clocked Sequential Circuits with T F.F. Example: Detect 3 or more consecutive 1’s Synthesis using T Flip-Flops
TA = A x’ + A’ B x TB = A’ B + B x B 0 0 1 0 A 1 0 0 1 x
B 0 1 1 1 A 0 1 0 1 x 56 / 60
Homework Mano ● Chapter 5 ♦ 5-1 ♦ 5-3 ♦ 5-6 ♦ 5-8 ♦ 5-9
57 / 60
Homework 5-1
The D latch is constructed with four NAND gates and an inverter. Consider the following three other ways for obtaining a D latch. In each case, draw the logic diagram and the circuit operation. (a) Use NOR gates for the SR latch part and AND gates for the other two. An inverter may be needed. (b) Use NOR gates for all four gates. Inverters may be needed. (c) Use four NAND gates only (without an inverter). This can be done by connecting the output of the upper gate that goes to the SR latch to the input of the lower gate instead of the inverter output. 58 / 60
Homework 5-3
Show that the characteristic equation for the complement output of a JK flip-flop is Q’(t+1) = J’Q + K Q
5-6
A sequential circuit with two D flip-flops, A and B; two inputs, x and y; and one output, z, is specified by the following next-state and output equations: A(t+1) = x’ y + x A B(t+1) = x’ B + x A z=B (a) Draw the logic diagram of the circuit. (b) List the state table for the sequential circuit. (c) Draw the corresponding state diagram. 59 / 60
Homework 5-8
Derive the state table and the state diagram of the sequential shown circuit. Explain the function that the circuit performs.
60 / 60
Homework 5-9
A sequential circuit has two JK flip-flops A and B and one input x. The circuit is described by the following flip-flop input equations: JA = x
KA = B’
JB = x
KB = A
(a) Derive the state equations A(t+1) and B(t+1) by substituting the input equations for the J and K variables. (b) Draw the state diagram of the circuit.
61 / 60
Combinational Circuit
Outputs Memory Elements
Synchronous Inputs
Outputs
Combinational Circuit Clock
Flip-flops
2 / 60
Latches SR Latch
S R Q0 0 0 0
0
0
0
1
Q 0
Q’ 1
Q = Q0
Initial Value 3 / 60
Latches SR Latch
S R Q0 0 0 0 0 0 1
0
1
0
0
Q 0 1
Q’ 1 0
Q = Q0 Q = Q0
4 / 60
Latches SR Latch
1
0
0
1
S 0 0 0
R 0 0 1
Q0 0 1 0
Q 0 1 0
Q’ 1 0 1
Q = Q0 Q=0
5 / 60
Latches SR Latch
1
1
0
0
S 0 0 0 0
R 0 0 1 1
Q0 0 1 0 1
Q 0 1 0 0
Q’ 1 0 1 1
Q = Q0 Q=0 Q=0
6 / 60
Latches SR Latch
0
1
0
S 0 0 0 0 1
R 0 0 1 1 0
Q0 0 1 0 1 0
Q 0 1 0 0 1
Q’ 1 0 1 1 0
Q = Q0 Q=0 Q=1
1
7 / 60
Latches SR Latch
0
1
1
S 0 0 0 0 1 1
R 0 0 1 1 0 0
Q0 0 1 0 1 0 1
Q 0 1 0 0 1 1
Q’ 1 0 1 1 0 0
Q = Q0 Q=0 Q=1 Q=1
0
8 / 60
Latches SR Latch
1
1
0
S 0 0 0 0 1 1 1
R 0 0 1 1 0 0 1
Q0 0 1 0 1 0 1 0
Q 0 1 0 0 1 1 0
Q’ 1 0 1 1 0 0 0
Q = Q0 Q=0 Q=1 Q = Q’
10
9 / 60
Latches SR Latch
1
1
10
S 0 0 0 0 1 1 1 1
R 0 0 1 1 0 0 1 1
Q0 0 1 0 1 0 1 0 1
Q 0 1 0 0 1 1 0 0
Q’ 1 0 1 1 0 0 0 0
Q = Q0 Q=0 Q=1 Q = Q’ Q = Q’
0
10 / 60
Latches SR Latch
S R
Q Q0
No change Reset 0 Set 1 Invalid Q=Q’=0
0 0 1 1
0 1 0 1
S 0 0 1 1
Q R Invalid 0 Q=Q’=1 Set 1 1 Reset 0 0 Q0 No change 1 11 / 60
Latches SR Latch
S R 0 0 1 1
0 1 0 1
S’ R’ 0 0 1 1
0 1 0 1
Q Q0
No change Reset 0 Set 1 Invalid Q=Q’=0 Q Invalid Q=Q’=1 Set 1 Reset 0 Q0 No change 12 / 60
Controlled Latches SR Latch with Control Input
C S R 0 1 1 1 1
x 0 0 1 1
x 0 1 0 1
Q Q0 Q0 0 1 Q=Q’
No change No change Reset Set Invalid
13 / 60
Controlled Latches D Latch (D = Data)
Timing Diagram C D Q
C D 0 x 1 0 1 1
Q Q0 0 1
t No change Reset Set
Output may change
14 / 60
Controlled Latches D Latch (D = Data)
Timing Diagram C D Q
C D 0 x 1 0 1 1
Q Q0 0 1
No change Reset Set
Output may change
15 / 60
Flip-Flops Controlled latches are level-triggered C
Flip-Flops are edge-triggered CLK
Positive Edge
CLK
Negative Edge
16 / 60
Flip-Flops Master-Slave D Flip-Flop D
D C
D Latch (Master)
Q
D C
D Latch (Slave) Master
CLK
Q
Q
Slave
CLK D
Looks like it is negative edge-triggered
QMaster QSlave 17 / 60
Flip-Flops Edge-Triggered D Flip-Flop D
Q Q
Positive Edge D
Q Q
Negative Edge
18 / 60
Flip-Flops JK Flip-Flop
J
Q
K
Q
D = JQ’ + K’Q 19 / 60
Flip-Flops T Flip-Flop
T
J K
Q
D
T
Q
Q
D = JQ’ + K’Q D = TQ’ + T’Q = T Q
Q
T
Q Q 20 / 60
Flip-Flop Characteristic Tables D
Q Q
J
Q
K
Q
T
Q Q
D 0 1 J 0 0 1 1 T 0 1
Q(t+1) 0 1
Reset Set
K Q(t+1) 0 Q(t) 1 0 0 1 1 Q’(t)
No change Reset Set Toggle
Q(t+1) Q(t) Q’(t)
No change Toggle 21 / 60
Flip-Flop Characteristic Equations D
Q Q
J
Q
K
Q
T
Q Q
D 0 1 J 0 0 1 1
Q(t+1) 0 1 K Q(t+1) 0 Q(t) 1 0 0 1 1 Q’(t)
T 0 1
Q(t+1) Q(t) Q’(t)
Q(t+1) = D
Q(t+1) = JQ’ + K’Q
Q(t+1) = T Q 22 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 1 1 0 0 0 1 1 0 1 1
No change Reset Set Toggle
23 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 1 0 1 1
No change Reset Set Toggle
24 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1
No change Reset Set Toggle
25 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J K
Q Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0
No change Reset Set Toggle
26 / 60
Flip-Flop Characteristic Equations Analysis / Derivation
J
Q
K
Q
J 0 0 0 0 1 1 1 1
K Q(t) Q(t+1) 0 0 0 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0
K 0 J 1
1 1
0 0
0 1
Q
Q(t+1) = JQ’ + K’Q 27 / 60
Flip-Flops with Direct Inputs Asynchronous Reset
D
Q
R’ 0
D CLK Q(t+1) x x 0
Q R Reset
28 / 60
Flip-Flops with Direct Inputs Asynchronous Reset
D
Q Q R
R’ 0 1 1
D CLK Q(t+1) x x 0 ↑ 0 0 ↑ 1 1
Reset
29 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0
Q CLR Reset
30 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0 x 0 1 x 1
Q CLR Reset
31 / 60
Flip-Flops with Direct Inputs Asynchronous Preset and Clear Preset PR D Q Q CLR
PR’ CLR’ D CLK Q(t+1) 1 0 x x 0 x 0 1 x 1 ↑ 1 1 0 0 ↑ 1 1 1 1
Reset
32 / 60
Analysis of Clocked Sequential Circuits The State ● State = Values of all Flip-Flops
Example AB=00
33 / 60
Analysis of Clocked Sequential Circuits State Equations
A(t+1) = DA = A(t) x(t)+B(t) x(t) =Ax+Bx B(t+1) = DB = A’(t) x(t) = A’ x y(t) = [A(t)+ B(t)] x’(t) = (A + B) x’
34 / 60
Analysis of Clocked Sequential Circuits State Table (Transition Table) Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1 t
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1 t+1
B 0 1 0 1 0 0 0 0
Output
y 0 0 1 0 1 0 1 0 t
A(t+1) = A x + B x B(t+1) = A’ x y(t) = (A + B) x’
35 / 60
Analysis of Clocked Sequential Circuits State Table (Transition Table) Present State
A 0 0 1 1 t
B 0 1 0 1
Next State
Output
x=0
x=1 x=0 x=1
A 0 0 0 0
A 0 1 1 1
B 0 0 0 0
t+1
B 1 1 0 0
y 0 1 1 1
y 0 0 0 0 t
A(t+1) = A x + B x B(t+1) = A’ x y(t) = (A + B) x’
36 / 60
Analysis of Clocked Sequential Circuits State Diagram
Present State
input/output
AB
0/0
1/0 0/1
00
Next State x=0
Output
x=1
x=0
x=1
A B 0 0
A B A B 0 0 0 1
y
y
0
0
0 1 1 0 1 1
0 0 0
1 1 1
0 0 0
0 0 0
1 1 1
1 0 0
10
0/1 1/0
0/1 01
1/0 11
1/0
37 / 60
Analysis of Clocked Sequential Circuits D Flip-Flops
Example: Present Input State
A 0 0 0 0 1 1 1 1
x 0 0 1 1 0 0 1 1
y 0 1 0 1 0 1 0 1
Next State
A 0 1 1 0 1 0 0 1
x y
D CLK
Q
A
Q
A(t+1) = DA = A x y 01,10 00,11
0
1
00,11
01,10 38 / 60
Analysis of Clocked Sequential Circuits JK Flip-Flops
Example: Present Next I/P State State A B x A B 0 0 0 0 1
Flip-Flop Inputs JA KA JB KB 0 0 1 0
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
1
1
0
0
1
1
1
0
1
0
0
1
1
0
0
1
1
0
0
1
1
1
0
1
1
0
0
0
0
0
1
1
0
0
0
1
1
1
1
1
1
1
1
1
1
0
0
0
JA = B JB = x’
KA = B x’ KB = A x
A(t+1) = JA Q’A + K’A QA = A’B + AB’ + Ax B(t+1) = JB Q’B + K’B QB = B’x’ + ABx + A’Bx’
39 / 60
Analysis of Clocked Sequential Circuits JK Flip-Flops
Example: Present Next I/P State State A B x A B 0 0 0 0 1
Flip-Flop Inputs JA KA JB KB 0 0 1 0
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
1
1
0
0
1
1
1
0
1
0
0
1
1
0
0
1
1
1
0
0
1
1
0
1
1
0
0
0
0
0
1
1
0
0
0
1
1
1
1
1
1
1
1
1
1
0
0
0
1
0
1 11
00
0
0 01
0 10
1
1 40 / 60
Analysis of Clocked Sequential Circuits T Flip-Flops
Example: Present Next F.F I/P O/P State State Inputs A B x A B TA TB y 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
1
0
0
1
0
0
0
0
1
1
1
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
1
TA = B x TB = x y =AB A(t+1) = TA Q’A + T’A QA = AB’ + Ax’ + A’Bx B(t+1) = TB Q’B + T’B QB =xB
41 / 60
Analysis of Clocked Sequential Circuits T Flip-Flops
Example: Present Next F.F I/P O/P State State Inputs A B x A B TA TB y 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
1
0
0
1
0
0
0
0
1
1
1
0
1
1
0
1
0
0
1
0
0
0
0
1
0
1
1
1
0
1
0
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
1
0/0
0/0 00
1/0
01
1/1 0/1
1/0 11
10
1/0
0/0 42 / 60
Mealy and Moore Models Mealy Present State A B 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
I/P x 0 1 0 1 0 1 0 1
Next O/P State A B y 0 0 0 0 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 1 0 0
For the same state, state the output changes with the input
Moore Present State A B 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1
I/P x 0 1 0 1 0 1 0 1
Next O/P State A B y 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1
For the same state, state the output does not change with the input 43 / 60
Moore State Diagram State / Output 0
0 1 00/0
01/0
1
1
11/1
10/0 1
0
0 44 / 60
Timing Diagram 0 A x
y
1 00/0 0
B
01/0 1
0
11/1 No effect
0
10/0
1
1
CLK x State
A
0
0
1
0
0
0
B
0
1
0
0
1
1
y
45 / 60
Timing Diagram 1/0
0/0
00
A x
y B
0/0
01
1/1
1/0
0/0 11
0/0
10 1/1
CLK x State
A
1
B
0
y
46 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s 1
0 S0 / 0
S1 / 0 0
0
1
0 S3 / 1
1
S2 / 0
State A B S0 0 0 S1 0 1 S2
1 0
S3
1 1
1 47 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
Output
y 0 0 0 0 0 0 1 1
1
0 S0 / 0
S1 / 0 0
0
0
S3 / 1 1
1 S2 / 0
1 48 / 60
Design of Clocked Sequential Circuits Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
Output
y 0 0 0 0 0 0 1 1
Synthesis using D Flip-Flops
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)
49 / 60
Design of Clocked Sequential Circuits with D F.F. Example: Detect 3 or more consecutive 1’s Synthesis using D Flip-Flops
DA (A, B, x) = ∑ (3, 5, 7) = Ax + Bx
B 0 0 1 0 A 0 1 1 0 x
DB (A, B, x) = ∑ (1, 5, 7) = A x + B’ x y (A, B, x) = ∑ (6, 7) = AB
B 0 0 0 0 A 0 0 1 1 x
B 0 1 0 0 A 0 1 1 0 x
50 / 60
Design of Clocked Sequential Circuits with D F.F. Example: Detect 3 or more consecutive 1’s Synthesis using D Flip-Flops
DA = A x + B x DB = A x + B’ x y = AB
51 / 60
Flip-Flop Excitation Tables Present Next State State
Q(t) Q(t+1) 0 0 0 1 1 0 1 1
F.F. Input
D 0 1 0 1
Present Next State State
F.F. Input
Q(t) Q(t+1) J K 0 x 0 0 0 1 1 x 1 0 x 1 1 1 x 0 Q(t) Q(t+1) 0 0 0 1 1 0 1 1
0 0 (No change) 0 1 (Reset) 1 0 (Set) 1 1 (Toggle) 0 1 (Reset) 1 1 (Toggle) 0 0 (No change) 1 0 (Set)
T 0 1 1 0 52 / 60
Design of Clocked Sequential Circuits with JK F.F. Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
Flip-Flop Inputs
B JA KA JB KB 0 0 x 0 x 1 0 x 1 x 0 0 x x 1 0 1 x x 1 0 x 1 0 x 1 x 0 1 x 0 x 1 x 1 1 x 0 x 0
Synthesis using JK F.F. 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) 53 / 60 dKB (A, B, x) = ∑ (0,1,4,5)
Design of Clocked Sequential Circuits with JK F.F. Example: Detect 3 or more consecutive 1’s Synthesis using JK Flip-Flops
JA = B x
KA = x’
JB = x
KB = A’ + x’
B
B
0 0 1 0 A x x x x x B
x x x x A 1 0 0 1 x B
0 1 x x A 0 1 x x x
x x 1 1 A x x 0 1 x
54 / 60
Design of Clocked Sequential Circuits with T F.F. Example: Detect 3 or more consecutive 1’s Present Input State
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
x 0 1 0 1 0 1 0 1
Next State
A 0 0 0 1 0 1 0 1
B 0 1 0 0 0 1 0 1
F.F. Input
TA 0 0 0 1 1 0 1 0
TB 0 1 1 1 0 1 1 0
Synthesis using T Flip-Flops TA (A, B, x) = ∑ (3, 4, 6) TB (A, B, x) = ∑ (1, 2, 3, 5, 6)
55 / 60
Design of Clocked Sequential Circuits with T F.F. Example: Detect 3 or more consecutive 1’s Synthesis using T Flip-Flops
TA = A x’ + A’ B x TB = A’ B + B x B 0 0 1 0 A 1 0 0 1 x
B 0 1 1 1 A 0 1 0 1 x 56 / 60
Homework Mano ● Chapter 5 ♦ 5-1 ♦ 5-3 ♦ 5-6 ♦ 5-8 ♦ 5-9
57 / 60
Homework 5-1
The D latch is constructed with four NAND gates and an inverter. Consider the following three other ways for obtaining a D latch. In each case, draw the logic diagram and the circuit operation. (a) Use NOR gates for the SR latch part and AND gates for the other two. An inverter may be needed. (b) Use NOR gates for all four gates. Inverters may be needed. (c) Use four NAND gates only (without an inverter). This can be done by connecting the output of the upper gate that goes to the SR latch to the input of the lower gate instead of the inverter output. 58 / 60
Homework 5-3
Show that the characteristic equation for the complement output of a JK flip-flop is Q’(t+1) = J’Q + K Q
5-6
A sequential circuit with two D flip-flops, A and B; two inputs, x and y; and one output, z, is specified by the following next-state and output equations: A(t+1) = x’ y + x A B(t+1) = x’ B + x A z=B (a) Draw the logic diagram of the circuit. (b) List the state table for the sequential circuit. (c) Draw the corresponding state diagram. 59 / 60
Homework 5-8
Derive the state table and the state diagram of the sequential shown circuit. Explain the function that the circuit performs.
60 / 60
Homework 5-9
A sequential circuit has two JK flip-flops A and B and one input x. The circuit is described by the following flip-flop input equations: JA = x
KA = B’
JB = x
KB = A
(a) Derive the state equations A(t+1) and B(t+1) by substituting the input equations for the J and K variables. (b) Draw the state diagram of the circuit.
61 / 60
Related Documents 3h463d
Digital Design Synchronous Sequential Logic 5j195q
July 2021 0
Digital Logic Design 5f53u
December 2019 62
Digital Logic Design Notes 616s4
November 2019 68
Digital Logic Design Lab Manual g5043
November 2021 0
Digital Logic Design Question Bank.pdf 3wr2o
November 2021 0
Synchronous Digital Hierarchy (sdh) 3h6i22
February 2022 0More Documents from "Pavan Kumar" 2pa4c
Digital Design Synchronous Sequential Logic 5j195q
July 2021 0
Nda And Anda 2q196b
February 2022 0
Respaper Avm School Prelim Paper Physics Icse 49384j
November 2019 92
Abb Acs550 Vfd Manual 4a325m
November 2019 101
Global Nutrition Philosophy Pdf g6k40
November 2021 0