Digital System Design-Synchronous Sequential Circuits

MATRUSRI ENGINEERING COLLEGE
DEPARTMENT OF ELECTRONICS AND COMMUNICATION
ENGINEERING
SUBJECT NAME: DIGITAL SYSTEM DESIGN WITH VERILOG
FACULTY NAME: Mrs. B. Indira Priyadarshini
MATRUSRI
ENGINEERING COLLEGE

INTRODUCTION:
Gives a detailed presentation of synchronous sequential circuits (finite state
machines). It explains the behavior of these circuits and develops practical
design techniques for both manual and automated design.Deals with a general
class of circuits in which the outputs depend on the past behavior of the circuit,
as well as on the present values of inputs. They are called sequential circuits. In
most cases a clock signal is used to control the operation of a sequential circuit;
such a circuit is called a synchronous sequential circuit.
UNIT-III
OUTCOMES:
After successful completion of this Unit students should be able to
Analyze, design and implement sequential logic circuits in terms of state
machines.
Implement sequential circuits using Verilog code
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CONTENTS:
Latches
Flip Flops
Counters
Shift registers applications
OUTCOMES:
Students will be able to design sequential logic modules in behavioral
modeling.
MODULE-I: Behavioural modeling of
sequential logic modules
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Latches
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Clock d qn q n+1
0 x x 0
1 0 0 0
1 1 0 1
1 0 1 0
1 1 1 1
module D latch (D, Clk, Q);
input D, Clk;
output Q;
reg Q;
always @(D or Clk)
if (Clk)
Q <= D;
endmodule
module dtest_v;
reg d;
reg clk;
wire q;
D latch uut(.d(D), .clk(Clk), .q(Q));
always
#5 clk=~clk;
initial begin
#100 $finish;
end
initial begin
clk=1; d=1'b0;
#20 d=1'b1;
#20 d=1'b0;
end
endmodule

D Flipflops
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module D_FF (Q, D, Clk);
output reg Q;
input D, Clk;
always@ ( posedge Clk)
Q <= D;
endmodule
module dtest_v;
reg d;
reg clk;
wire q;
D_FF uut(.d(D), .clk(Clk), .q(Q));
always
#5 clk=~clk;
initial begin
#100 $finish;
end
initial begin
clk=1; d=1'b0;
#20 d=1'b1;
#20 d=1'b0;
end
endmodule
Resetn clock d qn q n+1
1 x x 0
0 0 0 0
0 1 0 1
0 0 1 0
0 1 1 1

D Flipflops
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D flip-flop with synchronous reset
module D_FF (Q, D, Clk, rst);
output reg Q;
input D, Clk, rst;
always@ ( posedge Clk)
if (!rst)
Q <= 0;
else
Q <= D;
endmodule
D flip-flop with asynchronous reset
module DFF ( output reg Q, input D, Clk, rst);
always@ ( posedge Clk or negedge rst)
if (!rst)
Q <= 1'b0; // Same as: if (rst == 0)
else
Q <= D;
endmodule
Testbench:
module dtest_v;
reg d,clk,rst
wire q;
DFF uut(
.d(D), .clk(Clk), .rst(rst), .q(Q)
);
always #5 clk=~clk;
initial begin
#100 $finish;
end
initial begin
clk=1; rst=0; d=1'b0;
#20 rst n=1; d=1'b1;
#20 rst=0; d=1'b0;
end
endmodule

T Flipflops
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T Flipflop
module tff_behv(t, clk, rst, q);
input t, clk, rst;
output reg q;
always@(posedge clk)
if(rst == 1)
q<=1'b0;
else
begin
if(t==1)
q<= ~q;
else
q<= q;
end
endmodule
Testbench:
module tfftest_v;
reg t, clk,rst;
wire q;
tffuut( .t(t), .clk(clk), .rst(rst), .q
(q));
always
#5 clk = ~clk;
initial begin
#100 $finish;
end
initial begin
clk= 1; rst = 1; t=0;
#5 t=0;
#10 rst=0; t=1;
#30 t=0;
end
endmodule
Resetn clock t qn q n+1
1 x x 0
0 0 0 0
0 1 0 1
0 0 1 1
0 1 1 0

JK Flipflop
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JK Flipflop
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module jkff_behv(j, k, clk, rst, q,qb);
input j, k, clk, rst;
output reg q;
output qb;
assign qb=~q;
always@( posedge clk)
begin
if(rst == 1'b1)
q<= 1'b0;
else
case({j,k})
2'b00: q<=q;
2'b01: q<=1'b0;
2'b10: q<=1'b1;
2'b11: q<= ~q;
endcase
end
endmodule
Testbench:
module jkfftest1_v;
reg j, k, clk, rst;
wire q, qb;
jkffuut(.j(j),.k(k), .clk(clk), .rst(rst),.
q(q),.qb(qb));
always #5 clk = ~clk;
always #5 rst = ~ rst;
initial begin
#100 $finish;
end
initial begin
rst = 0; j=1; k=0; clk=1;
#5 j=0; k=1;
#5 j=1; k=1;
#5 j=1; k=0;
end
endmodule

SR Flipflop
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rst clk s r qn q n+1
1 x x x 0
0 0 0 0 0
0 0 0 1 1
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 1
0 1 1 x x
0 1 1 x x

SR Flipflop
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module srff_behv(s, r, clk, rst, q);
input s, r ;
output reg q;
always@( posedgeclk)
begin
if(rst == 1'b1)
q<= 1'b0;
else
casex({s,r})
2'b00: q<=q;
2'b01: q<=1'b0;
2'b10: q<=1'b1;
2'b11: q<= 1’bx;
endcase
end
endmodule
Testbench:
module srfftest1_v;
reg j, k, clk, rst;
wire q;
srffuut(.s(s),.r(r), .clk(clk), .rst(rst),.
q(q));
always #5 clk = ~clk;
always #5 rst = ~ rst;
initial begin
#100 $finish;
end
initial begin
rst = 0; s=1; r=0; clk=1;
#5 s=0; r=1;
#5 s=1; r=1;
#5 s=0; r=0;
end
endmodule

Counter
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Clock
Pluse
Q3 Q2 Q1 Q0
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1

Counter
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module updowncount (
R, clk, L, E, updwn, Q
);
parameter n = 8;
input [n-1]R;
input clk, L, E, updwn;
output [n-1] Q;
reg [n-1]Q;
integer direction;
always @(posedge clk)
begin
if (updwn)
direction = 1;
else
direction = −1;
if (L)
Q <= R;
else if (E)
Q <= Q + direction;
end
endmodule
Testbench:
module updowncounttest_v;
reg clk, L, E, updwn;
reg [n-1]R
wire [n-1:0] Q;
Updowncount uut(R, clk, L, E, updwn, Q);
always #5 clk= ~clk;
initial begin
#400 $finish;
end
initial
begin
L=1;clk=1;updwn=0; R=8’o23;E=1;
#10 L=0;
#150 L=1;
#20 L=0; updwn=1;
end
endmodule

Shift Register
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Right Shift:
module shift4 (R, L, w, Clock, Q);
input [3:0] R;
input L, w, Clock;
output reg [3:0] Q;
always @(posedge Clock)
if (L)
Q <= R;
else
Q[3:0]={w, Q[3: 1]};
endmodule
Left Shift:
module shift4 (R, L, w, Clock, Q);
input [3:0] R;
input L, w, Clock;
output reg[3:0] Q;
if (L)
Q <= R;
else
Q[3:0]={Q[2: 0],w};
endmodule
Testbench:
module shift4_tst_v;
reg L,w,Clock;
reg [3:0]R;
wire [3:0]Q;
shift4 uut(R, L, w, Clock, Q);
initial begin
R = 4’b1011;
L = 1;
w = 1;
#20; L = 0;
end
always #5 Clock = ~ Clock;
endmodule

Shift Register
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Serial In Serial Out:
module shift (C, SI, SO);
input C,SI;
output SO;
reg [7:0] tmp;
always @(posedge C)
begin
tmp = tmp << 1;
tmp[0] = SI;
end
assign SO = tmp[7];
endmodule
Parallel In Parallel Out:
module pipo(pin, clk, reset, pout);
input [3:0] pin;
input clk, reset;
output reg [3:0] pout;
always @ (posedgeclk or posedge reset)
begin
if (reset)
pout <= 4'b0000;
else
pout <= pin;
end
endmodule

Shift Register
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Parallel In Serial Out:
module Shiftregister_PISO(
Clk, Parallel_In,load,
Serial_Out);
input Clk,load;
input [3:0]Parallel_In;
output reg Serial_Out;
reg [3:0]tmp;
always @(posedge Clk)
begin
if(load)
tmp<=Parallel_In;
else
begin
Serial_Out<=tmp[3];
tmp<={tmp[2:0],1'b0};
end
end
endmodule
Serial In Parallel Out:
module ShiftRegister_SIPO(Clk, SI, PO);
input Clk,SI;
output [7:0] PO;
reg [7:0] tmp;
always @(posedge Clk)
begin
tmp = {tmp[6:0], SI};
end
assign PO = tmp;
endmodule

1. The output of latches will remain in set/reset until the trigger pulse is
given to change the state.
2. The purpose of the clock input to a flip-flop is to cause the output to
assume a state dependent on the controlling inputs.
3. How much storage capacity does each stage in a shift register represent?
Ans: One bit
4. Binary counter that counts reversely is called down counter.
Questions & Answers
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CONTENTS:
Design Procedure
Moore state Model
Mealy state Model
OUTCOMES:
Students will be able to design Mealy and Moore FSM models for completely
and incompletely specified circuits.
MODULE-II: Synchronous Sequential
Circuits
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Synchronous Sequential Circuits
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A circuit whose output(s) depend on past behaviour, and present inputs
•Clock is used => synchronous sequential circuits
•No clock => asynchronous sequence circuits
Also called Finite state machine (FSM)
State elements in synchronous sequential circuits are edge triggered
•To ensure state changes only once in a single cycle.

Synchronous Sequential Circuits
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Synchronous sequential circuits are of two types:
•Moore output depends only on state
•Mealy: output depends on state and inputs

Design Procedure
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The procedure for designing synchronous sequential circuits can be
summarized by a list of recommended steps.
1. From the word description and specifications of the desired operation,
derive a state diagram for the circuit.
2. Reduce the number of states if necessary.
3. Assign binary values to the states.
4. Obtain the binary-coded state table.
5. Choose the type of flip-flops to be used.
6. Derive the simplified flip-flop input equations and output equations.
7. Draw the logic diagram

Designing Sequential Circuit
Moore State Model
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 Suppose that we want a circuit with the following characteristics:
• One input w, and one output z
• Positive-edge-triggered design
• z = 1, if w = 1 during two consecutive clock cycles
Notes: using only input, we can not find an expression for output
•Hence need a state information – FSM
Clock cycle: t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10
w: 0 1 0 1 1 0 1 1 1 0 1
z: 0 0 0 0 0 1 0 1 1 0 0

Develop State Diagram
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The conceptually simplest method is to use a pictorial representation in the
form of a state diagram.
Optional to develop
One form to represent a FSM:
• How many states: States are circles
• Transitions between states: Transitions are directed edges
• Starting state: i.e. after reset/clear
Note in figure, reset is not treated as input: To simplify figure.

Develop State Table
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Another way to describe a FSM
When implemented in a logic circuit, each state is represented by a particular
valuation (combination of values) of state variables.
It contains information on:
• States of the machine
• Transitions from all states, for all possible inputs
• Output values
• Reset information ignored: State A is assumed to be “start” state
Present
state
Next state Output
Z
w = 0 w = 1
A
B
C
A B
A C
A C
0
0
1

Develop State Assignment
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Find number of flip/flops needed to represent state
•No. of FFs = log2(no. of states)
Assign each state a combination of values of state variables
• “State assigned table”
• All unused variable combination are normally used as don’t cares
Below is the resulting table after state assignment
Notice that:
•Output depends on current state only - Moore type
•2 state variables are sufficient to represent 3 states
•Y1 & Y2 are next-state variables, y1&y2 are present-state variables
Need to decide type of FF to use as state element
Use D-FF since it is easiest
D1 = Y1, and D2 = Y2
For every next state and output, derive their function from present state and
input

Develop State Assignment
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Y1 = w.y1y2
Y2 = w(y1+y2 )
z = y2
•State assignments has direct relation to the cost of derived implementation
Some state assignments are better than others
•Using the new state assignment a more cost effective realization in possible
Y1 = w, cheaper
Y2 = wy1, cheaper
z = y2 , same cost
Present state
y2y1
Next state Output
Z
w = 0 w = 1
Y2Y1 Y2Y1
A
B
C
D
00
01
10
11
00 01
00 10
00 10
d d
0
0
1
d
Present state
y2y1
Next state Output
Z
w = 0 w = 1
Y2Y1 Y2Y1
A
B
C
D
00
01
11
10
00 01
00 10
00 10
d d
0
0
1
d

Function Realization
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Resulting Logic Circuit
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Timing Diagram of Realization
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Mealy Machine Implementation
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Output values are generated using state & present inputs
State diagram State Table
State Assigned Table Logic Diagram
Y = D = w z = wy
Clock cycle: t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10
w: 0 1 0 1 1 0 1 1 1 0 1
z: 0 0 0 0 1 0 0 1 1 0 0
Present
state
Next state Output z
w = 0 w = 1 w = 0 w = 1
A
B
A B
A B
0 0
0 1
Present
state
Next state Output z
w = 0 w = 1 w = 0 w = 1
y Y Y z z
A
B
0
1
0 1
0 1
0 0
0 1

Timing Diagram of Mealy Machine
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Mealy implementation is more cost effective than Moore implementation
•However, circuit can be modified so that it behaves like a Moore machine
Note how output change based on state and input

1. Moore machine produces an output over the change of transition states.
2. In mealy machine, the O/P depends upon present states and inputs.
3. The relationship that exists among the inputs, outputs, present states and
next states can be specified by either the state table or the state diagram.
4. A state-transition table is a table showing what state a finite-state
machine will move to, based on the current state and other inputs.
Questions & Answers
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CONTENTS:
State Minimization
Partitioning Minimization Procedure
OUTCOMES:
Students will be able to design a more complex FSM with fewer flips-flops.
MODULE-III: State Minimization
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State Minimization
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Two states Si and Sj are said to be equivalent if and only if for every possible
input sequence, the same output sequence will be produced regardless of
whether Si or Sj is the initial state.
Lower no. of states => lower no. of FFs
Solved using “partitioning minimization procedure”
Partition: A set of states
A partition consists of one or more blocks, where each block comprises a
subset of states that may be equivalent, but the states in a given block are
definitely not equivalent to the states in other blocks.
States in a partition may be equivalent
Not equivalent to states in other partitions

State Minimization
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•P1 = (ABCDEFG)
Partition based on output z
•P2 = (ABD)(CEFG),
Partition based on 0- & 1-successor for
block (ABD) & (CEFG)
•P3 = (ABD)(CEG)(F),
block (ABD) & (CEG),
•P4 = (AD)(B)(CEG)(F)
block (AD) & (CEG), => Final
•Final Partitions: P5 = (AD)(B)(CEG)(F)
2 FFs are sufficient after state minimization instead of 3
Present
state
Next state Output
z
w = 0 w = 1
A
B
C
D
E
F
G
B C
D F
F E
B G
F C
E D
F G
1
1
0
1
0
0
0
Present
state
Next state Output
z
w = 0 w = 1
A
B
C
F
B C
A F
F C
C A
1
1
0
0

Incompletely Specified FSMs
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The partitioning scheme for minimization of states works well when all
entries in the state table are specified. FSMs of this type are said to be
completely specified.
If one or more entries in the state table are not specified, corresponding to
don’t-care conditions, then the FSM is said to be incompletely specified.
Affects the number of minimized states
Assume x’s are zeros:
P1 = (ABCDEFG)
P2 = (ABDG)(CEF),
P3 = (AB)(D)(G)(CE)(F),
P4 = (A)(B)(D)(G)(CE)(F),
P5 = P4 => 6 states
Assume x’s are ones:
P1 = (ABCDEFG)
P2 = (AD)(BCEFG),
P3 = (AD)(B)(CEFG),
P4 = (AD)(B)(CEG)(F),
P5 = P4 => 4 states
Present
state
Next state Output z
w = 0 w = 1 w = 0 w = 1
A
B
C
D
E
F
G
B C
D -
F E
B G
F C
E D
F -
0 0
0 -
0 1
0 0
0 1
0 1
0 -

1. State Minimizing reduces the number of flips-flops used in the FSM.
2. State Minimizing reduces the complexity of the combinational circuit
needed in the FSM.
3. By state minimization, two different FSMs may exhibit identical behavior
in terms of the outputs produced in response to all possible inputs.
4. If one or more entries in the state table are not specified, corresponding to
don’t-care conditions, then the FSM is said to be incompletely specified.
Questions & Answers
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CONTENTS:
Moore FSM Model
Mealy FSM Model
OUTCOMES:
Students will be able to write a verilog code for any sequence detector.
MODULE-IV: Sequence detector with
verilog HDL modeling
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module simple (Clock, Resetn, w, z);
input Clock, Resetn, w;
output z;
reg [2:1] y, Y;
parameter [2:1] A = 2'b00, B = 2'b01, C = 2'b10;
always @(w or y)
case (y)
A: if (w) Y = B;
else Y = A;
B: if (w) Y = C;
else Y = A;
C: if (w) Y = C;
else Y = A;
default: Y = 2'bxx;
endcase
always @(negedge Resetn or posedge Clock)
if (Resetn == 0)
y <= A;
else
y <= Y;
assign z = (y == C);
endmodule
Moore FSM Model
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module mealy (Clock, Resetn, w, z);
input Clock, Resetn, w;
output z;
reg y, Y, z;
parameter A = 0, B = 1;
always @(w or y)
case (y)
A: if (w) begin z = 0; Y = B; end
else begin z = 0; Y = A; end
B: if (w) begin z = 1; Y = B; end
else begin z = 0; Y = A; end
endcase
always @(negedge Resetn or posedge Clock)
if (Resetn == 0)
y <= A;
else
y <= Y;
endmodule
Mealy FSM Model

1. Mealy machine will have same or fewer states than Moore machine.
2. In Moore Machine both output and state change synchronous to the clock
edge.
3. Moore Machine requires more hardware to design than mealy machine.
4. In Mealy machine, asynchronous output generation through the state
changes synchronous to the clock.
Questions & Answers
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CONTENTS:
Modulo – 8 Counter
D Flip-flop
JK Flip-flop
Verilog code
OUTCOMES:
Students will be able to design different counters using sequential ciruit
approach.
MODULE-V: Modulo-8 Counter Using the
Sequential Circuit Approach
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Modulo-8 Counter
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Mod 8 counter (0, 1, .. 7, 0, 1 ..) {Moore Design}
State Diagram:
State asigned table
State table
Present
state
Next state Output
w = 0 w = 1
A
B
C
D
E
F
G
H
A B
B C
C D
D E
E F
F G
G H
H A
1
2
3
4
5
6
7
8
Present
state
y2y1y0
Next state Output
z2z1z0
w = 0 w = 1
Y2Y1Y0 Y2Y1Y0
A
B
C
D
E
F
G
H
000
001
010
011
100
101
110
111
0 0 0 0 0 1
0 0 1 0 1 0
0 1 0 0 1 1
0 1 1 1 0 0
1 0 0 1 0 1
1 0 1 1 1 0
1 1 0 1 1 1
1 1 1 0 0 0
000
001
010
011
100
101
110
111

Implementation Using D-Type Flip-Flops
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 
  2
1
0
1
0
2
1
0
2
1
0
2
1
0
2
1
0
2
1
2
0
2
2
1
0
1
0
1
0
1
0
1
0
1
0
0
1
1
1
0
0
0
0
y
y
wy
y
y
wy
y
y
y
w
y
y
wy
y
y
y
w
y
y
y
w
y
y
y
y
y
w
D
y
wy
y
wy
y
y
w
y
wy
y
y
w
y
wy
y
y
y
w
D
y
w
y
w
y
w
D



























Implementation Using D-Type Flip-Flops
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Implementation Using JK Flip-Flops
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Excitation table
Present
state
y2y1y0
Flip-flop inputs Count
z2z1z0
w = 0 w = 1
Y2Y1Y0 J2K2 J1K1 J0K0 Y2Y1Y0 J2K2 J1K1 J0K0
000
001
010
011
100
101
110
111
000
001
010
011
100
101
110
111
0d 0d 0d
0d 0d d0
0d d0 0d
0d d0 d0
d0 0d 0d
d0 0d d0
d0 d0 0d
d0 d0 d0
001
010
011
100
101
110
111
000
0d 0d 1d
0d 1d d1
0d d0 1d
1d d1 d1
d0 0d 1d
d0 1d d1
d0 d0 1d
d1 d1 d1
000
001
010
011
100
101
110
111

Implementation Using JK Flip-Flops
MATRUSRI
ENGINEERING COLLEGE

Verilog Implementation
MATRUSRI
ENGINEERING COLLEGE
module updowncount (
q, reset, s, clk,m
);
output reg [2:0]q;
input reset, clk;
always@(posedge clk)
begin
if(reset==1)
q<=0;
else
case(m)
0: q=q+1;
1: q=q-1;
endcase
end
endmodule
Testbench:
module updowncounttest_v;
reg reset,clk,m;
wire [2:0] q;
updowncountuut(
.q(q),.reset(reset),.s(s),.clk(clk),.m(m)
);
always #5 clk= ~clk;
initial begin
#400 $finish;
end
initial begin
reset=1;clk=1; m=0;
#10 reset=0;
#150 reset=1;
#20 reset=0; m=1;
end
endmodule

1. A counter is a Moore machine.
2. The output signals are specified as depending only on the state of the
counter at a given time.
3. The flip-flops must be edge triggered to ensure that only one transition
takes place during a single clock cycle.
4. Can attach a combinational circuit to a D flip-flop to convert it into JK flip-
flop.
Questions & Answers
MATRUSRI
ENGINEERING COLLEGE

CONTENTS:
One-Hot Encoding
OUTCOMES:
Students will be able to design FSM models for completely and incompletely
specified circuits.
MODULE-VI: One-Hot Encoding
MATRUSRI
ENGINEERING COLLEGE

One-Hot Encoding
MATRUSRI
ENGINEERING COLLEGE
• As an alternative state assignment, we can use one-hot encoding:
Use as many FFs as states
•In one-hot encoding method, for each state all but one of the state variables
are equal to 0. The variable whose value is 1 is deemed to be “hot.”
•Usually leads to simpler output expressions, faster circuits
Y1 = w͞
Y2 = wy1
Y3 = wy͞1
z = y3
Present state
y3y2y1
Next state Output
Z
w = 0 w = 1
Y3Y2Y1 Y3Y2Y1
001
010
100
001 010
001 100
001 100
0
0
1
Present
state
Next state Output
Z
w = 0 w = 1
A
B
C
A B
A C
A C
0
0
1

1. The variable whose value is 1 is deemed to be hot.
2. One-hot state assignment leads to simpler output expressions than do
assignments with the minimal number of state variables.
3. Simpler output expressions may lead to a faster circuit.
Questions & Answers
MATRUSRI
ENGINEERING COLLEGE

CONTENTS:
Analysis of Synchronous Circuits
OUTCOMES:
Students will be able to analysis of various Synchronous Circuits.
MODULE-VII: Analysis of Synchronous
Circuits
MATRUSRI
ENGINEERING COLLEGE

MATRUSRI
ENGINEERING COLLEGE
Analysis is the reverse of synthesis: Given a circuit, find out what does it do
Reverse synthesis steps:
• Construct state table (FFs’ type is a factor)
• Symbolic state table
• State diagram
State assigned table
State table
2
1
2
1
2
2
1
1
y
y
z
wy
wy
Y
wy
y
w
Y





Present
state
Next state Output
Z
w = 0 w = 1
A
B
C
D
A C
A D
A B
A D
0
0
0
1
Present
state
y1y2
Next state Output
z
w = 0 w = 1
Y1Y2 Y1Y2
00
01
10
11
00 10
00 11
00 01
00 11
0
0
0
1

MATRUSRI
ENGINEERING COLLEGE
Excitation table State table
w
K
wy
J
y
w
K
w
J





2
1
2
2
1
1
Present
state
y1y2
Next state Output
z
w = 0 w = 1
J2K2 J1K1 J2K2 J1K1
00
01
10
11
01 01 00 11
01 01 00 10
01 01 10 11
01 01 10 10
0
0
0
1
Present
state
Next state Output
z
w = 0 w = 1
A
B
C
D
B B A D
B B A C
B B C D
B B C C
0
0
0
1

1. The analysis task is much simpler than the synthesis task.
2. To analyze a circuit, reverse the steps of the synthesis process.
3. To design a synchronous sequential circuit, the designer has to be able to
analyze the behavior of an existing circuit.
Questions & Answers
MATRUSRI
ENGINEERING COLLEGE

CONTENTS:
Serial Adder
Mealy-Type FSM
Moore-Type FSM
Verilog Code
OUTCOMES:
Student will able to design and implement a FSM for serial adder
MODULE-VIII: Additional Topic
MATRUSRI
ENGINEERING COLLEGE

Mealy-Type FSM
MATRUSRI
ENGINEERING COLLEGE
State Diagram State Table
Present
state
Next state Output
ab = 00 01 10 11 00 01 10 11
G
H
G G G H
G H H H
0 1 1 0
1 0 0 1

Mealy-Type FSM
MATRUSRI
ENGINEERING COLLEGE
State Assigned Table Output equations:
Circuit Diagram
Present
state
Next state Output s
ab = 00 01 10 11 00 01 10 11
y Y S
0
1
0 0 0 1
0 1 1 1
0 1 1 0
1 0 0 1
y
b
a
S
by
ay
ab
Y







Moore-Type FSM
MATRUSRI
ENGINEERING COLLEGE
State Diagram State Table
State Assigned Table
Circuit Diagram
Present
state
Next state Output
S
ab = 00 01 10 11
G0
G1
H0
H1
G0 G1 G1 H0
G0 G1 G1 H0
G1 H0 H0 H1
G1 H0 H0 H1
0
1
0
1
Present
state
y2y1
Next state Output
s
ab = 00 01 10 11
Y2Y1
00
01
10
11
00 01 01 10
00 01 01 10
01 10 10 11
01 10 10 11
0
1
0
1
1
2
2
2
2
1
y
s
by
ay
ab
Y
y
b
a
Y








module serialadder (A, B, Reset, Clock, Sum);
input [3:0] A, B;
input Reset, Clock;
output wire [3:0] Sum;
reg [3:0] Count;
reg s, y, Y;
wire [3:0] QA, QB;
wire Run;
parameter G = 1'b0, H = 1'b1;
shift shift_A (A, Reset, 1'b1, 1'b0, Clock, QA);
shift shift_B (B, Reset, 1'b1, 1'b0, Clock, QB);
shift shift_S (4'b0, Reset, Run, s, Clock, Sum);
always @(QA, QB, y)
case (y)
G: begin
s = QA[0]^QB[0];
if (QA[0] & QB[0]) Y = H;
else Y = G;
end
Verilog Code
MATRUSRI
ENGINEERING COLLEGE
H: begin
s = QA[0]~^QB[0];
if (~QA[0] & ~QB[0])
Y = G;
else
Y = H;
end
default: Y = G;
endcase
if (Reset)
y <= G;
else
y <= Y;
if (Reset) Count = 4;
else if (Run) Count = Count-1;
assign Run =|Count;
endmodule

1. In a serial adder, bits are added a pair at a time.
2. Fast adders are more complex and more expensive.
3. A serial adder is a cost-effective and less speed.
4. Moore-type circuit is delayed by one clock cycle with respect to the Mealy-
type sequential circuit.
Questions & Answers
MATRUSRI
ENGINEERING COLLEGE

Question Bank
MATRUSRI
ENGINEERING COLLEGE
Short Answer Question
S.No Question
Blooms
Taxonomy
Level
Course
Outcome
1 Write Verilog model for JK flip flop. L2 CO3
2 Analyze the Gated-D latch circuit. L2 CO3
3 Explain about State assignment problem L1 CO3
4 Briefly explain One hot encoding. L2 CO3
5 Brief out the basic design steps for designing
Synchronous Sequential Circuits.
L1 CO3
6 Write short notes on incompletely specified FSM
model.
L1 CO3
7 Differentiate between latch and flipflop. L1 CO3
8 Explain Partitioning Minimization Procedure. L1 CO3
9 Explain difference between mealy and moore
model.
L1 CO3

Question Bank
MATRUSRI
ENGINEERING COLLEGE
Long Answer Question
S.No Question
Blooms
Taxonomy
Level
Course
Outcome
1 Briefly explain about shift register and write a verilog code for
4 bit shift right and left register.
L2 CO3
2 Explain the state minimization process in synchronous
sequential circuits.
L5 CO3
3 Design a synchronous 3 bit up-down counter. Write verilog
code with its test bench and waveforms.
L1 CO3
4 Design and write verilog code for modulo-8 counter using
sequential approach use T flipflop memory element..
L4 CO3
5 Explain serial adder with a neat diagram and implement in
Verilog language using Mealy and moore type models.
L2 CO3
6 Design sequential circuit for given state table and write verilog
code in behavioural modeling
L1 CO3
X
0 1
A
B
C
D
E
B/0 E/0
A/1 C/1
B/0 C/1
C/0 E/0
D/1 A/0

Question Bank
MATRUSRI
ENGINEERING COLLEGE
Long Answer Question
S.No Question
Blooms
Taxonomy
Level
Course
Outcome
7 Perform the partitioning procedure on the state table shown
below
L2 CO3
8 Analyser the given synchronous sequential circuit and write its
veilog code.
L5 CO3
Present
state
Next State output
input=0 input = 1
A B C 1
B D F 1
C F E 0
D B G 1
E F C 0
F E D 0
G F G 0

Assignment Questions
MATRUSRI
ENGINEERING COLLEGE
1. Design a counter which moves through the sequence 0,4,2,6,1,5,3,7,0,4
and so on using synchronous sequential circuit approach.
2. Design a Mealy and Moore type Sequence detector for detecting the
sequence 1010 and Implement in verilog HDL.
3. Reduce the state table given below and draw the minimized state diagram.
Present
State
Next State Output (z)
Input (x) Input (x)
X = 0 X = 1 X = 0 X = 1
A A B 0 0
B D C 0 1
C F E 0 0
D D F 0 0
E B G 0 0
F G C 0 1
G A F 0 0

Assignment Questions
MATRUSRI
ENGINEERING COLLEGE
4. Analyser the given synchronous sequential circuit and write its veilog code.
5. Design synchronous sequential circuit of state machine M1 shown using D
flip –flop. Assume state assignment as A = 00, B = 01and C= 10.
PS Next State , z
x = 0 x = 1
A B, 0 A, 1
B C, 0 A, 1
C A,1 B, 0

Digital System Design-Synchronous Sequential Circuits

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