EECS 465: Digital Systems
Lecture Notes # 5
Sequential Circuit (Finite-State Machine) Design
SHANTANU DUTT
Department of Electrical Engineering and Computer Science
University of Illinois, Chicago
Phone: (312) 355-1314: e-mail: dutt@eecs.uic.edu
URL: http://www.eecs.uic.edu/~dutt
1
Finite State Machine (FSM) Design
• FSMs are different from counters in the sense that they have external
I/Ps, and state transitions are dependent on these I/Ps and the current
state.
• Example : Problem Statement
There is a bit-serial I/P line. Design an FSM that outputs a ‘0’
if an even # of 1’s have been received on the I/P line and the
outputs a ‘1’ otherwise.
Note : If a synchronous sequential circuit is being designed, the counting
of the # of 1s occur every clock cycle.
x
O/p y
FSM
CLK
CLK
x
# of
1s
even
(0)
odd
(1)
even
(2)
odd
(3)
odd
(3)
2
• First determine the # of useful information classes required to solve
the problem.
• In this case, only 2 classes of information is required: whether an
even # of 1s have been received, or an odd # of 1s have been received
Solution 1: (Mealy)
Solution 2: (Moore)
0
0/0
Reset
Even
Odd
Even
Input
1/1
1/0
Transition
Arc
Reset
Output
Output
[0]
O/P is dependent
on current state and
input in Mealy
0/1
Mealy Machine: Output is associated
with the state transition, and appears
before the state transition is completed
(by the next clock pulse).
1
1
Odd
[1]
0
Input
Output is
dependent only
on current state
Moore Machine: Output is associated
with the state and hence appears
after the state transition take place.
3
External I/Ps
External O/Ps
m1
Comb.
Logic
n even

odd
n
Output
Logic
n
n
CLK
External Outputs
Moore Machine Model
Mealy Machine Model
Time t : Even I/P
FFs
m2
CLK
t+
Next State
Comb.
Logic
m2
FFs
t
External
I/Ps
m1
 = propagation delay of logic of Mealy M/C
t+TCLK t+TCLK+2
Even
Odd O/P=1
O/P=1
x=1
(Moore)
(Mealy)
O/P=0
2 = propagation delay of O/P
logic unit of Moore M/C
4
State Transition Table
(Even-Parity Checker)
Even State: 0 ;
Present
State
Input
Odd State: 1;
Next
State
State Variable A
Moore
O/P
Mealy
O/P
D-FF
Excit.
y1
y2
DA
T-FF
Excit.
A
x
A+
0
0
1
0
1
0
0
1
1
0
0
1
0
1
1
0
1
1
0
1
0
1
1
0
1
0
0
1
Input variables
to comb. logic
DA= Ax ; TA= x
y1 = A for Moore
y2 = Ax for Mealy
TA
Output functions
y2
x
N.S. & O/P
Logic
Q
A FF
x
N.S.
Logic
Or
Q
DA
CLK
A
O/P
Logic
FFs
DA
y1
5
Reset
State=0
Even
0
0/0
Reset
State=0
Even
[0]
1/1
1/0
1
1
State=1
Odd
N.S.
Logic
x
State=1
Odd
[1]
Q
0/1
Q
Mealy
0
D-
FF D
Moore
CLK
S.T. is complete.
Assume single bit state information stored in a D-FF
State Transition State Transition
is occurring
is occurring
S.T. is complete.
CLK
x
D
even
Q
odd
even
even
odd
odd
(state)
y2
(Mealy O/P)
6
y1
Moore
O/P)
Moore M/C Implementation
a) D-FF
b) T-FF
x
0
D
x=1
Q
R
CLK
A
y2
T
CLK
Q
R
Q
A
y2
Q
Reset
Reset
Moore O/P is synchronized with clock.
Mealy M/C Implementation
y1
0
1
D
x=1
CLK
Q
R
x
T
A
CLK
Q
R
y1
Q
Q
Reset
a) D-FF
Reset
b) T-FF
Mealy O/P is not synchronized with clock.
7
Difference Between Mealy and Moore Machine
Mealy
(1) O/Ps depend on the present
state and present I/Ps
(2) The O/P change asyn
-chronously with the
enabling clock edge
(3)
(4)
A counter is not a Mealy
machine
A Mealy machine will have
the same # or fewer states
than a Moore machine
Moore
O/Ps depend only on the
present state
Since the O/Ps change
when the state changes,
and the state change is
synchronous with the
enabling clock edge, O/Ps
change synchronously
with this clock edge
A counter is a Moore
machine
8
Another example: A simple vending machine
Here is how the control is supposed to work. The vending machine delivers
a package of gum after it has received 15 cents in coins. The machine has a single
coin slot that accepts nickels and dimes, one coin at a time. A mechanical sensor
indicates to the control whether a dime or a nickel has been inserted into the coin slot.
The controller’s output causes a single package of gum to be released down a chute
to the customer.
One further specification: We will design our machine so it does not give
change. A customer who pays with two dimes is out 5 cents!
Coin
Sensor
Reset
Vending
Machine
FSM
Open
Gum
Release
Mechanism
CLK
Vending Machine block diagram
States:
0C
5C
10C
15C
9
— The figure below show the Moore and Mealy machine state transition diagrams.
( N D  Reset )/0
0 cent
[0]
Reset
ND
0 cent
Reset / 0
N D /0
N/0
N
5 cent
[0]
D
( N D  Reset )/0
Reset / 0
Reset
D/0
D
N
10 cent
[0]
N+D
15 cent
[1]
Moore machine
5 cent
D/1
N/0
ND
10 cent
N D /0
N+D/1
15 cent
Mealy machine
Moore and Mealy machine state diagrams for the vending machine FSM
10
—State transition table for Moore and Mealy M/C.(Next state also gives D-FF
excitation).
Present State
Q1 Q2
0 0
0 1
1 0
1 1
Inputs
Next State
D
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
Q 1+
0
0
1
x
0
1
1
x
1
1
1
x
1
1
1
x
N
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Q2+
0
1
0
x
1
0
1
x
0
1
1
x
1
1
1
x
Moore Output Mealy Output
Open
0
0
0
x
0
0
0
x
0
0
0
x
1
1
1
x
Open
0
0
0
x
0
0
1
x
0
1
1
x
1
1
1
x
Encoded vending machine state transition table.
Q+ = D
Q Q+
0 0
0 1
1 0
1 1
D
0
1
0
1
11
Implementation using D-FFs
Q1Q0
Q1Q0
00
01
11
10
00
0
0
1
1
01
0
1
1
11
x
x
10
1
1
DN
Q1Q0
00
01
11
10
00
0
1
1
0
1
01
1
0
1
x
x
11
x
x
1
1
10
0
1
K-map for D1
DN
00
01
11
10
00
0
0
1
0
1
01
0
0
1
0
x
x
11
x
x
x
x
1
1
10
0
0
1
0
DN
K-map for Open (Moore)
K-map for D0
Q1Q0
D1 = Q1 + D + Q0·N
D0  N  Q0  Q0  N  Q1  N  Q1  D
OPEN = Q1·Q0
Moore OPEN = Q1·Q0 + D·Q0 + D·Q1 + N·Q1
00
01
11
10
00
0
0
1
0
01
0
0
1
1
DN
11
x
x
x
x
10
0
1
1
1
Mealy
K-map for Open (Mealy)
12
Q1
D1
D
D
Q0
N
Q1
CLK
Q
R
Q1
Q
OPEN
Reset
N
Q0
Q0
D0
N
Q1
D
CLK
Q
R
Q
Similarly,
a Mealy
implementation;
only the OPEN
function changes.
Q0
Q0
N
Reset
Q1
D
Vending machine FSM implementation based on D flip-flops(Moore).
13
Implementation using J-K FFS
J-K Excitation
D N
Q 1+ Q2+
J1 K 1 J0 K 0
Q Q+
0 0
0
0
0 x
0 x
0 1
0
1
0 x
1 x
0 0
1 0
1
0
1 x
0 x
0 1
1 1
x
x
x x
x x
1 0
0 1
0 0
0
1
0 x
x 0
1 1
0 1
1
0
1 x
x 1
1 0
1
1
1 x
x 0
1 1
x
x
x x
x x
1 0
0 0
1
0
x 0
0 x
0 1
1
1
x 0
1 x
1 0
1
1
x 0
1 x
1 1
x
x
x x
x x
1 1
0 0
1
1
x 0
x 0
0 1
1
1
x 0
x 0
1 0
1
1
x 0
x 0
1 1
x
x
x x
x x
Remapped next-state functions for the vending machine example.
Q1 Q2
0 0
J
0
1
x
x
14
K
x
x
1
0
Q1Q0
00
01
11
10
00
0
0
x
x
01
0
1
x
11
x
x
10
1
1
DN
Q1Q0
00
01
11
10
00
x
x
0
0
x
01
x
x
0
0
x
x
11
x
x
x
x
x
x
10
x
x
0
0
DN
K-map for J1
K-map for K1
Q1Q0
Q1Q0
00
DN
00
01
0
1
01
11
10
x
x
0
x
x
1
00
01
11
10
00
x
0
0
x
01
x
1
0
x
DN
11
x
x
x
x
11
x
x
x
x
10
0
x
x
1
10
x
0
0
x
K-map for J0
K-map for K0
K-maps for J-K flip-flop implementation of vending machine.
J1 = D + Q0·N
J 0  Q0  N  Q1  D
K1 = 0
K 0  Q1  N
15
N
Q0
CLK
D
Q0
J
Q
K R
Q
Q1
Q1
OPEN
N
Q1
D
CLK
Q1
J
Q
K R
Q
Q0
Q0
N
Reset
J-K flip-flop implementation for the vending machine example (Moore).
Similarly, a Mealy implementation; only the OPEN function changes.
16
Basic Steps in the FSM Design Procedure
1. Understand the problem and the different information classes
(minimal number) required to solve it.
2. Convert these information classes into distinct states, and
determine the state transition diagram of the FSM.
3. Encode states in binary, and obtain state transition table and FF
excitation for desired FF type.
4. Minimize the FF input functions (using K-Maps, for example)
and implement the FSM using these FFs and logic gates (or MUXes)
that implement the FF’s input functions.
17
FSM Word Problem 1:
• Design a system that outputs a ‘1’ whenever it receives a multiple
of 3 # of 1’s (i.e., 0, 3, 6, 9, etc. # of 1’s) on a serial input line x.
— Relevant information classes needed to solve the problem:
(A) A multiple of 3 # is received.
(B) A non-multiple of 3 # is received.
Questions to consider:
(1) How do we go from (A)(B)
Ans.: If a ‘1’ is received
(2) How do we go from (B)(A)
Ans.: Not clear. Need to split up (B) further into
(B1): 3y+1 # of 1’s received.
Where y is an integer  0.
(B2): 3y+2 # of 1’s received.
18
Note: (A): is 3y+0 = 3y # of 1’s received.
• Now the transitions between the3 classes of information is clear:
(A)  (B1)  (B2)  (A)
1 received
1 received
1 received
• Hence these classes of information can be considered states of
the required as states of the required FSM:
These 3 states can be represented by 3y+I, i = 0,1,2
00
Reset
0/1
i=0
0
Output
Input
Reset
i=0
[1]
0/0
1/0
1/1
10
01
i=1
1/0
i=2
0/0
Mealy Machine
1
1
0
i=1
[0]
i=2
[0]
1
0
19
Moore Machine
FSM Word Problem 2:
• Design a system that outputs a ‘1’ whenever it receives:
(a) A multiple of 3 # of 1’s AND (b) A non-zero even # of 0’s
# of 1’s
E.g., ((0,2) , (3,2) , (3,4) , (6,2) ,···)
# of 0’s
— Relevant classes of information:
- For # of 1’s: 3y+i, i = 0,1,2
[3 classes]
- For # of 0’s: 2z+j, j = 0,1
For j = 0, we need to distinguish between zero (z = 0)
and non-zero (z > 0) # of 0’s
- Thus we have 3 classes:
2z+0, z = 0 ( 0 )
2z+0, z > 0 ( non-zero even )
2z+1
( odd )
20
The relevant # of 1’s can be represented by i = { 0, 1, 2 }
( # of 1’s = 3y+i )
— The relevant # of 0’s can be represented by j= { 00 , 0>0 , 1 }
( # of 0’s = 2z+j ) where the subscript of the 0 indicates whether
z=0 or z>0.
— Since at any point time, a certain # of 1’s and # of 0’s will have
been received, the state of the system will be given by
a combination of relevant # of 1’s and # of 0’s.
— There are 9 combinations:
{ 0, 1, 2, } X { 00, 0>0, 1 } = (0,00), (0,0>0), (0,1), (1,00), (1,0>0),
(1,1), (2,00), (2,0>0), (2,1) 
# of 1’s
# of 0’s
Cartesian
Product
21
(0,00)
(0,0>0)
(1,00)
(0,1)
(2,00)
(1,1)
(2,1)
(1,0>0)
(2,0>0)
22
Note: 0>0  2z+j, j = 0
z>0
Reset
(0,00)
1/0
0/0
1/0
(0,0>0)
(1,00)
1/0
0/0
0/0
0/1
1/0
(0,1)
(2,00)
0/0
(1,1)
0/0
1/0
1/0
0/0
(2,1)
(1,0>0)
1/1
0/0
1/0
0/0
(2,0>0)
1/0
23