sequential circuits - kth · 2017-03-01 · sequential circuits william sandqvist william@kth.se if...

Sequential circuits

William Sandqvist william@kth.se

If the same input may produce

different output signal, we

have a sequential logic circuit.

It must then have an internal

memory that allows the

output to be affected by both

the current and previous

inputs!Logic circuit

Same input can produce

different output

Moore-machine

NEXT STATEDECODER STATE REGISTER

OUTPUTDECODER

Input-signals

Output-signals

For Moore machine the outputs depends on the inputs and

the internal state. The internal memory is the state register

consisting of D flip-flops.

State register D-flip-flops

State register D flip-flops slows down the race between signals

until the value is stable. (Compare with the tollbooth).

NEXT STATEDECODER

STATEREGISTER

OUTPUTDECODER

Output-signals

Input-signals

NEXT STATEDECODER

STATEREGISTER

OUTPUTDECODER

Output-signals

Input-signals

NEXT STATEDECODER

STATEREGISTER

OUTPUTDECODER

Output-signals

Input-signals

Quickie Question … flip-flopWhich of the following timing diagram is valid for a

edge-triggered D flip-flop?

Alt: A Alt: B Alt: C

Quickie Question … flip-flop

Alt: B Alt: C

Alt: A

D is copied to output at the edge, when

Clock goes from 0 to 1

Which of the following timing diagram is valid for a

edge-triggered D flip-flop?

Quickie Question … Latch

Which of the following timing diagram is valid

for a D Latch?

Enable

D is connected to output when Enable is 1,

and is locked when Enable is 0

Quickie Question … Latch

Which of the following timing diagram is valid

for a D Latch?

Design: ”two consecutive”

Sequence Detector. If w has been 1 in two

(or more) consecutive clock then z = 1.w w z

Specification

Sequence circuit has an input w and an output z

If input w has been 1 in during the current and

previous clock cycle then z will be set to 1

Use a Moore machine with D-flip-flops to realizing

the design.

(Reset)

Statediagram ”two consecutive”

C z 1 =

B z 0 = A z 0 = w 0 =

w 0 = w 1 =

State transition

State name

Output value

A zero

B single one

C two or moore

consecutive

State table

Present Next state Outputstate w = 0 w = 1 z

A A B 0

B A C 0

C A C 1

Three states – two flip-flops needed to

hold state numbers!

”two consecutive” as Moore-

machine

Combinationalcircuit

Designdecissions

The designer must decide which flip-flops?

to be used

D-, T-, or JK-flip-flop

The designer must choose the code for each

Design decision

This time:

D-flip-flop

State code A = 00, B = 01, C = 10

The code word 11 should not occur.

We choose don’t care.

Coded state table

Present Next state

state w = 0 w = 1 Output

A 00 00 01 0

B 01 00 10 0

C 10 00 10 1

11 dd dd d

12YY12 yy12YY

2 1 2 1 2 1( ) ( )Y Y f y y w z f y y

A = 00

B = 01

C = 10

Next state decoder

)()()(

1211221212

wyyfYwyyfYwyyfYY

The Next state decoder consists of two logic networks

available as input network to the two flip-flops.

In order to minimize logic networks, we enter the truth

tables in the form of Karnaugh maps.

From coded statetable to

Karnaughmap

w 00 01 11 10

Y1 wy 1y 2

w 00 01 11 10

)()()( 1211221212 wyyfYwyyfYwyyfYY

Output decoder

)( 12 yyfz

Implementation

Y1 wy 1y 2

Timing diagram ”two consecutive”

t 0 t 1 t 2 t 3 t 4 t 5 t 6 t 7 t 8 t 9 t 10

transitions

occurs only on

the positive

clock edge!

In other terms

)()()( 1211221212 wyyfywyyfywyyfyy

Exercises and Hemert-book:

yyPresent state Next state

With another lineup

)()()( 1211221212 wyyfywyyfywyyfyy

You could directly write down the codet state table as a

”Karnaugh map”.

In exercises

we use this

method

Mealy-machine

NEXT STATEDECODER STATE REGISTER

OUTPUTDECODER

Input-signals

Output-signals

In a Mealy Machine output signals depends on

both the current state and the inputs.

Moore machine

Present Next state Outputstate w = 0 w = 1 z

A A B 0

B A C 0

C A C 1

Mealy machine

Present Next state Output

state w = 0 w = 1

A A B 0

B A C 0

C A C 1

w = 0 w = 1

Two states that differs only in output can be merged

– input can be used to distinguish between outputs!

Mealy machine

state w = 0 w = 1

A A BC 0

BC A BC 0

w = 0 w = 1

w 0 = z 0 =

w 1 = z 1 = BC w 0 = z 0 =

w 1 = z 0 =

State diagram Mealy

w 0 = z 0 =

w 1 = z 1 = B w 0 = z 0 =

w 1 = z 0 =

Value of output signalState name

Value of input signal

”Two consecutive”

The state diagram for the Mealy-machine only

needs two states

Output signal depends booth on states and input.

State table

Present Next state Output z

state w = 0 w = 1 w = 0 w = 1

A A B 0 0

B A B 0 1

w 0 = z 0 =

w 1 = z 1 = B w 0 = z 0 =

w 1 = z 0 =

Two states – only one flip-flop is needed!

Coded state table

state w = 0 w = 1 w = 0 w = 1

y Y Y z z

A 0 0 1 0 0

B 1 0 1 0 1

)()( wyfzwyfY

ywzwY Directly from the table:

Implementation

Resetn

Timing diagram

t 0 t 1 t 2 t 3 t 4 t 5 t 6 t 7 t 8 t 9 t 10

The output may change during the clock period, since it

is a function of the input signal

Compared to Moore machine the Mealy machine is

moore 'responsive' (bit sequence is detected in t4

compared to t5 for the Moore-machine)

Mealy with output register

Resetn

The disadvantage of the Mealy machine is that the

output can be changed during the entire clock period

You can add a register (flip-flop) at the end so to

synchronize the output with the clock edge

Timing diagram with output register

t 0 t 1 t 2 t 3 t 4 t 5 t 6 t 7 t 8 t 9 t 10

With an output register there are no longer

any differences between the timing diagrams!

Moore vs Mealy

Moore-machine output values depend only

on the current state

Mealy-machine output values depend on the

current state and the values of the input signals

Mealy-machine often uses fewer states

Mealy-machine output signals are not inte

synchronized with the clock, why you often

has to add an output register

Selection of state encoding

Selection of state encoding can play a

major role in implementation as it affects

the logic for

Next-state-decoder

Output decoder

”two consecutive” state diagram

C z 1 =

B z 0 = A z 0 = w 0 =

w 0 = w 1 =

State code = Binary

Present Next state

y 2 y 1 Y 2 Y 1 Y 2 Y 1 z

A 00 00 01 0

B 01 00 10 0

C 10 00 10 1

11 dd dd d

A = 00

B = 01

C = 10

Realization (Binary code)

2 D-flip-flops

2 AND-gates

1 OR-gate

State code = Gray code

Present Next state

y 2 y 1 Y 2 Y 1 Y 2 Y 1

A 00 00 01 0

B 01 00 11 0

C 11 00 11 1

10 dd dd d

In the Gray-code only one bit at a time is changed, eg. 00,

01, 11, 10

Gray-code is good for counters

A = 00

B = 01

C = 11

Realization (Gray code)

Resetn

2 D-flip-flops

1 AND-gate

As the Gray code did

well in this case this

sequential circuit has

obviously has

“counting properties”

One-Hot-encoding

One-hot-encoding uses one flop-flop

per state

For each state one bit is ‘hot’ (1), all

other bits are 0,

eg. 0001, 0010, 0100, 1000

One-hot encoding minimizes the

combinatorial logic, but increases the

number of flip-flops!

What code should be chosen?

There is not a code that is the best in every

situation, it all depends on the state diagram

You can also have your 'own codes' that fits

into the design, eg. 00, 11, 10, 01

Example. - wait - cw - wait - ccw -

qZwait

With this state encoding

(Graycode) we need no output

decoder!

Can you can set up the encoded state table and

Karnaugh maps by your self … ?

211212 qiqiqqiqiq

- wait - cw - wait - ccw -

211212 qiqiqqiqiq

With NAND-gates

sequential circuits - kth · 2017-03-01 · sequential circuits william sandqvist william@kth.se if...

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