1. Sequential Systems with Random Inputs
Examples: Lift control systems, alarm systems.
In these systems, inputs, i.e. from sensors, limit switches, pushbuttons, etc, may be completely random and can come at any time and in any order.
Huffman Method can be readily employed for the design of such systems.
In designing such systems, all possible/feasible input combinations will need to be taken into account.
Introduction

2. Example: Alarm Annunciator system
X1=1 represents a danger condition.
System specifications:
If x1 = 1, system must produce z1 = 1, which actuates an alarm siren.
The operator can shut off the siren by pressing x2. This shuts off the siren (z1 = 0), replacing it with a flashing light (z2 = 1), which must continue to flash as long as x1 = 1.
As soon as x1 = 0 again, then z2 = 0 returning to the normal state.

3. Generating the Flow Diagram
The first step in the design is to draw the flow diagram.
In the Flow Diagram, the convention is to use a circle to represent a stable state, with inputs on the left separated from outputs on the right by a slash.
Example:
X1X2/Z1Z2

4. Generating the Flow Diagram
System specifications:
If x1 = 1, system must produce z1 = 1, which actuates an alarm siren.
The operator can shut off the siren by pressing x2. This shuts off the siren (z1 = 0), replacing it with a flashing light (z2 = 1), which must continue to flash as long as x1 = 1.
As soon as x1 = 0 again, then z2 = 0 returning to the normal state.
Example
For a completely random system, if there are n inputs, there should be n arrows leaving every stable state.

5. Derive the Primitive Flow Table
From the Flow Diagram, the Primitive Flow Table is drawn.
Construct the table with columns for all possible input combinations and columns for the outputs. X 1 2
Row 00 10 11 01 Z 3 4 5 6
Example
The number of rows should correspond to the number of stable states(circles) in the Flow Diagram.

6. Derive the Primitive Flow Table
Fill in the stables states under the proper columns and rows. X 1 2
Row 00 10 11 01 Z 3 4 5 6
Example

7. Derive the Primitive Flow Table
Fill in the output states for every stable state. X 1 2
Row 00 10 11 01 Z 3 4 5 6
Example

8. Derive the Primitive Flow Table
Fill in the unstable states. X 1 2
Row 00 10 11 01 Z 5 6 3 4
Example
You will need to refer to the Flow Diagram to determine the transitions between states.

9. Derive the Primitive Flow Table
Fill out the rest of the cells with “don’t cares”. X 1 2
Row 00 10 11 01 Z - 5 6 3 4
Example

10. Derive the Merged Flow TableX 1 2
Row 00 10 11 01 Z - 5 6 3 4
Primitive
Flow Table
Example X 1 2 00 10 11 01
1,5
2,6
3,4 -
Merged
Flow table
Determine which rows can be merged.

11. Derive the Merged Flow TableX 1 2
Row 00 10 11 01 Z - 5 6 3 4
Primitive
Flow Table
Example X 1 2 00 10 11 01
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Fill out the cells accordingly.

12. Assign the states. Example Primitive Flow Table Merged Flow table1 2
Row 00 10 11 01 Z - 5 6 3 4
Primitive
Flow Table
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
For three rows, we need at least two flip-flops.
Draw in the transition arrows and assign the states.

13. Assign the states. Example Primitive Flow Table Merged Flow table1 2
Row 00 10 11 01 Z - 5 6 3 4
Primitive
Flow Table
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Row 1&2 and rows 1&3 have greater numbers of transitions. They are thus given priority. One race problem(blue arrow).

14. Assign the states. Example
Rearranging the rows for K-map representation, and taking care of the race.
Merged
Flow table X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Example

15. Draw the excitation maps.
Fill out the excitation maps. X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Example x 1 2 00 01 11 10 y S1 R1

16. Draw the excitation maps.
For stable states, flip-flops should remain unchanged. X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Example - x 1 2 00 01 11 10 y S1 R1

17. Draw the excitation maps.
For unstable states, move to the respective stable states. X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Example 1 - x 2 00 01 11 10 y S1 R1

18. Draw the excitation maps.
Don’t cares get don’t cares. X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -
Merged
Flow table
Example

19. Get the Boolean expressions for Set/Reset inputs.
Example

20. Set/Reset inputs for Y2.
Example

21. The maps for the outputs.X 1 2
Row 00 10 11 01 Z - 5 6 3 4 x 1 2 00 01 11 10 y Z1 Z2
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -

22. The maps for the outputs.
Cells representing stable states get the outputs values shown in the primitive flow table. X 1 2
Row 00 10 11 01 Z - 5 6 3 4 1 x 2 00 01 11 10 y Z1 Z2
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -

23. The maps for the outputs.
In moving from stable state m to stable state n, if the output remains the same, we enter this value in the cell for unstable state n. X 1 2
Row 00 10 11 01 Z - 5 6 3 4 1 x 2 00 01 11 10 y Z1 Z2
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -

24. The maps for the outputs.
In moving from stable state m to stable state n, if the output changes, we enter a “don’t care” in the cell for unstable state n. X 1 2
Row 00 10 11 01 Z - 5 6 3 4 y 1 - x 2 00 01 11 10 Z1 Z2
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -

25. The maps for the outputs.
Don’t’ care cells get don’t cares. X 1 2
Row 00 10 11 01 Z - 5 6 3 4 x x 1 2 - - 00 1 1 - - -
Example 01 y y 1 2 11 - - - - - - - - X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 - - 1 1 - 10

26. Derive the Boolean expressions for the outputs.1 2
Row 00 10 11 01 Z - 5 6 3 4 1 - x 2 00 01 11 10 y
Example X 1 2 00 10 11 01 Y
1,5 3 5
2,6 6
3,4 4 -

27. Review of the START Pushbutton
The START pushbutton is required only for single- cycle fixed sequences which needs to be initiated.
Example:
START A+, B+, B-, C+, C-, A-

28. Review of the START Pushbutton
For other cases, for example systems with random inputs, there is no need for a START pushbutton.
Example:
Alarm Annunciator
System

29. Incorporating the START Pushbutton
Consider the example discussed earlier...
Example
START, A+, A-, A+, A-

30. Incorporating the START Pushbutton1 2 10 00 01 11 Y1 Y2 ‚ 3 - 5 „ ƒ … † 7 ˆ ‡
The merged Flow Table obtained was...
START
And we incorporated the START pushbutton by ANDing it with the S1 signal.

31. Incorporating the START Pushbutton
But it is not always the case that the START pushbutton is ANDed with the S1 signal.
The function of incorporating the START pushbutton is to force the sequence to stop at the end of the cycle.
In the example given, it is stopped in the Unstable 1 state. Pushing the START button will allow the system to jump to Stable State 1, thereby starting another sequence.

32. Incorporating the START Pushbutton1 2 10 00 01 11 Y1 Y2 ‚ 3 - 5 „ ƒ … † 7 ˆ ‡
If the states of the flip-flops were assigned as...
Example
Then START will need to be ANDed with the R2 signal to stop the sequence in Unstable 1. Actuation of START will cause the system to jump to Stable 1 . - 1
START

33. Examples of Systems with Random Inputs

34. Example of System with Random Inputs
A machine runs as long as it receives a T=1 signal. When a fault occurs, a sensor sends an F=1 signal, which shuts off T. After the fault has been cleared (F=0), the machine can be restarted only by pressing a reset button twice in a row. This provision is intended to prevent a false restarting of the machine due to accidental touching of the reset button.
Design a logic system to fulfill the above requirements.
Example 1

35. Example of System with Random Inputs
A machine runs as long as it receives a T=1 signal. When a fault occurs, a sensor sends an F=1 signal, which shuts off T. After the fault has been cleared (F=0), the machine can be restarted only by pressing a reset button twice in a row. 1
00/ 1 2
10/ 0 N
FR/T 6
01/ 1 8
11/ 0 7
01/ 0 3
00/ 0 5
00/ 0 4
01/ 0

36. Constructing the Primitive Flow Table
Row 00 01 11 10 T 1 2 ‚ 3 ƒ 4 „ 5 … 6 † 7 ‡ 8 - ˆ
Primitive Flow Table
Construct the table and enter first the stable states and the outputs.

37. Constructing the Primitive Flow Table
Row 00 01 11 10 T 1 6 2 3 8 ‚ ƒ 4 5 „ … † 7 ‡ ˆ
Primitive Flow Table
For each stable state, determine which other stable states it moves to and fill in the unstable states.

38. Constructing the Primitive Flow Table
Row 00 01 11 10 T 1 6 - 2 3 8 ‚ ƒ 4 5 „ … † 7 ‡ ˆ
Primitive Flow Table
Fill in the don’t cares.

39. Constructing the Merged Flow Table
F R
Rows 00 01 11 10
2,3 ƒ 4 8 ‚ 5 „ - … 6 2
1,6 †
7,8 3 ‡ ˆ
Merged Flow Table F R
Row 00 01 11 10 T 1 6 - 2 3 8 ‚ ƒ 4 5 „ … † 7 ‡ ˆ
Primitive Flow Table

40. Assigning the States ƒ ‚ „ …  † ‡ ˆ
Draw a Transition Diagram indicating movement among the merged rows.
F R
Rows 00 01 11 10
2,3 ƒ 4 8 ‚ 5 „ - … 6 2
1,6 †
7,8 3 ‡ ˆ
Merged Flow Table

41. Assigning the States ƒ ‚ „ …  † ‡ ˆ
Assign the states such that races are held to a minimum.
F R
Rows 00 01 11 10
2,3 ƒ 4 8 ‚ 5 „ - … 6 2
1,6 †
7,8 3 ‡ ˆ
Merged Flow Table
000
010
100
011
110

42. Construct the full Merged Flow Table
Construct the merged flow table accordingly, indicating the races.
Merged Flow Table
F R
Rows 00 01 11 10 Y 1 2 3
2,3 ƒ 4 8 ‚
0 1 0 5 „ - … 6
1 0
1,6 †
7,8 ‡ ˆ
0 1
1 0 1
We need to find alternative routes for the three races.

43. Construct the full Merged Flow Table
Replace the races with two-step movements.
F R
Rows 00 01 11 10 Y 1 2 3
2,3 ƒ 4 8 ‚
0 1 0 5 „ - … 6
1 0
1,6 †
7,8 ‡ ˆ
0 1
1 0 1
Note that individual steps of these two-step movements must themselves not be races.
Merged Flow Table

44. Construct the K-map for the S and R inputs
For cells with stable states… and the don’t cares...
Merged Flow Table
F R
Rows 00 01 11 10 Y 1 2 3
2,3 ƒ 4 8 ‚
0 1 0 5 „ - … 6
1 0
1,6 †
7,8 ‡ ˆ
0 1
1 0 1 - S1 R1

45. Construct the K-map for the S and R inputs
For cells with unstable states…
Merged Flow Table
F R
Rows 00 01 11 10 Y 1 2 3
2,3 ƒ 4 8 ‚
0 1 0 5 „ - … 6
1 0
1,6 †
7,8 ‡ ˆ
0 1
1 0 1 1 - - 1 S1 R1

46. And the others...
Similarly the K-maps for the S and R inputs for the flip-flops Y2 and Y3 are derived.
The K-map for the output T is then derived using both the merged flow table and the primitive flow table.

47. Systems with Inputs which are not completely randomX1 X2
Parts with a rectangular cross-section move on a conveyor belt. As shown above, two photoelectric sensor, x1 and x2, detect whether a part passing through is “standing up” or “lying down”. The sensor output is a logic 1 if a part is detected. The parts are required to be in the sequence (DOWN, DOWN, UP) with this sequence repeated continuously. Any incoming part which breaks this required sequence will cause an output signal T=1 is to be produced. This is used to actuate a cylinder which pushes these parts off the conveyor belt.
Example 2

48. Systems with Inputs which are not completely random
Note that the inputs X1 and X2 are not completely randomly actuated. As a “lying down” part and a “standing up” part passes through, they change as shown.
Example 2
When a “down” part passes through, X1X2 go through 00, 01, 00.
When an “up” part passes through, X1X2 go through 00, 01, 11, 10, 00.

49. When a “down” part passes through, X1X2 go through 00, 01, 00.
When an “up” part passes through, X1X2 go through 00, 01, 11, 10, 00. 10
10/1 9
11/0 12
10/1 11
11/0 1
00/0 2
01/0 3
00/0 4
01/0 8
10/0 7
11/0 6
01/0 5
00/0 13
00/1

50. End of Random Inputs