itacr_e311 multiprog exercises and solutions festo plc double side

ISaGRAF

Learning system for automation and communications

Programmable Logic Controller

Fundamentals and Applications of

Programmable Logic Controllers

Exercises

Solutions

Exercise 1

The YES Function

Problem Description

Actuation of a push button (S1) is to cause a lamp (H1) to be switched on. The lamp is to be illuminated as long as the push button is actuated.

Draw the circuit diagram and assemble the equipment.

Formulate the PLC program using MultiProg.

You can use any one or all of the five languages available.

Electrical Circuit Diagram

Solution 1


Variables

For Festo PLC

NameAttribTypeComment

Input0_0[Input]BOOLPush Button S1

Output0_0[Output]BOOLLamp H1

Ladder Programming

IL Programming

ST Programming

FBD Programming

Exercise 2

The NOT Function

Problem Description

A thin wire has been stretched behind a display window, which breaks if a burglary is attempted. A close circuit is interrupted as a result of this and a buzzer is sounded. A normally closed switch is used to simulate the unbroken wire.





Solution 2


Variables

For Festo PLC


Input0_0[Input]BOOLSensor S1

Output0_0[Output]BOOLSiren H1

Ladder Programming

IL Programming

ST Programming

FBD Programming

OR

Exercise 3

The AND Function

Problem Description

A press stamp 1.0 is to advance only if a push button is actuated and a protective guard is closed. If one of these conditions is not met, the press tool is to return immediately.

The position of the closed protective guard B1 is monitored by a proximity switch B1. The press tool is advanced or retracted by means of a spring return solenoid valve (Coil Y1).

Draw the electrical circuit diagram and Electro-pneumatic circuit diagram.

Assemble the equipment.

Formulate the PLC program using MultiProg


Electro-Pneumatic Circuit Diagram


Solution 3



Variables

For Festo PLC


Input0_0[Input]BOOLStart Switch S1

Input0_1[Input]BOOLSafety Sensor B1

Output0_0[Output]BOOLCylinder Extends Y1

Ladder Programming

IL Programming

ST Programming

FBD Programming

Exercise 4

The OR Function

Problem Description

An apartment bell is to ring if bell button S1 at the front door is pressed or bell button S2 at the apartment door.





Solution 4


Variables

For Festo PLC




Output0_0[Output]BOOLBell H1

Ladder Programming

IL Programming

ST Programming

FBD Programming for Festo PLC

Exercise 5

Combination of AND/OR/NOT Functions

Problem Description

A stamping device can be operated from three sides. A workpiece is inserted via a guide, whereby it touches two of the three proximity switches B1, B2 and B3. This causes a pneumatic cylinder 1.0 to extend via a solenoid valve (Coil Y1), whereby a recess is to be stamped into the workpiece.

The stamping cycle is to be triggered only if two signal generators are Typeesses. For reasons of safety the cylinder must be prevented from advancing, if all three proximity sensors are contacted.







Solution 5



Variables

For Festo PLC


Input0_0[Input]BOOLProximity Sensor B1



Flag1[Internal]BOOLInternal Relay 1



Output0_0[Output]BOOLCylinder Extends Y1

Ladder Programming

Instruction List Programming

Structured Text Programming

OR

FBD Programming

Exercise 6

Logic Control System with Branching.

Problem Description

The control of a bulk material mixing system consists of a push button (S1) and a selector switch (S2).

Once the appropriate silo has been selected using the selector switch (S2), pressing the push button (S1) opens the silo. Use single-solenoid 5/2-way directional control valves to control both cylinders. The cylinders are extended in the initial position.







Solution 6



Variables

For Festo PLC


Input0_0[Input]BOOLStart Button S1

Input0_1[Input]BOOLSelector Switch S2

Output0_0[Output]BOOLCylinder A Retracts Y1

Output0_1[Output]BOOLCylinder B Retracts Y2

Ladder Programming

IL Programming

ST Programming

FBD Programming

Exercise 7

Setting and Resetting of an Output

Technical Knowledge

Storage operations form part of the elementary PLC operations and apply in cases where a briefly occurring signal status is permanently stored. Typical examples of storage operations are the permanent setting or resetting of output signals. The standard function blocks SR and RS are available for the realisation of storage functions.

Function block SR, dominant setting flip-flop

Function block SR contains a dominant setting flip-flop.

The typical behaviour of the SR function block is therefore as follows:

A 1-signal at set input S1 sets the flip-flop, i.e. the value of Q1 becomes 1.

A 1 signal at reset input R sets the value of Q1 at 0 only if a 0-signal simultaneously applies at the S1 input. If a 1-signal applies both at the S1 and the R input output Q1 is set.Function block RS, dominant resetting flip-flop

Function block RS contains a dominant resetting flip-flop.

The behaviour of the block represented is as follows:

A 1-signal at reset input R1 sets the value of Q1 to 0, irrespective of which value applies at input S.

A 1-signal at set input S sets output Q1 to 1 only if a 0-signal simultaneously applies at the R1 input.

If 1-signals apply both at inputs S and R1, output Q1 is reset.

Problem Description

The drill on a drilling unit is monitored by means of a drill breakage sensor (B1).

If the drill is broken, the sensor interrupts the circuit. A buzzer (H1) is to sound in this event. The buzzer can only be switched off via push button S1. Draw the electrical circuit diagram.





Solution 7


Variables

For Festo PLC



Input0_1[Input]BOOLSensor B1

Output0_1[Output]BOOLBuzzer H1

NameFunction or Function BlockComment

RS_1RSRS Function Block 1

Ladder Diagram

IL Programming

OR

ST Programming

Remember to declare RS_1 in the Variables under Function Block.

FBD Programming

Using FBD in Ladder Diagrams

Exercise 8

Transfer of a Part

Problem Description

Using a double-acting cylinder, a part is to be moved from one roller track to another. The operation is to be started using a push button. An acknowledgement signal is required when the piston is fully extended to ensure that the part has been transferred. The return stroke is to be automatic. Use a double solenoid valve.







Solution 8



Variables

For Festo PLC



Input0_1[Input]BOOLCylinder A Back B1

Input0_2[Input]BOOLCylinder A Forward B2

Output0_0[Output]BOOLCylinder A Extends Y1

Output0_1[Output]BOOLCylinder A Retracts Y2

Ladder Diagram

IL Programming

OR

ST Programming

FBD Programming

OR

Exercise 9

Activating a Cylinder

Signal Edges

Technical Knowledge

PLC applications frequently require the detection and evaluation not of signal itself, but of the point of change of a signal. These signal changes are described as edges.

Rising (positive) edges mark the instant, during which a signal change takes place from 0 to 1.

Falling (negative) edges mark the instant, during which a signal change takes place from 1 to 0.

Function block R_TRIG for rising edge detection

This standard function block is used for the detection of a rising edge.

If a signal change takes place from 0 to 1 at input CLK, output Q assumes the value 1 during a program cycle.

Function block F_TRIG for falling edge detection

This standard function block is used for the detection of a falling edge.

If a signal change takes place from 1 to 0 at input CLK, output Q carries a 1-signal during a program cycle.Problem Description

A cylinder is actuated by means of a spring-returned solenoid valve (coil Y1). Two proximity switches signal the positions "extended" (B2) and "retracted" (B1). Push button (S1) is used to actuate the cylinder in such a way that it advances from the retracted end position into the opposite direction. The cylinder must advance only once per push button actuation. To trigger a second movement of the cylinder, the push button must be released and actuated afresh. Draw the electrical circuit diagram and Electro-pneumatic circuit diagram.






Solution 9



Variables

For Festo PLC






Output0_0[Output]BOOLCylinder Solenoid Y1


R_TRIG_1R_TRIGRising Edge 1

RS_1RSReset-Set 1

Ladder Diagram

IL Programming for Festo PLC

ST Programming

FBD Programming

Exercise 10

Bonding of Components

Pulse Timer

Technical Knowledge

A large number of control tasks require the programming of time. Through IEC61131-3, standard function blocks are available for timers with different time response.

Timers are available for the realisation of a pulse time response, a switch-on signal delay and a switch-off signal delay.

Function block TP, pulse timer

Standard function block TP is a pulse timer

The response of function block TP is as follows:

Function block TP is started via a short or long signal at input IN.

Once the timer has started, a 1-signal applies at output Q for the time specified at input PT.

The current timer value (the time, which has elapsed since the start) is available at output ET.

The timer can only be started again once it has expired.

Problem Description

Two components are to be bonded together with the help of a pneumatic cylinder 1.0. To do this, the bonding surfaces are pressed together with a defined force for 5 seconds. The time is commenced once the cylinder advances from its retracted end position (sensor B1).

The cylinder is controlled by single solenoid directional control valves (Y1). Once the 5 seconds have expired, the cylinder is to return to the initial position. The bonding process is started by a push button S1.







Solution 10



Variables

For Festo PLC






TP_1TPTimer Preset 1

Ladder Diagram

IL Programming

ST Programming


Exercise 11

Embossing Device

Switch-on Signal Delay

Technical Knowledge

Function block TON, Switch-on signal delay

The standard function block TON is used to generate a switch-on signal delay.

The behaviour of function block TON is as follows:

Function block TON is started by means of a 1-signal at input IN.

Upon expiry of the time specified at input PT, output Q carries a 1-signal. The 1-signal at output Q applies until the input signal IN reverts to the value 0.

If the duration of the input signal IN is shorter than the specified time PT, the value of output Q remains a constant 0.

The current timer value (the time, which has elapsed since the start) is available at output ET.

Problem Description

A workpiece is to be embossed by activating a start button (S1). In order to ensure that the embossing cycle is not triggered inadvertently, the embossing cycle is to be triggered only after 3 seconds have expired. During this time the start button must be permanently actuated. The position of the cylinder 1.0 is established by means of the proximity switches B1 (retracted) and B2 (extended). The cylinder is controlled by single solenoid directional control valves (Y1). Draw the electrical circuit diagram and Electro-pneumatic circuit diagram.






Solution 11



Variables

For Festo PLC







TON_1TONTimer Delay On 1

RS_1RSReset-Set 1

Ladder Diagram

IL Programming

ST Programming


Exercise 12

Clamping Device

Switch-off Signal Delay

Technical Knowledge

Function block TOF, switch-off signal delay

The standard function block TOF is used to generate a switch-off signal delay.

The behaviour response of function block TOF is as follows:

Function block TOF is started via a 1-signal at input IN. Output Q simultaneously receives the value 1.

After the input signal IN has reverted to the value 0, the 1 signal continues to be applied at output Q for the time specified at the PT input and then returns to the value 0.

Problem Description

A workpiece is to be clamped by activating the start button S1. When the workpiece is clamped by cylinder 1.0, cylinder 2.0 extends and embosses the workpiece. Since the workpiece requires time to cool down, it remains clamped for a period of 3 seconds. This time is started with the advancing of cylinder 1.0.

Use single solenoid DCV for cylinder 1.0 and double solenoid DCV for cylinder 2.0.







Solution 12



Variables

For Festo PLC





Input0_3[Input]BOOLCylinder B Back B3

Input0_4[Input]BOOLCylinder B Forward B4


Output0_1[Output]BOOLCylinder B Extends Y2

Output0_2[Output]BOOLCylinder B Retracts Y3


TOF_1TOFTimer Delay Off 1


RS_1RSReset-Set 1

RS_2RSReset-Set 2

Ladder Diagram

IL Programming

ST Programming

FBD Programming

Exercise 13

Bending Device with Counter

Counters

Technical Background

Function block CTU, incremental counter

Function block CTU realises an incremental counter. Its interface is defined by means of three input and two output parameters.

The characteristic behaviour of an incremental counter is as follows:

A 1-signal at reset input R sets the current counter status CV at 0.

Thereafter, the value CV of the counter is increased by 1 with each rising edge at input CU.

A 1-signal applies at output Q as soon as the current value CV is equal or greater than the preselect value PV. Output Q carries a 0-signal as long as current counter status CV is less than the preselect value PV.

Function block CTD, Decremental counter

Function block CTD being a decremental counter is counterpart to function block CTU.

The behaviour of a decremental counter is as follows:

A 1-signal at the LD input sets the current counter status CV equal to the preselect value PV.

Each rising edge at the CD input decreases the current counter status CV by 1.

Output Q carries a 0-signal as long as the current counter status CV is greater than 0. Only if the current value is less or equal to 0, does a 1-signal apply at output Q.

Problem Description

A double-acting cylinder is used for a bending process and is to operate for 5 continuous cycle and then stop. The cycle is started by a push button. The cylinder is controlled via single solenoid valves. Draw the electrical circuit diagram and Electro-pneumatic circuit diagram.



Use the Ladder or FBD language.



Solution 13



Variables

For Festo PLC









CTU_1CTUCounter Up 1



RS_1RSReset-Set 1

RS_2RSReset-Set 2

Ladder Diagram

IL Programming

ST Programming

Function Block Diagram

Exercise 14

Lifting Device for Packages

Linear Sequence

Technical Knowledge

Sequence control systems

Sequence control systems describe processes, which proceed in several clearly separated steps.

The transition from one step to the next is dependent on the process statuses. It is possible for the process to be branched into partial processes in relation to the process statuses established.

The program of a sequence control system must therefore fulfill three basic exercises:

Steps of a sequential function chart

Example of double divergence and convergence.

Note that if it is an OR logic, the line would be a single line.

Problem Description

A roller conveyor is monitored by proximity switch B1 as to whether a package is present. If a package is present, the package is picked up by a Cylinder 1.0 (Lifting cylinder) and then transferred to another conveyor by means of Cylinder 2.0 (Transfer cylinder).

Cylinder 1.0 is to retract first, followed by Cylinder 2.0. The cylinders are retracted and advanced by means of solenoid valves (Coil Y1 and Y2). The cylinder positions are monitored by means of proximity switches B2 to B5.

On the feed side, packages have been arranged in such a way that they reach the lifting devices individually.




You can use all the programming languages but should try the Sequential Function Chart.



Solution 14



Variables

For Festo PLC







Output0_0[Output]BOOLCylinder A Solenoid Y1

Output0_1[Output]BOOLCylinder B Solenoid Y2







SR_1SRSet-Reset 1

SR_2SRSet-Reset 2

SR_3SRSet-Reset 3

SR_4SRSet-Reset 4

SR_5SRSet-Reset 5

SR_6SRSet-Reset 6

SR_7SRSet-Reset 7

Ladder Diagram

Ladder with Function Blocks

Structured Text

Function Block Diagram

Instruction List

Sequential Function Chart for Festo PLC

Sequential Function Chart

Exercise 15

Distributing Device for Boxes

Problem Description

Metal strips are placed manually in a fixture. Cylinder 1.0 clamps the metal strip when a push button is pressed. Cylinder 2.0 starts the bend and retracts. Cylinder 3.0 completes the bending operation. After Cylinder 3.0 has returned to its initial position, Cylinder 1.0 releases the part.

Sequence abbreviation: A+ B+ B- C+ C- A-

Reed switches sense cylinder positions.




You should use either the Ladder Diagram or the Sequential Function Chart.



Solution 15

For Festo PLC







Input0_5[Input]BOOLCylinder C Back B6

Input0_6[Input]BOOLCylinder C Forward B7

Output0_0[Output]BOOLCylinder A Solenoid Y1

Output0_1[Output]BOOLCylinder B Solenoid Y2

Output0_2[Output]BOOLCylinder C Solenoid Y3








Ladder Diagram

Sequential Function Chart

IN

Input0_1

Input0_0

TP

Q1

RESET1

SET

RS

RS_1

Q

CLK

R_TRIG

AND

R_TRIG_1

AND

Input0_2

Input0_0

Input0_1

Output0_0

R_TRIG_1(CLK:=Input0_0);

RS_1(SET:=R_TRIG1_1.Q AND Input0_1,RESET1:=R_TRIG_1.Q AND Input0_2);

Output0_0:= RS_1.Q1;

LD Input0_0

ST R_TRIG_1.CLK

CAL R_TRIG_1

LD R_TRIG_1.Q

AND Input0_1

S Output0_0

LD R_TRIG_1.Q

AND Input0_2

R Output0_0

R

Output0_0

Input0_2

CLK

Output0_0

Input0_1

Q

S

Input0_0

Input0_2

Output0_1

Input0_1

Output0_0

Input0_0

AND

Output0_0:= Input0_0 AND Input0_1;

Output0_1:= Input0_2;

LD Input0_0

AND Input0_1

S Output0_0

R Output0_1

LD Input0_2

S Output0_1

R Output0_0

LD Input0_0

AND Input0_1

ST Output0_0

LD Input0_2

ST Output0_1

R

R

Output0_0

Output0_1

Output0_1

Output0_0

Input0_2

Input0_1

S

S

Input0_0

Output0_0

Q1

RESET1

SET

Input0_0

Input0_1

RS

Q1

RESET1

SET

Input0_0

Input0_1

RS

Output0_0

RS_1(SET:=Not Input0_1, RESET1:= Input0_0);


LDN Input0_1

S Output0_0

LD Input0_0

R Output0_0

R

Output0_0

Input0_0

S

Input0_1

Output0_0

Output0_0

Input0_1

Input0_0

AND

Input0_1

Input0_0

AND

Output0_1

Output0_0:= Input0_0 AND NOT Input0_1;


LD Input0_0

ANDN Input0_1

ST Output0_0

LD Input0_0

AND Input0_1

ST Output0_1

Output0_1

Output0_0

Input0_1

Input0_1

Input0_0

Input0_0

AND

AND

AND

OR

Input0_1

Input0_0

Input0_2

Input0_1

Input0_0

Output0_0

Input0_2

Input0_1

Input0_0

LDN Input0_1

ST RS_1.SET

LD Input0_0

ST RS_1.RESET1

CAL RS_1

LD RS_1.Q1

ST Output0_0

Flag1:= Input0_0 AND Input0_1 AND NOT Input0_2;

Flag2:=Input0_0 AND Input0_2 AND NOT Input0_1;

Flag3:=Input0_1 AND Input0_2 AND NOT Input0_0;

Output0_0:=Flag1 OR Flag2 OR Flag3;

LD Input0_0

AND Input0_1

ANDN Input0_2

ST Flag1

LD Input0_0

AND Input0_2

ANDN Input0_1

ST Flag2

LD Input0_1

AND Input0_2

ANDN Input0_0

ST Flag3

LD Flag1

OR Flag2

OR Flag3

ST Output0_0

Output0_0

Input0_2

Input0_2

Input0_1

Input0_1

Input0_2

Input0_1

Input0_0

Input0_0

Input0_0

Input0_1

OR

Input0_0

Output0_0

Output0_0:= Input0_0 OR Input0_1;

LD Input0_0

OR Input0_1

ST Output0_0

Input0_1

Output0_0

Input0_0

LD Input0_1

OR Input0_2

ST Outuput0

Input0_1

AND

Input0_0

Output0_0


LD Input0_0

AND Input0_1

ST Output0_0

Output0_0

Input0_1

Input0_0

Flag5

Flag3

SR_5

Output0_0:= NOT Input0_0;

LDN Input0_0

ST Output0_0

Output0_0

Input0_0

H1

S1

EMBED Word.Picture.8

0V

0V

24V

0.3

0.2

0.1

0.0

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

0V

24V

PLC

O

I

SET1

Output0_0:=Input0_0;

LD Input0_0

ST Output0_0

Output0_0

Input0_0

H1

S1

Output0_0

Output0_0

0.6

0.5

TON_1(IN:=Input0_0 AND Input0_1,PT:= T#3s);

RS_1(SET:=TON_1.Q,RESET1:= Input0_2);


LD Input0_0

AND Input0_1

ST TON_1.IN

LD T#3s

ST TON_1.PT

CAL TON_1

LD TON_1.Q

S Output0_0

LD Input0_2

R Output0_0

R

Output0_0

Input0_2

S

T#3s

ET

Output0_0

Q

PT

IN

Input0_1

Input0_0

TON

ET

T#5s

Q

PT

IN

TP

AND

Input0_1

Input0_0

Output0_0

TP_1(IN:=Input0_0 AND Input0_1,PT:= T#5s);

Output0_0:= TP_1.Q;

LD Input0_0

AND Input0_1

ST TP_1.IN

LD T#5s

ST TP_1.PT

CAL TP_1

LD TP_1.Q

ST Output0_0

T#5s

ET

Output0_0

Q

PT


0.4

0.3

0.2

0.1

0.0

0V

24V

PLC

0V

0.7

O

0.0

I

0.1

0V

24V

0.3

0.2

Input0_1

AND

TON

IN

PT

Q

T#3s

ET

RS

SET

RESET1

Q1

Input0_2

TOF

Input0_0

Input0_1

IN

PT

Q

Output0_0

ET

T#3s

Input0_2

Output0_1

S

Input0_3

R_TRIG_1

Input0_3

Input0_4

Output0_1

R

Output0_2

S

Input0_3

Output0_2

R

LD Input0_0

AND Input0_1

AND Input0_3

ST TOF_1.IN

LD T#3s

ST TOF_1.PT

CAL TOF_1

LD TOF_1.Q

ST Output0_0

LD Input0_2

ST R_Trig_1.CLK

CAL R_Trig_1

LD R_Trig_1.Q

AND Input0_3

S Output0_1

LD Input0_4

R Output0_1

S Output0_2

LD Input0_3

R Output0_2

TOF_1 (IN:=Input0_0 AND Input0_1 AND Input0_3,PT:= T#3s);

Output0_0:= TOF_1.Q;

R_Trig_1 (Input0_2);

RS_1 (SET:=R_Trig_1.Q AND Input0_3,RESET1:= Input0_4);


RS_2 (SET:=Input0_4,RESET1:= Input0_3);


CTU_1

Input0_0

Input0_1

AND

TOF

IN

PT

Q

T#3s

ET

Input0_3

Output0_1

Input0_2

Input0_3

AND

R_TRIG

CLK

Q

RS

SET

RESET1

Q1

Input0_4

Output0_2

Input0_0

Input0_1

RS

SET

RESET1

Q1

Input0_2

Output0_0

CTU

Input0_2

CU

PV

Q

Flag2

CV

Flag2

Input0_1

Flag1

Output0_0

Input0_0

Flag1

Flag2

Flag1

5

RESET

S

S

R

R

Output0_0

R_TRIG_1

Flag2

Input0_1

Input0_1

Flag1

AND

R_TRIG

CLK

Q

RS

SET

RESET1

Q1

Input0_2

Flag2

Input0_2

Flag2

CTU

CU

PV

Q

RS

SET

RESET1

Q1

Q

CLK

R_TRIG

5

RESET

CV

Input0_0

Flag5

Flag1

Flag1

Input0_2

Flag1

Flag2

Flag2

Input0_4

Flag2

Flag3

Flag3

Input0_1

Output0_0

R

S007

Input0_5

Flag2

Flag1

Input0_3

Input0_3

Output0_0

Input0_0

Input0_2

Input0_0

CTU_1

R_TRIG_2

Input0_0

RS_1

Output0_1

R

S005

Output0_2

R

S006

Input0_6

Output0_2

S

S005

Input0_3

Input0_1

Output0_0

R

S004

Input0_4

Output0_1

S

S003

Input0_2

Flag1

RS_2

TP_1

TP_1

RS-1

RS_1

TON_1

TON_1

RS_1

TOF_1

R_TRIG_1

TOF_1

RS_1

RS_2

LD Input0_0

S Flag1

LD Input0_1

AND Flag1

S Output0_0

LD Input0_2

ST R_TRIG_1.CLK

CAL R_TRIG_1

LD R_TRIG_1.Q

R Output0_0

LD Input0_2

ST R_TRIG_2.CLK

CAL R_TRIG_2

LD R_TRIG_2.Q

ST CTU_1.CU

LD Flag2

ST CTU_1.Reset

LD 5

ST CTU_1.PV

CAL CTU_1

LD CTU_1.Q

ST Flag2

LD Flag2

R Flag1

RS_1 (SET:=Input0_0,RESET1:= Flag2);

RS_2 (SET:=Input0_1 AND Flag1,RESET1:=R_TRIG_1.Q);


CTU_1 (CU:=R_TRIG_2.Q,RESET:= Flag2,PV:= 5);


Flag1:= RS_1.Q1;


Flag2:= CTU_1.Q;

Output0_1

Flag4

Flag2

Output0_0

Flag3

Flag1

Flag5

Flag4

Input0_1

Flag4

Flag4

Flag3

Input0_3

Output0_1

R

S004

Input0_4

Output0_1

Output0_1

2

R

1

Action Step

3

Initial Step

2

Action Step

1

Output0_0

S

S002

Output0_0

R

Input0_0

S001

S

S003

Input0_2

Output0_2

R

Output0_1

R

Output0_0

S

S002

Flag2

Flag2

Output0_0

R

Input0_0

S001

Flag1

Input0_2

Flag1

Flag1

Flag7

Input0_0

Transition Step

2

4

Action Step

Transition Step

3

Transition Step

4

2

101

End of Process 2

Process2

1

true

Wait for Process 2

End of Process 1

Process1

Run

3

2

1

101

3

2

1

Transition Step

Initialize

Wait for Process 2

102

Input0_4

Flag2

Flag3

Flag3

Input0_3

Flag3

Flag4

Flag4

Input0_6

Flag4

Flag5

Flag5

Input0_5

Flag5

Flag6

Flag6

Input0_1

Flag6

Flag7

Flag1

Flag6

Output0_0

Flag2

Flag3

Output0_1

Flag4

Flag5

Output0_2

RESET

Flag1

Input0_1

Flag4

Q1

SR

AND

Flag4

Flag2

SR_4

SET1

RESET

Flag5

Input0_4

Flag3

Q1

SR

AND

Flag1

Flag3

SR_3

SET1

RESET

Flag4

Input0_2

Flag2

SR_1 (SET1:=Input0_1 AND Input0_3, RESET:=Flag2);

Flag1:=SR_1.Q1;

SR_2 (SET1:=Input0_0 AND Flag1, RESET:=Flag3);

Flag2:=SR_2.Q1;

SR_3 (SET1:=Input0_2 AND Flag2 AND NOT Flag1, RESET:=Flag4);

Flag3:=SR_3.Q1;


Flag4:=SR_4.Q1;


Flag5:=SR_5.Q1;

SR_6 (SET1:=Flag2, RESET:=Flag4);

Output0_0:=SR_6.Q1;

SR_7 (SET1:=Flag3, RESET:=Flag5);

Output0_1:=SR_7.Q1;

Q1

SR

AND

Flag2

SR_7

Q1

RESET

SET1

RS

Flag5

Output0_1

SR_2

Flag3

SR_6

Q1

RESET

Input0_1

SR_5

Q1

RESET

SET1

RS

Flag1

Flag5

Flag4

Flag3

Input0_4

SR_4

Q1

RESET

SET1

RS

Flag5

Flag4

Flag3

Flag2

Input0_2

SR_3

Q1

RESET

SR_1

Q1

RESET

SET1

RS

SET1

RS

Flag4

Output0_0

SET1

Flag2

SET1

RS

Flag4

Flag3

Flag2

Flag1

SR_2

Q1

RESET

SET1

RS

Flag3

Flag2

Input0_0

Flag1

RESET

Flag3

Input0_0

Input0_0

Output0_0

Input0_0

Flag1

Q1

SR

AND

Flag1

SR_1

SET1

Output0_0:= (Input0_0 AND Input0_1 AND NOT Input0_2)

OR (Input0_0 AND Input0_2 AND NOT Input0_1)

OR(Input0_1 AND Input0_2 AND NOT Input0_0);

RESET

NOT

Output0_0


Flag2

Input0_3

Input0_1

Q1

SR

AND

Output0_1

SR_7

SET1

RESET

Flag5

Flag3

LD Input0_2

ANDN Flag1

AND Flag2

ST SR_3.SET1

LD Flag4

ST SR_3.RESET

CAL SR_3

LD SR_3.Q1

ST Flag3

LD Input0_4

ANDN Flag2

AND Flag3

ST SR_4.SET1

LD Flag5

ST SR_4.RESET

CAL SR_4

LD SR_4.Q1

ST Flag4

LD Input0_1

ANDN Flag3

AND Flag4

ST SR_5.SET1

LD Flag1

ST SR_5.RESET

CAL SR_5

LD SR_5.Q1

ST Flag5

LD Flag2

ST SR_6.SET1

LD Flag4

ST SR_6.RESET

CAL SR_6

ST Output0_0

LD Flag3

ST SR_7.SET1

LD Flag5

ST SR_7.RESET

CAL SR_7

ST Output0_1

Q1

SR

Output0_0

SR_6

SET1

RESET

Flag4

Flag2

LD Input0_1

AND Input0_3

ST SR_1.SET1

LD Flag2

ST SR_1.RESET

CAL SR_1

LD SR_1.Q1

ST Flag1

LD Input0_0

AND Flag1

ST SR_2.SET1

LD Flag3

ST SR_2.RESET

CAL SR_2

LD SR_2.Q1

ST Flag2

Q1

SR

RS_1

Q1

RESET1

SET

RS

Input0_2

Output0_1

Input0_2

Input0_1

Input0_0

AND

Output0_0

Q

CLK

R_TRIG

R_TRIG_2

Q

CLK

R_TRIG

R_TRIG_1

Q

CLK

R_TRIG

B1

B2

Y2

B2

B1

B2

B1

B1

B1 B2

B2

B1

B4

B3

B2

B1

R_TRIG

R_TRIG_1

E311 Fundamentals and Applications of Programmable Logic ControllersFesto Didactic

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