Control Lab
(0904419)
Short Lab Report
Faculty of Engineering and Technology
The University of Jordan, Amman-Jordan
Process Control Simulator
By
Mohammad S. Al-Dweik (0116983)
Omar F. Abu-Znait (0114619)
Zaid S. Fahed (0110507)
Mohammad O. Qasem (0110529)
Section #: 1
March 2015
Abstract
The behavior of various systems with different modifiers such as lag, integral and
inverter were studied in the process control Experiment by building different systems and
stabilizing them using different parameters.
In the first case the output was equal to the input so there was no obvious problem in the
system, in the second case the output was 2.5 and it was corrected by adding Ki or by
increasing the value of Kp. In the third case the output went to infinity and it was
stabilized by adding a feedback to the system to give the desired output. In the fourth
case the output of system was oscillating around the value of 5 due to undamped system
so a damping block was added and then the system was stabilized.
Objectives
The main objectives of this experiment are:
1. Build different systems and observe the output in each case.
2. Modify the output using external controllers.
Experimental setup
The front panel of the process control simulator is shown in Fig (1) where each of the
blocks in the panel has its own function.
11
8
12
9
2
10
0
7
6
3
5
4
1
Figure (1): Front panel of the process control simulator
1. Set Value Block: Enables the user to set a range of input values from -10V up to
+10 V.
2. INTEGRAL ACTION block: it is used for adding integral control in the forward
loop of the system according to the transfer function:
𝑉𝑜(𝑆)
1
𝐻(𝑆) =
=
𝑉𝑖(𝑆)
𝑇𝑖 𝑆
3. DERIVATIVE ACTION block: it is used for adding derivative integral control in
the forward loop of the system according to the transfer function:
𝑉𝑜(𝑆)
𝐻(𝑆) =
= 𝑇𝑑𝑆
𝑉𝑖(𝑆)
4. PROPORTIONAL BAND block: it is used to add the gain required in the forward
loop with a scale that is calibrated with p (proportional band).
The value of the gain is given by:
100
𝑝
P varies in the range [4,200], the corresponding variation in the gain K is in the
range [0.5, 25].
5. Linear/ Nonlinear Control Switch: this switch allows the user to determine the
mode of control desired for implementation. The modes that can be selected are:
Linear with two steps, Nonlinear with two steps, Nonlinear with hysteresis,
Nonlinear with multi-step (staircase V-V characteristics).
6. LAG/INTEG Block:
When the mode of the switch is set to LAG, the transfer function is:
𝑉𝑜(𝑆)
1
𝐻(𝑆) =
= −
𝑉𝑖(𝑆)
𝑇𝑆 + 1
When the mode of the switch is set to INTEG, the transfer function is:
𝑉𝑜(𝑆)
1
𝐻(𝑆) =
= −
𝑉𝑖(𝑆)
𝑇𝑆
If the speed switch is set to SLOW, T= 1.0
If the speed switch is set to FAST, T = 0.01
7. LAG Block:
The transfer function for the lag block is
𝑉𝑜(𝑆)
1
𝐻(𝑆) =
= −
𝑉𝑖(𝑆)
𝑇𝑆 + 1
𝐾=
If T assumed as (1.0) which means the speed switch is set to slow this will give
𝑉𝑜(𝑆)
−1
−1
𝐻(𝑆) =
=
=
𝑉𝑖(𝑆)
(1)𝑆 + 1 𝑆 + 1
And if T is assumed as (0.1) which means the speed switch is set to fast this will
give
𝑉𝑜(𝑆)
−1
−100
𝐻(𝑆) =
=
=
𝑉𝑖(𝑆)
(0.1)𝑆 + 1 𝑆 + 100
8. Inverter Block:
The Transfer function of the Inverter Block is given as:
𝑉𝑜(𝑆)
𝐻(𝑆) =
= −1
𝑉𝑖(𝑆)
9. Measured Value: the measured value is the output of the process measured at the
inverter output.
10. Meters: there are two meters for different purposes, the lower one measures the
set value in the range [-10V, +10V].
The upper one measures the output value in the range [-10V, +10V].
11. MEASURED VALUE/ DEVIATION Switch: this switch allows the user to
measure the output value or the difference between the input and the output
values.
12. DISTANCE-VELOCITY Block: the transfer function of the DISTANCE –
VELOCITY Block is:
𝑉𝑜(𝑆)
𝐻(𝑆) =
= 𝑒 −𝑆𝑇𝑑
𝑉𝑖(𝑆)
Where the time delay (Td) is controlled according to the speed switch as follows:
S = j2*pi*f
Speed Switch
Valid frequency range
(Hz)
SLOW: Td = 1
0<= f <= 0.3
FAST: Td = 0.01
0 <= f <= 0.3
Procedure
Case 1
Lag-Lag-Inverter, Open Loop
1. Connect two lags and inverter in cascade
2. Set the proportional band to 100% (K=1)
3. Set the value for K = 5 using the set value knob
4. Record the measured value
Case 2
Lag-Lag-Inverter, Closed Loop
1.
2.
3.
4.
5.
Connect two lags and inverter in cascade
Set the proportional band to 100% (K=1)
Connect the Feedback wire between the two knobs to close the loop
Set the value for K = 5 using the set value knob
Record the measured value
Case 3
Lag-Integral-Inverter, Closed Loop
1.
2.
3.
4.
5.
Connect a lag block , integral block and inverter in cascade
Set the proportional band to 100% (K=1)
Connect the feedback wire between the two knobs to close the loop
Set the value for K = 5 using the set value knob
Record the measured value
Case 4
Integral-Integral-Inverter, Closed Loop
1.
2.
3.
4.
5.
Connect a Integral block with another integral block and inverter in cascade
Set the proportional band to 100% (K=1)
Connect The Feedback wire between the two knobs to make the loop closed
Set the value for K = 5 using the set value knob
Record the measured value
Results and sample of calculations
Case 1:
Lag + Lag + Open loop + Inverter
1
Transfer Function = 𝑆+1 ∗
1
1
= 𝑆2 +2𝑆+1
𝑆+1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
1
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆2 +2𝑆+1 = 5
𝑆→0
Case 2:
Lag + Lag + Feedback + Inverter
Transfer Function =
1
𝑆2 +2𝑆+1
1
1+ 2
𝑆 +2𝑆+1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
1
= 𝑆2 +2𝑆+2
5
1
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆2 +2𝑆+2 = 2.5
𝑆→0
The output is 2.5 which is not the desired value, so the problem is solved by:
1. Adding ki:
Transfer Function =
1
𝑆3 +2𝑆2 +𝑆
1
1+ 3
𝑆 +2𝑆2 +𝑆
=
1
𝑆 3 +2𝑆 2 +𝑆+1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
Yss = lim ∗ 𝑆 ∗
𝑆→0 𝑆
1
𝑆 3 +2𝑆 2 +𝑆+1
=5
2. Setting the value of Kp to 200 ( large value)
Transfer Function =
200
𝑆2 +2𝑆+1
200
1+ 2
𝑆 +2𝑆+1
200
= 𝑆2 +2𝑆+201
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
200
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆2 +2𝑆+201 ≅ 5
𝑆→0
Case 3:
Lag + Integral + Open loop + Inverter
1
Transfer Function = 𝑆+1 ∗
1
𝑆
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
1
5
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆2 +𝑆 = 0 = ∞
𝑆→0
1
= 𝑆2 +𝑆
When adding Feedback:
1
𝑆2 +𝑆
1
1+ 2
𝑆 +𝑆
Transfer Function =
=
1
𝑆 2 +2𝑆+1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
1
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆2 +2𝑆+1 = 5
𝑆→0
Case 4:
Integral + Integral + Feedback + Inverter
Transfer Function =
1
𝑆2
1
1+ 2
𝑆
1
= 𝑆2 +1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
Yss = lim ∗ 𝑆 ∗
𝑆→0 𝑆
1
𝑆 2 +1
=5
Kd is added to remove the oscillation:
Transfer Function =
1
𝑆
1
1+
𝑆
1
= 𝑆+1
Yss = lim 𝑠𝑡𝑒𝑝 𝑖𝑛𝑝𝑢𝑡 ∗ 𝑆 ∗ 𝑇. 𝐹.
𝑆→0
5
1
Yss = lim 𝑆 ∗ 𝑆 ∗ 𝑆+1 = 5
𝑆→0
Discussion
The system was built in four different ways such that each case represents a real life
problem. In the first case a system was built so that it contained two lag blocks and an
inverter without feedback and it has been noticed that the output is equal to 5. In the
second case the system was built so that it contained two lag blocks and an inverter with
feedback and it was noticed that the output is equal to 2.5, the solution for this kind of
problems is simply by adding Ki or by setting Kp to a large value. In the third part a
system was built so that it contained lag block, an integral block and an inverter without
feedback and it has been noticed that the output will lead to infinity, this is because there
is no feedback to the system so the system continuously increases the value of the output
until it goes to the infinity. The solution for this kind of problems is simply adding a
feedback sensor to the system which will stop the system from going to infinity. In the
fourth part a system was built so that it contained two integral blocks and an inverter with
feedback and it has been noticed that the reading was oscillating around 5 V (Undamped
System) and it was stabilized by adding a damping block (Kd).
References
[1] Dr. Musa Abdalla control lab notes.
[2] Dr. Musa Abdalla Control Systems course notes.