Dynamics of Open-Loop Systems (Simulations)
Arboleda, G.P. | Busico, B.D. | Lonzaga, J.P.
CHE180P – Process Dynamics and Control (Paired)
3rd term of A.Y. 2021-2022
College of Engineering and Architecture
Malayan Colleges Mindanao
ABSTRACT
MATLAB is an engineering software that can be
used to simulate process dynamics and process
control using Simulink. Through Simulink, one can
conveniently find out and model a process control
system and graphically show how the system reacts
to certain factors like the time-constant 𝜏 and the
damping factor 𝜁. In this experiment, the students
had found out about the differences between a firstorder response to a second-order response by the
time-constant
𝜏
and
the
damping
factor
𝜁.
Simulations done through Simulink also showed
what would happen if the time constant and damping
factor values were increased and how it affected the
different systems present. Overall, this experiment
also showed how a non-interacting, two-tank system
reaches a faster steady-state value as compared to
an interacting two-tank system.
INTRODUCTION
From the image above, the control action applied to
the system is the input. The response of the system
to a given input is the output. Some disturbances
that interact with the system can also alter the
output.
With the use of MATLAB, it’s simple to represent
and then simulate a mathematical model of a
physical system with Simulink. Simulink models are
represented graphically as block diagrams. The user
can choose from a variety of blocks in the offered
libraries to represent various phenomena and
models in a variety of ways. One of the key benefits
of using Simulink (and simulation in general) for
dynamic system analysis is that it allows us to swiftly
study the behavior of sophisticated systems that
The input or control action in an open loop control
system is independent of the system's output. The
input signal is independent of the system's reaction,
regardless of how it acts.
would
otherwise
be
impossible
to
analyze
analytically. Simulink can numerically approximate
the solutions of mathematical models that we can't
solve "by hand" or don't want to.
Simulink can mimic the entire control system,
including the control algorithm. Simulink is notably
useful for generating approximate solutions to
mathematical models that are prohibitively difficult to
solve. Consider the situation when you have a
Lab Exercise 2. Dynamics of Open-Loop Systems (Simulations)
CHE180P
By: Group 5
BS in Chemical Engineering
nonlinear plant. A popular method is to create a
the second tank does not influence the liquid level of
linearized model of the plant and then use analytical
the first tank. As for an interacting system, the flow
approaches to develop a controller using the
rate Q1 is derived from the height difference of the
linearized model. Simulink may then be used to
first tank and the second tank (H1-H2) divided by the
replicate your controller's performance when applied
resistance R1.
to the entire nonlinear model. Simulink can be used
to create the linearized model, while MATLAB can
A transfer function was derived from the first tank by
relating its liquid level H1 and the input function Q0.
be used to create the controller.
For this experiment, liquid-level systems with
variable flow input and output flows that depend
This results to a first order transfer function.
Similarly, the liquid level of the second tank H2 was
related to the input function Q0 which derives a
linearly on resistance will be simulated.
second order transfer function.
OBJECTIVES
You will be given the task of designing the flow of
the exercise itself. In doing so, make sure you
address the following objectives. Upon completing
this exercise, you should be able to:
A, 𝑚2
R, 𝑚𝑖𝑛
Tank 1
10
0.1
Tank 2
15
0.1
1. understand how first-order responses
differ from second-order responses
2. understand how the parameters 𝜏 and 𝜁
influence the dynamic responses of
I.
For non-interacting system:
First Order Transfer Function:
𝑯𝟏 (𝒔)
𝑹𝟏
𝟎. 𝟏
=
=
𝑸𝟎 (𝒔) 𝛕𝟏 𝒔 + 𝟏 𝒔 + 𝟏
firsthand second-order systems
3. differentiate
the
behavior
of
an
interacting second-order system from
Second Order Transfer Function:
that of a non-interacting second-order
𝑯𝟐 (𝒔)
𝑹𝟐
=
𝟐
𝑸𝟎 (𝒔) (𝛕𝟏 𝛕𝟐 )𝒔 + (𝛕𝟏 + 𝛕𝟐 )𝒔 + 𝟏
𝟎. 𝟏
=
𝟐
𝟏. 𝟓𝒔 + 𝟐. 𝟓𝒔 + 𝟏
system
4. simulate and visualize the responses of
open-loop processes to different input
functions in Simulink
II.
For interacting system:
First Tank:
METHODOLOGY
The two-tank system composes of two tanks with
different areas A1 and A2. On the first tank, a liquid
𝐻1 (𝑠) − 𝐻2 (𝑠)
= 𝑄0 (𝑠) − 𝐴1 𝑠𝐻1 (𝑠)
𝑅1
Second Tank:
enters with a flow rate Q0 and acts as the source of
the input function for the transfer functions of the two
tanks. There is a resistance R1 for the first tank and
𝐻1 (𝑠) =
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2 (𝑠) + 𝐻2 (𝑠)
𝑅2 2 2
Second Order Transfer Function:
a resistance R2 for the second tank. The flow rate
that enters the second tank is Q1, which is the
output of the first tank. For a non-interacting system,
2
𝑯𝟐 (𝒔)
𝑹𝟐
=
𝟐
𝑸𝟎 (𝒔) (𝛕𝟏 𝛕𝟐 )𝒔 + (𝛕𝟏 + 𝛕𝟐 + 𝑹𝟐 𝑨𝟏 )𝒔 + 𝟏
𝟎. 𝟏
=
𝟏. 𝟓𝒔𝟐 + 𝟑. 𝟓𝒔 + 𝟏
Lastly, the impulse function value was changed
To simulate the graphs of the two transfer
system. In the interacting setup, the liquid level
functions, MATLAB Simulink was utilized. The
of the second tank affects the liquid level of the
setup of the simulation includes a source (input
first tank. This results in an extra term (𝑅2 𝐴1 ) on
function), transfer function block(s), a mux, and
the s coefficient for the second order transfer
a sink (scope). The source can be any input
function. The responses of the interacting and
function such as a step function, impulse
non-interacting setup were compared by using
function, or a sinusoidal function. The purpose
a step and impulse function.
in an increasing manner.
The last comparison involves the interacting
and non-interacting setup of the two-tank
of using MATLAB Simulink is to observe the
changes on the output graphs by varying the
time constant τ and the damping factor ζ.
III.
Changing Time Constant Coefficient (𝜏)
Input: Step Function
The first input function used was a step
function. With a step function of 1, the value of τ
was varied by adjusting its value for the first
and second order transfer function. A similar
process was done for the second order transfer
function by changing the ζ value by increasing
and decreasing the coefficient of s on the
denominator. Afterwards, the step function
value was varied to test how the transfer
functions would respond.
To simulate an impulse function, a derivative
block was used to filter the step function. The
starting time must be non-zero for the step
function, but it should be close to zero. For the
simulations involving an impulse function, a
value of 0.00001 was chosen as the starting
time. In the same manner as the step function
process, the value of τ was varied for the first
order transfer function and the second order
transfer function. Then, for the second order
transfer function, the ζ value was also varied.
3
Input: Impulse Function
V.
Changing Input Value
Input: Step Function
IV.
Changing Damping Value (ζ)
Input: Step Function
4
Input: Impulse Function
DATA AND RESULTS
I.
Changing Time Constant Coefficient (𝜏)
Input: Step Function
Transfer Function
(First Order)
1
2
3
VI.
Coefficient of τ
1
2
3
Two-tank Interacting System vs. NonInteracting System
Input: Step Function
Input: Impulse Function
5
Transfer Function
(Second Order)
1
Coefficient of τ
2
3
2.5
3.5
1.5
Transfer Function
(Second Order)
1
Coefficient of τ
2
2.5
3
3.5
Input: Impulse Function
1.5
There is a trend that can be observed by changing
the time constant τ. For a first order transfer
function, increasing the time constant causes a
delay on its response to the input function. As
observed from Graph 3.1 and 3.3, the transfer
function with the least time constant value reached
the steady state value fastest. For the second order
transfer function, a lesser time constant reduces the
fluctuations of the response. The first order transfer
function reaches the steady state value faster than
the second order transfer function.
II.
Changing Damping Value (ζ)
Input: Step Function
Transfer Function
(Second Order)
1
2
3
Coefficient of s
Damping Value
2.5
1.5
3.5
1.02
0.61
1.43
Transfer Function
(First Order)
1
Coefficient of τ
2
3
2
3
1
6
a value of 2.5 on the coefficient of s results to a
damping value of 1.02. A damping value equal to 1
indicates a critically damped response from the
transfer function. An underdamped response occurs
when the damping value is less than 1, and an
overdamped response occurs when the damping
value is greater than 1. The 2nd transfer function is
an example of an underdamped response with a
damping value of 0.6, and this results in more
oscillations to reach the steady state value. The 3rd
transfer function is overdamped and when compared
to the 1st transfer function, it takes more time to
reach the steady state value.
III.
Changing Input Value
Input: Step Function
Transfer Function
(Second Order)
1
2
3
Input Value
1
2
3
Input: Impulse Function
Transfer Function (First
Order)
1
2
3
Input Value
1
2
3
The change in the damping value dictates the
oscillation of the response. Transfer function 1 had a
smoother transition to the steady state value
compared to the other transfer functions. This is
because for the calculated second transfer function,
7
Transfer Function
(Second Order)
1
Input Value
2
3
2
3
Input: Impulse Function
1
Transfer Function
(Second Order)
1
2
Coefficient of s
2.5
3.5
Increasing the input value for the first order transfer
function results in a higher gain (𝐾𝑐 ). When the
value of 𝐾𝑐 increases, the amount of time constants
it takes to reach the new steady state value also
increases. This is also evident from Graph 3.10 with
a higher peak value as 𝐾𝑐 increases caused by an
impulse input.
IV.
Two-tank Interacting System vs. NonInteracting System
8
As observed from Graphs 3.11 and 3.12, the
Input: Step Function
response of an interacting system is delayed
compared to the response of a non-interacting
system. It also reaches the new steady state value
slower than the non-interacting system. This is due
to the addition of the term 𝑅2 𝐴1 on the coefficient of
s in the second order transfer function. The extra
term contributes to a sluggish response on the
interacting system caused by the difference in liquid
level height from the two tanks.
CONCLUSION
The dynamics of open-loop systems can be
mathematically shown using first and second order
transfer functions. Open-loop systems can be
affected by factors such as the time-constant 𝜏, the
damping factor, 𝜁 and the gain, 𝐾𝑐 . In this
experiment, the students found out of the following:
Input: Impulse Function
•
First-order responses reach a faster steady
state value when compared to second-order
transfer function responses.
•
The time-constant 𝜏 and the damping factor
𝜁 affect the outcomes and the trend of the
responses of first order and second order
transfer functions. For first order transfer
functions, the time constant 𝜏 affects the
system if it is increased, as this will result to
the increase in time to achieve a steadystate value. For second order transfer
functions, the higher the value of the time
constant
𝜏
increases
the
number
of
fluctuations of the responses. If the damping
Transfer Function (First
Order)
1
Input Value
2
2
3
3
factor 𝜁 has a value close to 1, the system is
considered critically damped. This means
1
that a second order system transitions to a
steady state value smoothly. Increasing or
decreasing the value of the damping factor 𝜁
makes
the
system
overdamped
and
underdamped, respectively.
9
•
A two-tank, non-interacting system reaches
a new steady state value faster than an
interacting system. The transfer functions for
both systems differ; the two-tank interacting
𝐻1′ (𝑠)
1
𝑅1
=
=
→ 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑎𝑛𝑘
′
𝑄𝑜 (𝑠) 𝐴𝑠 + 1 𝐴1 𝑅1 𝑠 + 1
𝑅
′ (𝑠)
𝐻2
𝑅2
→ 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑎𝑛𝑘
′ (𝑠) =
𝑄1
𝐴2 𝑅2 𝑠 + 1
system has an additional 𝑅2 𝐴1 term on its
denominator, resulting in a more sluggish
𝑅𝑒𝑙𝑎𝑡𝑖𝑛𝑔 𝐻2′ (𝑠) 𝑡𝑜 𝑄𝑜′ (𝑠):
response to a new steady-state value. The
height of the liquid level is also a factor on
𝐻2′ (𝑠) 𝐻2 (𝑠) 𝑄1′ (𝑠) 𝐻1 (𝑠) 𝑄1
1
=
×
×
;
=
𝑄𝑜′ (𝑠) 𝑄1′ (𝑠) 𝐻1 (𝑠) 𝑄𝑜′ (𝑠) 𝐻1 𝑅1
why the response of an interacting system is
=
more sluggish compared to a non-interacting
system.
•
MATLAB and Simulink were used for the
𝑅2
1
𝑅1
× ×
𝐴2 𝑅2 𝑠 + 1 𝑅1 𝐴1 𝑅1 𝑠 + 1
𝐻2′ (𝑠)
𝑅2
𝑅2
=
=
′
𝑄𝑜 (𝑠) (𝐴2 𝑅2 𝑠 + 1)(𝐴1 𝑅1 𝑠 + 1) (𝜏2 𝑠 + 1)(𝜏1 𝑠 + 1)
𝑅2
(𝜏2 𝑠 × 𝜏1 𝑠) + 𝜏2 𝑠 + 𝜏1 𝑠 + 1
collection of data and results. The students
=
had used the step-function as well as the
impulse-function and compared the results
𝐻2′ (𝑠)
𝑅2
=
′
2
𝑄𝑜 (𝑠) 𝜏2 𝜏1 𝑠 + (𝜏2 + 𝜏1 )𝑠 + 1
through graphs.
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝜏 𝑎𝑛𝑑 𝜁:
𝐻2′ (𝑠)
0.1
=
𝑄𝑜′ (𝑠) (𝑠 + 1) (𝑠 + 2)
3
APPENDICES
I.
For Non-interacting system
=(
𝑎
𝑏
2
+
) (𝑠 + 1) (𝑠 + )
2
𝑠+1 𝑠+
3
3
2
0.1 = 𝑎 (𝑠 + ) + 𝑏(𝑠 + 1)
3
2
2
@𝑠 = − : 0.1 = 𝑏 (− + 1)
3
3
1
0.1 = 𝑏 ∴ 𝑏 = 0.3
3
2
@𝑠 = −1: 0.1 = 𝑎 (−1 + )
3
1
0.1 = − a ∴ 𝑎 = −0.3
3
𝑑(𝜌𝑉)
= 𝜌𝑞(𝑡) − 𝜌𝑞1 (𝑡)
𝑑𝑡
𝑑ℎ
ℎ
𝐴
= 𝑞(𝑡) − 𝑞1 (𝑡) = 𝑞
𝑑𝑡
𝑅
ℒ {𝐴1
𝐻𝑅′
𝑑𝐻
= 𝑄𝑜′ − }
𝑑𝑡
𝑅
1
𝐴𝑠𝐻(𝑠) = 𝑄(𝑠) − 𝐻(𝑠)
𝑅
𝐻2′ (𝑠)
0.1
0.3
0.3
=
=−
+
𝑄𝑜′ (𝑠) (𝑠 + 1) (𝑠 + 2)
𝑠+1 𝑠+2
3
3
ℒ −1 {0.3 (
1
𝑠+
2
3
−
1
)}
𝑠+1
2
= 0.3 (𝑒 −3𝑡 − 𝑒 −𝑡 )
𝐻2′ (𝑠)
0.1
0.1
2
=
=(
)÷
′
2
2
𝑄𝑜 (𝑠) (𝑠 + 1) (𝑠 + 2)
3
𝑠2 + 𝑠 + 𝑠 +
3
3
3
10
=
𝐻1′ (𝑠) − 𝐻2′ (𝑠)
1
= 𝐻2′ (𝑠) (𝐴2 𝑠 + )
𝑅1
𝑅2
0.1
3 2 5
𝑠 + 𝑠+1
2
2
𝐻1′ (𝑠) − 𝐻2′ (𝑠) =
3
∴ 𝜏 = √ = 1.22;
2
𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝜁: 2𝜁𝜏 =
II.
5
∴ 𝜁 = 1.02
2
Interacting system:
𝐻1′ (𝑠) =
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2′ (𝑠)
𝑅2 2 2
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2′ (𝑠) + 𝐻2′ (𝑠) → 2
𝑅2 2 2
𝐻1′ (𝑠) − 𝐻2′ (𝑠)
= 𝑄𝑜′ (𝑠) − 𝐴1 𝑠𝐻1′ (𝑠)
𝑅1
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2′ (𝑠) + 𝐻2′ (𝑠) − 𝐻2′ (𝑠)
𝑅2 2 2
𝑅1
= 𝑄𝑜 (𝑠)
− 𝐴1 𝑠 (
+ 𝐻2′ (𝑠))
𝐹𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 𝑡𝑎𝑛𝑘:
𝑞𝑜 − 𝑞1 = 𝐴1
𝑞𝑜 −
𝑑ℎ1
ℎ1 − ℎ2
; 𝑞1 =
𝑑𝑡
𝑅1
(𝑅2 𝐴2 𝑠 + 1)𝐻2′ (𝑠)
𝑅1
ℎ1 − ℎ2
𝑑ℎ1
= 𝐴1
𝑅1
𝑑𝑡
= 𝑄𝑜 (𝑠) − 𝐴1 𝑠
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2′ (𝑠)
𝑅2 2 2
+ 𝐴1 𝑠𝐻2′ (𝑠)
𝑆. 𝑠.
𝑞𝑜 𝑠 −
𝑅1
(𝑅 𝐴 𝑠 + 1)𝐻2′ (𝑠)
𝑅2 2 2
ℎ1 𝑠 − ℎ2 𝑠
𝑑ℎ1 𝑠
= 𝐴1
𝑅1
𝑑𝑡
(𝑅2 𝐴2 𝑠 + 1)𝐻2′ (𝑠)
𝑅1
+ 𝐴1 𝑠 (𝑅2 𝐴2 𝑠 + 1)𝐻2′ (𝑠)
𝑅2
𝑅2
+ 𝐴1 𝐻2′ (𝑠) = 𝑄𝑜 (𝑠)
𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑓𝑜𝑟𝑚:
ℒ {𝑄𝑜′ −
𝐻1′ − 𝐻2′
𝑑𝐻1′
= 𝐴1
}
𝑅1
𝑑𝑡
𝑄𝑜 (𝑠) −
𝐻1′ (𝑠) − 𝐻2′ (𝑠)
= 𝐴1 𝑠𝐻1′ (𝑠)
𝑅1
𝐻1′ (𝑠) − 𝐻2′ (𝑠)
𝑅1
= 𝑄𝑜′ − 𝐴1 𝐻1′ (𝑠) → 1
𝐹𝑜𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑎𝑛𝑘:
𝑞1 − 𝑞2 = 𝐴2
𝑑ℎ2
ℎ1 − ℎ2
ℎ2
; 𝑞1 =
; 𝑞2 =
𝑑𝑡
𝑅1
𝑅2
𝑆. 𝑠.
ℎ1 𝑠 − ℎ2 𝑠 ℎ2 𝑠 𝐴2 𝑑ℎ2 𝑠
−
=
𝑅1
𝑅2
𝑑𝑡
𝑅2 𝐴2 𝑠 + 1
𝑅1
𝐻2′ (𝑠) [
+ 𝐴1 𝑠 (𝑅2 𝐴2 𝑠 + 1) + 𝐴2 𝑠]
𝑅2
𝑅2
= 𝑄𝑜 (𝑠)
𝐻2′ [
𝑅2 𝐴2 𝑠 + 1 + 𝐴1 𝑠𝑅1 (𝑅2 𝐴2 𝑠 + 1) + 𝑅2 𝐴1 𝑠
] = 𝑄𝑜 (𝑠)
𝑅2
𝐻2′ (𝑠)
𝑅2
=
𝑄0′ (𝑠) 𝜏2 𝑠 + 1 + 𝜏1 𝑠(𝜏2 𝑠 + 1) + 𝑅2 𝐴1 𝑠
𝐻2′ (𝑠)
𝑅2
=
𝑄0′ (𝑠) 𝜏1 𝑠 + 1 + 𝜏1 𝜏2 𝑠 2 + 𝜏1 𝑠 + 𝑅2 𝐴1 𝑆
𝐻2′ (𝑠)
𝑅2
=
𝑄0′ (𝑠) 𝜏1 𝜏2 𝑠 2 + (𝜏1 𝜏2 + 𝑅2 𝐴1 )𝑠 + 1
𝑑𝑒𝑣𝑖𝑎𝑡𝑜𝑛 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:
ℒ{
𝐻1′ − 𝐻2′ 𝐻2′
𝑑𝐻2′
−
= 𝐴2
}
𝑅1
𝑅2
𝑑𝑡
𝐻1′ (𝑠) − 𝐻2′ (𝑠) 𝐻2′ (𝑠)
−
= 𝐴2 𝑠𝐻2′ (𝑠)
𝑅1
𝑅2
𝐻1′ (𝑠) − 𝐻2′ (𝑠)
𝐻2′ (𝑠)
= 𝐴2 𝑠𝐻2′ (𝑠) +
𝑅1
𝑅2
11