Process Dynamics and Control - Solved Exam Questions

PROCESS DYNAMICS

and CONTROL

QUESTIONS

1)​ Feedback vs. Feedforward Control: Explain the fundamental differences between

feedback and feedforward control strategies. What are the primary advantages of

feedforward control, and why is a "feedback trim" usually added to it in industrial

applications?

Feedback control is based on the principle that when a disturbance enters the process (or we

change the set point of a process variable) the process variable (deviation) changes and the

controller tries to adjust it by modifying a manipulated variable.

The feedforward instead, since the disturbance is the first cause of the process variables

changes, as soon as the disturbance is detected, the feedforward controller tries to modify the

manipulated variable to act on the process variable, before the actual deviation from the set

point of the controlled variable occurs. This is possible only if the disturbance is measurable

or there are variables that indicate the presence of a disturbance. Moreover if the disturbance

or process dynamics (TFs) are not well modeled, the feedforward controller doesn’t work

very well. To conclude, feedforward control would also fail if there are other disturbances

that have not been studied. For these many reasons, a feedback trim is always needed.

2)​ Deviation Variables: Why are "deviation variables" used when developing dynamic

models for process control? Explain how they simplify the process of solving

differential equations using Laplace transforms.

Deviation variables are defined as the difference between the actual value of a process

variable and its value at a steady-state. They are used because:

●​ allow engineers to ignore the absolute "background" values of a process and focus

entirely on the disturbance or change.

●​ facilitate the linearization of complex systems. By shifting the frame of reference to

the steady-state, the initial value of the deviation variable at t = 0 is, by definition,

zero.

●​ With deviation variables, the system starts at the "baseline," so y(0) = 0. This

()

simplifies the transform of a derivative .

⎡ ⎤

= () − (0) = ()

⎣ ⎦

●​ the transfer functions are independent on the initial system configuration

3)​ First-Order System Dynamics: Define the "time constant" (τ) and "steady-state

gain" (K) for a first-order system. If a step change of magnitude M is applied to the

input, describe the output response and how much of the final change is reached

after a time t = τ. 1

A first order system dynamics is defined by the following transfer function: .

() = τ+1

The time constant τ is defined as the characteristic time for the system and could be intended

as the time to reach 63% of the final steady state value.

The steady state gain K measures the final change of the output with respect to the change of

the input, when the steady state is reached.

We could say that τ defines how the output will change while K defines how much the output

will change.

If a step change of magnitude M is applied to the input of a first order system, we can find the

output response by .

() = ()() =

(

τ+1

)

Thinking about the graph for such a response we would see an asymptotical exponential rise

from the first steady state value to the final one (defined by the steady state gain).

The larger the K is and the larger will be the difference in steady state values while the longer

will be the time constant and the slower will be the response.

4)​ Second-Order System Damping: Describe how the damping factor (ζ) influences

the dynamic behavior of a second-order system. Specifically, explain the

characteristics of overdamped, underdamped, and critically damped responses

when the system is perturbed by a step change.

The second order transfer function looks like this: () = 2 2

τ +2ζτ+1

If a step change is applied to the input of the process, the response would be

.

() = 2 2

( )

τ +2ζτ+1

The characteristics of the response change as the value of the damping factor ζ changes. For

example:

●​ if ζ is higher (overdamped) or equal to 1 (critically damped) the response will be very

similar to the one of a first order system (asymptotic exponential rise) but it will take

a longer time and at the beginning there is a latency since the derivative is 0

●​ if ζ is lower than 1 (underdamped) the response will show some oscillations around

the steady state value. The lower the damping factor and the larger will be the

oscillations (the poles are complex)

●​ if ζ is smaller than -1 the response is unstable and it won’t settle to a steady state

because not all the poles are negative ⍵

5)​ If a sinusoidal change of amplitude A and angular frequency is applied to the

input of a first order system, describe the output response.

By applying a sinusoidal profile to the input of a first order system we would obtain a

response with two contributes:

●​ a first exponential contribute that dies out quickly

●​ an oscillation contribute that have: 2

○​ an amplitude that depends on the steady state gain of the TF, the frequency of

the oscillations and the time constant

■​ the larger the K and the larger the amplitude

■​ the bigger the frequency the smaller the amplitude

■​ the bigger the time constant the smaller the amplitude

○​ the same frequency of the input

○​ a phase angle (basically a delay) that depends on the frequency

6)​ Second-Order System Damping: Describe how the damping factor (ζ) influences

the dynamic behavior of a second-order system perturbed by a sinusoidal input

profile.

As for the first order system, the response is made of two contributions: one dies out quickly

while the other one is oscillating continuously.

The frequency of the oscillating response is the same of the input, but the amplitude depends

on: ●​ the frequency

●​ the time constant (the higher the time constant and the smaller the amplitude)

●​ the damping factor

The dependence from frequency and damping factor is more complicated than the one of the

first order system:

●​ if the damping factor is higher than 0.707 the amplitude is dampened for every

possible frequency (always smaller than the amplitude of the input)

●​ if the damping factor is lower than 0.707 the amplitude could be both bigger or

smaller than the amplitude of the input, depending on the frequency value

○​ for low frequency value the amplitude is amplified

○​ for high frequency value the amplitude is dampened

A resonance frequency exists, at which the amplification of the amplitude is maximum.

7)​ Describe the step response of a pure integrator. Where is the pole of a pure

integrator positioned inside the complex plane of the solutions?

The step response of a pure integrator is characterized by a linear divergence from the starting

steady state value. The system won’t reach another steady state. The transfer function of a

'

pure integrator is .

() =

The pure integrator is positioned exactly on the origin of the complex plane, since its solution

is s = 0.

8)​ How can we predict the system behaviour by looking at the complex plane of poles?

The position of the poles will exactly tell how the system works:

●​ if the pole is in the right plane, the response will diverge

●​ if the pole is in the left plane, the response in stable 3

●​ if the pole is in the origin, the system acts as a pure integrator

●​ if the poles are complex and conjugated, the system will show some oscillations

●​ if the complex poles are on the y-axis, the oscillations will remain continuously

●​ the farthest the solution from the y-axis and the faster is the response

9)​ Describe the step response of a lead-lag system.

A lead lag system is a first order system with numerator dynamics, whose TF looks like this:

τ +1 . Since the relative order is 0, this system doesn’t exist in nature.

() = τ +1

1

The step response of a lead lag system depends both on the lead time and on the lag time.

The final value depends only on the steady state value.

Let’s assume that we fix the value of the lag time.

The trend of the response will depend on the corresponding value of the lead time:

●​ if the lead time is longer than the lag time, the response will start from above the

steady state value and will eventually reach it monotonically

●​ if the lead time is shorter than the lag time, the response will start from below the

steady state value and will eventually reach it monotonically

Let’s note that also the initial value of the response depends on the values of lead and lag

time.

10)​Describe the step response of a second order system with numerator dynamics.

In this case we have two “lag” times (time constants):

τ +1

() = ( )

( )

τ +1

τ +1

1 2

The trend of the response depends mostly on the value of the lead time with respect to the

bigger lag time (the one that controls the overall time needed to reach the steady state). After

setting the two “lag” times and understanding which one is the bigger, the response will

depend on the lead time value: 4

●​ if the lead time is zero we fall into the transfer function of a normal second order

system

●​ if the lead time is between 0 and the bigger lag time the response is faster, meaning

that it approaches the steady state value earlier than a second order system

●​ if the lead time is larger than the bigger lag time, the response will overcome the

steady state value, reach a peak and then decreasing towards the steady state value

(this is called overshoot)

●​ if the lead time is negative (positive zero), the response is called “inverse” because at

first it moves in the opposite direction with respect to the steady state value, but

eventually it will reach it

11)​How can a system respond to an input change with an inverse response? Give an

example on a system like that.

We can predict the inverse response by looking at the TF of the system. If such TF has a

positive zero, there will be an inverse response.

A system with an inverse response could be a system made by two TF in parallel:

●​ one TF has a negative steady state gain and a faster response (small time constant)

●​ the other TF has a positive steady state gain, bigger than the other and a slower

response

At the beginning the faster response will prevail, moving the output variable towards the

negative direction and then, when some time has passed, the response with the higher steady

state value will prevail, moving the output towards the other direction.

12)​Describe the effect of a dead time inside a system.

When a system is characterized by a dead time it shows a time interval after the manifestation

of the input change where there is no sign of response. The dead time is represented by an

−θ

exponential at the numerator of the TF: .

() =

The dead time is an irrational function so, if we need to factor it, we need to approximate it.

The most used approximations are the Taylor expansion and the Padè approximation.

13)​What happens when more than one first order system is put in series? In that case,

can we simplify the final TF? 5

When systems are put in series, the overall transfer function is the product of all the transfer

functions so the order of such a final function will be higher than the single one. The

consequence from a physical point of view is that the response will be slower and slower as

we increase the number of systems.

The final TF can be simplified by approximation: −τ

( )

●​ the numerator dynamics are usually approximated as dead times

τ + 1 =

●​ we assume as time constant the largest time constant of the denominator −τ

1

●​ the other contributions are approximated as dead times (for example )

2

≃

( )

τ +1

2

The steady state gain is the same as the one of the original TF.

The time constant is the largest between the singular time constants.

The dead time is the sum of the lead time and the lag times that have been discarded

−θ

:.

() = τ+1

14)​How is a model for a specific system estimated?

A model is estimated through this procedure:

1.​ a perturbation on the input is made (usually a rectangular one to see the response both

of a positive and negative change)

2.​ the data of the response is recorded

3.​ assumption of a possible system, the most common are first order system with time

delay and second order system with time delay

4.​ regression analysis (usually made by softwares with numerical procedures to find the

correct parameters of the model to make the estimated values fit the best the real

values

5.​ if more than one model can fit reasonably the response, the one with the lowest order

is chosen

15)​How does a feedback controller work?

A feedback controller is a device that receives a signal from a sensor which records variation

in a process variable (output). The controller will calculate an action to make on the final

control element (usually a valve) to modify a manipulated variable (input) that will cause

(hopefully) the adjustment of the process variable. Basically, it is activated by a variation in

the output. 6

16)​How should we decide to design the type of final control element (fail open or fail

closed)?

The final control element is usually a control valve. The main choice that must be done is

about how the valve opens or closes. The main reason for that is safety during energy failure.

When energy fails, air stops being sent to the valve and the type of valve will define if it will

close or remain open.

The choice is therefore:

●​ fail open valve if we need the valve to remain open during failures

●​ fail closed valve if we need the valve to remain closed during failures

This design choice will also define the sign of the steady state gain of the valve.

17)​How does an on-off controller work?

An on-off controller is a controller that has only two possible output signals:

●​ one when the error is positive

●​ the other one when the error is negative

The manipulated variable will subject continuous oscillations since the controller is not able

to quantify the response as a function of the value of error. To avoid wear-and-tear problems,

a neutral dead band is usually defined, where the controller does not intervene. This type of

controller is used very rarely since it produces wear and is inaccurate.

18)​How does a proportional controller work? What are its limitations and advantages?

Would it be possible to remedy the main disadvantage of the controller?

The control law for a proportional control is that correspond to the TF

() = + ()

'() . Its action is proportional to the error that is detected on the process

() = =

()

variable.

Its main limitation is that it can’t remove the steady state offset, meaning that if a perturbation

is sustained over time, the controller will make the system sit in a different steady state value

with respect to the desired one.

The controller is designed by deciding the value of the controller steady state gain K . The

C

higher the gain, the more sensitive (or aggressive) the controller becomes and the new steady

state will be closer to the desired set point.

We could remedy the steady state offset by changing manually the bias of the controller.

The limits of the controller action are defined by the maximum and minimum signal that the

controller can send (fully open and fully closed valve).

Its advantages are that it reacts as soon as the error is detected and its design is very simple

(only one parameter to set). 7

19)​How does an integral controller work? '() 1

The TF of an integral controller is and its control law is

() = =

() τ

* *

1 . The advantage of such a controller is that it doesn’t stop changing

() = + ∫ ( )

τ

0

the manipulated variable until the error is zero so it can remove any steady state offset.

By the way, its response is based on the integral of the error over the time so it could take

some time before seeing a non-negligible action of the controller (slow action).

It is designed by choosing the integral time constant: the smaller the constant and the more

aggressive the action. Due to its slowness, it is usually implemented together with a

proportional action.

20)​How does a proportional integral controller work?

⎡ ⎤

* *

1

The control law of the controller is and its transfer

⎢ ⎥

() = + () + ∫ ( )

⎢ ⎥

τ

⎣ ⎦

0

( ) ( )

τ +1

'() 1

function is .

1

() = = 1 + =

() τ τ

1 1

Combining proportional and integral action we obtain:

●​ a faster response than the integral action thanks to the proportional contribution

●​ the possibility of removing the steady state offset thanks to the integral action

By the way, this controller needs the choice of two parameters:

●​ the controller gain that if increased lead to a faster response but the possibility of

oscillations or even an unstable response

●​ the integral time constant that if increased lead to slower response but dampened

21)​How does a derivative controller work? ()

The control law of a derivative controller is and its transfer function is

() = + τ

'() .

= τ

()

The advantage of the derivative controller is the fast response, since it depends on the

derivative of the error (meaning its tendency). The larger the time constant the more

aggressive the action.

The disadvantage are:

●​ it can’t remove the steady state offset: for example, if the error remains constant, but

the steady state value is not at the set point, the controller won’t act since its

derivative is zero.

●​ it is highly disturbed by noise, because a very fast noise could activate the controller

even if there is no need to (wear and tear problem)

●​ when implemented in a PID controller, if the set point is changed stepwise (for

example manually) there will be a quick change in the controller output (since it will

8

measure a large derivative of the error). To avoid this problem, it is convenient to

measure the derivative of the measured output (process variable) instead of its error

We can also see from its transfer function that it is physically unrealizable so we should add a

first order system to both making it realizable and dampen the noise:

τ τ

'() ⍺

. Usually is chosen a 0.1.

= =

() τ +1 ατ +1

The derivative action should not be used when the system is fast, because there won’t be any

advantage.

22)​How does a PID controller work? Which are the main issues to be solved and how

could we solve them?

PID controllers combine the action of proportional, integral and derivative controllers.

The derivative action is dominant at the beginning, and the larger the derivative time constant

the more aggressive the action will be. The derivative action will help to avoid oscillations in

the response, but if the time constant is increased too much it could lead to an unstable

response.

The derivative action should not be used when the system is fast, because there won’t be any

advantage.

The integral action instead will dominate after some time in order to correct a possible steady

state offset.

The main disadvantage is the difficulty of tuning, since there are three parameters to be

chosen.

There are some cases in which the PID controller could fail in efficiency:

●​ derivative kick: when we change stepwise the set point, the controller will measure an

instantaneous variation in the error, leading to a very fast response of the derivative

action, leading to fast variation in the controlled variable. To avoid this problem,

instead of measuring the derivative of the error, we should measure the derivative of

the process variable (measured)

●​ reset windup: in some cases, when a disturbance leads to very strange and variable

outside their usual values, the proportional and derivative action could not be able to

make the controlled variable return to the set point. When this happens, the integral

control will continue to increase its contribution (since the integral continues to grow)

so even when the disturbance disappears, the integral action is still going. At that

point, the controlled variable needs to create a sufficiently large integral on the other

side of the error to balance the integral action. To avoid such a problem, we should set

a reset windup, a procedure in which the integration stops when the valve is fully

open or closed (depending on the system).

●​ wear and tear due to derivative kick in response to noise 9

23)​How can we choose between direct acting and reverse acting controller?

A good habit is to make the product of all the steady state gain inside the feedback loop

positive. So, after deciding the steady state gain of the valve (fail closed or fail open) which is

usually a safety concern, the steady state of the controller is chosen.

A direct acting controller will send a higher signal when the process variable increases while

a reverse acting controller will send a higher signal when the process variable decreases.

Let’s assume that the product of the gains inside the feedback loop apart from controller and

valve are positive:

●​ if the valve is fail closed (positive K ) the controller is a reverse acting controller

V

(positive gain K )

C

●​ if the valve is fail open (negative K ) the controller is a direct acting controller

V

(negative gain K )

C

24)​Which are the main topics to discuss when deciding the type of controller to use?

The main rules to decide the type of controller to use are:

●​ use the derivative control (in PID controllers) only if the loop is s