Introduction
Okay, this is a continuation of the previous lecture (the one I didn’t make flashcards about), where we ended up getting stuck when we had non-linear differential equations
Here, we will see how we can check if a system is linear or not, and if not linear, we will see how to derive the linear approximate model (using Taylor expansion) and check see stability criteria of the model based on root locus
How can we determine if a system is linear or not?
To check whether a system is linear or not, we take a combination of inputs and see if the combinations of outputs remain the same as when we have only one input.
This is known as the superposition model, where:
F(ax) = aF(x)
The representation and illustration of the superposition model is shown in digital notes
How can we derive linear approximate models
An approximate linear model can be obtained by local linearization of the process behaviour in the operating point; each non-linear term is approximated by a Taylor series expansion
In digital notes, you can see the Taylor expansion depending on one variable and on two variables. For the two-variable case, we have to take into account the partial derivative of the function with respect to both variables.
In digital notes, there is also an illustration of linearization up to 1st order
What are deviation variables?
Deviation variables represent perturbations from the normal operating point (or reference point). From a control perspective, it is highly desirable to express functions in terms of these deviation variables because they simplify analysis and design by eliminating constant terms, thereby making the system equations linearised around the operating point and easier to handle in control design.
An example of this is shown in digital notes notes
Second order modelling examples
The example is shown in digital notes - She flew through it so just look and try to understand - What is important however is in image 3 where we highlight the importance of chosing which varible to chose as a distrubance. This could be either your disturbance variable of Ca0, or your feed F. If you chose to only vary your Ca0 and keep your Feed constant you woud only have one nonelinear term (that of CA^2) while FCa0 and FCa would be constant. However if you chose to varry your F then you will end up with 3 nonlinear terms (since CA0 is chaning either or):
Therefore when we want to find the linear approximate model of the example it is best to chose the distrubution varible to be CA,0
How to obtain transfer functions from non-linear process models (summary)?
Follow the steps below:
Keep in mind this proccess requires you to get rid of constant terms —> most often tern then into deviaition varibles or take as zero
Break moment
Okay we saw how to check whether a system is first order ot not and we saw how to derive the linear approximate model of a none linear system. Now we move on to a INTEGRAL part of this leture - stability and the root locus
What are the certia for a stable dynamic system?
Bounded means that the input and output have a magnitude (absolute value) that never exceeds some finite, fixed number M for all time. so the system doesnt blow-up (go to inifinity) BIBO stability is a fundamental, and generally desirable, property because it guarantees that a controlled system will remain predictable and physically manageable. If a system were not BIBO stable, a common and harmless input (like a step change or a sine wave) could cause a critical variable (like temperature or concentration) to escalate infinitely, leading to equipment damage or failure.
In digital notes there is an example that better illustrates these stability creteria
Can you give an example of a chemical system that can be unstable? (Setup for control and rootlocus)
The most famous example is the of a CSTR with an irreversible reation A –> B (Shown in digital notes).
We an see 2 normalized differentail equations representing such system. The first corresponds to the material balance and the second to the energy balance.
We can also see that the output y = x2 and k comes from the arrehniius equation.
Now depending on your input equation you will end up in one of 3 steady states. This means that you might have 3 opertating conditions which for mosr chemical reactions of ofcourse not desrible.
When numerical solver are implemetned to solve such system we end up seeing that they go to either the top or bottom steady state depending on the starting conditions(/positions) and they never go to the middle steady state due to its unstability. However we have seen that economic profitablitiy comes from the operating conditions being an the unstable middle point. to achieve this we then need a feedback controllerto stabilize the reactor at that middle, most profitable point. To design your feedback controller you use your the root locus analysis (#connections we will see what root lous is one and next lecutre we can see how we can use it to design controllers)
break moment
we will not define the root locus method then follow it up with a bunch of examples to lock it in with each example more insight will be given!
What is the root locus?
The root locus is a powerful graphical method in control theory that shows how the closed-loop poles of a system move in the complex s-plane as a specific system parameter, typically the gain (K), is varied from zero to infinity.
A simple example in digital notes is shown to illustrate this. Where we see that as we varry our gain from zero to infinity our system will never become unstable, you can however make your system faster.
Note that yes you make your system faster but if you have such a large value of k then you will require a large input to have any insightfull change so chose your k to have a nice margin for your input varible. We disccused this peroiusly so ig a reminder ;)
Example 2
Shown in digital notes. This is a typical example of a lead-lag system. We see that as K approahes infinity the pole will then approch the zero of your open loop transfer fucntion , H(s, which in this case is -b.
This behavour occours for lead-lag systems or systems with defined zeros and we will use this trick to make systems stable (as seen in this lecture) and to design controllers (as will be seen in the next lecture)
Example 3
Shown in digital notes. This is a second order system with complex poles. Her as you spam your root locus (i.e vary K) you will only have a change in your imagenary part the real part (the one that defines stability) remains the same. This means that if you have a stable systne you cannot make it unstable, however by increasing your k you will allow for more dominant oscillations.
This again shows how the root lous helps you chose the appropraite value of k
Example 4
Shown in digital notes. This is a third-order system with repeated poles in the open-loop transfer function and three poles in the closed-loop transfer function: one real pole and two complex conjugate poles. As the gain K varies and the root locus is traced, the complex conjugate poles move from the negative real axis region toward the positive real axis. This behavior limits the allowable range of K to maintain system stability.
The limitation on K can be related to the gain margin obtained from the Bode plot of the closed-loop transfer function. The gain margin indicates the maximum gain that can be applied before the system becomes unstable.
What are the general rules for rootloci?
Digital notes
Note that she stated that we wont have to extensively use rootlocus due to the limitations of python.
Example 5
Shown on digital notes. Here you can then see that if you muliply your open loop transfer function of example 4 by (s+10)(S+50) you will effectively be introducing zeros into the system, then due to the behavour of the root locus this will cause your system, instead of increasing to unstability to rather stay witin the sable region.
Therefore by introducing zeros to the system you have changed root locus –> you are designing a controller!
Note that there is still aregion og instability so you are still sort of limited with your gain values
Example 6
In this example you can then see that your pole is positive and there is no way you can make the system stable for any value of K. However, what you can do is instead of just having only a k block you can introduce system dynamics via a control block C(s) = K(s+10)(s+50).
Controller design is all about placing the poles on the LHS of the complex plane and by using system dynamics (also similar with exmaple 5 but not directly) we are able to do such.
Not that this example is not possible the physical world
summary
In digital notes