Welcome back to this class. We will continue from the last class. So, in the last class we have seen the state space modeling in modern control theory.
We have discussed the advantages of modern control theory and we have taken the example of a spinum system and I have tried to explain you the what do you mean by the state space model. So, in this class what we will cover we will discuss something more about the control modern control theory and in state space modeling also we will discuss something more. I will try to explain the state space model with the block diagram and we will in the last class we have taken the spring mass system example but in this example we will take the LRC circuit example because this for the starting lecture this course is all about to build the general foundation of a control theory that can be used for a other branches also.
So, we will take the LRC example and in next upcoming lectures we will move to the core aerospace controls. We will discuss today the full state feedback or pole placement. What do you mean by the pole placement? We will discuss with the examples. So, first we will discuss the modern control theory in little more.
As we have seen that it has some advantage when you in time domain actually you can see the other response also. If you give the input for example like elevator you can see the example of all other aircraft states also. And if the system is very highly non-linear then from the time domain it is good to design the controller using the time domain. And also when you need some kind of optimization for example like you want to optimize the control power or if you minimize the fuel requirement. And depending upon the cost function you can actually.
use the optimal control technique in a modern control theory in order to design your good controller. So, in last example, you are aware of this, right? BU and Y equal to CX plus DU.
So, how will you draw the block diagram of this equation X dot equal to AX plus BU and Y equal to CX plus DU? Because from the block diagram it is okay to visualize like what signal is going out and what signal is going in and how you are taking the feedback. Feedback is multiplying by what gain.
So all the things you can actually easily see in the block diagram. So now you can say this is my x dot. So if you integrate x dot what you will get?
x right. So take a integration block. This is 1 by S S integration you will get X. Now what is written here AX right. So you will take the feedback and you multiply by the A that will become the AX and you will sum up here.
So AX. But now the second term is BU. So what you will do you will take the signal u which is a input and u sum up here. So, now, here you what you will get?
a x plus b u first equation. Now, when you multiply by x to c, you will get c x. Now, you take u From this side, you take this side, this side, multiply by d and you sum up here.
So, Cx plus du will be your y. So, x dot equal to Ax plus bu. And y equal to cx plus du. c multiplied by x you will get cx.
And then you take u and multiply by the d. And sum here you will get y equal to cx plus du. So, easily you can draw the broad diagram of any state space model right. Now, we will take the LRC circuit example and again I will try to give you some feel for a state space modeling. In C easily you can see how you are integrating getting the x multiplying by c getting the output and what is d actually d is feed forward that means even if my, so Let us say x equal to 0, you are not getting the any dynamical response from the model, but see when you apply the input, you are getting the output, right.
So, input and output are directly related. So, we will take the LRC circuit example. So, you are very much familiar with this, right. This is what?
Resistance, right. And this is what? Inductance, right. and this is what capacitor right and this is ground and this is voltage difference v1 and current direction is this and here you can consider this v2 or vc and this is a voltage v1 now so in a spinning mass system what we have considered First state is position, second state was velocity, but an input was f, right.
Take the, you can stretch the spring, that was the input, right. But here what is the input? What you are changing at the beginning? V1, right. So, your V1 is what?
V1 you can consider as a u, input u, x dot equal to x plus v u, that u is v1. And now, What you are changing? By changing this V1, what you are changing? I, right? So, you can consider first state is I.
Like in spring mass system, we have considered the position. And what will be the second state? Second state you consider the voltage, V2 or Vc.
So, two states you have defined and input also you have defined. Now, you write the equation. What you write from this? You can write from this diagram, you can write V1 is VL, VR plus VL plus VC.
I am summing all the voltage. VR, VL plus VC is V1. Now, what is VR from Ohm's law?
IR, right? And what is VL? L di by dt, right?
And what is VC? 1 by C integral i dt, right? What is V1?
You will write. U, right? I have defined U, right? What is I?
X1, right? What is d i by dt? See, if you differentiate this x1 dot will be that d i by dt, right? So, I can write x1 dot, right? And what is integral i dt?
See. Vc is 1 by c integral i dt right. So, if you take Vc dot will become i by c right and Vc dot is what x2 dot right x2 dot equal to i by c.
So, you can write x2 dot equal to x1 by c. So, Vc you can write directly you can write Vc as a x2 simple. Directly you can add VCSX2 right. So now take X1 dot because I want X1 dot and X2 dot.
So take X1 dot one side. So what you will get? I will take LX1 RX1 minus X2 plus U to divide by L. So what you will get R by L X1 minus 1 by L X2. and 1 by l u this you call equation 1. And what is x2 dot?
x2 dot you have written what? x2 dot you have already found out. x1 by c.
So, I can write x1 and I can write 0x2. Okay. So, can I write x1 dot x2 dot equal to what?
I can write R by L minus 1 by L, 1 by C, 0, x1, x2 plus 1 by L, 0, u, right? So, how I have written? You can see, right?
See, x1 dot x2 dot I have written one column. Now, what is the coefficient of x1? Minus r by l I have written.
What is the coefficient of x2? Minus l by l, minus 1 by l I have written. What is the coefficient of x1 here in this second equation? 1 by c I have written.
What is the coefficient of x2? 0 I have written. So, again I will write this equation x1 dot x1 dot x2 dot what is our equation minus r by l minus 1 by l 1 by c 0 and this is x1 x2 and this is 1 by l this is 0 this is u. Please see this, okay.
Minus R by L minus 1 by C. So, what is this? What is this matrix? A. What is this?
X. What is this? B.
And this is U already, right? And this is what? What is this?
X dot. X dot equal to AX plus BU. from the LRC circuit also you can actually write in a state space form and x1 dot and x2 dot is what first order right. So, you can also comment on that the second order system you can write rewrite and say two first order system.
So, when I will tell you in the later lectures aircraft is also highly non-linear model. Not the first order, but you can write in the terms of first order u dot, b dot, w dot, p dot, q dot, r dot. So, this is all about the LRC circuit. LRC circuit is over. Now, we will see the full state feedback or pole placement.
So, now, the name itself implies full state feedback means all the states who are taking feedback, whether it is 1 or 12. Doesn't matter at all. All the states you are taking feedback. That is full state feedback.
Or it is also called pole placement. Why it is called pole placement? I already told you what do you mean by pole, right? But the name itself implies that you want to place at a correct position in order to make your system stable.
For example, let's say I have written this one. x dot equal to ax plus bu. There are two possibilities.
Based on A matrix, you can comment on my system, is it stable or unstable. But in both the cases, you want to control in a more effective way. For example, if the system is stable in this case, in the first case, but you want to try, you want to control with some other criteria. So, we will try to change the A matrix.
And the second case is, the system is unstable like a quadrotor model. So, eigenvalue is positive. So, let us say some root was here.
right half the plane but all the root that means poles you have to bring this on a left half plane in order to make the stable. So that means something you will do with u in order to make this in order to make the system stable. For example let us say in x dot equal to ax plus bu you put u equal to minus kx.
So what will be the x dot? x dot will be AX minus BKX. So you can write A minus. BK into X so simply only a is a pole for a open loop right when you have not applied any contours but when you take u equal to KX so that a matrix will become a minus BK so depending upon the P you can actually make the system stable to more stable or less stable depending upon the k and for unstable case you can make the stable. So, A minus B k what will be the Eigen value of this characteristic matrix will represent the closed loop Eigen value.
So, in reality whether your system is unstable or unstable in open loop case, but in a closed loop case it has to be stable because talking about the closed loop unstable is meaningless. Now, we always have a habit to take some example in order to try to understand what so far we are talking. So, now, a minus bk into u, I will take this transfer function, let us say transfer function 1 upon s minus 1, 1 upon s minus 1. So, if you locate at the complex plane, you will be here somewhere, right? 1, right? x equal to 1, right?
So, it is a stable and unstable. Unstable, right? Because roots are right half plane, right?
So, now, then the question is how will you bring this pole to this side? That means you are placing something, right? That is why it is called pole placement. How will you achieve? Using this k.
Now, simple. So, I can write, I can write, see, s equal to 1, right? That means root lambda equal to 1, right?
So, I can write this matrix as a x dot equal to, a is 1 you can consider and b also you can consider 1. And this is k into k. And this is u. So, x dot equal to 1 minus k u, right?
So, What is the solution? x equal to x0 e to the power a minus bk into t. h dot equal to a minus bk into x, right? So, this is x. x dot equal to a minus bk into x because you have substitute u equal to kx right x dot equal to a minus bk what is the solution of this solution of this is this right x not equal to a minus bk into t now what i said here you take a equal to 1 b equal to 1 So, what will be your solution?
x equal to x naught e to the power 1 minus k into t. Now, when k is less than 0, so this part is positive, right? So, e to the power positive diverging, right? Let us say k equal to 0. So, e to the power t again diverging, okay?
Now, if you put k greater than 1. So, what is the value? Let us say k equal to 2. So, x equal to x naught e to the power minus 1 into t, right. So, that means converging, right.
So, when k was not there, 0, your system was diverging by changing the k value only. you make the system from diverging to converging that means unstable to stable. Why unstable?
Because I have considered the transfer function 1 upon s minus 1 where the root is right half the plane, right half plane. Now, if you increase k, then this coefficient will be more negative. So, most faster convergence.
So, that means, and what is k? k is gain. So, now, you have a feel for gain, right?
High gain, low gain. correct again, if you take k value negative, for all k less than 0 will be diverging case, diverging case, right. And all k greater than will be converging case, right.
So, in the first case, you can see when there is a some kind of, for example, let us say if you put t equal to 0 here. So, it is x naught, right. So, I will mark as a x naught.
And when you increase this t then you will try to as t tends to infinity as t tends to infinity it will try to x will try to reach 0 right from this region right and for an unstable case it will try to So, by changing the gate, you have placed the pole at a correct position, right? So, now, here it is minus 1, right? A minus BK, right?
So, I can put this root, new root here, right? New root here. This is my new root. Okay? And if you increase the K value more, this root will try to shift here.
So, that is the feel for how will you place the pole. And each location you will get some response. So, if you want to get some desired response like this optimal response I want.
So, place the eigenvalue at that position. So, see how you can do the optimization. So, that is the advantage of steady space.
This A matrix can be more in order. For example, in a spring mass system and LRC circuit, what is the order of a matrix? 2 cross 2, right?
Okay, so in aircraft case, it can be more than 2. For example, in aircraft case, 1 is velocity, x direction, y direction and z direction. So, 3 equation is there, u dot, b dot, w dot and 3 rotational, p dot, q dot, r dot. So, 6 equation is there. And if you want to include phi dot, theta dot, psi dot, that is 3 more. and if you want to include x dot, y dot, z dot, three more.
So, total 12. So, there is a 12 states in aircraft. So, by considering this 12 states, what will be the A matrix if you include both lateral and longitudinal and direction R together? It will be 12 cross 12 matrix. But see, I have tried to, for example, then in that case, there will be, if you want to decide full feedback, there will be 12 feedback.
But generally it happens that if you want to control one state then other states will be automatically controlled. So you will see that in aircraft case I will show you that no need to take complete 12 states. Because if it is 12 independent states then you can take.
So you can take 2 cross 2 matrix or 3 cross 3 matrix and you can actually demonstrate. If it is 2 cross 2 matrix then what will be the gain value? 2 gain value. value right because here you will take two feedback right here you will take two feedback and if it is a 3 cross 3 then you will take three feedback so 3 gain is there okay or is it depending upon the gain value you can actually locate the appropriate location you can actually locate the root at a appropriate So now you have a some kind of foundation of a control system design in both classical approach and both modern control theory approach.
So what I will try to cover in the next two three lecture maybe I will take two three examples I will design the examples and then I will try to tell you how to design the controller and I will also try to implement the same. in the MATLAB. So, we will discuss these things.
and you should also have some kind of feel if this is far from this origin then you have a more stability and it is a if you are close to the origin you have a less stability so if you want to for example like if you want to design highly maneuverable aircraft so what you what you will do you will try to place near the origin or far from the origin near the origin. So, that is a feeling for like for example, like if you see if I tell you this is aircraft case for example, I am just saying that one route is this and another route is this. So, which one is a commercial case and which one is a fighter case just example. So, this one is a fighter case and this one is a commercial case. That is why when a fighter aircraft give the small definition of Elrond is certainly rule.
or pitch or yeah so the design the one actually actually where actually you want to place the pole for example in a commercial aircraft that is the criteria is different and for a fighter aircraft the criteria is different but see how the modern control theory is good right here so actually you can use both modern control theory and classical control theory to see some advantage is there in the classical control theory and some advantage is there in modern so, parallely you can design using both the approach some information you can extract from that approach some information you can extract from this approach and actually usually you can design the controller but here i you can also see when when i write x dot equal to ax plus bu and y equal to cx plus du so is a is changing with the time No, right. So, that means it is a linear time invariant, right. But in aircraft case, when aircraft go from one stage to another stage, the things will change, right. So, you have to linearize every time. So, linearization also I will tell you how to linearize.
That means here is a very simple. Even spring mass case system, the A matrix is constant. LRC circuit also the A matrix is constant. Also B matrix is constant. But, for example, even in spring mass system case also, depending upon the condition to condition, the K value will change and C value will change.
So, that will no longer the constant value. So, where the condition you are flying on, where the condition in your spring mass system, you have to each and every time you have to linearize. So, that is the importance of linearization. And how will you linearize?
using the Taylor series. You will ignore the higher order term. You will consider the only first order term. So, in next class also, I will try to explain how will you linearize the system. For example, like this is a very simple case x dot equal to x plus bu where I have taken a is constant and b is constant, c is constant, d is constant.
But this can change depending on the flight conditions. So, using the Taylor series, you can linearize and using the Jacobian, Please note down this for using the Jacobian. You can actually find out the A matrix and B matrix.
So in this I will design some example to give you the feel for the control system design. And I will also implement the same in the MATLAB so that you can have a feel for the gain while changing the gain, how the response is coming. So actually you can see what I'm saying here.
If it is close to the origin, it is a more. controllable and far from the origin it is a less controllable so all kind of actually feel or visualization you can actually get from that design example so we will meet in the next class