Introduction to Full State Feedback Control

hello everyone and welcome to another video today I'd like to talk about full state feedback we're gonna see that full state feedback is an incredibly powerful yet ridiculously simple control system and I know that sounds a little bit like an oxymoron right you usually have simple or powerful not both at the same time it almost seems like a paradox like if Cinderella's slipper fits so perfectly how did it fall off in the first place well in this case if this controller is so simple how is it able to be so powerful well that's exactly what I'd like to investigate today so if all this sounds like more fun than a box full of GI Joes and I guarantee you will actually play with a bunch of GI Joes before the video is over now let's jump over to the whiteboard and get started alright so let's start with our good old friend the linear state space representation of the system okay so if you remember that's just saying that we've got our model of the system as X dot is equal to ax plus bu and in this case Y is equal to CX plus D you okay all right so graphically again if we want to look at this as a block diagram you've got your system here so this is your plant so it's got our state space representation here of X dot is equal to ax plus bu and now let's deal actually with this output here so what I would like to consider now is uh what it would happen if C was the identity matrix and D was 0 so if C was the identity matrix and D is a big fat 0 matrix what we're basically saying here is that the output of the system is the entire state of the system right so that's one of the assumptions we're gonna have to make for this full state feedback discussion here we're gonna see later on that that might come back too I don't want to say to come back to buy this here but it is a bit of a restriction but for now let's make this assumption here so in this case we have y is equal to X right so coming in to the system is obviously you but coming out is the entire state vector okay so if we have this system the control law that we would like to consider is very simple for a full state feedback so let's consider a control law that is very very simple and this is where the simplicity comes in so we're gonna say that the control law here is that the input of the system is just the state vector but it is gonna be scaled by some matrix K and typically by convention we're also going to put a minus sign in front of it the minus sign it could be consumed into the K sometimes depending on who you're talking to but most places will see this as a controller U is equal to minus KX okay so let's box this up because this is actually the control law for full state feedback and I don't have to keep writing a full state feedback so let's call this FS FB okay so full state feedback is basically saying the control it's just the state's scaled by a matrix here so if I wanted to draw this up in my block diagram we got to see that the only thing you have to do is make u equal to the state vector and I just need it to come back through and I'm gonna put a little block here just for fun and you'll see why in just a second this goes through a matrix K okay and then to pick up the negative sign here maybe I'll just draw it like such okay so you can see clearly that in this diagram U is equal to negative K X all right so this K here and it can be thought of as your full state feedback and I don't want to call it a controller really because if you look at this picture here maybe a better name for this would be regulator because what this thing does is that it's going to try to regulate the states to zero there's no input signal there's no reference signal there's no R of T coming in anywhere in this picture right later on down the road maybe we'll incorporate that but for now let's keep it simple and say that this is the overall system so this is a again let's just write this down that this might be considered a regulator because what it tries to do here is regulate nonzero initial conditions back to the origin again so this is the concept of there's no reference signal for it to track all it does is if that the system starts at some nonzero initial conditions this control law here or this full state feedback controller or regulator will try to regulate the states back to zero so just a little side note that I think we I wanted to point out okay so um why don't we take another step with this in mind we can almost ask now how does this overall system behave so maybe I will maybe I'll draw a little dashed box around this thing in green all right so this green you could kind of think of it as the closed loop system right so actually what's interesting about this is that I'm just gonna extend this arrow out a little bit right there again we see that there's no inputs to the green box alright there's no reference signal there's no disturbance torques there's no nothing like that the only thing that comes out of this is X so it's almost like it's an autonomous system by itself right it just responds to initial conditions and it spits out what the state vector does under this control us so why don't we try to write down what are the closed loop dynamics of how does this green box how does the overall system the plant that is being controlled by this full state feedback control law how does that evolve so it's actually pretty simple so let's go ahead and write down the dynamics so we know the dynamics of the plant here right our X dot is equal to ax plus bu right okay now we're gonna make a note that well what is U well that our control law said you was minus KX right so just substitute that in for u so we get a X plus B negative K X right okay so let's distribute this B through so we like what a X minus B okay and again make sure you're not flip-flopping these are all matrices right these are matrix multiplication so you can't just flip-flop the order where are we oh here we go oh yeah sorry my minus BK x right so let's pull X out on the other side so you have a minus B K X right and now if you look at this you have X dot is equal to some matrix X why don't we call this a closed loop or ACL okay so the overall dynamics of the green block are basically X dot is equal to a closed loop times X and again this back and I maybe we shouldn't we should make a very explicit note here so a closed loop is a minus BK right so again this backs up our discussion here of the green box there there are no inputs to the system right it's just an autonomous system or you can think of it from from a differential equation perspective it's a homogeneous system here right there are no external inputs of the system and you see the only way that you get stay trajectories are in response to a nonzero initial condition in your state vector right okay so this is beautiful and basically the next thing that we should ask now and we see from this disk from this this formulation here is that we know that the way that this thing is going to evolve is basically determined by this ACL matrix right so we should say here that the behavior of the closed loop system is governed by ACL right and more specifically the thing that we're usually caring about here is the eigenvalues right so more specifically by the eigenvalues of a Co right that's what we care about and let me erase this because it's gotta getting in our way or a discussion okay so your job as a control engineer might be to ask you should probably be thinking ken is there some knob or can I change the eigenvalues of a closed loop right so your job or your goal here is can we move again values of a closed loop right well why didn't here do you get to play with right do you get to play with a no typically that's part of the plant right this is like the the dynamics your aircraft has already been built so you don't get to change the a matrix similarly B the B matrix was the beat was the control matrix or the control input matrix of your original open loop plant so again that has something to do with like how many engines are on the aircraft where are they mounted what kind of control surfaces do you have all that kind of stuff again you usually don't get to mess with B right however K right the K from your control law that you do get to mess around with so the question maybe instead of saying goal we we should have sittin pose this as a question here of can we move the eigenvalues of a CL well if you look at this the way you can do this is by changing the K matrix you can hopefully change the closed-loop a matrix and now hopefully you should be able to change the eigenvalues of that that closed-loop a matrix okay so yeah maybe this again that and the goal is the wrong one let's put this as a question right so the question is can we move I can values of a CL and the answer is yes I can values of a CL can be placed anywhere if a and B are controllable okay so let's let's let's box this up because this is is is huge and there very interesting although we might need to take us take a step back here okay so first off this concept of controllable here make sure that you've watched our other video right this one over here that talks about controllability right because that talked about how to test of a system was controllable and you can get to that video again there's a link to the URL in the description of this video or click on the card right up here okay so if your system is controllable let's get back to maybe the ramifications and the discussion of this right if you're consistent is controllable what this claims here is that yes you're able to manipulate the eigenvalues of the closed-loop system to anywhere right okay that is incredibly powerful right you're basically saying that I can make this system here respond or behave any way I want I can put it's an eigen values anywhere okay and furthermore let's look at the control law alright how did we do that how are you gonna do this you're gonna do this with this ridiculously simple control alright I don't see any integrals derivatives zeroes polls any complicated behavior here there's nothing this is almost like if you think about this this is almost like a stupid proportional controller right because the control law right here that you can shoot computing at the current time T is just dependent on the current state right now and all you're doing is you're you're you're multiplying n right this is just a big fat matrix multiplied so the control is just directly proportional to the current state so this is what we were talking about that you get you get the best of both worlds you get power you get incredible power to move the poles anywhere but it's ridiculously simple here right there's there's got to be some saying here that goes along with you know ultimate power and all that kind of stuff but this is this is this is the gist of it right so here's the the large-scale picture of you get incredible amount of power with a very simple controller okay so tell you what let's talk about now the next thing is now that we've got all this power we got to we see that the way you get that power is by intelligently choosing this major k-o-k so that's the whole game with full state feedback is how do we pick this matrix K to get us the eigenvalue locations that we want so give me a second I'll pause the video erase the board and let's jump into an example of how we're going to design this K alright so I've erased the board and I've moved the critical information up to the upper left side of the board write the control law again was really really simple and building the architecture to implement that control law was just as simple okay so the first example I want to look at is a controllable system okay so the idea like we said earlier is I need to be able to pick this matrix K here and I want to pick that K such that I put the eigenvalues or the poles of this closed loop system at desired locations this is what's sometimes referred to as pole placement and we're going to see that there are various tools and methods that allow us to perform this pole placement operation okay so let's go ahead and start with an example system so the example system I like to look at is let's go ahead and use our good old friend again our linear system X dot is equal to a which is just a 2 by 2 that looks something like this plus B which looks something like how about minus 2 1 you okay so here's your system okay so here's your a matrix here's your B matrix now you remember we said that okay if this pair a B is controllable the world is our oyster we can move these item days of the closed-loop system to anywhere we want so let's go ahead and first check controllability okay so check controllability so you could go ahead and test but bility all right you can go ahead and use the controllability matrix or the pbh test I think the controllability matrix is probably easier so let's go and do that so PC in this case is going to be B and then a B so I think if you plug all that in I think the B matrix obviously is pretty easy to one if you do the matrix multiplication of a times be I think you're gonna end up with three zero okay so here's your controllability matrix and if you went and check the rank why I guess you don't even need to check the rank you can easily see that this first column is linearly independent of the second column here at they point in two separate directions so we see that the rank of PC is equal to two which is the number of states so yeh system is controllable we should be good to go okay so now what we need to do here is we need to go ahead and try to solve for K okay well we need some way to pick K the first thing we should probably do is what is the dimension of K what does K look like No maybe we should have talked about this a little bit earlier so we said that maybe maybe we should've did let's let's do it over here I get sorry I we probably should have talked about this a little bit earlier but just looking at the control all right U is equal to minus KX okay so if you look at this X that is your number of states this is an N by 1 right so X here is an N by 1 vector u what is U U is the number of control inputs that you have in your system right so U is going to be an M by 1 vector here so again just to make sure n is number of states right and M is number of controls right so just for to make this happen you can easily see that K is actually it's actually got to be a matrix of size what M by n right that's what K has to be in order for the dimensions to work out right so we see K is going to be a matrix it's going to have as many rows as you have control inputs and it's going to have as many columns as you have States ok so perfect so in this case we see that K is actually gonna be okay how many controls you have there's only one control so it's only gonna be one row how many states do you have there's two states so it's going to be a one by two matrix so there's gonna be like a k1 here and a k2 okay so K again keeping the back your head K is a matrix game right okay so let's keep going alright now we know that this is what our K matrix is gonna look like we also know from our previous discussion that we know that the closed loop system of this right it's closed over a matrix is gonna look like maybe we should write this down so we know closed loop a matrix is we called it ACL I think earlier right and that was gonna be a minus B times K all right okay so your a matrix is pretty simple it is a going to be a zero 3 - 4 - what's your B matrix so it's a minus 2 1 and now we know our K matrix here is a 1 by 2 it's k1 k2 so again we see the we see the dimensions work out so this is boring algebra let me skip it you can go order mathematic and do this or heck do it by hand it's it's pretty easy to do this matrix multiplication and then the subtraction but you would basically get that at the end of the day your ACL this is gonna look like again it's a 2 by 2 all right so it's gonna look like 2 K 1 3 plus 2 K - 2 minus K 1 and then great so you see that the case weasel their way into the a matrix right so again you should start thinking that oh wow you can easily see that if I start screwing around with K 1 and K 2 I change the a matrix therefore I hopefully change the eigen values so we want to directly change the eigen values then right so what we should maybe do next is we know that the eigen values come from solving basically the the characteristic equation of s Si minus a alright so why don't we write this down here that the closed loop I ghen values right if you want to find the eigen values of any matrix right you're basically solving lambda I minus the matrix right and you want to solve this thing or you actually you want you want this to be to be singular right so the eigen values are the values that make the ACL singular so in other words I want the determinant of this thing to be equal to zero all right so the determinant of this thing should equal zero right so those will give me all the eigenvalues of the ACL matrix right okay so this is pretty simple so again let's go ahead and just do this so lambda I so this this looks like really simple it's a lambda 0 0 lambda minus ACL which is this thing up here right 2k 1 3 plus 2 K 2 2 minus K 1 and then 4 minus K 2 right and I still got to take the determinant of this whole thing and set that equal to 0 so again you can see that it's not complicated it's just algebra at this point right subtract these two take the determinant right do the do the diagonals and you'll basically end up with and again I'll skip this the the nitty-gritty is here in my notes but I think you can see that it's very simple matrix operations at this point you could get this thing to look like you'll basically see you'll get a second-order polynomial in lambda right so what I'm gonna do is I'm going to write this in the form of what do we end up here we end up with lambda squared plus negative 4 minus 2 k1 plus k2 times lambda plus 11 k1 minus 6 minus 4 K 2 ok there we go okay why don't we call this this is like your characteristic equation right because this is what I need this to be equal to 0 all right you solve for the lambda values that gives you the eigenvalues of the closed-loop system okay so let's call this how about P of lambda all right so let's box this up okay and maybe we'll make a note here that this is the characteristic equation of the closed-loop system using full state feedback of k1 k2 right so this is the characteristic equation of closed-loop system under full state feedback right okay and actually you know what let's let's erase some of this because I want to isolate this thing by itself because it's this is the important bit okay okay so all right we've got ourselves the characteristic equation of the closed-loop system then we use full state feedback is going to look like that so again you see that basically by manipulating K 1 and K 2 you can change the characteristic equation therefore you can change the eigenvalues of the system okay so now the question is where do we want those eigenvalues to go okay this is where full state feedback is so powerful we can now choose where do we want the poles to go so for example let's say our suppose that you want closed-loop eigenvalues at a specific location let's make something up how much I really would like the closed-loop eigenvalue the first one to be here at I'll make something up how about -5 plus 2i and then I would really like the second closed-loop eigenvalue to be a minus 5 minus 2i right that's where I want these this system to have I gain values I want this overall system here by the time I close the loop and do everything I want it to behave like it's got pole or eigen values at these locations right well okay if you are gonna be this location what is the characteristic equation that is going to yield these locking values right so the other way to think about that is if you want these eigen value locations you better make sure that the characteristic equation looks like this right so you better make sure the characteristic equation is well it better be lambda what plus 5 minus 2i right and then lambda plus 5 plus 2i right is equal to 0 right this better be the characteristic equation right let's call this thing maybe P desired of Lee right you really want the characteristic equation to look like this because if you do if you go if you get a characteristic equation to look like this I can tell you right now yeah the roots of this characteristic equation are exactly where you want it to be right okay so tell you what expand this thing out so you could also write this thing to look like it would be lambda squared plus 10 lambda plus 29 by the time you just foil this thing out right you do the expansion and I think that's what you would end up with okay so tell you it again let's erase the junk that we don't care about any longer okay so here's P desired of lambda right and let's box this up all right and again let's make a note here that this is the desired characteristic equation in order to place poles / eigenvalues right at desired locations right okay so here's the characteristic equation that we want here's the characteristic equation you get under full state feedback control so I just need to make these two equal to each other right so in other words all I need to do is you know what compare the coefficient of lambda here to the coefficient of lambda over here those two better match right so we get one equation here so if I compare the coefficient of lambda maybe I'll just write it like this right we basically get an expression here that alright you get minus 4 minus 2 k1 plus k2 had better equal the coefficient of lambda over here had better equal 10 right okay and then their second equation here is the constant right let's do it in another color the constant of what I want had better equal the constant of what I've got right so if I look at the constant term right again we end up with what do we end it's 11 k1 minus 6 minus 4 k2 had better equal the constant here of 29 right great so if you look at this all this thing is it's two equations two unknowns right and in fact this is even easier they're two linear equations and since the system was controllable you can pretty much guarantee that this is going to be solvable right and if furthermore there will be in this case there gonna is gonna be a unique solution here so run to your favorite package rate go ahead and write this in matrix form and row reduced do whatever you want to do to solve two equations for two unknowns and you'll end up with what do we end up with here so I think you'll end up with k1 has got to be equal to 91 thirds about 30 0.33 and k2 has got to equal 224 over three which is again this is approximately seventy four point six seven okay so here we go this went ahead we solved now for the the K matrix that will get us the eigen values to be placed in exactly these locations right and you saw that it's actually not that bad a procedure at least in this case all right all it's doing here all you need to do is look at what is the characteristic equation you get under full state feedback control compare it to the characteristic equation that you want alright and just make the two of them equal to each other okay now to support this workflow again you talked about this whole workflow is referred to as pole placement right because that's exactly what we're doing we're placing the poles of this closed-loop I get of the closed-loop system to wherever we want it to be matlab's got a nice routine that will allow us to do this called place okay so tell you what let's run over to MATLAB and do this exact example and see if we can just use the place command to to automate this process for us all right so this is actually really really simple in MATLAB so let's go ahead and first define our a matrix so I think this is what we ended up using and also let's define our B matrix and now what we should probably do here is check controllability right we talked about how that is a critical component so let's check the rank of the controllability matrix and again if you would like refresher on rank or CTR B please go ahead and check out our controllability lecture so let's go and run this and we see yep full rank so this system should be controllable so what I can do now is I would like to perform pole placement okay so to do that like we said MATLAB has this nice command called place and place will as it says compute the feedback matrix K to put the eigenvalues of the closed-loop system where we want so what I need to do here is make a let's call P desired right this is the location of where I want to put the poles and I think we said it was -5 plus 2i and -5 minus 2i right that's the locations where I want to put the poles so now it's really really simple I go ahead and say place a whoops a B and then P desired right and what this will return for me is the matrix K that should place the poles at that location so if I run this here we go these are the gains that we computed on the board and maybe what we should just do one last double-check before we leave MATLAB is check that the closed-loop I ghen values are in the desired location all right so to do that all we need to do is first let's make a closed loop right I think we said that was a minus B times K right and then let's just look at the eigenvalues of the closed-loop system so if I run this yay look at that that is exactly where we want it to go okay before we leave this maybe what I will mention is there's actually another command you could also use in MATLAB besides the the place command and that is a cur a cker so this is using Ackerman's formula to do a similar operation so for example we could come up here and instead of calling place let's go ahead and I will copy this and let's paste it below you could also call a cur which should hopefully you similar results okay so if we run this again yet we get the exact same results just a quick note here I think if you read the help on Acker it's says that it's not reliable for larger order systems so place might be a better way to go but again I just wanted to point out that there's more than one way to ride this donkey you can either use place or you could use Acker okay so this worked quite well for a controllable system let's jump back to the white board and look at a system that maybe is uncontrollable all right so let's look at another example because I bet you're just dying to know does this what happens if you have a system that's uncontrollable does it still work and the the the answer is actually no but let's go ahead and walk through this just to show where this breaks down so let's consider the area similar system right we'll keep the exact same a matrix we had earlier but now what I'm gonna do is let's choose a B matrix that is uncontrollable here so this uh B matrix is going to be oh gosh I need a little bit more room actually because it's actually a little bit involved here it's still a two by one here but this is now 1/2 times negative 2 plus square root 10 and then one you okay and this I think if you wanted to write it out numerically I think this comes out to about zero point five eight one one and then a one okay interestingly for I guess there's a little side note for those of you who have watched the controllability lecture this is actually an eigen vector right corresponding to one of the eigen values and I think we showed last time that that if you have a basic your B matrix that's exactly an eigen vector right you end up with like an uncontrollable system so this system we claim is uncontrollable we'll check the controllability in a second well actually you know what maybe we should do this right now so again let's check controllability PC here you do the exact same thing be a B right and what you end up with error is you end up with ok the B matrix again is pretty simple it's one negative two plus root ten and then one and then a times B when you do that multiplication I think you get three and then 2 plus root ten okay and you check the rank of this thing and it's only one right so it's ranked efficient so the system uncontrollable okay so you could do the exact same operations we did earlier and let's see where this is going to be great a breakdown so you perform the same pole placement so if you remember all we did last time was we computed a CL right as being a minus BK right and we know here that K is again a k1 and a k2 right it's just a one by two matrix so you do that math alright and let me see is it ugly or not I have it here in the notes I guess it's not horribly ugly tell you what let's all just write it down for completeness just so everyone can follow along here so a closed loop what you would end up with still a two by two here so this is going to be minus one half times negative two plus root ten times k1 and then you'd end up with 2 minus k1 and then in this entry up here you get 3 minus 1/2 times negative 2 plus root 10 times k2 and then you'd end up with 4 minus k2 okay so that's close a loop now what we would do is we would compute the characteristic equation for this system here right I'd go ahead and do determinant of lambda I minus a closed loop is equal to zero okay so again that is something ugly but basically what you can do is after a bunch of algebraic manipulation you can get this thing to look like lambda squared plus this quantity here so it's minus 4 plus UPS that's a minus minus k1 plus root 5 over 2 K 1 plus K 2 this is times lambda plus gotcha constant I ran out of room so the constant so it's a second-order polynomial so the constant here or the zeroth order this is what - 6 + 7 K 1 minus 2 root 10 K 1 minus 2 K 2 plus root 10 K 2 again something right this year is P of lambda here right so again this is the closed loop characteristic equation under full state feedback right and now we need to just compare with this with what we want so let's use the exact same poles that we wanted to earlier that we considered earlier so let's just gonna say I want to place this system at P desired of lambda at lambda minus 5 minus 2i lambda minus 5 plus 2i right so I have poles at it's wait a second what did it do hole is that a plus I think it's a plus didn't we want the poles at positive I have to make sure did I mess this up in the last one we wanted the poles at s and lambda is equal to minus 5 plus or minus 2i so it's got to look like this so I'll have to double check the video to see did I did I drop this plus sign in the earlier one if I did on all I'll correct it okay anyway again this is the exact same characteristic equation we were we were considering earlier right so that is if you expand it out it's lambda squared come on where do I put this thing here oh yeah plus ten lambda plus 29 right so again you do the exact same thing you compare the coefficient of S or the coefficient of lambda with the coefficient of lambda here right okay so what we would end up with is do to do to do so that's two equations and two unknowns so we end up with minus four minus K 1 plus square root of 5 over 2 K 1 plus K 2 has got to equal 10 all right and then we're going to compare the constant here or the 0th term with this term right so then we end up with minus 6 plus 7 K - to root 10 K 1 minus 2 K 2 + root 10 K 2 is gotta equal to 29 right okay so now we just need to solve these two equations for these two unknowns K 1 and K 2 and it might be easier if we write this in matrix form alright so I think if you want to write this in matrix form you could write and I let's pull this out so you got this okay K 1 K 2 and actually you can see there's actually this constant over here there's a there's a minus 4 and then a minus 6 here right has got to equal 10 and 29 right and then this entry right over here I think if you had gone ahead and done this you would end up with root 5 over 2 minus 1 1 7 minus 2 root n root 10 minus 2 right okay so let's move this to the other side so we can get this to look like an ax is equal to B so what you would end up with then is you'd end up with a matrix expression that looks something like okay that that whoops that's probably a little bit big we don't need it that big okay so this this so this is the same thing root 5 over 2 minus 1 1 7 minus 2 root 10 root 10 minus 2 times k1 k2 it's equal to once at 14 and then 35 right okay so now this looks like a big ax is equal to B problem right so here's your your a here's your X in here B right this looks like ax is equal to B but if you solve for or if you try to compute the rank of a you find that it's only rank 1 here so you see that rank of a is equal to 1 so the inverse doesn't exist so now we're into one of two situations there's either infinite solutions or there's no solutions so the way we can figure that out here is let's create our augmented matrix by putting the B matrix over here so I'm going to consider our augmented matrix a tilde to be a B right and then I'm going to row reduce that thing so again all that that entails is maybe let's just do one more step to make sure we're all on the same page all right so I'm just gonna augment this thing wrote 5 over 2 minus 1 1 7 minus 2 root 10 and then root 10 minus 2 so here's the a and then I'm gonna miss with the B matricide okay so here's my a till this is the Augmented matrix and what I can now do is run over to any mathematical package or if you want to do the elementary row operations yourself and row reduce this thing you're free to but you know I did this in Mathematica and what I ended up with was this came out looking like 1 2 over negative 2 plus root 10 0 and then 0 0 1 right so clearly this is inconsistent right because if you look at this bottom line this bottom line basically reads K 1 or basically it reads 0 is equal to 1 right it's 0 times K 1 plus 0 times K 2 is equal to 1 so this is clearly inconsistent right so there's no solutions so in this case the system was uncontrollable so I can't find a K matrix that is going to move the eigenvalues to the locations that I desire and if you try this in MATLAB it's also gonna choke em just gonna yell at you and say hey systems uncontrollable you know and if you don't believe me here tell you what let's run over there those do it real quick alright so let's go ahead and start example 2 which was an uncontrollable system ok so we said the only difference here was the B matrix so the B matrix now is this weird 0.5 times negative 2 plus square root of 10 and then a 1 right that was our B matrix so now let's again check controllability right so I'm gonna look at rank CTR B of a B and this should be deficient and indeed it is ranked efficient so now if we go ahead and try to perform pole placement basically this command up here let's just go ahead and copy that exact command because that's all I'm trying to do is the exact same thing again is gonna not like this and you see error right it couldn't place these because it's uncontrollable okay so let's look at one more example so example 3 so this one is going to be again let's revisit our controllable system but now let's use multiple controls okay and see what happens so let's go and revisit the system we had in example 1 right so I think the example 1 we said was something like this is 0 3 2 4 X plus negative 2 1 you okay but now what I want to do here is instead of just having one input let's add another column to this B matrix and do something like this right so now in this case you see we've got two control inputs okay so in this case now where I have two control inputs again we could check controllability and and you know what table let's save until we get over to MATLAB because Mallove's a lot quicker than me just waiting out the controllability matrix and check hey Frank take my word that it is controllable well actually you know what you should know for a fact that it is controlled because this is the B matrix we checked already like this first column back in example one that was controllable it's adding more this is obviously it's not then make it little if anything it's get more controllable so this is still controllable so what we can now do here is let's go ahead and compute K so again make a note here that K in this case remember we said K was a matrix with as many rows as there are controls so now this thing has got two rows and now it's actually got two columns as well so this thing is actually a two by two so now I've got a k1 one ak-12 a k2 one and a k2 two right okay so we do the exact same thing we did before so let's go ahead and get a closed loop right a closed loop for the overall system it's still a minus BK so go ahead and do that matrix math here take here's your a here's your B here's your K go ahead and do that you're gonna get something let me see it's not it's not ugly so tell you what I will go ahead and write it out just for completeness it's not it's not ridiculous you still get a 2x2 its 2k maybe I need a little more room hold on makes a little bit bigger ok so it's 2 K 1 1 minus K 2 1 and then it's 3 plus 2 K 1 2 minus K 2 2 and then you get a 2 minus K 1 1 minus K 2 1 and then 4 minus K 1 2 minus K 2 2 ok there we go so now let's go ahead and get the characteristic equation of this closed-loop system by basically doing our good old friend determinant of lambda I minus a closed-loop is equal to 0 right so this is a little bit more complicated but still not horribly ugly I'll write it out for just completeness so we get lambda squared plus minus 4 minus 2 K 1 1 plus K 1 2 plus K 2 1 plus K 2 2 times lambda plus the constant term here is minus 6 plus 11 K 1 1 minus 4 K 1 2 plus 3 K 1 2 times K 2 1 plus 2 K 2 2 minus 3 K 1 1 K 2 2 yuck a little bit uglier okay so again box this sucker up here's your characteristic equation under full state feedback right and then let's go ahead and compare this with what we want let's again use the exact same desired characteristic equation as before so this was this guy which put our poles at plus from minus 5 plus or minus 2i so here we go so here's the characteristic equation we want so it's the exact same game right compare the coefficient of lambda with the coefficient lambda compare the constant with the constant but now we see we get a little bit more freedom because we've got now two equations and four unknowns right so there's an additional degree of freedom here so for example what we could do is we could solve what did I do in my example let's solve k2 1 & 2 2 let's solve let's solve the bottom row of this K matrix with respect to the top row so if you do that let me go ahead and find someplace to erase all right it's just algebra at this point so what we can say is again you got 4 would start two equations right two equations four unknowns so there's infinite solutions in this case so what you could do is you could get basically K - 1 is gonna equal something ugly I don't know if I really want to write this all up okay tell you what I will write this up once for completeness okay so you get minus 4 minus 2 we're sorry whoops no I'm looking at the wrong line sorry K - 1 is it's a negative in front of this whole thing and then you get negative 7 minus 27 K 1 1 minus 6 K 1 1 squared minus 6 K 1 2 plus 3 K 1 1 K 1 2 ok all over 3 times negative 1 plus K 1 1 plus K 1 2 ok so that's K 2 1 and then K 2 2 is going to be something also kind of ugly you get a negative and then the fraction looks like 49 minus 9 K 1 1 minus 39 K 1 2 minus 6 K 1 1 K 1 2 plus 3 K 1 2 squared all of this thing over 3 times the quantity negative 1 plus K 1 1 plus K 1 2 yuck ok so a couple of degrees of freedom we're now free to choose K 1 1 and K 1 2 to be pretty much anything we want and then that will dictate what K 2 1 and K 2 2 have to be in order to get the eigen values where we want but this additional degree of freedom is actually nice because it allows us to do other or try to meet other requirements or optimize the system in other fashions so for example you know one thing you could do is what if we wanted the first control to not be used as much as the second control if you remember this system here be now there are two inputs there's a u one and a u - right and we said that the control law is U is equal to minus K X all right so again if you write this thing out this is a u1 a u2 is equal to K 1 1 K 1 2 K 2 1 K 2 2 times X 1 X 2 right so the first control you can see that if we if I make K 1 1 and K 1 2 small relative to these two it would mean that the magnitude of U 1 is less than the magnitude of U 1 relatively speaking here right so in this case example what if you one was expensive like you one was the was was was thrust from your satellite or something like that it was something you know super expensive it's like jet fuel or something like that or I guess jet fuel is not really that expensive but you get though you get what I'm trying to say all right you won 1 how about you 1 1 is is you're shooting babies out the back of your rocket that's how you're gonna generate thrust right and that's much worse than u 2 u 2 is like just using you know traditional hydrazine or something like that as a propellant here so if I don't if I want to save the babies right if I don't want you want to be large what I would then want to do is make sure that K 1 1 and K 1 2 are small so for example you know now what we could say is let's go ahead and choose you since we're free to choose I could say how about K 1 1 is like 1/10 and a cave sorry 1 1 and K 1 2 is also 1/10 something small okay so now what I would do then is plug these numbers back into these expressions to get what is K 2 1 and K 2 2 have to be so if you do that you would end up with K 2 1 is being what was this something like minus 10 33 over 240 which is about negative 4 three and then k 2 2 would be 4 4 1 7 over 240 which is about eighteen point four more four oh right okay great so this K matrix now so now we see that the control law right so u 1 and u 2 it's going to be 0.1 0.1 and then negative four point three and then eighteen point four times x1 x2 right so we can see that for a given X right if x1 and x2 are let's just say both of the both of these are 1 right so this vectors on what is ones right so out here what what you one would would evaluate to is just one point two right so you only shoot point two babies out the back of your rocket that's less that's like what a fifth of a baby I guess that's okay I guess I can live with that right but down here how much hydrogen are you burning it's actually what four point three minus eight so it's like 14 units of hydrogen so you can see in this case that u1 is small relative to you two right so this is nice so the pole placement technique and and what's great about this is this K matrix still will put the eigen values of the system the closed-loop system you know this whole thing it will still put it at the desired locations of minus 5 plus or minus 2i right so this is this is great it shows us there's a there's a little bit of wiggle room here so and the pole placement technique is still great well I don't I don't know if I want to say gracefully but you can still take it into account if you manually do this this sort of sets the stage for our discussion later on when we're gonna start talking about linear quadratic regulator zor-el QR and this is another way that you can use to tune which one of these controls do you care about more and use pole placement a full state feedback in this entire framework but I think I'm getting a little bit ahead of myself maybe what we should do though is let's run over to my lab and see that how is MATLAB gonna handle this situation where we have more than one control input and we're trying to design a K matrix for this system so again keep this in mind that this is the case matrix we designed ourselves manually because we wanted to tune how many babies are we shooting out of the back of this rocket and let's go see what does MATLAB do automatically alright so let's go ahead and start example three this was our controllable system with multiple inputs okay so actually before we get started let's comment out line 34 because we saw that this is going to choke right line 34 was where we had you're trying to place an uncontrollable system so this is not going to help us so let's comment that out okay so now what we said is all we're going to do is augment the B matrix so I think our new B matrix looks like this right this is the B matrix with two controls so again let's go ahead and rank check this thing of the controllability matrix to make sure we are fully controllable and y8 here we go so now I can well you know what it's actually just this command alright it's our place command so again let's copy this paste it down here and we should be good to go and here here you go so interestingly look at this since place doesn't have any knowledge of how much do we want to penalize one control versus the other or any of these other ways to narrow down the degrees of freedom rank we said there's infinite solutions so here's one that MATLAB chooses for us right it decides that here here's a game that should work let's just verify that maybe so let's come back over here and let's store this to a variable K and just go ahead and check that matlab's computed K yields acceptable or desired eigen values alright so I want to look at AG of a minus B times K all right so run this sucker and yay it totally works right to put it where we want it to be um however let's go and check that our whiteboard designed K's yields desired eigen values right so K let's call it K how about white board right this is what did up on the whiteboard what did we say this was this was 1/10 1/10 and then I think it was negative 10 30 3 over 2 40 and 44 17 over 2 40 right so now let's go ahead and check that the eigenvalues of a minus K whiteboard get us where we want as well and there you go look at this it also works so again there's multiple ways to skin this cat in the situation where you've got multiple inputs because we've got more degrees of freedom than we have constraints all right so I want to leave with a couple of closing thoughts so the first thought is if you look at the control law here which is equal to minus KX we see that full state feedback is approximately it's basically proportional control right in the sense that to compute the control right now all you do is you take the current state and you multiply or you proportionate it and what that's actually not a verb isn't here but anyway the control is directly proportional to the current state right so you don't need anything any information about the past all right there's like like an integrator right you with PID control we saw that you needed to add components that would have some idea of the past and you also had things that needed had rough ideas of the future here but you don't need any of that stuff here in full state feedback control it's really simple I can compute the control server control input now based on the information I know now I don't need anything else the other thing to think about here is what we're talking about maybe comparing it to PID control if you remember here there's when we were talking about PID control we looked at designing them via root locus in fact we looked at designing any linear controller using the root locus technique with full state feedback control there there's no concept of root locus right so full state feedback you know has no root locus in the sense that you are no longer constrained to move the closed-loop pole along given locus lines right so if you plot out you know your real imaginary axis and let's say your open-loop system had pulled Raman blue here right had poles here here here you know it had it had dynamics that looked like this right and with root locus we saw that by adding a controller these poles would move in specific lines or specific locations and and half the half of your time was just wrestling with making sure that we placed the controller poles and zeros and locations that would drag the locus lines to where we want and we spent an incredible amount of time just fighting with the system but with full state feedback control there's no idea there's no lines here right you can move them anywhere you want right so if you don't like these locations and you want the poles to go somewhere else like you know here here and here right you don't have to drag them along lines you can directly move them a full state feedback right full state feedback says well we'll move them wha-bam right there okay and and if you think about this like this has huge ramifications right for example consider what if one of these blue poles were like your open-loop system here so like we said this is open-loop system this is your natural system right so again I I'm I'm an aerospace so I like to think of aircraft examples let's say you've got this slow Piper Cub right this is a low performance aircraft and it it has open-loop poles like like this like these these blue X's right well if I slap a full state feedback controller on this I can move it to anywhere like what if what you know here let's let's call this a Piper Cub right what if these are like f-14 dynamics right so I don't want this I want it to look more like an f-14 right so I can get rid of this full state feedback can make my Piper Cub behave like an f-14 all right so now it can fly around it can be doing Mach 1 it can be turnin and burnin doing all this kind of cool stuff and it doesn't stop there right there's nothing to stop it let's say let's say you've got other aircraft like you you you want this thing to behave in and yet another fashion maybe you want it even faster you want all the poles even further to the left right you can move it here to here to here right you want it to now behave like some forward-swept you know other ridiculous fighter jets so I can now turn my Piper Cobra into my f-14 then I can turn it into into something faster or you can even there's no limit to this right you can keep going you can turn this if you don't like this and you want your experimental crazy prototype hypersonic vehicle that can fly around and and and do everything but nothing is stopping you with full state feedback control right full state feedback control says you can go anywhere you want right you can be gosh it sounds so uh so this you know my daughter's in kindergarten it sounds like one of those things they tell you in kindergarten right you can be anything you want to be right that's exactly what full state feedback control is telling you it says you can be anything you want I can change the dynamics of my system I can put the eigenvalues anywhere so in summary in this short lecture we saw how full state feedback was an incredibly simple yet incredibly powerful control architecture as long as your system is linear and controllable full state feedback is going to allow you to move the closed-loop eigenvalues to your system to pretty much anywhere you want so what this means is you can make any system behave like any other system as long as they have the same number of eigenvalues so you can literally transform your system to be whatever you want so if I wanted to act like a human with a dog's head on top full state feedback would make this possible now obviously this is a little ridiculous so you have to ask yourself where does full state feedback break down I mean if full state feedback is so great why doesn't anyone use it well I'm glad you asked that question because in our very next video we're going to take a look at some practical implementation issues that you're going to encounter with full state feedback so with that being said I hope you enjoyed the discussion today and if you're already a subscriber thanks for sticking with it with us if you're new to the channel please consider subscribing as we'll have lots of other discussions on Controls Engineering's in the future and I hope you'll join us at a future video now I don't know about you but I'm pretty hungry after all this talking so I'm gonna grab a bite to eat until our next time I hope to see you later talk to you later by

Channel: Christopher Lum

Views: 18,251

Keywords: Full state feedback control, Full state feedback controller, pole placement, place, acker, Ackermann’s formula

Id: 1zIIcYfp5QA

Length: 62min 50sec (3770 seconds)

Published: Mon May 27 2019